*Article* **Note on Type 2 Degenerate** *q***-Bernoulli Polynomials**

### **Dae San Kim 1, Dmitry V. Dolgy 2, Jongkyum Kwon 3,\* and Taekyun Kim <sup>4</sup>**


Received: 17 June 2019; Accepted: 12 July 2019; Published: 13 July 2019

**Abstract:** The purpose of this paper is to introduce and study type 2 degenerate *q*-Bernoulli polynomials and numbers by virtue of the bosonic *p*-adic *q*-integrals. The obtained results are, among other things, several expressions for those polynomials, identities involving those numbers, identities regarding Carlitz's *q*-Bernoulli numbers, identities concerning degenerate *q*-Bernoulli numbers, and the representations of the fully degenerate type 2 Bernoulli numbers in terms of moments of certain random variables, created from random variables with Laplace distributions. It is expected that, as was done in the case of type 2 degenerate Bernoulli polynomials and numbers, we will be able to find some identities of symmetry for those polynomials and numbers.

**Keywords:** type 2 degenerate *q*-Bernoulli polynomials; *p*-adic *q*-integral

**MSC:** 11S80; 11B83

### **1. Introduction**

There are various ways of studying special polynomials and numbers, including generating functions, combinatorial methods, umbral calculus techniques, matrix theory, probability theory, *p*-adic analysis, differential equations, and so on.

In [1], it was shown that odd integer power sums (alternating odd integer power sums) can be represented in terms of some values of the type 2 Bernoulli polynomials (the type 2 Euler polynomials). In addition, some identities of symmetry, involving the type 2 Bernoulli polynomials, odd integer power sums, the type 2 Euler polynomials, and alternating odd integer power sums, were obtained by introducing appropriate quotients of bosonic and fermionic *p*-adic integrals on Z*p*. Furthermore, in [1], it was shown that the moments of two random variables, constructed from random variables with Laplace distributions, are closely connected with the type 2 Bernoulli numbers and the type 2 Euler numbers.

In recent years, studying degenerate versions of various special polynomials and numbers, which began with the paper by Carlitz in [2], has attracted the interest of many mathematicians. For example, in [3], the degenerate type 2 Bernoulli and Euler polynomials, and their corresponding numbers were introduced and some properties of them, which include distribution relations, Witt type formulas, and analogues for the interpretation of integer power sums in terms of Bernoulli polynomials, were investigated by means of both types of *p*-adic integrals.

As a *q*-analogue of the Volkenborn integrals for uniformly differentiable functions, the bosonic *p*-adic *q*-integrals were introduced in [4] by Kim. These integrals, together with the fermionic *p*-adic integrals and the fermionic *p*-adic *q*-integrals, have proven to be very useful tools in studying many problems arising from number theory and combinatorics. For instance, in [5], the type 2 *q*-Bernoulli

(*q*-Euler) polynomials were introduced by virtue of the bosonic (fermionic) *p*-adic *q*-integrals. Then, it was noted, among other things, that the odd *q*-integer (alternating odd *q*-integer) power sums are expressed in terms of the type 2 *q*-Bernoulli (*q*-Euler) polynomials.

In this short paper, we would like to introduce the type 2 degenerate *q*-Bernoulli polynomials and the corresponding numbers by making use of the bosonic *p*-adic *q*-integrals, as a degenerate version of and also as a *q*-analogue of the type 2 Bernoulli polynomials, and derive several basic results for them. The obtained results are several expressions for those polynomials, identities involving those numbers, identities regarding Carlitz's *q*-Bernoulli numbers, identities concerning degenerate *q*-Bernoulli numbers, and the representations of the fully degenerate type 2 Bernoulli numbers (*q* = 1 and *x* = 1 cases of the type 2 degenerate *q*-Bernoulli polynomials) in terms of moments of certain random variables, created from random variables with Laplace distributions.

The motivation for introducing the type 2 degenerate *q*-Bernoulli polynomials and numbers is to study their number-theoretic and combinatorial properties, and their applications in mathematics and other sciences in general. One novelty of this paper is that they arise naturally by means of the bosonic *p*-adic *q*-integrals so that it is possible to easily find some identities of symmetry for those polynomials and numbers, as it was done, for example, in [1]. In the rest of this section, we recall what is needed in the latter part of the paper.

Throughout this paper, *p* is a fixed odd prime number. We use the standard notations Z*p*, Q*p*, and C*<sup>p</sup>* to denote the ring of *p*-adic integers, the field of *p*-adic rational numbers, and the completion of the algebraic closure of <sup>Q</sup>*p*, respectively. The *<sup>p</sup>*-adic norm on <sup>C</sup>*<sup>p</sup>* is normalized as <sup>|</sup>*p*|*<sup>p</sup>* <sup>=</sup> <sup>1</sup> *p* .

As is well known, the Bernoulli numbers are given by the recurrence relation

$$B\_0 = 1, \quad (B+1)^n = \begin{cases} 1, & \text{if } n = 1, \\ 0, & \text{if } n > 1, \end{cases} \tag{1}$$

where, as usual, *B<sup>n</sup>* are to be replaced by *Bn* (see [2,6,7]).

Additionally, the Bernoulli polynomials of degree *n* are given by

$$B\_{\rm II}(\mathbf{x}) = \sum\_{l=0}^{n} \binom{n}{l} B\_{l} \mathbf{x}^{n-l},\tag{2}$$

(see [3,8,9]).

Let *q* be an indeterminate in C*p*. For *q* ∈ C*p*, we assume that |1 − *q*|*<sup>p</sup>* < *p* <sup>−</sup> <sup>1</sup> *<sup>p</sup>*−<sup>1</sup> .

In [7], Carlitz considered the *q*-Bernoulli numbers which are given by the recurrence relation:

$$\beta\_{0,q} = 1, \quad q(q\beta\_q + 1)^n - \beta\_{n,q} = \begin{cases} 1, & \text{if } n = 1, \\ 0, & \text{if } n > 1, \end{cases} \tag{3}$$

where *β<sup>n</sup> <sup>q</sup>* are to be replaced by *βn*,*q*, as usual.

In addition, he defined the *q*-Bernoulli polynomials as

$$\beta\_{n,q}(\mathbf{x}) = \sum\_{l=0}^{n} \binom{n}{l} [\mathbf{x}]\_q^{n-l} q^{l\mathbf{x}} \beta\_{l,q} \quad (n \ge 0), \tag{4}$$

where [*x*]*<sup>q</sup>* <sup>=</sup> <sup>1</sup>−*q<sup>x</sup>* <sup>1</sup>−*<sup>q</sup>* , (see [7]).

Recently, the type 2 Bernoulli polynomials have been defined as

$$\frac{t}{e^t - e^{-t}} e^{xt} = \sum\_{n=0}^{\infty} b\_n(x) \frac{t^n}{n!} \tag{5}$$

(see [1,3,8]).

When *x* = 0, *bn* = *bn*(0) are called the type 2 Bernoulli numbers.

From (5), we note that

$$\sum\_{l=0}^{n-1} (2l+1)^k = \frac{1}{k+1} (b\_{k+1}(2n) - b\_{k+1}), \ (k \ge 0). \tag{6}$$

Let *f* be a uniformly differentiable function on Z*p*. Then, Kim defined the *p*-adic *q*-integral of *f* on Z*<sup>p</sup>* as

$$\begin{split} I\_q(f) &= \int\_{\mathbb{Z}\_p} f(\mathbf{x}) d\mu\_q(\mathbf{x}) \\ &= \lim\_{N \to \infty} \frac{1}{[p^N]\_q} \sum\_{\mathbf{x}=0}^{p^N-1} f(\mathbf{x}) q^{\mathbf{x}} \\ &= \lim\_{N \to \infty} \sum\_{\mathbf{x}=0}^{p^N-1} f(\mathbf{x}) \mu\_q(\mathbf{x} + p^N \mathbb{Z}\_p), \end{split} \tag{7}$$

(see [4]). Here, we note that *<sup>μ</sup>q*(*<sup>x</sup>* <sup>+</sup> *<sup>p</sup>N*Z*p*) = *<sup>q</sup><sup>x</sup>* [*pN*]*<sup>q</sup>* is a distribution but not a measure. The details on the existence of the *p*-adic *q*-integrals for uniformly differentiable functions *f* on Z*<sup>p</sup>* can be found in [4,10].

From (7), we note that

$$qI\_q(f\_1) = I\_q(f) + (q-1)f(0) + \frac{q-1}{\log q}f'(0),\tag{8}$$

where *f*1(*x*) = *f*(*x* + 1).

By virtue of (8) and induction, we get

$$q^n I\_\emptyset(f\_n) = I\_\emptyset(f) + (q - 1) \sum\_{l=0}^{n-1} q^l f(l) + \frac{q - 1}{\log q} \sum\_{l=0}^{n-1} q^l f'(l),\tag{9}$$

where *fn*(*x*) = *f*(*x* + *n*), (*n* ≥ 1).

The degenerate exponential function is defined by

$$
\sigma\_{\lambda}^{x}(t) = (1 + \lambda t)^{\frac{\tilde{\lambda}}{\lambda}},
\tag{10}
$$

(see [11]), where *λ* ∈ C*<sup>p</sup>* with |*λ*|*<sup>p</sup>* < *p* <sup>−</sup> <sup>1</sup> *<sup>p</sup>*−<sup>1</sup> . For brevity, we also set

$$e\_{\lambda}(t) = e\_{\lambda}^{1}(t) = (1 + \lambda t)^{\frac{1}{\lambda}}.\tag{11}$$

Carlitz defined the degenerate Bernoulli polynomials as

$$\frac{t}{\varepsilon\_{\lambda}(t) - 1} e\_{\lambda}^{x}(t) = \sum\_{n=0}^{\infty} B\_{n, \lambda}(x) \frac{t^{n}}{n!} \tag{12}$$

where *Bn*,*<sup>λ</sup>* = *Bn*,*λ*(0) are called the degenerate Bernoulli numbers.

From (12), we note that

$$\sum\_{l=0}^{n-1} (l)\_{k,\lambda} = \frac{1}{k+1} (B\_{k+1,\lambda}(n) - B\_{k+1,\lambda}), \quad (n \ge 0), \tag{13}$$

where (*l*)0,*<sup>λ</sup>* = 1, (*l*)*k*,*<sup>λ</sup>* = *l*(*l* − *λ*)···(*l* − (*k* − 1)*λ*), (*k* ≥ 1).

In the special case of *λ* = 1, the falling factorial sequence (also called the Pochammer symbol) is given by

$$(l)\_0 = 1,\\
(l)\_k = l(l-1)\cdots(l-(k-1)),\ \ (k \ge 1). \tag{14}$$

In this paper, we study type 2 degenerate *q*-Bernoulli polynomials and investigate some identities and properties for these polynomials.

### **2. Type 2 Degenerate** *q***-Bernoulli Polynomials**

Throughout this section, we assume that *q* ∈ C*<sup>p</sup>* with |1 − *q*|*<sup>p</sup>* < *p* <sup>−</sup> <sup>1</sup> *<sup>p</sup>*−<sup>1</sup> and *λ* ∈ C*p*. Now, we define the *type 2 degenerate q-Bernoulli polynomials* by

$$\sum\_{n=0}^{\infty} b\_{n,q}(x \mid \lambda) \frac{t^n}{n!} = \frac{1}{2} \int\_{\mathbb{Z}\_p} e^{[x+2y]\_q}\_{\lambda}(t) d\mu\_q(y). \tag{15}$$

By (15), we get

$$\frac{1}{2} \int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + 2\mathbf{y}]\_\theta \right)\_{\mathfrak{n},\lambda} d\mu\_\theta(\mathbf{y}) = b\_{\mathfrak{n},\mathfrak{q}}(\mathbf{x} \mid \lambda), \ (\mathfrak{n} \ge 0). \tag{16}$$

When *x* = 1, *bn*,*q*(*λ*) = *bn*,*q*(1 | *λ*) are called the type 2 degenerate *q*-Bernoulli numbers. We observe here that

$$\begin{aligned} &\lim\_{q\to 1} \lim\_{\lambda \to 0} \frac{1}{2} \int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + 2y]\_q \right)\_{n,\lambda} d\mu\_q(y) \\ &= \lim\_{q\to 1} \lim\_{\lambda \to 0} \frac{1}{2} b\_{\mathbb{N},q}(\mathbf{x} \mid \lambda) = b\_{\mathbb{N}}(\mathbf{x} - 1), \ (\mathbf{n} \ge 0). \end{aligned} \tag{17}$$

The degenerate Stirling numbers of the first kind appear as the coefficients in the expansion

$$\mathbf{y}(\mathbf{x})\_{n,\lambda} = \sum\_{l=0}^{n} S\_{1,\lambda}(n,l)\mathbf{x}^{l}, \ (n \ge 0), \tag{18}$$

(see [12]).

Thus, by (18), we get

$$\begin{aligned} &\frac{1}{2}\int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + 2y]\_q \right)\_{n,\lambda} d\mu\_q(y) \\ &= \frac{1}{2}\sum\_{l=0}^n S\_{1,\lambda}(n,l) \int\_{\mathbb{Z}\_p} [\mathbf{x} + 2y]\_q^l d\mu\_q(y) \\ &= \sum\_{l=0}^n S\_{1,\lambda}(n,l) b\_{l,\emptyset}(\mathbf{x})\_l \end{aligned} \tag{19}$$

where *bl*,*q*(*x*) is the type 2 *q*-Bernoulli polynomials given by <sup>1</sup> 2 " Z*p* [*x* + 2*y*] *n <sup>q</sup> dμq*(*y*) = *bn*,*q*(*x*), (*n* ≥ 0), (see [5]).

Therefore, by (16) and (19), we obtain the following theorem.

**Theorem 1.** *For n* ≥ 0*, we have*

$$b\_{n,q}(\mathbf{x} \mid \boldsymbol{\lambda}) = \sum\_{l=0}^{n} S\_{1,\lambda}(n,l) b\_{l,q}(\mathbf{x}). \tag{20}$$

Now, we observe that

$$\begin{split} &\frac{1}{2}\int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + 2\mathbf{y}]\_q \right)\_{n,\lambda} d\mu\_q(\mathbf{y}) \\ &= \frac{1}{2}\sum\_{l=0}^n S\_{1,\lambda}(n,l) \int\_{\mathbb{Z}\_p} [\mathbf{x} + 2\mathbf{y}]\_q^l d\mu\_q(\mathbf{y}) \\ &= \frac{1}{2}\sum\_{l=0}^n S\_{1,\lambda}(n,l) \frac{1}{(1-q)^l} \sum\_{m=0}^l \binom{l}{m} (-q^x)^m \frac{2m+1}{[2m+1]\_q}. \end{split} \tag{21}$$

Therefore, by (21), we obtain the following theorem.

**Theorem 2.** *For n* ≥ 0*, we have*

$$b\_{n,q}(\mathbf{x} \mid \boldsymbol{\lambda}) = \frac{1}{2} \sum\_{l=0}^{n} \frac{S\_{1,\boldsymbol{\lambda}}(n,l)}{(1-q)^{l}} \sum\_{m=0}^{l} \binom{l}{m} (-q^{x})^{m} \frac{2m+1}{[2m+1]\_{q}}.\tag{22}$$

From (16), we note that

$$\begin{split} \sum\_{n=0}^{\infty} b\_{n,q}(\mathbf{x} \mid \boldsymbol{\lambda}) \frac{t^n}{n!} &= \frac{1}{2} \sum\_{n=0}^{\infty} \int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + 2\mathbf{y}]\_q \right)\_{n,\lambda} d\mu\_q(\mathbf{y}) \frac{t^n}{n!} \\ &= \frac{1}{2} \int\_{\mathbb{Z}\_p} \left( 1 + \lambda t \right)^{\frac{[x+2q]\_q}{\lambda}} d\mu\_q(\mathbf{y}) \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^n \sum\_{m=0}^k \binom{k}{m} q^{mx} [\mathbf{x}]\_q^{k-m} \mathcal{S}\_1(n,k) \lambda^{n-k} b\_{m,q} \right) \frac{t^n}{n!} \end{split} \tag{23}$$

where *S*1(*n*, *k*) are the Stirling numbers of the first kind and *bn*,*<sup>q</sup>* are the type 2 *q*-Bernoulli numbers. Therefore, by (23), we get the following theorem.

**Theorem 3.** *For n* ≥ 0*, we have*

$$b\_{\boldsymbol{n},\boldsymbol{q}}(\boldsymbol{x}\mid\boldsymbol{\lambda}) = \sum\_{k=0}^{n} \sum\_{m=0}^{k} \binom{k}{m} q^{mx} [\boldsymbol{x}]\_{q}^{k-m} \mathbf{S}\_{1}(\boldsymbol{n},\boldsymbol{k}) \boldsymbol{\lambda}^{n-k} b\_{\boldsymbol{m},\boldsymbol{q}}.\tag{24}$$

*In particular,*

$$b\_{n,q}(\lambda) = \sum\_{k=0}^{n} q^{k\chi} S\_1(n,k) \lambda^{n-k} b\_{k,q}.$$

In [4], Kim expressed Carlitz's *q*-Bernoulli polynomials in terms of the following *p*-adic *q*-integrals on Z*p*:

$$\int\_{\mathbb{Z}\_p} [\mathbf{x} + y]\_q^n d\mu\_q(y) = \beta\_{n,q}(\mathbf{x}), \ (n \ge 0). \tag{25}$$

From (9) and (25), we have

$$\begin{split} q^n \beta\_{m,q}(n) &= \int\_{\mathbb{Z}\_p} q^n [\mathbf{x} + n]\_q^m d\mu\_q(\mathbf{x}) \\ &= \int\_{\mathbb{Z}\_p} [\mathbf{x}]\_q^m d\mu\_q(\mathbf{x}) + (q-1) \sum\_{l=0}^{n-1} q^l [l]\_q^m + m \sum\_{l=0}^{n-1} [l]\_q^{m-1} q^{2l} \\ &= \beta\_{m,q} + (q-1) \sum\_{l=0}^{n-1} q^l [l]\_q^m + m \sum\_{l=0}^{n-1} [l]\_q^{m-1} q^{2l} \\ &= \beta\_{m,q} + (m+1) \sum\_{l=0}^{n-1} q^{2l} [l]\_q^{m-1} - \sum\_{l=0}^{n-1} q^l [l]\_q^{m-1}, \end{split} \tag{26}$$

where *n* is a positive integer.

Therefore, we obtain the following theorem.

**Theorem 4.** *For n* ≥ 0*, we have*

$$q^n \beta\_{m,q}(n) - \beta\_{m,q} = (m+1) \sum\_{l=0}^{n-1} q^{2l} [l]\_q^{m-1} - \sum\_{l=0}^{n-1} q^l [l]\_q^{m-1}.\tag{27}$$

$$\begin{split} \text{Let us take } f(\mathbf{x}) = \left( [\mathbf{x}]\_q \right)\_{m,\lambda} \text{ } (m \ge 1). \text{ From (9), we have} \\ \\ q^n \int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + \mathbf{n}]\_q \right)\_{m,\lambda} d\mu\_q(\mathbf{x}) \\ = \int\_{\mathbb{Z}\_p} \left( [\mathbf{x}]\_q \right)\_{m,\lambda} d\mu\_q(\mathbf{x}) + (q-1) \sum\_{l=0}^{n-1} q^l \binom{[l]\_q}{[l]\_q}\_{m,\lambda} \\ + \sum\_{l=0}^{n-1} \left( \sum\_{k=0}^{m-1} \frac{1}{[l]\_q - k\lambda} \right) \left( [l]\_q \right)\_{m,\lambda} q^{2l} . \end{split} \tag{28}$$

In [13], the degenerate *q*-Bernoulli polynomials are defined by Kim as

$$\int\_{\mathbb{Z}\_p} e^{[x+y]\_{\mathfrak{J}}}\_{\lambda}(t) d\mu\_{\mathfrak{q}}(y) = \sum\_{n=0}^{\infty} \beta\_{n,\lambda,\mathfrak{q}}(x) \frac{t^n}{n!}. \tag{29}$$

In particular, the degenerate *q*-Bernoulli numbers are given by *βn*,*λ*,*<sup>q</sup>* = *βn*,*λ*,*q*(0).

From (29), we have

$$\int\_{\mathbb{Z}\_p} \left( [\mathbf{x} + y]\_q \right)\_{n,\lambda} d\mu\_q(y) = \beta\_{n,\lambda,q}(\mathbf{x}), \ (n \ge 0). \tag{30}$$

By (28) and (30), this completes the proof for the next theorem.

**Theorem 5.** *For m*, *n* ∈ N*, we have*

$$\begin{split} &q^n \beta\_{m,\lambda,q}(n) - \beta\_{m,\lambda,q} \\ = &(q-1) \sum\_{l=0}^{n-1} q^l \binom{[l]\_q}{[l]\_q}\_{m,\lambda} + \sum\_{l=0}^{n-1} \left( \sum\_{k=0}^{m-1} \frac{1}{[l]\_q - k\lambda} \right) \binom{[l]\_q}{[l]\_q}\_{m,\lambda} q^{2l}. \end{split} \tag{31}$$

$$\begin{split} \text{Let us take } f(\mathbf{x}) &= \left( [2\mathbf{x} + 1]\_q \right)\_{m, \lambda} \text{ } (m \ge 1). \text{ From (9), we have} \\ \\ q^n \int\_{\mathbb{Z}\_p} \left( [2\mathbf{x} + 2\mathbf{n} + 1]\_q \right)\_{m, \lambda} d\mu\_q(\mathbf{x}) \\ &= \int\_{\mathbb{Z}\_p} \left( [2\mathbf{x} + 1]\_q \right)\_{m, \lambda} d\mu\_q(\mathbf{x}) + (q - 1) \sum\_{l=0}^{n-1} q^l \binom{[2l + 1]\_q}{[2l + 1]\_q}\_{m, \lambda} \\ &+ 2 \sum\_{l=0}^{n-1} \left( \sum\_{k=0}^{m-1} \frac{1}{[2l + 1]\_q - k\lambda} \right) \left( [2l + 1]\_q \right)\_{m, \lambda} q^{3l + 1} . \end{split} \tag{32}$$

From (16) and (32), we have

$$\begin{split} &q^n b\_{m,q}(2n+1\mid\lambda) - b\_{m,q}(\lambda) \\ &= \frac{q-1}{2} \sum\_{l=0}^{n-1} q^l \left( [2l+1]\_q \right)\_{m,\lambda} + \sum\_{l=0}^{n-1} \left( \sum\_{k=0}^{m-1} \frac{1}{[2l+1]\_q - k\lambda} \right) \left( [2l+1]\_q \right)\_{m,\lambda} q^{3l+1} . \end{split} \tag{33}$$

Therefore, by (33), we obtain the following theorem.

**Theorem 6.** *For m*, *n* ∈ N*, we have*

$$\begin{aligned} \frac{q-1}{2} \sum\_{l=0}^{n-1} q^l \left( [2l+1]\_q \right)\_{m,\lambda} + \sum\_{l=0}^{n-1} \left( \sum\_{k=0}^{m-1} \frac{1}{[2l+1]\_q - k\lambda} \right) \left( [2l+1]\_q \right)\_{m,\lambda} q^{3l+1} \\ = q^n b\_{m,q} (2n+1 \mid \lambda) - b\_{m,q}(\lambda) . \end{aligned} \tag{34}$$

From (7), we can derive the following integral equation:

$$\begin{split} \int\_{\mathbb{Z}\_T} f(x) d\mu\_q(x) &= \lim\_{N \to \infty} \frac{1}{[p^N]\_q} \sum\_{x=0}^{p^N - 1} f(x) q^x \\ &= \lim\_{N \to \infty} \frac{1}{[dp^N]\_q} \sum\_{x=0}^{dp^N - 1} f(x) q^x \\ &= \lim\_{N \to \infty} \frac{1}{[dp^N]\_q} \sum\_{a=0}^{d-1} \sum\_{x=0}^{p^N - 1} f(a+dx) q^{a+dx} \\ &= \sum\_{a=0}^{d-1} q^a \frac{1}{[d]\_q} \lim\_{N \to \infty} \frac{1}{[p^N]\_{q^d}} \sum\_{x=0}^{p^N - 1} f(a+dx) q^{dx} \\ &= \frac{1}{[d]\_q} \sum\_{a=0}^{d-1} q^a \int\_{\mathbb{Z}\_T} f(a+dx) d\mu\_{q^d}(x), \end{split} \tag{35}$$

where *d* is a positive integer.

**Lemma 1.** *For d* ∈ N*, we have*

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x})d\mu\_q(\mathbf{x}) = \frac{1}{[d]\_q} \sum\_{a=0}^{d-1} q^a \int\_{\mathbb{Z}\_p} f(a+dx)d\mu\_{q^d}(\mathbf{x}).\tag{36}$$

We obtain the following theorem from Lemma 1.

**Theorem 7.** *For n*, *d* ∈ N*, we have*

$$b\_{n,q}(\lambda) = [d]\_q^{n-1} \sum\_{a=0}^{d-1} q^a b\_{n,q^d} \left( \frac{2a+1}{d} \mid \frac{\lambda}{[d]\_q} \right). \tag{37}$$

**Proof.** Let us apply Lemma 1 with *<sup>f</sup>*(*x*) = -[2*x* + 1]*<sup>q</sup> n*,*λ* , (*n* ∈ N). Then, by virtue of (16), we have

$$\begin{split} &\int\_{\mathbb{Z}\_p} \left( [2\mathbf{x} + \mathbf{1}]\_q \right)\_{n,\lambda} d\mu\_q(\mathbf{x}) = \frac{1}{[d]\_q} \sum\_{a=0}^{d-1} q^a \int\_{\mathbb{Z}\_p} \left( [2(a + dx) + \mathbf{1}]\_q \right)\_{n,\lambda} d\mu\_{q^d}(\mathbf{x}) \\ &= \frac{1}{[d]\_q} \sum\_{a=0}^{d-1} q^a [d]\_q^n \int\_{\mathbb{Z}\_p} \left( \left[ \frac{2a + 1}{d} + 2\mathbf{x} \right]\_{q^d} \right)\_{n,\frac{\lambda}{[d]\_q}} d\mu\_{q^d}(\mathbf{x}) \\ &= [d]\_q^{n-1} \sum\_{a=0}^{d-1} q^a \int\_{\mathbb{Z}\_p} \left( \left[ \frac{2a + 1}{d} + 2\mathbf{x} \right]\_{q^d} \right)\_{n,\frac{\lambda}{[d]\_q}} d\mu\_{q^d}(\mathbf{x}) \\ &= 2[d]\_q^{n-1} \sum\_{a=0}^{d-1} q^a b\_{n,q^d} \left( \frac{2a + 1}{d} \mid \frac{\lambda}{[d]\_q} \right). \end{split}$$

### **3. Further Remarks**

Assume that *X*1, *X*2, *X*3, ··· are independent random variables, each of which has the Laplace distribution with parameters 0 and 1. Namely, each of them has the probability density function given by <sup>1</sup> <sup>2</sup> exp(−|*x*|).

Let *Z* be the random variable given by *Z* = ∑<sup>∞</sup> *k*=1 *Xk* <sup>2</sup>*k<sup>π</sup>* . In addition, let *bn* be the type 2 Bernoulli numbers defined by

$$\frac{t}{e^t - e^{-t}} = \sum\_{n=0}^{\infty} b\_n \frac{t^n}{n!} \,. \tag{38}$$

Then, it was shown in [1] that

$$\begin{split} \sum\_{n=0}^{\infty} E[Z^n] \frac{(it)^n}{n!} &= \frac{t}{e^{\frac{t}{2}} - e^{-\frac{t}{2}}} \\ &= \sum\_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n-1} b\_n \frac{t^n}{n!} . \end{split} \tag{39}$$

Thereby, it was obtained that

$$i^n E[Z^n] = \left(\frac{1}{2}\right)^{n-1} b\_n. \tag{40}$$

Before proceeding further, we recall that the Volkenborn integral (also called the *p*-adic invariant integral) for a uniformly differentiable function *f* on Z*<sup>p</sup>* is given by

$$\int\_{\mathbb{Z}\_p} f(y)d\mu(y) = \lim\_{N \to \infty} \frac{1}{p^N} \sum\_{y=0}^{p^N - 1} f(y). \tag{41}$$

Then, it is well known (see [14]) that this integral satisfies the following integral equation:

$$\int\_{\mathbb{Z}\_p} f(y+1)d\mu(y) = \int\_{\mathbb{Z}\_p} f(y)d\mu(y) + f'(0). \tag{42}$$

When *q* = 1 and *x* = 1, by virtue of (41), (15) becomes

$$\begin{split} \sum\_{n=0}^{\infty} b\_n(\lambda) \frac{t^n}{n!} &= \frac{1}{2} \int\_{\mathbb{Z}\_p} \mathfrak{e}\_{\lambda}^{1+2y}(t) d\mu(y) \\ &= \frac{\frac{1}{\lambda} \log(1+\lambda t)}{\mathfrak{e}\_{\lambda}(t) - \mathfrak{e}\_{\lambda}^{-1}(t)}. \end{split} \tag{43}$$

Here, *bn*(*λ*) may be called the fully degenerate type 2 Bernoulli numbers, even though they were defined slightly differently in [3]. Replacing *t* with <sup>2</sup> *<sup>λ</sup>* log(1 + *λt*) in (39) and by making use of (43), we have

$$\begin{split} \sum\_{m=0}^{\infty} & E[Z^{m}] (2i)^{m} \frac{1}{m!} \left( \frac{\log(1+\lambda t)}{\lambda} \right)^{m} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^{n} S\_{1,\lambda}(n,m) (2i)^{m} E[Z^{m}] \right) \frac{t^{n}}{n!} \\ &= 2 \sum\_{n=0}^{\infty} b\_{n}(\lambda) \frac{t^{n}}{n!} . \end{split} \tag{44}$$

Here, *S*1,*λ*(*n*, *k*) are the degenerate Stirling numbers of the first kind (see [12]) either given by

$$\frac{1}{m} \left( \frac{\log(1 + \lambda t)}{\lambda} \right)^m = \sum\_{n=m}^{\infty} S\_{1,\lambda}(n, m) \frac{t^n}{n!} \tag{45}$$

or given by

$$\mathbf{x}(\mathbf{x})\_{\mathbf{n},\lambda} = \sum\_{m=0}^{n} \mathbf{S}\_{1,\lambda}(n,m)\mathbf{x}^{\mathbf{m}} = \sum\_{m=0}^{n} \mathbf{S}\_{1}(n,m)\lambda^{n-m}\mathbf{x}^{\mathbf{m}}.\tag{46}$$

Thus, by (44), we have shown that

$$2b\_{\mathfrak{n}}(\lambda) = \sum\_{m=0}^{n} \mathcal{S}\_{1,\lambda}(n,m) \left(2i\right)^{m} \mathcal{E}[Z^{m}].$$

Here, we remark that we only considered *q* = 1 and *x* = 1 cases of (15), namely the fully degenerate type 2 Bernoulli numbers. This is because we do not see how to express type 2 degenerate *q*-Bernoulli polynomials or type 2 degenerate *q*-Bernoulli numbers in terms of the moments of some suitable random variables, constructed from random variables with Laplace distributions. We leave this as an open problem to the interested reader.

### **4. Conclusions**

Studies on various special polynomials and numbers have been preformed using several different methods, such as generating functions, combinatorial methods, umbral calculus techniques, matrix theory, probability theory, *p*-adic analysis, differential equations, and so on.

One way of introducing new special polynomials and numbers is to study various degenerate versions of some known special polynomials and numbers, which began with Carlitz's paper in [2]. Actually, degenerate versions were investigated not only for some polynomials but also for a transcendental function, namely the gamma function. For this, we refer the reader to [11]. Another way of introducing new special polynomials and numbers is to study various *q*-analogues of some known special polynomials and numbers. The bosonic *p*-adic *q*-integrals, together with the fermionic *p*-adic *q*-integrals, turned out to be very powerful and fruitful tools for naturally constructing such *q*-analogues. They were introduced by Kim in [4] and have been widely used ever since their invention.

In this paper, the type 2 degenerate *q*-Bernoulli polynomials and the corresponding numbers were introduced and investigated as a degenerate version of and also as a *q*-analogue of type 2 Bernoulli polynomials by making use of the bosonic *p*-adic *q*-integrals [1,3,5]. Here, as an introductory paper on the subject, only very basic results were obtained. The obtained results are several expressions for those polynomials, identities involving those numbers, identities regarding Carlitz's *q*-Bernoulli numbers, identities concerning degenerate *q*-Bernoulli numbers, and the representations of the fully degenerate type 2 Bernoulli numbers (*q* = 1 and *x* = 1 cases of the type 2 degenerate *q*-Bernoulli polynomials) in terms of moments of certain random variables, created from random variables with Laplace distributions. We are planning to study more detailed results relating to these polynomials and numbers in a forthcoming paper.

**Author Contributions:** All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

**Funding:** This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No. 2017R1E1A1A03070882).

**Acknowledgments:** The authors would like to thank the referees for their valuable comments and suggestions. **Conflicts of Interest:** The authors declare no conflict of interest.

### **References**


© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **Some Identities of Ordinary and Degenerate Bernoulli Numbers and Polynomials**

**Dmitry V. Dolgy 1, Dae San Kim 2, Jongkyum Kwon 3,\* and Taekyun Kim <sup>4</sup>**


Received: 28 May 2019; Accepted: 26 June 2019; Published: 1 July 2019

**Abstract:** In this paper, we investigate some identities on Bernoulli numbers and polynomials and those on degenerate Bernoulli numbers and polynomials arising from certain *p*-adic invariant integrals on Z*p*. In particular, we derive various expressions for the polynomials associated with integer power sums, called integer power sum polynomials and also for their degenerate versions. Further, we compute the expectations of an infinite family of random variables which involve the degenerate Stirling polynomials of the second and some value of higher-order Bernoulli polynomials.

**Keywords:** Bernoulli polynomials; degenerate Bernoulli polynomials; random variables; *p*-adic invariant integral on Z*p*; integer power sums polynomials; Stirling polynomials of the second kind; degenerate Stirling polynomials of the second kind

### **1. Introduction**

We begin this section by reviewing some known facts. In more detail, we recall the integral equation for the *p*-adic invariant integral of a uniformly differentiable function on Z*<sup>p</sup>* and its generalizations, the expression in terms of some values of Bernoulli polynomials for the integer power sums, and the *p*-adic integral representaions of Bernoulli polynomials and of their generating functions.

Throughout this paper, Z*p*, Q*<sup>p</sup>* and C*<sup>p</sup>* will denote the ring of *p*-adic integers, the field of *p*-adic rational numbers and the completion of the algebraic closure of Q*p*, respectively. The *p*-adic norm is normalized as <sup>|</sup>*p*|*<sup>p</sup>* <sup>=</sup> <sup>1</sup> *<sup>p</sup>* . Let *f* be a uniformly differentiable function on Z*p*. Then the *p*-adic invariant integral of *f* (also called the Volkenborn integral of *f*) on Z*<sup>p</sup>* is defined by

$$\begin{split} I\_0(f) &= \int\_{\mathbb{Z}\_p} f(\mathbf{x}) d\mu\_0(\mathbf{x}) = \lim\_{N \to \infty} \frac{1}{p^N} \sum\_{\mathbf{x}=0}^{p^N - 1} f(\mathbf{x}) \\ &= \lim\_{N \to \infty} \sum\_{\mathbf{x}=0}^{p^N - 1} f(\mathbf{x}) \mu\_0(\mathbf{x} + p^N \mathbb{Z}\_p). \end{split} \tag{1}$$

Here we note that *μ*0(*x* + *pN*Z*p*) = <sup>1</sup> *<sup>p</sup><sup>N</sup>* is a distribution but not a measure. The existence of such integrals for uniformly differentiable functions on Z*<sup>p</sup>* is detailed in [1,2]. It can be seen from (1) that

$$I\_0(f\_1) = I\_0(f) + f'(0),\tag{2}$$

where *f*1(*x*) = *f*(*x* + 1), and *f* (0) = *d f*(*x*) *dx* |*x*=0, (see [1,2]). In general, by induction and with *fn*(*x*) = *f*(*x* + *n*), we can show that

$$I\_0(f\_{\mathbb{N}}) = I\_0(f) + \sum\_{k=0}^{n-1} f'(k), \ (n \in \mathbb{N}), \tag{3}$$

As is well known, the Bernoulli polynomials are given by the generating function (see [3–5])

$$\frac{t}{e^t - 1} e^{\mathbf{x}t} = \sum\_{n=0}^{\infty} B\_n(\mathbf{x}) \frac{t^n}{n!} \,\tag{4}$$

When *x* = 0, *Bn* = *Bn*(0) are called the Bernoulli numbers. From (4), we note that (see [3–5])

$$B\_{\rm tr}(\mathbf{x}) = \sum\_{l=0}^{n} \binom{n}{l} B\_{l} \mathbf{x}^{n-l}, \quad (n \ge 0), \tag{5}$$

and

$$B\_0 = 1, \quad \sum\_{k=0}^n \binom{n}{k} B\_k - B\_n = \begin{cases} 1, & \text{if } n = 1, \\\ 0, & \text{if } n > 1, \end{cases}$$

Let (see [6–13])

$$S\_p(n) = \sum\_{k=1}^n k^p, \ (n, p \in \mathbb{N}).\tag{6}$$

The generating function of *Sp*(*n*) is given by

$$\begin{split} \sum\_{p=0}^{\infty} S\_p(n) \frac{t^p}{p!} &= \sum\_{k=1}^n e^{kt} = \frac{1}{t} \left( \frac{t}{e^t - 1} \left( e^{(n+1)t} - e^t \right) \right) \\ &= \sum\_{p=0}^{\infty} \left( \frac{B\_{p+1}(n+1) - B\_{p+1}(1)}{p+1} \right) \frac{t^p}{p!} . \end{split} \tag{7}$$

Thus, by (7), we get

$$S\_p(n) = \frac{B\_{p+1}(n+1) - B\_{p+1}(1)}{p+1}, \ (n, p \in \mathbb{N}).\tag{8}$$

From (2), we have

$$\int\_{\mathbb{Z}\_p} e^{(x+y)t} d\mu\_0(y) = \frac{t}{e^t - 1} e^{xt} = \sum\_{n=0}^{\infty} B\_n(x) \frac{t^n}{n!}.\tag{9}$$

By (9), we get (see [11,12])

$$\int\_{\mathbb{Z}\_p} (\mathbf{x} + \mathbf{y})^n d\mu\_0(\mathbf{y}) = B\_n(\mathbf{x}), \ (n \ge 0), \tag{10}$$

From (8) and (10), we can derive the following equation.

$$\int\_{\mathbb{Z}\_p} (\mathbf{x} + k + 1)^{p+1} d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} \mathbf{x}^{p+1} d\mu\_0(\mathbf{x}) = (p+1) \sum\_{n=1}^k n^p, \ (p \in \mathbb{N}).\tag{11}$$

Thus, by (6) and (11), and for *p* ∈ N, we get

$$S\_{\mathcal{P}}(k) = \frac{1}{p+1} \left\{ \int\_{\mathbb{Z}\_p} (\mathbf{x} + k + 1)^{p+1} d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} \mathbf{x}^{p+1} d\mu\_0(\mathbf{x}) \right\}. \tag{12}$$

The purpose of this paper is to investigate some identities on Bernoulli numbers and polynomials and those on degenerate Bernoulli numbers and polynomials arising from certain *p*-adic invariant integrals on Z*p*.

The outline of this paper is as in the following. After reviewing well- known necessary results in Section 1, we will derive some identities on Bernoulli polynomials and numbers in Section 2. In particular, we will introduce the integer power sum polynomials and derive several expressions for them. In Section 3, we will obtain some identities on degenerate Bernoulli numbers and polynomials. Especially, we will introduce the degenerate integer power sum polynomials, a degenerate version of the integer power sum polynomials and deduce various representations of them. In the final Section 4, we will consider an infinite family of random variables and compute their expectations to see that they involve the degenerate Stirling polynomials of the second and some value of higher-order Bernoulli polynomials.

### **2. Some Identities of Bernoulli Numbers and Polynomials**

For *p* ∈ N, we observe that

$$\begin{split} (j+1)^{p+1} - j^{p+1} &= \sum\_{i=0}^{p+1} \binom{p+1}{i} j^i - j^{p+1} \\ &= (p+1)j^p + \sum\_{i=1}^{p-1} \binom{p+1}{i} j^i + 1. \end{split} \tag{13}$$

Thus, we get

$$\mathbb{P}(n+1)^{p+1} = \sum\_{j=0}^{n} \left\{ (j+1)^{p+1} - j^{p+1} \right\} = (p+1) \sum\_{j=0}^{n} j^p + \sum\_{i=1}^{p-1} \binom{p+1}{i} \sum\_{j=0}^{n} j^i + (n+1). \tag{14}$$

From (14), we have

$$S\_p(n) = \frac{1}{p+1} \left\{ (n+1)^{p+1} - (n+1) - \sum\_{i=1}^{p-1} \binom{p+1}{i} S\_i(n) \right\}.\tag{15}$$

Therefore, by (15), we obtain the following lemma.

**Lemma 1.** *For n*, *p* ∈ N*, we have*

$$\int\_{\mathbb{Z}\_p} (\mathbf{x} + n + 1)^{p+1} d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} \mathbf{x}^{p+1} d\mu\_0(\mathbf{x})$$

$$= (n+1)^{p+1} - (n+1) - \sum\_{i=1}^{p-1} \binom{p+1}{i} \frac{1}{i+1} \tag{16}$$

$$\times \left\{ \int\_{\mathbb{Z}\_p} (\mathbf{x} + n + 1)^{i+1} d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} \mathbf{x}^{i+1} d\mu\_0(\mathbf{x}) \right\}.$$

From Lemma 1, we note the following.

**Corollary 1.** *For n*, *p* ∈ N*, we have*

$$B\_{p+1}(n+1) - B\_{p+1} = (n+1)^{p+1} - (n+1) - \sum\_{i=1}^{p-1} \binom{p+1}{i} \frac{1}{i+1} \left( B\_{i+1}(n+1) - B\_{i+1} \right). \tag{17}$$

For *n* ∈ N<sup>0</sup> = N ∪ {0}, by (1), we get

$$\int\_{\mathbb{Z}\_p} \left( y + 1 - \mathbf{x} \right)^n d\mu\_0(y) = (-1)^n \int\_{\mathbb{Z}\_p} (y + \mathbf{x})^n d\mu\_0(y). \tag{18}$$

From (18), we note that

$$B\_n(1-\mathbf{x}) = (-1)^n B\_n(\mathbf{x}), \ (n \ge 0). \tag{19}$$

Now, we observe that, for *n* ≥ 1,

$$\begin{split} B\_n(2) &= \sum\_{l=0}^n \binom{n}{l} B\_l(1) = B\_0 + \binom{n}{1} B\_1(1) + \sum\_{l=2}^n \binom{n}{l} B\_l(1) \\ &= B\_0 + \binom{n}{1} B\_1 + n + \sum\_{l=2}^n \binom{n}{l} B\_l = n + \sum\_{l=0}^n \binom{n}{l} B\_l \\ &= n + B\_n(1) . \end{split} \tag{20}$$

Thus we have completed the proof for the next lemma.

**Lemma 2.** *For any n* ∈ N0*, the following identity is valid:*

$$B\_n(2) = n + B\_n + \delta\_{n,1\prime} \tag{21}$$

where *δn*,1 is the Kronecker's delta. For any *n*, *m* ∈ N with *n*, *m* ≥ 2, we have

$$\begin{split} \int\_{\mathbb{Z}\_p} \mathbf{x}^m (-1 + \mathbf{x})^n d\mu\_0(\mathbf{x}) &= \sum\_{i=0}^n \binom{n}{i} (-1)^{n-i} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m+i} d\mu\_0(\mathbf{x}) \\ &= \sum\_{i=0}^n \binom{n}{i} (-1)^{n-i} B\_{m+i} \\ &= (-1)^{n-m} \sum\_{i=0}^n \binom{n}{i} B\_{m+i} .\end{split} \tag{22}$$

On the other hand,

$$\begin{split} \int\_{\mathbb{Z}\_p} \mathbf{x}^m (\mathbf{x} - \mathbf{1})^n d\mu\_0(\mathbf{x}) &= \sum\_{i=0}^m \binom{m}{i} \int\_{\mathbb{Z}\_p} (\mathbf{x} - \mathbf{1})^{n+i} d\mu\_0(\mathbf{x}) \\ &= \sum\_{i=0}^m \binom{m}{i} (-1)^{n+i} \int\_{\mathbb{Z}\_p} (\mathbf{x} + \mathbf{2})^{n+i} d\mu\_0(\mathbf{x}) \\ &= \sum\_{i=0}^m \binom{m}{i} (-1)^{n+i} \binom{B\_{n+i} + n + i}{} \\ &= \sum\_{i=0}^m \binom{m}{i} (-1)^{n+i} B\_{n+i} \\ &= \sum\_{i=0}^m \binom{m}{i} B\_{n+i} .\end{split} \tag{23}$$

Therefore, by (22) and (23), we obtain the following theorem.

**Theorem 1.** *For any m*, *n* ∈ N *with m*, *n* ≥ 2*, the following symmetric identity holds:*

$$(-1)^{n} \sum\_{i=0}^{n} \binom{n}{i} B\_{m+i} = (-1)^{m} \sum\_{i=0}^{m} \binom{m}{i} B\_{n+i}.\tag{24}$$

From (5), we note that

$$B\_n(1) = \sum\_{l=0}^n \binom{n}{l} B\_{l\prime} \quad (n \ge 0).$$

For *n* ≥ 2, we have

$$B\_{ll} = B\_{ll}(1) = \sum\_{l=0}^{n} \binom{n}{l} B\_{l} = \sum\_{l=0}^{n} \binom{n}{l} B\_{n-l}.\tag{25}$$

Now, we define the *integer power sum polynomials* by

$$S\_P(n|\mathbf{x}) = \sum\_{k=0}^n (k+\mathbf{x})^p, \ (n, p \in \mathbb{N}\_0). \tag{26}$$

Note that *Sp*(*n*|0) = *Sp*(*n*), (*n* ∈ N0, *p* ∈ N). For *N* ∈ N0, we have

$$t\sum\_{k=0}^{N} e^{(k+x)t} = \int\_{\mathbb{Z}\_p} e^{(N+1+x+y)t} d\mu\_0(y) - \int\_{\mathbb{Z}\_p} e^{(x+y)t} d\mu\_0(y). \tag{27}$$

Then it is immediate to see from (27) that we have

$$\sum\_{k=0}^{N} \varepsilon^{(k+\mathbf{x})t} = \sum\_{n=0}^{\infty} \frac{1}{n+1} \left\{ \int\_{\mathbb{Z}\_p} (N+1+\mathbf{x}+\mathbf{y})^{n+1} d\mu\_0(\mathbf{y}) - \int\_{\mathbb{Z}\_p} (\mathbf{x}+\mathbf{y})^{n+1} d\mu\_0(\mathbf{y}) \right\} \frac{t^n}{n!}.\tag{28}$$

Now, we see that (28) is equivalent to the next theorem.

**Theorem 2.** *For n*, *N* ∈ N0*, we have*

$$S\_n(N|\mathbf{x}) = \frac{1}{n+1} \left\{ B\_{n+1}(\mathbf{x} + N + 1) - B\_{n+1}(\mathbf{x}) \right\}.\tag{29}$$

Let denote the difference operator given by

$$
\triangle f(\mathbf{x}) = f(\mathbf{x} + \mathbf{1}) - f(\mathbf{x}).\tag{30}
$$

Then, by (30) and induction, we get

$$
\triangle^n f(\mathbf{x}) = \sum\_{k=0}^n \binom{n}{k} (-1)^{n-k} f(\mathbf{x} + k) \,\,\, (n \ge 0). \tag{31}
$$

Now, we can deduce the Equation (32) from (27) as in the following:

$$\begin{split} \sum\_{k=0}^{N} \epsilon^{(k+x)t} &= \frac{1}{t} \epsilon^{xt} (\epsilon^{(N+1)t} - 1) \int\_{\mathbb{Z}p} \epsilon^{yt} d\mu\_0(y) \\ &= \frac{1}{\epsilon^t - 1} \left( \sum\_{m=0}^{N+1} \binom{N+1}{m} (\epsilon^t - 1)^m - 1 \right) \epsilon^{xt} \\ &= \frac{1}{\epsilon^t - 1} \sum\_{m=0}^{N+1} \binom{N+1}{m} (\epsilon^t - 1)^m \epsilon^{xt} \\ &= \sum\_{m=0}^{N} \left( \sum\_{m=0}^{N+1} \binom{N+1}{m+1} \sum\_{k=0}^m \binom{m}{k} (-1)^{m-k} (k+x)^n \right) \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \left\{ \sum\_{k=0}^{N} \sum\_{m=k}^{N} \binom{N+1}{m+1} \binom{m}{k} (-1)^{m-k} (k+x)^n \right\} \frac{t^n}{n!} . \end{split} \tag{32}$$

Therefore, (31) and (32) together yield the next theorem.

**Theorem 3.** *For n*, *N* ≥ 0*, we have*

$$S\_{\mathbb{H}}(N|\mathbf{x}) = \sum\_{m=0}^{N} \binom{N+1}{m+1} \triangle^{m} \mathbf{x}^{\mathbf{n}} = \sum\_{k=0}^{N} (k+\mathbf{x})^{\mathbf{n}} T(N,k),\tag{33}$$

*where T*(*N*, *k*) = ∑*<sup>N</sup> <sup>m</sup>*=*<sup>k</sup>* ( *N*+1 *m*+1)(*<sup>m</sup> <sup>k</sup>* )(−1)*m*−*k*.

*In particular, we have*

$$S\_0(N|x) = \sum\_{k=0}^{N} T(N,k) = N+1.$$

We recall here that the Stirling polynomials of the second kind *S*2(*n*, *k*|*x*) are given by (see [14])

$$\frac{1}{k!} (e^t - 1)^k e^{xt} = \sum\_{n=k}^{\infty} S\_2(n, k|x) \frac{t^n}{n!}. \tag{34}$$

Note here that *S*2(*n*, *k*|0) = *S*2(*n*, *k*) are Stirling numbers of the second kind. Then, we can show that, for integers *n*, *m* ≥ 0, we have

$$\frac{1}{m!} \triangle^m \mathbf{x}^n = \begin{cases} S\_2(n, m|\mathbf{x}), & \text{if } n \ge m, \\\ 0, & \text{if } n < m. \end{cases} \tag{35}$$

We can see this, for example, by taking *λ* → 0 in (51).

**Remark 1.** *Combing* (33) *and* (35)*, we obtain*

$$S\_{\Pi}(N|\mathbf{x}) = \sum\_{m=0}^{\min\{N,n\}} \binom{N+1}{m+1} m! S\_2(n,m|\mathbf{x}).$$

For any *m*, *k* ∈ N with *m* − *k* ≥ 2, we observe that

$$\begin{split} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m-k} d\mu\_0(\mathbf{x}) &= \int\_{\mathbb{Z}\_p} (\mathbf{x}+1)^{m-k} d\mu\_0(\mathbf{x}) \\ &= \sum\_{j=0}^{m-k} \binom{m-k}{m-k-j} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m-k-j} d\mu\_0(\mathbf{x}) \\ &= \sum\_{j=k}^m \binom{m-k}{m-j} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m-j} d\mu\_0(\mathbf{x}) \\ &= \frac{1}{\binom{m}{k}} \sum\_{j=k}^m \binom{m}{j} \binom{j}{k} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m-j} d\mu\_0(\mathbf{x}). \end{split} \tag{36}$$

Thus we have shown the following result.

**Theorem 4.** *For any m*, *k* ∈ N *with m* − *k* ≥ 2*, the following holds true:*

$$
\binom{m}{k} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m-k} d\mu\_0(\mathbf{x}) = \sum\_{j=k}^m \binom{m}{j} \binom{j}{k} \int\_{\mathbb{Z}\_p} \mathbf{x}^{m-j} d\mu\_0(\mathbf{x}).\tag{37}
$$

From (10) and (37), we derive the following corollary.

**Corollary 2.** *For m*, *k* ∈ N *with m* − *k* ≥ 2*, we have*

$$
\binom{m}{k} B\_{m-k} = \sum\_{j=k}^{m} \binom{m}{j} \binom{j}{k} B\_{m-j}.\tag{38}
$$

### **3. Some Identities of Degenerate Bernoulli Numbers and Polynomials**

In this section, we assume that 0 = *λ* ∈ C*<sup>p</sup>* with |*λ*|*<sup>p</sup>* < *p* <sup>−</sup> <sup>1</sup> *<sup>p</sup>*−<sup>1</sup> . The degenerate exponential function is defined as (see [3,13]) *x*

$$e^x\_\lambda(t) = (1 + \lambda t)^{\frac{\lambda}{\lambda}} \text{ }$$

Note that lim*λ*→<sup>0</sup> *<sup>e</sup><sup>x</sup> <sup>λ</sup>*(*t*) = *<sup>e</sup>xt*. In addition, we denote (<sup>1</sup> + *<sup>λ</sup>t*) 1 *<sup>λ</sup>* = *e*<sup>1</sup> *<sup>λ</sup>*(*t*) simply by *eλ*(*t*). As is well known, the degenerate Bernoulli polynomials are defined by Carlitz as

$$\frac{t}{\varepsilon\_{\lambda}(t) - 1} e\_{\lambda}^{x}(t) = \frac{t}{(1 + \lambda t)^{\frac{1}{\lambda}} - 1} (1 + \lambda t)^{\frac{x}{\lambda}} = \sum\_{n=0}^{\infty} \beta\_{n,\lambda}(x) \frac{t^{n}}{n!}. \tag{39}$$

When *x* = 0, *βn*,*<sup>λ</sup>* = *βn*,*λ*(0) are called the degenerate Bernoulli numbers, (see [3,15]). From (39), we note that (see [3])

$$\beta\_{n,\lambda}(\mathbf{x}) = \sum\_{l=0}^{n} \binom{n}{l} (\mathbf{x})\_{n-l,\lambda} \beta\_{l,\lambda'} \tag{40}$$

where (*x*)0,*<sup>λ</sup>* = 1, (*x*)*n*,*<sup>λ</sup>* = *x*(*x* − *λ*)···(*x* − (*n* − 1)*λ*), (*n* ≥ 1).

$$\text{By (39) and (40), we get}$$

$$
\beta\_{n,\lambda}(1) - \beta\_{n,\lambda} = \delta\_{n,1} \,. \tag{41}
$$

Now, we observe that

$$\begin{split} \sum\_{k=0}^{N} \epsilon\_{\lambda}^{k+1}(t) &= \frac{\epsilon\_{\lambda}^{N+1}(t) - 1}{\epsilon\_{\lambda}(t) - 1} e\_{\lambda}^{x}(t) = \frac{1}{t} \left\{ \frac{t}{\epsilon\_{\lambda}(t) - 1} \left( e\_{\lambda}^{N+1+x}(t) - e\_{\lambda}^{x}(t) \right) \right\} \\ &= \frac{1}{t} \sum\_{n=0}^{\infty} \left( \beta\_{n,\lambda}(N+1+x) - \beta\_{n,\lambda}(x) \right) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \frac{\beta\_{n+1,\lambda}(N+1+x) - \beta\_{n+1,\lambda}(x)}{n+1} \right) \frac{t^{n}}{n!} \quad (n \in \mathbb{N}\_{0}). \end{split} \tag{42}$$

On the other hand,

$$\sum\_{k=0}^{N} \varepsilon\_{\lambda}^{k+\chi}(t) = \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{N} (k+\chi)\_{n,\lambda} \right) \frac{t^n}{n!}. \tag{43}$$

Let us define a degenerate version of the integer power sum polynomials, called the *degenerate integer power sum polynomials*, by

$$S\_{p,\lambda}(n|x) = \sum\_{k=0}^{n} (k+x)\_{p,\lambda\prime} \ (n \ge 0). \tag{44}$$

Note that lim*λ*→<sup>0</sup> *Sp*,*λ*(*n*|*x*) = *Sp*(*n*|*x*), (*n* ≥ 0). Therefore, by (42) and (43), we obtain the following theorem.

**Theorem 5.** *For n*, *N* ∈ N0*, we have*

$$S\_{n,\lambda}(N|\mathbf{x}) = \frac{1}{n+1} \left(\beta\_{n+1,\lambda}(N+1+\mathbf{x}) - \beta\_{n+1,\lambda}(\mathbf{x})\right). \tag{45}$$

Now, we observe that

$$\begin{split} \sum\_{k=0}^{N} \varepsilon\_{\lambda}^{x+k}(t) &= \frac{1}{\varepsilon\_{\lambda}(t) - 1} \left( \varepsilon\_{\lambda}^{N+1}(t) - 1 \right) \varepsilon\_{\lambda}^{x}(t) \\ &= \frac{1}{\varepsilon\_{\lambda}(t) - 1} \left( (\varepsilon\_{\lambda}(t) - 1 + 1)^{N+1} - 1 \right) \varepsilon\_{\lambda}^{x}(t) \\ &= \frac{1}{\varepsilon\_{\lambda}(t) - 1} \sum\_{m=0}^{N+1} \binom{N+1}{m} (\varepsilon\_{\lambda}(t) - 1)^{m} \varepsilon\_{\lambda}^{x}(t) \\ &= \sum\_{m=0}^{N} \binom{N+1}{m+1} (\varepsilon\_{\lambda}(t) - 1)^{m} \varepsilon\_{\lambda}^{x}(t) \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^{N} \binom{N+1}{m+1} \sum\_{k=0}^{m} \binom{m}{k} (-1)^{m-k} (k + x)\_{n, \lambda} \right) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{N} \sum\_{m=k}^{N} \binom{N+1}{m+1} \binom{m}{k} (-1)^{m-k} (k + x)\_{n, \lambda} \right) \frac{t^{n}}{n!}. \end{split} \tag{46}$$

Therefore, (31) and (46) together give the next result.

**Theorem 6.** *For any n*, *N* ∈ N0*, the following identity holds:*

$$S\_{\boldsymbol{n},\boldsymbol{\lambda}}(\boldsymbol{N}|\boldsymbol{x}) = \sum\_{m=0}^{N} \binom{N+1}{m+1} \boldsymbol{\triangle}^{m}(\boldsymbol{x})\_{\boldsymbol{n},\boldsymbol{\lambda}} = \sum\_{k=0}^{N} (k+\mathbf{x})\_{\boldsymbol{n},\boldsymbol{\lambda}} T(\boldsymbol{N},k),\tag{47}$$

*where T*(*N*, *k*) = ∑*<sup>N</sup> <sup>m</sup>*=*<sup>k</sup>* ( *N*+1 *m*+1)(*<sup>m</sup> <sup>k</sup>* )(−1)*m*−*k*.

As is known, the degenerate Stirling polynomials of the second kind are defined by Kim as (see [14])

$$(x+y)\_{n,\lambda} = \sum\_{k=0}^{n} S\_{2,\lambda}(n,k|x)(y)\_{k\prime} \tag{48}$$

where (*x*)<sup>0</sup> = 1,(*x*)*<sup>n</sup>* = *x*(*x* − 1)···(*x* − *n* + 1), (*n* ≥ 1).

From (48), we can derive the generating function for *S*2,*λ*(*n*, *k*|*x*), (*n*, *k* ≥ 0), as follows:

$$\frac{1}{k!} (e\_\lambda(t) - 1)^k e\_\lambda^x(t) = \sum\_{n=k}^\infty S\_{2,\lambda}(n, k|x) \frac{t^n}{n!}.\tag{49}$$

When *x* = 0, *S*2,*λ*(*n*, *k*|0) = *S*2,*λ*(*n*, *k*) are called the degenerate Stirling numbers of the second kind.

By (49), we get

$$\begin{split} \sum\_{n=m}^{\infty} \mathcal{S}\_{2,\lambda}(n,m|\mathbf{x}) \frac{t^n}{n!} &= \frac{1}{m!} (\boldsymbol{\varepsilon}\_{\lambda}(t) - 1)^m \boldsymbol{\varepsilon}\_{\lambda}^{\boldsymbol{x}}(t) \\ &= \frac{1}{m!} \sum\_{k=0}^m \binom{m}{k} (-1)^{m-k} \boldsymbol{\varepsilon}\_{\lambda}^{k+\boldsymbol{x}}(t) \\ &= \sum\_{n=0}^{\infty} \left( \frac{1}{m!} \sum\_{k=0}^m \binom{m}{k} (-1)^{m-k} (\mathbf{x} + k)\_{n,\lambda} \right) \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \frac{1}{m!} \triangle^m (\mathbf{x})\_{n,\lambda} \right) \frac{t^n}{n!} . \end{split} \tag{50}$$

Now, comparison of the coefficients on both sides of (50) yield following theorem.

**Theorem 7.** *For any n*, *m* ≥ 0*, the following identity holds:*

$$\frac{1}{m!} \triangle^m (\mathbf{x})\_{n,\lambda} = \begin{cases} \ S\_{2,\lambda}(n, m|\mathbf{x}), & \text{if } n \ge m, \\\ 0, & \text{if } n < m. \end{cases} \tag{51}$$

**Remark 2.** *Combing* (47) *and* (51)*, we obtain*

$$S\_{\mathfrak{u},\lambda}(N|\mathfrak{x}) = \sum\_{m=0}^{\min\{N,n\}} \binom{N+1}{m+1} m! S\_{2,\lambda}(n, m|\mathfrak{x}) .$$

From (30) and proceeding by induction, we have

$$(1+\triangle)^{k}f(\mathbf{x}) = \sum\_{m=0}^{k} \binom{k}{m} \triangle^{m} f(\mathbf{x}) = f(\mathbf{x}+k), \ (k \ge 0). \tag{52}$$

By (52), we get

$$\sum\_{k=0}^{N} (\mathbf{x} + k)\_{n,\lambda} = \sum\_{k=0}^{N} (\mathbf{1} + \triangle)^{k} (\mathbf{x})\_{n,\lambda}. \tag{53}$$

It is known that Daehee numbers are given by the generating function

$$\frac{\log(1+t)}{t} = \sum\_{n=0}^{\infty} D\_n \frac{t^n}{n!} \quad \text{(see [1,4,6])}.\tag{54}$$

From (2), we have

$$\begin{split} \int\_{Z\_{p}} \boldsymbol{e}\_{\lambda}^{x+y}(t) d\mu\_{0}(y) &= \frac{\frac{1}{\lambda} \log(1+\lambda t)}{\boldsymbol{e}\_{\lambda}(t)-1} \boldsymbol{e}\_{\lambda}^{x}(t) \\ &= \frac{\log(1+\lambda t)}{\lambda t} \frac{t}{\boldsymbol{e}\_{\lambda}(t)-1} \boldsymbol{e}\_{\lambda}^{x}(t) \\ &= \sum\_{l=0}^{\infty} D\_{l} \frac{\lambda^{l} t^{l}}{l!} \sum\_{m=0}^{\infty} \beta\_{m,\lambda}(x) \frac{t^{m}}{m!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{l=0}^{n} \binom{n}{l} \lambda^{l} D\_{l} \beta\_{n-l,\lambda}(x) \right) \frac{t^{n}}{n!} .\end{split} \tag{55}$$

From (55), we have

$$\int\_{\mathbb{Z}\_p} (\mathbf{x} + \mathbf{y})\_{n,\lambda} d\mu\_0(\mathbf{y}) = \sum\_{l=0}^n \binom{n}{l} \lambda^l D\_l \pounds\_{n-l,\lambda}(\mathbf{x}), \ (n \ge 0).$$

### **4. Further Remark**

A random variable *X* is a real-valued function defined on a sample space. We say that *X* is a continuous random variable if there exists a nonnegative function *f* , defined on (−∞, ∞), having the property that for any set *B* of real numbers (see [16,17])

$$P\{X \in B\} = \int\_{B} f(\mathbf{x})d\mathbf{x}.\tag{56}$$

The function *f* is called the probability density function of random variable *X*.

Let *X* be a uniform random variable on the interval (*α*, *β*). Then the probability density function *f* of *X* is given by

$$f(\mathbf{x}) = \begin{cases} \frac{1}{\beta - \mathbf{a}^\*} & \text{if } \mathbf{a} < \mathbf{x} < \beta, \\ 0, & \text{otherwise.} \end{cases} \tag{57}$$

Let *X* be a continuous random variable with the probability density function *f* . Then the expectation of *X* is defined by

$$E[X] = \int\_{-\infty}^{\infty} x f(x) dx.$$

For any real-valued function *g*(*x*), we have (see [16])

$$E[\mathbf{g}(X)] = \int\_{-\infty}^{\infty} \mathbf{g}(\mathbf{x}) f(\mathbf{x}) d\mathbf{x}.\tag{58}$$

Assume that *X*1, *X*2, ··· , *Xk* are independent uniform random variables on (0, 1). Then we have

$$\begin{split} &E[\boldsymbol{\epsilon}\_{\lambda}^{x+X\_{1}+X\_{2}+\cdots+X\_{k}}(t)] = \boldsymbol{e}\_{\lambda}^{x}(t)E[\boldsymbol{e}\_{\lambda}^{X\_{1}}(t)]E[\boldsymbol{e}\_{\lambda}^{X\_{2}}(t)]\cdots E[\boldsymbol{e}\_{\lambda}^{X\_{k}}(t)] \\ &= \boldsymbol{e}\_{\lambda}^{x}(t)\underbrace{\frac{\lambda}{\log(1+\lambda t)}(\boldsymbol{e}\_{\lambda}(t)-1)\times\cdots\times\frac{\lambda}{\log(1+\lambda t)}(\boldsymbol{e}\_{\lambda}(t)-1)}\_{k-\text{times}} \\ &= \left(\frac{\lambda t}{\log(1+\lambda t)}\right)^{k}\frac{t!}{t^{k}}\frac{1}{k!}(\boldsymbol{e}\_{\lambda}(t)-1)^{k}\boldsymbol{e}\_{\lambda}^{x}(t) \\ &= \frac{k!}{t^{k}}\sum\_{l=0}^{\infty}B\_{l}^{(l-k+1)}(1)\lambda^{l}\frac{t^{l}}{l!}\sum\_{m=k}^{\infty}S\_{2,\lambda}(m,k\mid\mathbf{x})\frac{t^{m}}{m!} \\ &= \frac{k!}{t^{k}}\sum\_{m=k}^{\infty}\left(\sum\_{m=k}^{n}\binom{n}{m}S\_{2,\lambda}(m,k\mid\mathbf{x})B\_{n-m}^{(n-m-k+1)}(1)\lambda^{n-m}\right)\frac{t^{n}}{n!}, \end{split} \tag{59}$$

where *<sup>B</sup>*(*α*) *<sup>n</sup>* (*x*) are the Bernoulli polynomials of order *<sup>α</sup>*, given by (see [4,7,8])

$$\left(\frac{t}{e^t - 1}\right)^a e^{xt} = \sum\_{n=0}^\infty B\_n^{(a)}(x) \frac{t^n}{n!} \,\tag{60}$$

and we used the well-known formula

$$\left(\frac{t}{\log(1+t)}\right)^n (1+t)^{x-1} = \sum\_{k=0}^{\infty} B\_k^{(k-n+1)}(x) \frac{t^k}{k!}.\tag{61}$$

From (59), we note that

$$\begin{split} \binom{n}{k} E[(\mathbf{x} + X\_1 + X\_2 + \dots + X\_k)\_{n-k, \lambda}] \\ = \sum\_{m=k}^n \binom{n}{m} S\_{2, \lambda}(m, k \mid \mathbf{x}) B\_{n-m}^{(n-m-k+1)}(1) \lambda^{n-m}. \end{split} \tag{62}$$

### **5. Conclusions**

It is well-known and classical that the first *n* positive integer power sums can be given by an expression involving some values of Bernoulli polynomials. Here we investigated some identities on Bernoulli numbers and polynomials and those on degenerate Bernoulli numbers and polynomials, which can be deduced from certain *p*-adic invariant integrals on Z*p*.

In particular, we introduced the integer power sum polynomials associated with integer power sums and obtained various expressions of them. Namely, they can be given in terms of Bernoulli polynomials, difference operators, and of the Stirling polynomials of the second kind. In addition, we introduced a degenerate version of the integer power sum polynomials, called the degenerate integer power sum polynomials and were able to find several representations of them. In detail, they can be represented in terms of Carlitz degenerate Bernoulli polynomials, difference operators, and of the degenerate Stirling numbers of the second kind.

In the final section, we considered an infinite family of random variables and proved that the expectations of them are expressed in terms of the degenerate Stirling polynomials of the second and some value of higher-order Bernoulli polynomials.

Most of the results in Sections 1 and 2 are reviews of known results, other than that, we demonstrated the usefulness of the *p*-adic invariant integrals in the study of integer power sum polynomials. However, we emphasize that the results in Sections 3 and 4 are new. In particular, we showed that the degenerate Stirling polynomials of the second kind, introduced as a degenerate version of the Stirling polynomials of the second kind, appear naturally and meaningfully in the context of calculations of an infinite family of random variables (see (62)). We also showed that they appear in an expression of the degenerate integer power sum polynomials (Remark 2) which is a degenerate version of the integer power sum polynomials (see (26)).

We have witnessed in recent years that studying various degenerate versions of some old and new polynomials, initiated by Carlitz in the classical papers [3,15], is very productive and promising (see [3,5,14,15,18,19] and references therein). Lastly, we note that this idea of considering degenerate versions of some polynomials extended even to transcendental functions like the gamma functions (see [19]).

**Author Contributions:** All authors contributed equally to the manuscript, and typed, read and approved the final manuscript.

**Funding:** This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No. 2017R1E1A1A03070882).

**Conflicts of Interest:** The authors declare that they have no competing interests.

### **References**


© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **Some Convolution Formulae Related to the Second-Order Linear Recurrence Sequence**

### **Zhuoyu Chen <sup>1</sup> and Lan Qi 2,\***


Received: 9 May 2019; Accepted: 12 June 2019; Published: 13 June 2019

**Abstract:** The main aim of this paper is that for any second-order linear recurrence sequence, the generating function of which is *f*(*t*) = <sup>1</sup> <sup>1</sup>+*at*+*bt*<sup>2</sup> , we can give the exact coefficient expression of the power series expansion of *<sup>f</sup> <sup>x</sup>*(*t*) for *<sup>x</sup>* ∈ **<sup>R</sup>** with elementary methods and symmetry properties. On the other hand, if we take some special values for *a* and *b*, not only can we obtain the convolution formula of some important polynomials, but also we can establish the relationship between polynomials and themselves. For example, we can find relationship between the Chebyshev polynomials and Legendre polynomials.

**Keywords:** Fibonacci numbers; Lucas numbers; Chebyshev polynomials; Legendre polynomials; Jacobi polynomials; Gegenbauer polynomials; convolution formula

**MSC:** 11B83

### **1. Introduction**

For any integer *n* ≥ 1 and any real number *y*, the Fibonacci polynomials *Fn*(*y*) and the Lucas polynomials *Ln*(*y*) are defined by the second-order linear recurrence sequence

$$F\_{n+1}(y) = y F\_n(y) + F\_{n-1}(y)$$

and

$$L\_{n+1}(y) = yL\_n(y) + L\_{n-1}(y),$$

where the first two terms are *F*0(*y*) = 0, *F*1(*y*) = 1, *L*0(*y*) = 2 and *L*1(*y*) = *y*.

If we take *<sup>α</sup>* = *<sup>y</sup>*<sup>+</sup> √*y*2+<sup>4</sup> <sup>2</sup> , *<sup>β</sup>* <sup>=</sup> *<sup>y</sup>*<sup>−</sup> √*y*2+<sup>4</sup> <sup>2</sup> , according to the properties of the second-order linear recurrence sequence, we have

$$F\_n(y) = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$

and

$$L\_n(y) = \mathfrak{a}^n + \beta^n.$$

For any integer *n* ≥ 0, the Fibonacci numbers *Fn* = *Fn*(1) can be defined by the generating function

$$\frac{1}{1-t-t^2} = \sum\_{n=0}^{\infty} F\_n t^n.$$

For any integer *n* ≥ 0, the first and the second kind Chebyshev polynomials *Tn*(*y*) and *Un*(*y*) are defined by the second-order linear recurrence sequence

$$T\_{n+2}(y) = 2yT\_{n+1}(y) - T\_n(y)$$

and

$$\mathcal{U}\_{\mathfrak{n}+2}(y) = 2y\mathcal{U}\_{\mathfrak{n}+1}(y) - \mathcal{U}\_{\mathfrak{n}}(y),$$

where the first two terms are *T*0(*y*) = 1, *T*1(*y*) = *y*, *U*0(*y*) = 1 and *U*1(*y*) = 2*y*.

If we take *<sup>α</sup>* <sup>=</sup> *<sup>y</sup>* <sup>+</sup> *<sup>y</sup>*<sup>2</sup> <sup>−</sup> 1, *<sup>β</sup>* <sup>=</sup> *<sup>y</sup>* <sup>−</sup> *<sup>y</sup>*<sup>2</sup> <sup>−</sup> 1, according to the properties of the second-order linear recurrence sequence, we have

$$T\_n(y) = \frac{\alpha^n + \beta^n}{2}$$

and

$$\mathcal{U}\_n(y) = \frac{\mathfrak{a}^{n+1} - \beta^{n+1}}{\mathfrak{a} - \beta}.$$

On the other hand, the second kind Chebyshev polynomials *Un*(*y*) can be also defined by the generating function

$$\frac{1}{1 - 2yt + t^2} = \sum\_{n=0}^{\infty} \mathcal{U}\_n(y) t^n.$$

Besides Fibonacci polynomials, Lucas polynomials and Chebyshev polynomials, other orthogonal polynomials have also been studied by interested scholars.

For example, the Legendre polynomials *Pn*(*y*) are defined by the generating function

$$\left(\frac{1}{1-2yt+t^2}\right)^{\frac{1}{2}} = \sum\_{n=0}^{\infty} P\_n(y)t^n.$$

The Jacobi polynomials {*P*(*α*,*β*) *<sup>n</sup>* (*y*)}0≤*n*<<sup>∞</sup> are defined by the generating function

$$\left[ R(1+R-t)^{\alpha}(1+R+t)^{\beta} \right]^{-1} = \sum\_{k=0}^{\infty} 2^{-\alpha-\beta} P\_n^{(\alpha,\beta)}(y) t^n y^k$$

where *<sup>R</sup>* <sup>=</sup> <sup>1</sup> <sup>−</sup> <sup>2</sup>*yt* <sup>+</sup> *<sup>t</sup>*2, <sup>|</sup>*t*<sup>|</sup> <sup>&</sup>lt; 1, *<sup>α</sup>*, *<sup>β</sup>* <sup>&</sup>gt; <sup>−</sup>1.

The Gegenbauer polynomials {*C<sup>λ</sup> <sup>n</sup>* (*y*)}0≤*n*<<sup>∞</sup> are defined by the generating function

$$\left(\frac{1}{1-2yt+t^2}\right)^{\lambda} = \sum\_{n=0}^{\infty} \mathbb{C}\_n^{\lambda}(y)t^n, \left(\lambda > -\frac{1}{2}\right).$$

It is well know that polynomials and sequence occupy indispensable positions in the research of number theory. Especially, Fibonacci and Lucas numbers, Chebyshev and Legendre polynomials and others. These polynomials and numbers are closely related and there are a variety of meaningful results which have been researched by interested scholars until now. For example, the identities of Chebyshev polynomials can be found in [1–9], and the contents about Fibonacci and Lucas numbers in [10,11]. Some authors have a research which connects Chebyshev polynomials and Fibonacci or Lucas polynomials (see [12–14]).

In particular, we can find many significant results in the aspect of studying the calculating problem of one kind sums of some important polynomials. For example, Yuankui Ma and Wenpeng Zhang have calculated one kind sums of Fibonacci Polynomials (see [15]) as follows.

Let *h* be a positive integer, for any integer *n* ≥ 0, they proved

$$\sum\_{a\_1 + a\_2 + \cdots + a\_{h+1} = n} F\_{a\_1}(\mathbf{x}) F\_{a\_2}(\mathbf{x}) \cdot \cdots \cdot F\_{a\_{h+1}}(\mathbf{x}) = \frac{1}{h!} \cdot \sum\_{j=1}^{h} \frac{(-1)^{h-j} \cdot \mathcal{S}(h, j)}{\mathbf{x}^{2h-j}}$$

$$\times \left( \sum\_{i=0}^{n} \frac{(n-i+j)!}{(n-i)!} \cdot \binom{2h+i-j-1}{i} \cdot \frac{(-1)^i \cdot 2^i \cdot F\_{n-i+j}(\mathbf{x})}{\mathbf{x}^i} \right),$$

where the summation is over all *h* + 1-tuples with non-negative integer coordinates (*a*1, *a*2, ··· , *ah*<sup>+</sup><sup>1</sup> such that *a*<sup>1</sup> + *a*<sup>2</sup> + ··· + *ah*<sup>+</sup><sup>1</sup> = *n*, and *S*(*h*, *i*) is a second order non-linear recurrence sequence defined by *S*(*h*, 0) = 0, *S*(*h*, *h*) = 1, and *S*(*h* + 1, *i* + 1) = 2 · (2*h* − 1 − *i*) · *S*(*h*, *i* + 1) + *S*(*h*, *i*) for all positive integers 1 ≤ *i* ≤ *h* − 1.

Yixue Zhang and Zhuoyu Chen have researched the calculating problem of one kind sums of the second kind Chebyshev polynomials (see [16]) as follows.

Let *h* be a positive integer, for any integer *n* ≥ 0, they proved

$$\sum\_{a\_1 + a\_2 + \cdots + a\_{h+1} = n} \mathcal{U}\_{a\_1}(\mathbf{x}) \mathcal{U}\_{a\_2}(\mathbf{x}) \cdot \cdots \mathcal{U}\_{a\_{h+1}}(\mathbf{x})$$

$$= \frac{1}{2^{h} \cdot h!} \cdot \sum\_{j=1}^{h} \frac{\mathcal{C}(h, j)}{\mathbf{x}^{2h-j}} \sum\_{i=0}^{n} \frac{(n-i+j)!}{(n-i)!} \cdot \binom{2h+i-j-1}{i} \cdot \frac{\mathcal{U}\_{n-i+j}(\mathbf{x})}{\mathbf{x}^{i}},$$

where *C*(*h*, *i*) is a second order non-linear recurrence sequence defined by *C*(*h*, 0) = 0, *C*(*h*, *h*) = 1, *C*(*h* + 1, 1) = 1 · 3 · 5 ···(2*h* − 1)=(2*h* − 1)!! and *C*(*h* + 1, *i* + 1)=(2*h* − 1 − *i*) · *C*(*h*, *i* + 1) + *C*(*h*, *i*) for all 1 ≤ *i* ≤ *h* − 1.

Shimeng Shen and Li Chen have studied the calculating problem of one kind sums of Legendre Polynomials (see [17]) as follows.

For any positive integer *k* and integer *n* ≥ 0, they proved

$$(2k-1)!!\sum\_{a\_1+a\_2+\cdots+a\_{2k+1}=n}P\_{a\_1}(\mathbf{x})P\_{a\_2}(\mathbf{x})\cdots P\_{a\_k}(\mathbf{x})$$

$$=\sum\_{j=1}^k \mathbb{C}(k,j)\sum\_{i=0}^n \frac{(n+k+1-i-j)!}{(n-i)!} \cdot \frac{\binom{i+j+k-2}{i}}{\mathbf{x}^{k-1+i+j}} \cdot P\_{n+k+1-i-j}(\mathbf{x})$$

where(2*<sup>k</sup>* <sup>−</sup> <sup>1</sup>)!! <sup>=</sup> <sup>1</sup> <sup>×</sup> <sup>3</sup> <sup>×</sup> <sup>5</sup> ···(2*<sup>k</sup>* <sup>−</sup> <sup>1</sup>) = <sup>2</sup>*k*( <sup>1</sup> <sup>2</sup> )*k*, and *C*(*k*, *i*) is a recurrence sequence defined by *C*(*k*, 1) = 1, *C*(*k* + 1, *k* + 1)=(2*k* − 1)!! and *C*(*k* + 1, *i* + 1) = *C*(*k*, *i* + 1)+(*k* − 1 + *i*) · *C*(*k*, *i*) for all 1 ≤ *i* ≤ *k* − 1.

They have converted the complex sums of *Fn*(*x*) into a simple combination of *Fn*(*x*), the complex sums of *Un*(*x*) into a simple combination of *Un*(*x*), and the complex sums of *Pn*(*x*) into a simple combination of *Pn*(*x*).

Very recently, Taekyun Kim and other people researched the properties of Fibonacci numbers through introducing the convolved Fibonacci numbers *pn*(*x*) by generating function as follows (see [18]):

$$\left(\frac{1}{1-t-t^2}\right)^\chi = \sum\_{n=0}^\infty p\_n(x) \frac{t^n}{n!}, (x \in \mathbb{R}).$$

They researched some new and explicit identities of the convolved Fibonacci numbers for *x* ∈ **N**. For example, for *n* ≥ 0 and *r* ∈ **N**, they have proved the recurrence relationship of *pn*(*x*) (see [18]):

$$p\_n(\mathbf{x}) = \sum\_{l=0}^n p\_l(r) p\_{n-l}(\mathbf{x} - r) = \sum\_{l=0}^n p\_{n-l}(r) p\_l(\mathbf{x} - r).$$

The convolved Fibonacci numbers *pn*(*x*) seems to be only connected with the simple power square. In fact, it can establish the relationship between polynomials and themselves, so the further research of *pn*(*x*) is very significant. They have provided us a new perspective to study the properties of some vital polynomials. For example, Taekyun Kim and other people have proved the relationship between *pn*(*x*) and the combination sums about Fibonacci numbers:

$$\frac{p\_n(r+1)}{n!} = \sum\_{l\_1=0}^n \sum\_{l\_2=0}^{n-l\_1} \cdots \sum\_{l\_r=0}^{n-l\_1-\cdots-l\_{r-1}} F\_{I\_1} F\_{I\_2} \cdots F\_{I\_r} F\_{n-l\_1-l\_2-\cdots-l\_r} \cdots$$

They have converted the complex sums of *Fn*(*x*) into a calculation problem of *pn*(*x*) and the calculation method is easier and the expression is simpler.

Inspired by this article, in this paper, for any second-order linear recurrence sequence, the generating function of which is *f*(*t*) = <sup>1</sup> <sup>1</sup>+*at*+*bt*<sup>2</sup> , we can define

$$\left(\frac{1}{1+at+bt^2}\right)^x = \sum\_{n=0}^\infty p\_{ll}(\mathbf{x}) \frac{t^n}{n!}, (a,b,\mathbf{x}\in \mathbb{R}).\tag{1}$$

Firstly, we give a specific computational formula of *pn*(*x*) for *x* ∈ **R** using the elementary methods. After that for any polynomial or sequence, the generating function of which is *f*(*t*) = <sup>1</sup> <sup>1</sup>+*at*+*bt*<sup>2</sup> , we can obtain its convolved formula easily and directly.

Secondly, if we take some special values for *a*, *b* in *f*(*t*) and *x* in *pn*(*x*), we can find some relationship between special polynomials and themselves. For example, we will establish the relationship between the convolved Fibonacci numbers and Lucas numbers, the relationship between the convolved formula of the second kind Chebyshev polynomials and the first kind Chebyshev polynomials, and the relationship between Legendre polynomials and the first kind Chebyshev polynomials and others.

At last, through the computational formula of *pn*(*x*), especially for *x* ∈ **N**, we can also convert the complex sums of *Fn* into a liner combination of *Ln*; and express the complex sums of *Un*(*y*) as a liner combination of *Tn*(*y*). More importantly, the forms are more common and the calculations are easier than previous results.

We will prove the main results as follows:

**Theorem 1.** *Let f*(*t*) = <sup>1</sup> <sup>1</sup>−*t*−*t*<sup>2</sup> , *for any integer n* <sup>≥</sup> <sup>0</sup> *and x* <sup>∈</sup> **<sup>R</sup>***, we can obtain*

$$p\_n(\mathbf{x}) = \frac{1}{2} \sum\_{i=0}^n (-1)^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} L\_{n-2i} \mathbf{x}$$

*where x<sup>n</sup>* = *x*(*x* + 1)(*x* + 2)···(*x* + *n* − 1) *and* (*x*)<sup>0</sup> = 1*.*

**Theorem 2.** *Let f*(*t*) = <sup>1</sup> <sup>1</sup>−2*yt*+*t*<sup>2</sup> , *for any integer n* <sup>≥</sup> <sup>0</sup> *and x*, *<sup>y</sup>* <sup>∈</sup> **<sup>R</sup>***, we can obtain*

$$p\_n(\mathfrak{x}; y) = \sum\_{i=0}^n \binom{n}{i} \langle \mathfrak{x} \rangle\_i \langle \mathfrak{x} \rangle\_{n-i} T\_{n-2i}(y).$$

From Theorem 1 we can deduce the following:

**Corollary 1.** *For any positive integer k, we have the identity*

$$\sum\_{a\_1 + a\_2 + \cdots + a\_k = n} F\_{a\_1} F\_{a\_2} \cdots F\_{a\_k}$$

$$\xi = \frac{1}{2((k-1)!)^2} \sum\_{i=0}^n (-1)^i \frac{(k+i-1)!(k+n-i-1)!}{i!(n-i)!} \cdot L\_{n-2i-1}$$

From Theorem 2 we can deduce the following:

**Corollary 2.** *For any positive integer k, we have the identity*

$$\sum\_{a\_1 + a\_2 + \cdots + a\_k = n} \mathcal{U}\_{a\_1}(y) \cdot \mathcal{U}\_{a\_2}(y) \cdot \cdots \cdot \mathcal{U}\_{a\_k}(y)$$

$$= \frac{1}{((k-1)!)^2} \sum\_{i=0}^n \frac{(k+i-1)!(k+n-i-1)!}{i!(n-i)!} \cdot T\_{n-2i}(y).$$

**Corollary 3.** *If x* = <sup>1</sup> <sup>2</sup> *, we have the identity*

$$P\_n(y) = \frac{1}{2^n} \sum\_{i=0}^n \frac{(2i-1)!!(2n-2i-1)!!}{i!(n-i)!} \cdot T\_{n-2i}(y) \dots$$

**Corollary 4.** *If x* <sup>=</sup> <sup>−</sup><sup>1</sup> <sup>2</sup> *, we have the identity*

$$R = \sum\_{n=0}^{\infty} \frac{1}{2^n} \sum\_{i=0}^n \frac{(2i-3)!!(2n-2i-3)!!}{i!(n-i)!} \cdot T\_{n-2i}(y) \cdot t^n.$$

**Corollary 5.** *If x* <sup>=</sup> *<sup>λ</sup>* <sup>&</sup>gt; <sup>−</sup><sup>1</sup> <sup>2</sup> *, we have the identity*

$$\mathcal{C}\_n^{\lambda}(y) = \frac{1}{n!} \sum\_{i=0}^n \binom{n}{i} \langle \lambda \rangle\_i \langle \lambda \rangle\_{n-i} T\_{n-2i}(y).$$

Theorems 1 and 2 give the computational formula of *pn*(*x*) of some famous polynomials. Especially, we know that polynomials are closely connected and they can be converted to each other. According to these theorems, we can obtain the relationship between the polynomials easily. It cannot only extend the application of orthogonal polynomials, but also make replacement calculations according to its complexity. For example, if we make a calculation involving the Gegenbauer polynomials, for simple calculations, we can convert it into Chebyshev polynomials according to Corollary 5.

### **2. A Simple Lemma**

In order to prove our theorems, we are going to introduce a simple lemma.

**Lemma 1.** *For any integer n* ≥ 0 *and a*, *b*, *x* ∈ **R***, we can obtain the equation*

$$p\_n(\mathbf{x}) = \frac{1}{2} \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( \left( \frac{-a + \sqrt{a^2 - 4b}}{2} \right)^{n-2i} + \left( \frac{-a - \sqrt{a^2 - 4b}}{2} \right)^{n-2i} \right) \dots$$

**Proof.** Firstly, according Equation (1), we have

$$\sum\_{n=0}^{\infty} p\_n(\mathbf{x}) \frac{t^n}{n!} \quad = \quad \left(\frac{1}{1+at+bt^2}\right)^x = (1-at)^{-x}(1-\beta t)^{-x}.\tag{2}$$

We can easily know that *<sup>α</sup>* <sup>+</sup> *<sup>β</sup>* <sup>=</sup> <sup>−</sup>*a*, *αβ* <sup>=</sup> *<sup>b</sup>* and *<sup>α</sup>* <sup>=</sup> <sup>−</sup>*a*<sup>+</sup> <sup>√</sup>*a*2−4*<sup>b</sup>* <sup>2</sup> , *<sup>β</sup>* <sup>=</sup> <sup>−</sup>*a*<sup>−</sup> <sup>√</sup>*a*2−4*<sup>b</sup>* <sup>2</sup> are two roots of 1 + *at* + *bt*<sup>2</sup> = 0.

Then, applying the properties of power series, we obtain

$$(1 - at)^{-x} = \sum\_{n=0}^{\infty} \binom{-x}{n} (-1)^n (at)^n = \sum\_{n=0}^{\infty} \frac{(-x)\_n}{n!} (-1)^n a^n t^n \tag{3}$$

and

$$(1 - \beta t)^{-x} = \sum\_{n=0}^{\infty} \binom{-x}{n} (-1)^n (\beta t)^n = \sum\_{n=0}^{\infty} \frac{(-x)\_n}{n!} (-1)^n \beta^n t^n,\tag{4}$$

where (*x*)*<sup>n</sup>* = *x*(*x* − 1)(*x* − 2)···(*x* − *n* + 1) and (*x*)<sup>0</sup> = 1. Combining Equations (2)–(4), we get

$$\begin{split} \sum\_{n=0}^{\infty} p\_n(\mathbf{x}) \frac{t^n}{n!} &= \quad \left( \sum\_{n=0}^{\infty} \frac{(-\mathbf{x})\_n}{n!} (-1)^n \alpha^n t^n \right) \left( \sum\_{n=0}^{\infty} \frac{(-\mathbf{x})\_n}{n!} (-1)^n \beta^n t^n \right) \\ &= \quad \sum\_{n=0}^{\infty} \left( \sum\_{i=0}^n \frac{(-\mathbf{x})\_i (-1)^i \alpha^i t^i}{i!} \cdot \frac{(-\mathbf{x})\_{n-i} (-1)^{n-i} \beta^{n-i} t^{n-i}}{(n-i)!} \right) \\ &= \quad \sum\_{n=0}^{\infty} \frac{(-1)^n}{n!} \left( \sum\_{i=0}^n \binom{n}{i} (-\mathbf{x})\_i (-\mathbf{x})\_{n-i} \mathbf{a}^i \beta^{n-i} \right) t^n. \end{split} \tag{5}$$

Similarly, according the symmetry of *α* and *β*, we can easily obtain

$$\sum\_{n=0}^{\infty} p\_n(x) \frac{t^n}{n!} = \sum\_{n=0}^{\infty} \frac{(-1)^n}{n!} \left( \sum\_{i=0}^n \binom{n}{i} (-x)\_i (-x)\_{n-i} \beta^i a^{n-i} \right) t^n. \tag{6}$$

Then, combining Equations (5) and (6), we know that

$$\begin{split} \sum\_{n=0}^{\infty} p\_n(\mathbf{x}) \frac{t^n}{n!} &= \frac{1}{2} \sum\_{n=0}^{\infty} \frac{(-1)^n}{n!} \left( \sum\_{i=0}^n \binom{n}{i} (-\mathbf{x})\_i (-\mathbf{x})\_{n-i} (a\beta)^i \left( \beta^{n-2i} + a^{n-2i} \right) \right) t^n \\ &= \frac{1}{2} \sum\_{n=0}^{\infty} \frac{1}{n!} \left( \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( \beta^{n-2i} + a^{n-2i} \right) \right) t^n. \end{split} \tag{7}$$

Comparing the coefficients of *t <sup>n</sup>* in Equation (7), we get

$$\begin{aligned} p\_n(\mathbf{x}) &= \ & \frac{1}{2} \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( a^{n-2i} + \delta^{n-2i} \right) \\ &= & \frac{1}{2} \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( \left( \frac{-a + \sqrt{a^2 - 4b}}{2} \right)^{n-2i} + \left( \frac{-a - \sqrt{a^2 - 4b}}{2} \right)^{n-2i} \right) .\end{aligned}$$

Now we have completed the proof of the Lemma 1.

### **3. Proof of the Theorem**

**Proof of Theorem 1.** If we take *a* = −1 and *b* = −1 in Equation (1), we know that *f*(*t*) is the generating function of Fibonacci number. That is,

$$f(t) = \frac{1}{1 - t - t^2} = \sum\_{n=0}^{\infty} F\_n t^n.$$

The convolved Fibonacci numbers *pn*(*x*) are defined by the generating function as [18]

$$f^{x}(t) = \left(\frac{1}{1 - t - t^{2}}\right)^{x} = \sum\_{n=0}^{\infty} p\_{n}(x) \frac{t^{n}}{n!}.\tag{8}$$

In this time, *α* = <sup>1</sup><sup>+</sup> √5 <sup>2</sup> , *<sup>β</sup>* <sup>=</sup> <sup>1</sup><sup>−</sup> √5 <sup>2</sup> . According to the Lemma 1 and *Ln* = *α<sup>n</sup>* + *βn*, we can get

$$\begin{split} p\_n(\mathbf{x}) &= \quad \frac{1}{2} \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( \left( \frac{-a + \sqrt{a^2 - 4b}}{2} \right)^{n-2i} + \left( \frac{-a - \sqrt{a^2 - 4b}}{2} \right)^{n-2i} \right) \\ &= \quad \frac{1}{2} \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^{n-2i} + \left( \frac{1 - \sqrt{5}}{2} \right)^{n-2i} \right) \\ &= \quad \frac{1}{2} \sum\_{i=0}^n b^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} \left( a^{n-2i} + \beta^{n-2i} \right) \\ &= \quad \frac{1}{2} \sum\_{i=0}^n (-1)^i \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} L\_{n-2i} . \end{split} \tag{9}$$

In this equation, *pn*(*x*) is expressed as a combined forms of Lucas number. The Proof of Theorem 1 has finished.

About the convolved Fibonacci numbers *pn*(*x*), Taekyun Kim and others have obtained its some-recurrence formulae in reference [18]. Based on [18], we have given an exact computational formula of *pn*(*x*) for any arbitrary *x* in Theorem 1. Compared with the results in [18], Theorem 1 is more general and easier.

If we take *x* = *k* ∈ **N** in Equation (8), we get

$$\begin{aligned} &\sum\_{n=0}^{\infty} p\_n(k) \frac{t^n}{n!} = \left(\frac{1}{1 - t - t^2}\right)^k = \left(\sum\_{n=0}^{\infty} F\_n t^n\right)^k\\ &= \quad \left(\sum\_{a\_1=0}^{\infty} F\_{a\_1} \cdot t^{a\_1}\right) \left(\sum\_{a\_2=0}^{\infty} F\_{a\_2} \cdot t^{a\_2}\right) \cdots \left(\sum\_{a\_k=0}^{\infty} F\_{a\_k} \cdot t^{a\_k}\right)\\ &= \quad \left(\sum\_{a\_1=0}^{\infty} \sum\_{a\_2=0}^{\infty} \cdots \sum\_{a\_k=0}^{\infty} F\_{a\_1} \cdot F\_{a\_2} \cdot \cdots \cdot F\_{a\_k} \cdot t^{a\_1 + a\_2 + \cdots + a\_k}\right)\\ &= \quad \sum\_{n=0}^{\infty} \left(\sum\_{a\_1 + a\_2 + \cdots + a\_k = n} F\_{a\_1} \cdot F\_{a\_2} \cdot \cdots \cdot F\_{a\_k}\right) \cdot t^n,\end{aligned}$$

and then combining Equation (9), we can obtain

$$\begin{aligned} &\sum\_{a\_1+a\_2+\cdots+a\_k=n} F\_{a\_1} \cdot F\_{a\_2} \cdot \cdots \cdot F\_{a\_k} \\ &= \quad \frac{1}{2n!} \sum\_{i=0}^n (-1)^i \binom{n}{i} \langle k \rangle\_i \langle k \rangle\_{n-i} L\_{n-2i} \\ &= \quad \frac{1}{2((k-1)!)^2} \sum\_{i=0}^n (-1)^i \frac{(k+i-1)!(k+n-i-1)!}{i!(n-i)!} L\_{n-2i} .\end{aligned}$$

The proof of Corollary 1 has finished.

For every *Fal* (<sup>1</sup> ≤ *<sup>l</sup>* ≤ *<sup>k</sup>*), <sup>∑</sup>*a*1+*a*2+···+*ak*=*<sup>n</sup> Fa*<sup>1</sup> · *Fa*<sup>2</sup> ··· *Fak* is symmetry.

**Proof of Theorem 2.** If we take *a* = −2*y* and *b* = 1 in Equation (1), we all know *f*(*t*; *y*) is the generating function of the second-kind Chebyshev polynomials *Un*(*y*)

$$f(t;y) = \frac{1}{1 - 2yt + t^2} = \sum\_{n=0}^{\infty} \mathcal{U}\_n(y)t^n.$$

The convolved second-kind Chebyshev polynomials *pn*(*x*; *y*) are defined by the generating function as [18]

$$f^x(t;y) = \left(\frac{1}{1-2yt+t^2}\right)^x = \sum\_{n=0}^{\infty} p\_n(x;y)\frac{t^n}{n!}.\tag{10}$$

In this time, *<sup>α</sup>* <sup>=</sup> *<sup>y</sup>* <sup>+</sup> *<sup>y</sup>*<sup>2</sup> <sup>−</sup> 1, *<sup>β</sup>* <sup>=</sup> *<sup>y</sup>* <sup>−</sup> *<sup>y</sup>*<sup>2</sup> <sup>−</sup> 1. According to the Lemma 1 and *Tn*(*y*) = <sup>1</sup> <sup>2</sup> (*α<sup>n</sup>* + *<sup>β</sup>n*), we can get

$$p\_n(\mathbf{x}; y) = \sum\_{i=0}^n \binom{n}{i} \langle \mathbf{x} \rangle\_i \langle \mathbf{x} \rangle\_{n-i} T\_{n-2i}(y). \tag{11}$$

In this equation, *pn*(*x*; *y*) is expressed as a combined form of the first-kind Chebyshev polynomials *Tn*(*x*).

If we take *x* = *k* ∈ **N** in Equation (10), and combining Equation (11) we can easily prove the Corollary 2.

Take *x* = <sup>1</sup> <sup>2</sup> in Equation (10), we know *f* 1 <sup>2</sup> (*t*; *y*) is the generating function of the Legendre polynomials *Pn*(*x*) as follows:

$$f^{\frac{1}{2}}(t) = \left(\frac{1}{1 - 2yt + t^2}\right)^{\frac{1}{2}} = \sum\_{n=0}^{\infty} P\_n(y)t^n = \sum\_{n=0}^{\infty} p\_n\left(\frac{1}{2}; y\right)\frac{t^n}{n!}.$$

According to Theorem 2, we can easily obtain

$$\begin{aligned} p\_n\left(\frac{1}{2};y\right) &= \sum\_{i=0}^n \binom{n}{i} \left\langle \frac{1}{2} \right\rangle\_i \left\langle \frac{1}{2} \right\rangle\_{n-i} T\_{n-2i}(y) \\ &= \frac{1}{2^{2n}} \sum\_{i=0}^n \frac{n!(2i)!(2(n-i))!}{(i!)^2((n-i)!)^2} \cdot T\_{n-2i}(y) \\ &= \frac{n!}{2^n} \sum\_{i=0}^n \frac{(2i-1)!!(2n-2i-1)!!}{i!(n-i)!} \cdot T\_{n-2i}(y). \end{aligned}$$

In a word, we know the Legendre polynomials *Pn*(*x*) can be expressed as combined forms of the first kind Chebyshev polynomials *Tn*(*x*) as follows:

$$P\_n(y) = \frac{1}{2^n} \sum\_{i=0}^n \frac{(2i-1)!!(2n-2i-1)!!}{i!(n-i)!} \cdot T\_{n-2i}(y) \dots$$

The proof of Corollary 3 has finished.

If we take *<sup>x</sup>* <sup>=</sup> <sup>−</sup><sup>1</sup> <sup>2</sup> in (10), then we can easily obtain

$$\begin{aligned} R &=& \sum\_{n=0}^{\infty} p(-\frac{1}{2}; y) \frac{t^n}{n!} = \sum\_{n=0}^{\infty} \frac{1}{2^{2n-2}} \sum\_{i=0}^{n} \frac{(2(i-1))!(2(n-i-1))!}{i!(i-1)!(n-i)!(n-i-1)!} \cdot T\_{n-2i}(y) \cdot t^n \\ &=& \sum\_{n=0}^{\infty} \frac{1}{2^n} \sum\_{i=0}^{n} \frac{(2i-3)!!(2n-2i-3)!!}{i!(n-i)!} \cdot T\_{n-2i}(y) \cdot t^n .\end{aligned}$$

The proof of Corollary 4 has finished.

Taking *<sup>x</sup>* <sup>=</sup> *<sup>λ</sup>* <sup>&</sup>gt; <sup>−</sup><sup>1</sup> <sup>2</sup> in Equation (10), we know *<sup>f</sup> <sup>λ</sup>*(*t*; *<sup>y</sup>*) is the generating function of the Gegenbauer polynomials {*C<sup>λ</sup> <sup>n</sup>* (*y*)}0≤*n*<<sup>∞</sup> as follows:

$$\left(\frac{1}{1-2yt+t^2}\right)^\lambda = \sum\_{n=0}^\infty \mathbb{C}\_n^\lambda(y)t^n = f^\lambda(t;y) = \sum\_{n=0}^\infty p\_n\left(\lambda;y\right)\frac{t^n}{n!}.$$

According to Theorem 2, we can easily obtain

$$p\_n\left(\lambda;y\right) = \sum\_{i=0}^n \binom{n}{i} \langle \lambda \rangle\_i \langle \lambda \rangle\_{n-i} T\_{n-2i}(y).$$

The proof of Corollary 5 has finished.

**Author Contributions:** Writing—original draft: Z.C.; Writing—review and editing: L.Q.

**Funding:** This work is supported by the N. S. F. (11771351), (11826205), (11826203) of P. R. China and N.S.B.R.P in Shaanxi Province (2018JQ1093).

**Acknowledgments:** The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**


© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **On** *r***-Central Incomplete and Complete Bell Polynomials**

### **Dae San Kim 1, Han Young Kim 2, Dojin Kim 3,\* and Taekyun Kim <sup>2</sup>**


Received: 16 April 2019; Accepted:23 May 2019; Published: 27 May 2019

**Abstract:** Here we would like to introduce the extended *r*-central incomplete and complete Bell polynomials, as multivariate versions of the recently studied extended *r*-central factorial numbers of the second kind and the extended *r*-central Bell polynomials, and also as multivariate versions of the *r*-Stirling numbers of the second kind and the extended *r*-Bell polynomials. In this paper, we study several properties, some identities and various explicit formulas about these polynomials and their connections as well.

**Keywords:** extended *r*-central complete bell polynomials; extended *r*-central incomplete bell polynomials; complete *r*-Bell polynomials; incomplete *r*-bell polynomials

### **1. Introduction**

We begin this section by briefly recalling several definitions related to the central factorial numbers of the second kind and the central Bell polynomials and also to their generalizations of the extended *r*-central factorial numbers of the second kind and the extended *r*-central Bell polynomials (see [1]). The central factorial *x*[*n*] is given by the generating function

$$
\left(\frac{t}{2} + \sqrt{1 + \frac{t^2}{4}}\right)^{2x} = \sum\_{n=0}^{\infty} x^{\left[n\right]} \frac{t^n}{n!}.\tag{1}
$$

A proof of (1) can be found in [2], p. 215, Equations (27), (28) and (27a), (see also [1,3–5]). It is well known that Formula (1) shows that

$$\mathbf{x}^{[0]} = 1, \quad \mathbf{x}^{[n]} = \mathbf{x}(\mathbf{x} + \frac{n}{2} - 1) \cdots (\mathbf{x} - \frac{n}{2} + 1), \quad (n \ge 1), \tag{2}$$

where *x*[*n*] is of degree *n* in *x*.

The central factorial numbers of the second kind *T*(*n*, *k*) are the coefficients in the expansion of *x<sup>n</sup>* in terms of central factorials as follows:

$$\mathbf{x}^n = \sum\_{k=0}^n T(n,k)\mathbf{x}^{[k]},\tag{3}$$

(see [6–11]) and it is known that *T*(2*n*, 2*n* − 2*k*) enumerates the number of ways to place *k* rooks on a 3D-triangle board of size (*n* − 1) (see [12,13]). The generating function of *T*(*n*, *k*) is given by

*Symmetry* **2019**, *11*, 724; doi:10.3390/sym11050724 www.mdpi.com/journal/symmetry

$$\frac{1}{k!} \left( \varepsilon^{\frac{t}{2}} - \varepsilon^{-\frac{t}{2}} \right)^k = \sum\_{n=k}^{\infty} T(n,k) \frac{t^n}{n!} \tag{4}$$

which follows, for example, from (1) and (3).

Indeed, on the one hand by making use of (3) we have

$$\mathbf{x}^{\mathbf{x}\mathbf{t}} = \sum\_{n=0}^{\infty} \mathbf{x}^{\mathbf{n}} \frac{\mathbf{t}^{\mathbf{n}}}{n!} = \sum\_{k=0}^{\infty} \left( \sum\_{n=k}^{\infty} T(n,k) \frac{\mathbf{t}^{\mathbf{n}}}{n!} \right) \mathbf{x}^{[k]}.\tag{5}$$

On the other hand, by virtue of (1) we also have

$$\begin{split} x^{xt} &= \left( \frac{e^{\frac{\xi}{2}} - e^{-\frac{\xi}{2}}}{2} + \sqrt{1 + \frac{(e^{\frac{\xi}{2}} - e^{-\frac{\xi}{2}})}{4}} \right)^{2x} \\ &= \sum\_{k=0}^{\infty} \frac{1}{k!} (e^{\frac{\xi}{2}} - e^{-\frac{\xi}{2}})^k x^{[k]}. \end{split} \tag{6}$$

Now, it can be easily seen that Equation (4) follows from (5) and (6).

Kim-Kim in [11] introduced the central Bell polynomials by means of generating function as

$$\,\_2\varepsilon \left( \,^{x \left( \frac{\hbar}{2} - \varepsilon \right)} \frac{\,^{x} \mathbf{1}}{\,} \right) = \sum\_{n=0}^{\infty} B\_n^{(c)}(x) \frac{t^n}{n!} . \tag{7}$$

We note by making use of (4) that identity (7) implies (see [1,11])

$$B\_n^{(c)}(\mathbf{x}) = \sum\_{k=0}^n T(n,k)\mathbf{x}^k, \quad (n \ge 0).$$

For a nonnegative integer *r*, Kim-Dolgy-Kim-Kim in a recent work [1] introduced the extended *r*-central factorial numbers of the second kind given by the generating function:

$$\frac{1}{k!} \left( \varepsilon^{\frac{1}{2}} - \varepsilon^{-\frac{1}{2}} \right)^k \varepsilon^{r!} = \sum\_{n=k}^{\infty} T^{(r)}(n+r, k+r) \frac{t^n}{n!}. \tag{8}$$

From (8), it is noted that (see [1])

$$(\mathbf{x} + \mathbf{r})^n = \sum\_{k=0}^n T^{(r)}(n+r, k+r) \mathbf{x}^{[k]}.\tag{9}$$

The extended *r*-central Bell polynomials [1] are defined by

$$\left(\varepsilon^{\left(\frac{t}{\varepsilon^2} - \varepsilon^{-\frac{t}{2}}\right)}\varepsilon^{r\,t} = \sum\_{n=0}^{\infty} B\_n^{(\varepsilon,r)}(x) \frac{t^n}{n!} \right. \tag{10}$$

By definition (10), it is also known that (see [1])

$$B\_{il}^{(c,r)}(\mathbf{x}) = \sum\_{k=0}^{n} \mathbf{x}^{k} T^{(r)}(n+r, k+r), \quad (n \ge 0). \tag{11}$$

The purpose of this paper is to introduce and study the extended *r*-central incomplete and complete Bell polynomials, as multivariate versions of the recently studied the extended *r*-central factorial numbers of the second and the extended *r*-central Bell polynomials (see [1]), and also as multivariate versions of the *r*- Stirling numbers of the second kind and the extended *r*-Bell polynomials (see Section 2). Then we investigate their properties, some identities and various explicit formulas related to these polynomials and also their connections.

This paper is organized as follows. In Section 2, we introduce the incomplete and complete *r*-Bell polynomials and give some of their simple properties. We observe that these polynomials are multivariate versions of the *r*- Stirling numbers of the second kind and the extended *r*-Bell polynomials . In Section 3, we introduce our object of study, namely the extended *r*-central incomplete and complete Bell polynomials, and provide several properties, some identities and various explicit formulas for them. Finally, in Section 4, brief summaries for the obtained results about newly defined polynomials are provided.

#### **2. Preliminaries**

The *r*-Stirling numbers *S*(*r*) <sup>2</sup> (*n*, *k*) of the second kind are defined by the generating function (see [14–19])

$$\frac{1}{k!} \left( e^t - 1 \right)^k e^{rt} = \sum\_{n=k}^{\infty} S\_2^{(r)}(n+r, k+r) \frac{t^n}{n!} \tag{12}$$

and they enumerate the number of partitions of the set {1, 2, ··· , *n*} into *k* nonempty disjoint subsets in such a way that 1, 2, ··· ,*r* are in distinct subsets.

The extended *r*-Bell polynomials are given by (see [15])

$$\varepsilon^{rt}\varepsilon^{x(\varepsilon^t-1)} = \sum\_{n=0}^{\infty} B\_n^{(r)}(x)\frac{t^n}{n!}.\tag{13}$$

One can show that Equations (12) and (13) imply

$$\begin{split} B\_{n}^{(r)}(\mathbf{x}) &= \mathbf{e}^{-\mathbf{x}} \sum\_{k=0}^{\infty} \frac{(k+r)^{n}}{k!} \mathbf{x}^{k} \\ &= \sum\_{k=0}^{n} \mathbf{x}^{k} S\_{2}^{(r)}(n+r, k+r), \quad (n \ge 0). \end{split} \tag{14}$$

In particular *<sup>x</sup>* <sup>=</sup> 1, *<sup>B</sup>*(*c*,*r*) *<sup>n</sup>* <sup>=</sup> *<sup>B</sup>*(*c*,*r*) *<sup>n</sup>* (1) are called the extended *<sup>r</sup>*-Bell numbers.

The incomplete *r*-Bell polynomials are given by the generating function

$$\frac{1}{k!} \left( \sum\_{j=1}^{\infty} \mathbf{x}\_j \frac{t^j}{j!} \right)^k \left( \sum\_{j=0}^{\infty} y\_{j+1} \frac{t^j}{j!} \right)^r = \sum\_{n \ge k} B\_{n+r,k+r}^{(r)} (\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{x}\_k, y\_1, y\_2 \dots) \frac{t^n}{n!}. \tag{15}$$

Thus, we have

$$\begin{split} &B\_{n+r,k+r}^{(r)}(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots;y\_{1},y\_{2}\cdots) \\ &=\sum \left(\frac{n!}{k\_{1}!k\_{2}!\cdots} \left(\frac{\mathbf{x}\_{1}}{1!} \right)^{k\_{1}} \left(\frac{\mathbf{x}\_{2}}{2!} \right)^{k\_{2}}\cdots\right) \left(\frac{r!}{r\_{0}!r\_{1}!r\_{2}!\cdots} \left(\frac{y\_{1}}{0!} \right)^{r\_{0}} \left(\frac{y\_{2}}{1!} \right)^{r\_{1}} \left(\frac{y\_{3}}{2!} \right)^{r\_{2}}\cdots\right), \end{split} \tag{16}$$

where the summation is over all integers *k*1, *k*2, ···≥ 0 and *r*0,*r*1,*r*<sup>2</sup> ···≥ 0, such that

$$\sum\_{i\geq 1} k\_i = k\_r \quad \sum\_{j\geq 0} r\_j = r\_r \quad \text{and} \quad (k\_1 + r\_1) + 2(k\_2 + r\_2) + 3(k\_3 + r\_3) + \dots = n.$$

Let *a*1, *a*2, ··· , and *b*1, *b*2, ··· be any sequences of nonnegative integers. Then, as was noted in [20], *B*(*r*) *<sup>n</sup>*+*r*,*k*+*r*(*a*1, *a*2, ··· ; *b*1, *b*<sup>2</sup> ···) enumerates the number of partitions of a set with (*n* + *r*) elements into (*k* + *r*) blocks satisfying:


From (12) and (16), we note that

$$B\_{n+r,k+r}^{(r)}(1,1,\cdots;1,1,\cdots) = S\_2^{(r)}(n+k,k+r),\tag{17}$$

$$B\_{n+r,k+r}^{(r)}(a\mathbf{x}\_1, a\mathbf{x}\_2, \cdots; a\mathbf{y}\_1, a\mathbf{y}\_2 \cdot \cdots) = a^{k+r} B\_{n+k,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \cdots; \mathbf{y}\_1, \mathbf{y}\_2 \cdot \cdots),\tag{18}$$

and

$$B\_{n+r,k+r}^{(r)}(a\mathbf{x}\_1, a^2\mathbf{x}\_2, \dots, \mathbf{y}\_1, a\mathbf{y}\_2, a^2\mathbf{y}\_3, \dots) = a^n B\_{n+k,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{y}\_1, \mathbf{y}\_2, \dots),\tag{19}$$

where *α* is a real number.

By using (15), we get

$$\begin{split} \sum\_{n=k}^{\infty} B\_{n+r,k+r}^{(r)} (\mathbf{x}, 1, 0, 0, \cdots \; \vdots \; \mathbf{1}, 0, 0, \cdots \; \mathbf{1}) \frac{t^n}{n!} &= \frac{1}{k!} \left( \mathbf{x} t + \frac{t^2}{2} \right)^k \\ &= t^k \frac{1}{k!} \sum\_{n=0}^k \binom{k}{n} \left( \frac{t}{2} \right)^n \mathbf{x}^{k-n} \\ &= \sum\_{n=0}^k \frac{(n+k)!}{k!} \binom{k}{n} \left( \frac{1}{2} \right)^n \mathbf{x}^{k-n} \frac{t^{n+k}}{(n+k)!}. \end{split} \tag{20}$$

Also, it can be seen that

$$\sum\_{n=k}^{\infty} B\_{n+r,k+r}^{(r)}(\mathbf{x},1,0,0,\cdots;\mathbf{1},0,0,\cdots) \frac{t^n}{n!} = \sum\_{n=0}^{\infty} B\_{n+k+r,k+r}^{(r)}(\mathbf{x},1,0,0,\cdots;\mathbf{1},0,0,\cdots) \frac{t^{n+k}}{(n+k)!}.\tag{21}$$

Thus, by (20) and (21), we have the following equation given by

$$B\_{n+k+r,k+r}^{(r)}(\mathbf{x},1,0,0,\cdots;1,0,0,\cdots) = \begin{cases} \frac{(n+k)!}{k!} \binom{k}{n} \left(\frac{1}{2}\right)^n \mathbf{x}^{k-n}, & \text{if } 0 \le n \le k,\\ 0, & \text{if } n > k. \end{cases} \tag{22}$$

By replacing *n* by *n* − *k* in (22), we get

$$B\_{n+r,k+r}^{(r)}(\mathbf{x},1,0,0,\cdots;1,0,0,\cdots) = \frac{n!}{k!} \binom{k}{n-k} \mathbf{x}^{2k-n} \left(\frac{1}{2}\right)^{n-k}, \quad (k \le n \le 2k). \tag{23}$$

Now, we define the complete *r*-Bell polynomials by virtue of generating function as

$$\left(\exp\left(\sum\_{l=1}^{\infty} \mathbf{x}\_{l} \frac{t^{l}}{i!} \right) \left(\sum\_{j=0}^{\infty} y\_{j+1} \frac{t^{j}}{j!} \right)^{r} = \sum\_{n=0}^{\infty} B\_{n}^{(r)} (\mathbf{x}\_{1}, \mathbf{x}\_{2}, \cdots, \mathbf{y}\_{1}, y\_{2}, \cdots) \frac{t^{n}}{n!}. \tag{24}$$

From (15) and (24), we have

$$\begin{split} \sum\_{n=0}^{\infty} B\_{n}^{(r)}(\mathbf{x}\_{1}, \mathbf{x}\_{2}, \cdots; y\_{1}, y\_{2}, \cdots) \frac{t^{n}}{n!} &= \sum\_{k=0}^{\infty} \frac{1}{k!} \left( \sum\_{i=1}^{\infty} x\_{i} \frac{t^{i}}{i!} \right)^{k} \left( \sum\_{j=0}^{\infty} y\_{j+1} \frac{t^{j}}{j!} \right)^{r} \\ &= \sum\_{k=0}^{\infty} \sum\_{n=k}^{\infty} B\_{n+r,k+r}^{(r)}(x\_{1}, x\_{2}, \cdots; y\_{1}, y\_{2}, \cdots) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} B\_{n+r,k+r}^{(r)}(x\_{1}, x\_{2}, \cdots; y\_{1}, y\_{2}, \cdots) \frac{t^{n}}{n!} . \end{split} \tag{25}$$

Comparing both sides of (25) gives us the identity

$$B\_n^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{y}\_1, \mathbf{y}\_2, \dots \dots) = \sum\_{k=0}^n B\_{n+r,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{y}\_1, \mathbf{y}\_2, \dots \dots). \tag{26}$$

Now, we observe that

$$\begin{split} \mathbf{B}\_{n}^{(r)}(\mathbf{x}, \mathbf{x}, \cdots; \mathbf{1}, \mathbf{1}, \mathbf{1}, \cdots \text{ \cdot}) &= \sum\_{k=0}^{n} \mathbf{B}\_{n+r,k+r}^{(r)}(\mathbf{x}, \mathbf{x}, \cdots; \mathbf{1}, \mathbf{1}, \mathbf{1}, \cdots \text{ \cdot}) \\ &= \sum\_{k=0}^{n} \mathbf{x}^{k} \mathbf{B}\_{n+r,k+r}^{(r)}(\mathbf{1}, \mathbf{1}\_{\prime}, \cdots; \mathbf{1}, \mathbf{1}, \mathbf{1}, \cdots \text{ \cdot}) \\ &= \sum\_{k=0}^{n} \mathbf{x}^{k} \mathbf{S}\_{2}^{(r)}(n+r, k+r) \\ &= \mathbf{B}\_{n}^{(r)}(\mathbf{x}), \quad (n \ge 0). \end{split} \tag{27}$$

### **3. An Extended** *r***-Central Complete and Incomplete Bell Polynomials**

Recently, in [21], we initiated the study of central incomplete Bell polynomials *Tn*,*k*(*x*1, *<sup>x</sup>*2, ··· , *xn*−*k*+1) and the central complete Bell polynomials *B*(*c*) *<sup>n</sup>* (*x*|*x*1, *x*2, ··· , *xn*), respectively given by

$$\frac{1}{k!} \left( \sum\_{m=1}^{\infty} \frac{1}{2^m} (\mathbf{x}\_m - (-1)^m \mathbf{x}\_m) \frac{t^m}{m!} \right)^k = \sum\_{n=k}^{\infty} T\_{n,k}(\mathbf{x}\_1, \mathbf{x}\_2, \dots, \mathbf{x}\_{n-k+1}) \frac{t^n}{n!},$$

and

$$\exp\left(\mathbf{x}\sum\_{i=1}^{\infty}\frac{1}{2^i}(\mathbf{x}\_i-(-1)^i\mathbf{x}\_i)\frac{t^i}{i!}\right)=\sum\_{n=0}^{\infty}B\_n^{(c)}(\mathbf{x}|\mathbf{x}\_1,\mathbf{x}\_2,\cdots,\mathbf{x}\_n)\frac{t^n}{n!},$$

and studied some properties and identities concerning these polynomials. It was observed, in particular, that the number of partitioning a set with *n* elements into *k* blocks with odd sizes is given by the number of monomials appearing in *Tn*,*k*(*x*1, 2*x*2, ··· , 2*n*−*kxn*−*k*+1), and that the number of partitioning a set with *<sup>n</sup>* elements into a certain *k* blocks with odd sizes is the coefficient of the corresponding monomial appearing in *Tn*,*k*(*x*1, 2*x*2, ··· , 2*n*−*kxn*−*k*+1).

Here we will consider '*r*-extensions' of the central incomplete and complete Bell polynomials. In light of (15), we may define the extended *r*-central incomplete Bell polynomials by

$$\frac{1}{k!} \left( \sum\_{m=1}^{\infty} \left( \frac{1}{2} \right)^m (\mathbf{x}\_{\text{ill}} - (-1)^m \mathbf{x}\_{\text{ill}}) \frac{t^m}{m!} \right)^k \left( \sum\_{j=0}^{\infty} y\_{j+1} \frac{t^j}{j!} \right)^r = \sum\_{n=k}^{\infty} T\_{n+r,k+r}^{(r)} (\mathbf{x}\_1, \mathbf{x}\_2, \cdots, \mathbf{x}\_r; y\_1, y\_2, \cdots) \frac{t^n}{n!} \tag{28}$$

for any *k* ∈ N ∪ {0}. Then, for *n*, *k* ≥ 0 with *n* ≥ *k*, by (28), one can check that

$$\begin{split} T\_{n+r,k+r}^{(r)}(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots,\mathbf{y}\_{1},\mathbf{y}\_{2},\cdots \; \mathbf{y}) &= \sum \left( \frac{n!}{k\_{1}!k\_{3}!k\_{5}!\cdots} \left( \frac{\mathbf{x}\_{1}}{1!} \right)^{k\_{1}} \left( \frac{\mathbf{x}\_{3}}{2^{2}3!} \right)^{k\_{3}} \left( \frac{\mathbf{x}\_{5}}{2^{4}5!} \right)^{k\_{5}} \cdots \right) \\ &\quad \times \left( \frac{r!}{r\_{0}!r\_{1}!r\_{2}!\cdots} \left( \frac{\mathbf{y}\_{1}}{0!} \right)^{r\_{0}} \left( \frac{\mathbf{y}\_{2}}{1!} \right)^{r\_{1}} \left( \frac{\mathbf{y}\_{3}}{2!} \right)^{r\_{2}} \cdots \right), \end{split} \tag{29}$$

where the summation is over all integers *k*1, *k*3, *k*<sup>5</sup> ···≥ 0 and *r*0,*r*1,*r*<sup>2</sup> ···≥ 0, such that

$$\sum\_{i\geq 1} k\_{2i-1} = k\_\prime \sum\_{i\geq 0} r\_i = r\_\prime \text{ and } \sum\_{i\geq 1} (2i-1)k\_{2i-1} + \sum\_{i\geq 1} ir\_i = n. \tag{30}$$

The extended *r*-central incomplete Bell polynomials have the following combinatorial interpretation. This can be seen from (29). Let *a*1, *a*2, ··· , and *b*1, *b*2, ··· be any sequences of nonnegative integers. Then *T*(*r*) *<sup>n</sup>*+*r*,*k*+*r*(*a*1, 2*a*2, 2<sup>2</sup>*a*3, ··· ; *<sup>b</sup>*1, *<sup>b</sup>*2, *<sup>b</sup>*3, ···) enumerates the number of partitions of a set with (*<sup>n</sup>* <sup>+</sup> *<sup>r</sup>*) elements into *k* blocks of odd sizes and *r* blocks of any sizes satisfying:


From (15), (16) and (29), we note that

$$T\_{n+r,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \cdots, \mathbf{y}\_1, \mathbf{y}\_2, \cdots) = B\_{n+r,k+r}^{(r)}(\mathbf{x}\_1, 0, \frac{\mathbf{x}\_3}{2^2}, 0, \cdots, \mathbf{y}\_1, \mathbf{y}\_2, \mathbf{y}\_3, \cdots). \tag{31}$$

Therefore, we obtain the following theorem.

**Theorem 1.** *For n*, *k* ≥ 0*, with n* ≥ *k, we have*

$$T\_{n+r,k+r}^{(r)}(\mathbf{x}\_1,\mathbf{x}\_2,\cdots,\mathbf{y}\_1,\mathbf{y}\_2,\cdots) = B\_{n+r,k+r}^{(r)}(\mathbf{x}\_1,0,\frac{\mathbf{x}\_3}{2^2},0,\cdots,\mathbf{y}\_1,\mathbf{y}\_2,\mathbf{y}\_3\cdots).$$

From (28), we have

$$\begin{split} \sum\_{n=k}^{\infty} T\_{n+r,k+r}^{(r)} (1,1,\cdots;1,1,\cdots) \frac{t^n}{n!} &= \frac{1}{k!} \left( \sum\_{m=1}^{\infty} \left( \frac{1}{2} \right)^m (1-(-1)^m) \frac{t^m}{m!} \right)^k \left( \sum\_{j=0}^{\infty} \frac{t^j}{j!} \right)^r \\ &= \frac{1}{k!} \left( e^{\frac{t}{2}} - e^{-\frac{t}{2}} \right)^k e^{rt} \\ &= \sum\_{n=k}^{\infty} T^{(r)} (n+r, k+r) \frac{t^n}{n!} . \end{split} \tag{32}$$

Therefore, by (32), we obtain the following corollary.

**Corollary 1.** *For n*, *k* ≥ 0*, with n* ≥ *k, we have*

$$T\_{n+r,k+r}^{(r)}(1,1,\cdots \cdot;1,1,\cdots \cdot) = T^{(r)}(n+r,k+r), \quad (r \in \mathbb{N} \cup \{0\}).$$

Let *n*, *k* be nonnegative integers. Then, from (28), we get

$$\begin{split} \sum\_{n=k}^{\infty} T\_{n+r,k+r}^{(r)} (\mathbf{x}, \mathbf{x}^2, \mathbf{x}^3, \dots; \mathbf{1}, \mathbf{x}, \mathbf{x}^2, \dots) \frac{t^n}{n!} &= \frac{1}{k!} \left( \mathbf{x} t + \frac{\mathbf{x}^3}{2^2} \frac{t^3}{3!} + \frac{\mathbf{x}^5}{2^4} \frac{t^5}{5!} + \dotsb \right)^k \left( 1 + \mathbf{x} t + \frac{\mathbf{x}^2}{2} t^2 + \dotsb \right)^r \\ &= \frac{1}{k!} \left( e^{\frac{\mathbf{x}^2}{2}} - e^{-\frac{\mathbf{x}^3}{2}} \right)^k e^{\mathbf{x} \cdot \mathbf{t}} \\ &= \frac{1}{k!} \sum\_{l=0}^k \binom{k}{l} (-1)^{k-l} e^{(l+r-\frac{1}{2}) \mathbf{x} \cdot \mathbf{t}} \\ &= \sum\_{n=0}^\infty \frac{x^n}{k!} \sum\_{l=0}^k \binom{k}{l} (-1)^{k-l} \left( l + r - \frac{k}{2} \right)^n \frac{t^n}{n!} . \end{split} \tag{33}$$

Therefore, comparing both sides of (33) yields the following theorem.

**Theorem 2.** *For n*, *k* ≥ 0*, we have*

$$\frac{x^n}{k!} \sum\_{l=0}^k \binom{k}{l} (-1)^{k-l} \left(l + r - \frac{k}{2}\right)^n = \begin{cases} T\_{n+r,k+r}^{(r)}(x, x^2, x^3, \dots; 1, x, x^2, \dots; \dots), & \text{if } n \ge k, \\ 0, & \text{otherwise.} \end{cases}$$

In [10], Kim-Dolgy-Kim-Kim proved the following equation (34) given by

$$\frac{1}{k!} \sum\_{l=0}^{k} \binom{k}{l} (-1)^{k-l} \left( l + r - \frac{k}{2} \right)^n = \begin{cases} T^{(r)}(n + r, k + r), & \text{if } n \ge k, \\ 0, & \text{otherwise,} \end{cases} \tag{34}$$

where *n*, *k* ∈ Z with *n*, *k* ≥ 0. Therefore, by (34), the following corollary is established.

**Corollary 2.** *For n*, *k* ∈ N ∪ {0}*, with n* ≥ *k, we have*

$$T\_{n+r,k+r}^{(r)}(\mathbf{x},\mathbf{x}^2,\mathbf{x}^3,\cdots;\mathbf{1},\mathbf{x},\mathbf{x}^2,\cdots \ ) = \mathbf{x}^n T^{(r)}(n+r,k+r) \dots$$

From (29) and Corollary 2 , one can also have the following identity.

**Corollary 3.** *For n*, *k* ≥ 0*, with n* ≥ *k, we have*

$$T\_{n+r,k+r}^{(r)}(\mathbf{x},\mathbf{x}^2,\mathbf{x}^3,\cdots;\mathbf{1},\mathbf{x},\mathbf{x}^2,\cdots) = \mathbf{x}^{\mathbf{n}}T\_{n+r,k+r}^{(r)}(1,1,\cdots;\mathbf{1},1,\cdots\\\cdots)$$

*and*

$$\begin{split} T\_{n+r,k+r}^{(r)}(1,1,\cdots,\cdot;1,1,\cdots \; \cdot) &= T^{(r)}(n+r,k+r) \\ &= B\_{n+r,r}^{(r)}(1,0,\frac{1}{2^{2}},0,\frac{1}{2^{4}},0,\cdots \; \cdot;1,1,1,\cdots \; \cdot) \\ &= \sum \left( \frac{n!}{k\_{1}!k\_{3}!k\_{5}!\cdots} \left(\frac{1}{1!}\right)^{k\_{1}} \left(\frac{1}{2^{2}3!}\right)^{k\_{3}} \left(\frac{1}{2^{4}5!}\right)^{k\_{5}} \cdots \right) \\ &\quad \times \left( \frac{r!}{r\_{0}!r\_{1}!r\_{2}!\cdots} \left(\frac{1}{0!}\right)^{r\_{0}} \left(\frac{1}{1!}\right)^{r\_{1}} \left(\frac{1}{2!}\right)^{r\_{2}} \cdots \right), \end{split}$$

*where the summation is over all integers k*1, *k*3, *k*<sup>5</sup> ···≥ 0 *and r*0,*r*1,*r*<sup>2</sup> ···≥ 0*, satisfying the conditions in* (30)*.*

For *n*, *k* ≥ 0, we have

$$\sum\_{n=k}^{\infty} T\_{n+r,k+r}^{(r)}(\mathbf{x}, 1, 0, 0, \cdot, \cdots, \vdots, 1, 0, 0, \cdot, \cdots) \frac{t^n}{n!} = \frac{1}{k!} (\mathbf{x} t)^k. \tag{35}$$

By comparing the coefficients on both sides of (35), we have

$$T\_{n+r,k+r}^{(r)}(\mathbf{x},1,0,0,\cdots;1,0,0,\cdots) = \mathbf{x}^k \binom{0}{n-k}.\tag{36}$$

Also, by (29), one can obtain that

$$\begin{split} T\_{n+r,k+r}^{(r)}(\mathbf{x},\mathbf{x},\cdots,\mathbf{y},\mathbf{y},\cdots \mid \mathbf{y},\mathbf{y},\cdots \mid \mathbf{y} &= \mathbf{x}^k y^r T\_{n+r,k+r}^{(r)}(\mathbf{1},\mathbf{1},\cdots \mid \mathbf{1},\mathbf{1},\cdots \mid \mathbf{1},\mathbf{1},\cdots \mid \mathbf{y}) \\ &= \mathbf{x}^k y^r T^{(r)}(n+r,k+r), \end{split} \tag{37}$$

and

$$T\_{n+r,k+r}^{(r)}(\mathbf{a}\mathbf{x}\_1, \mathbf{a}\mathbf{x}\_2, \cdots \mathbf{ \cdot}; \mathbf{a}\mathbf{y}\_1, \mathbf{a}\mathbf{y}\_2, \cdots \mathbf{ \cdot}) = \mathbf{a}^{k+r} T\_{n+r,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \cdots, \mathbf{\cdot}; \mathbf{y}\_1, \mathbf{y}\_2, \cdots \mathbf{ \cdot}),\tag{38}$$

where *n*, *k* are nonnegative integers with *n* ≥ *k*.

Now, we observe that

$$\begin{split} \exp\left(\mathbf{x}\sum\_{i=1}^{\infty}\left(\frac{1}{2}\right)^{i}\left(\mathbf{x}\_{i}-(-1)^{i}\mathbf{x}\_{i}\right)\frac{t^{i}}{l!}\right)\left(\sum\_{j=0}^{\infty}y\_{j+1}\frac{t^{j}}{j!}\right)^{r} \\ &=\sum\_{k=0}^{\infty}\mathbf{x}^{k}\frac{1}{k!}\left(\sum\_{j=1}^{\infty}\left(\frac{1}{2}\right)^{i}\left(\mathbf{x}\_{i}-(-1)^{i}\mathbf{x}\_{i}\right)\frac{t^{i}}{l!}\right)^{k}\left(\sum\_{j=0}^{\infty}y\_{j+1}\frac{t^{j}}{j!}\right)^{r} \\ &=\sum\_{k=0}^{\infty}\mathbf{x}^{k}\sum\_{n=k}^{\infty}T\_{n+r,k+r}^{(r)}\left(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots,\mathbf{y}\_{1},\mathbf{y}\_{2},\cdots,\right)\frac{t^{n}}{n!} \\ &=\sum\_{k=0}^{\infty}\sum\_{k=0}^{n}\mathbf{x}^{k}T\_{n+r,k+r}^{(r)}\left(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots,\mathbf{y}\_{1},\mathbf{y}\_{2},\cdots,\right)\frac{t^{n}}{n!}. \end{split} \tag{39}$$

Taking (24) into account, we may define the extended *r*-central complete Bell polynomials by

$$\exp\left(\mathbf{x}\sum\_{i=1}^{\infty}\left(\frac{1}{2}\right)^{i}\left(\mathbf{x}\_{i}-(-1)^{i}\mathbf{x}\_{i}\right)\frac{t^{i}}{i!}\right)\left(\sum\_{j=0}^{\infty}y\_{j+1}\frac{t^{j}}{j!}\right)^{r}=\sum\_{k=0}^{\infty}B\_{n}^{(c,r)}(\mathbf{x}|\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots,\mathbf{y}\_{1},y\_{2},\cdots)\frac{t^{n}}{n!}.\tag{40}$$

In particular, when *<sup>x</sup>* <sup>=</sup> 1, *<sup>B</sup>*(*c*,*r*) *<sup>n</sup>* (1|*x*1, *<sup>x</sup>*2, ··· ; *<sup>y</sup>*1, *<sup>y</sup>*2, ···) = *<sup>B</sup>*(*c*,*r*) *<sup>n</sup>* (*x*1, *<sup>x</sup>*2, ··· ; *<sup>y</sup>*1, *<sup>y</sup>*2, ···) are called the extended *r*-central complete Bell numbers.

For *n* ≥ 0, by (39) and (40), we get

$$B\_n^{(c,r)}(\mathbf{x}\_1, \mathbf{x}\_2, \cdots, \mathbf{y}\_1, \mathbf{y}\_{2'}, \cdots) = \sum\_{k=0}^n T\_{n+r,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_{2'}, \cdots, \mathbf{y}\_{1'}, \mathbf{y}\_{2'}, \cdots) \tag{41}$$

and

$$B\_n^{(\varepsilon,r)}(\mathbf{x}|\mathbf{x}\_1,\mathbf{x}\_2,\cdots,\mathbf{y}\_1,\mathbf{y}\_2,\cdots) = \sum\_{k=0}^n \mathbf{x}^k T\_{n+r,k+r}^{(r)}(\mathbf{x}\_1,\mathbf{x}\_2,\cdots,\mathbf{y}\_1,\mathbf{y}\_2,\cdots).$$

It is easily noted that *B*(*c*,*r*) <sup>0</sup> (*x*1, *<sup>x</sup>*2, ··· ; *<sup>y</sup>*1, *<sup>y</sup>*2, ···) = *<sup>y</sup><sup>r</sup>* 1. Hence, one can have the following theorem.

**Theorem 3.** *For n* ≥ 0*, we have*

$$B\_n^{(c,r)}(\mathbf{x}|\mathbf{x}\_1, \mathbf{x}\_{2\prime}\cdots; \mathbf{y}\_1, \mathbf{y}\_{2\prime}\cdots) = \sum\_{k=0}^n \mathbf{x}^k T\_{n+r,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_{2\prime}\cdots; \mathbf{y}\_1, \mathbf{y}\_{2\prime}\cdots)$$

*and*

$$B\_{\mathfrak{n}}^{(c,r)}(\mathbf{x}\_1, \mathbf{x}\_2, \cdots, \mathbf{y}\_1, \mathbf{y}\_2, \cdots) = \sum\_{k=0}^{n} T\_{\mathfrak{n}+r,k+r}^{(r)}(\mathbf{x}\_1, \mathbf{x}\_2, \cdots, \mathbf{y}\_1, \mathbf{y}\_2, \cdots).$$

Please note that

$$\begin{aligned} B\_n^{(c,r)}(1,1,\cdots,\cdot;1,1,\cdots) &= \sum\_{k=0}^n T\_{n+r,k+r}^{(r)}(1,1,\cdots,\cdot;1,1,\cdots) \\ &= \sum\_{k=0}^n T^{(r)}(n+r,k+r) \\ &= B\_n^{(c,r)} \end{aligned}$$

and

$$B\_n^{(\circledast r)}(\mathbf{x}|1,1,\ldots,\cdots;1,1,\cdots) = \sum\_{k=0}^n \mathbf{x}^k T^{(r)}(n+r,k+r) = B\_n^{(\circledast r)}(\mathbf{x}), \quad (n \ge 0).$$

By (39), we get

$$\begin{split} & \exp\left(\sum\_{i=1}^{\infty} \left(\frac{1}{2}\right)^{i} \left(x\_{i} - (-1)^{i} x\_{i}\right) \frac{t^{i}}{i!}\right) \left(\sum\_{j=0}^{\infty} y\_{j+1} \frac{t^{j}}{j!}\right)^{r} \\ & \qquad = \sum\_{n=0}^{\infty} \left\{ \sum \left(\frac{n!}{k\_{1}! k\_{3}! k\_{5}! \cdots} \left(\frac{x\_{1}}{1!}\right)^{k\_{1}} \left(\frac{x\_{3}}{2^{2} 3!}\right)^{k\_{3}} \left(\frac{x\_{5}}{2^{4} 5!}\right)^{k\_{5}} \cdots \right) \right. \\ & \qquad \times \left(\frac{r!}{r\_{0}! r\_{1}! r\_{2}! \cdots} \left(\frac{y\_{1}}{0!}\right)^{r\_{0}} \left(\frac{y\_{2}}{1!}\right)^{r\_{1}} \left(\frac{y\_{3}}{2!}\right)^{r\_{2}} \cdots \right) \right\} \frac{t^{n}}{n!} \end{split} \tag{42}$$

where the inner sum runs over all integers *k*1, *k*3, *k*<sup>5</sup> ···≥ 0 and *r*0,*r*1,*r*<sup>2</sup> ···≥ 0, such that

$$\sum\_{i\geq 0} r\_i = r,\text{ and } \sum\_{i\geq 1} (2i - 1)k\_{2i - 1} + \sum\_{i\geq 1} ir\_i = n. \tag{43}$$

For *n* ≥ 0, we have

$$\begin{split} B\_{n}^{(c,r)}(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots;y\_{1},y\_{2},\cdots \; \mathbf{y}) &= \sum\_{k=0}^{n} T\_{n+r,k+r}^{(r)}(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots;y\_{1},y\_{2},\cdots \; \mathbf{y}) \\ &= \sum \left( \frac{n!}{k\_{1}!k\_{3}!k\_{3}!\cdots} \left( \frac{\mathbf{x}\_{1}}{1!} \right)^{k\_{1}} \left( \frac{\mathbf{x}\_{3}}{2^{2}3!} \right)^{k\_{3}} \left( \frac{\mathbf{x}\_{3}}{2^{4}5!} \right)^{k\_{3}} \cdots \right) \\ &\quad \times \left( \frac{r!}{r\_{0}!r\_{1}!r\_{2}!\cdots} \left( \frac{y\_{1}}{0!} \right)^{r\_{0}} \left( \frac{y\_{2}}{1!} \right)^{r\_{1}} \left( \frac{y\_{3}}{2!} \right)^{r\_{2}} \cdots \right), \end{split} \tag{44}$$

where the sum is over all integers *k*1, *k*3, *k*<sup>5</sup> ···≥ 0 and *r*0,*r*1,*r*<sup>2</sup> ···≥ 0, satisfying the conditions in (43). Thus, the following theorem is established.

**Theorem 4.** *For n* ≥ 0*, we have*

$$\begin{split} B\_{\boldsymbol{n}}^{(c,r)}(\mathbf{x}\_{1},\mathbf{x}\_{2},\cdots,\mathbf{y}\_{1},\mathbf{y}\_{2},\cdots) &= \sum\_{\mathbf{x}} \left( \frac{n!}{k\_{1}!k\_{3}!k\_{3}!\cdots} \left( \frac{\mathbf{x}\_{1}}{1!} \right)^{k\_{1}} \left( \frac{\mathbf{x}\_{3}}{2^{2}3!} \right)^{k\_{3}} \left( \frac{\mathbf{x}\_{5}}{2^{4}5!} \right)^{k\_{5}} \cdots \right) \\ &\quad \times \left( \frac{r!}{r\_{0}!r\_{1}!r\_{2}!\cdots} \left( \frac{\mathbf{y}\_{1}}{0!} \right)^{r\_{0}} \left( \frac{\mathbf{y}\_{2}}{1!} \right)^{r\_{1}} \left( \frac{\mathbf{y}\_{3}}{2!} \right)^{r\_{2}} \cdots \right), \end{split}$$

where the sum is over all integers *k*1, *k*3, *k*<sup>5</sup> ···≥ 0 and *r*0,*r*1,*r*<sup>2</sup> ···≥ 0, satisfying the conditions in (43). Now, we observe that

$$\begin{split} \exp\left(x\sum\_{i=1}^{\infty}\left(\frac{1}{2}\right)^{i}\left(1-(-1)^{i}\right)\frac{t^{i}}{i!}\right)\left(\sum\_{j=0}^{\infty}\frac{t^{j}}{j!}\right)^{r} &= \sum\_{k=0}^{\infty}x^{k}\frac{1}{k!}\left(\sum\_{i=1}^{\infty}\left(\frac{1}{2}\right)^{i}\left(1-(-1)^{i}\right)\frac{t^{i}}{i!}\right)^{k}\left(\sum\_{j=0}^{\infty}\frac{t^{j}}{j!}\right)^{r} \\ &= \sum\_{k=0}^{\infty}x^{k}\sum\_{n=k}^{\infty}T\_{n+r,k+r}^{(r)}(1,1,\cdots,\cdot;1,1,\cdots)\frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty}\sum\_{k=0}^{n}x^{k}T\_{n+r,k+r}^{(r)}(1,1,\cdots,\cdot;1,1,\cdots)\frac{t^{n}}{n!}. \end{split} \tag{45}$$

Alternatively, the left hand side of (45) can be simplified in the following way:

$$\exp\left(\mathbf{x}\sum\_{i=1}^{\infty}\left(\frac{1}{2}\right)^{i}\left(1-(-1)^{i}\right)\frac{t^{i}}{i!}\right)\left(\sum\_{j=0}^{\infty}\frac{t^{j}}{j!}\right)^{r} = e^{\mathbf{x}\left(\mathbf{e}^{\frac{t}{2}}-\mathbf{e}^{-\frac{t}{2}}\right)}\\e^{rt} = \sum\_{n=0}^{\infty}B\_{n}^{(c,r)}(\mathbf{x})\frac{t^{n}}{n!}.\tag{46}$$

Comparing the coefficients in (45) and (46) gives the following identity.

**Theorem 5.** *For n* ≥ 0*, we have*

$$\sum\_{k=0}^{n} \mathbf{x}^{k} T\_{n+r,k+r}^{(r)}(1,1,\cdots,\vdots;1,1,\cdots) = \mathcal{B}\_{n}^{(c,r)}(\mathbf{x})\,.$$

From (29), it is noted that

$$\begin{aligned} \sum\_{k=0}^{n} \mathbf{x}^{k+r} T\_{n+r,k+r}^{(r)} (\mathbf{1}, \mathbf{1}, \cdots, \mathbf{1}, \mathbf{1}, \cdots \cdot) &= \sum\_{k=0}^{n} T\_{n+r,k+r}^{(r)} (\mathbf{x}, \mathbf{x}, \cdots, \mathbf{x}, \mathbf{x}, \cdots \cdot) \\ &= B\_{n}^{(c,r)} (\mathbf{x}, \mathbf{x}, \cdots, \mathbf{x}, \mathbf{x}, \cdots \cdot), \end{aligned}$$

which yields the next corollary.

**Corollary 4.** *For n* ≥ 0*, we have*

$$B\_n^{(c,r)}(\mathfrak{x}, \mathfrak{x}, \cdot \cdot \cdot; \mathfrak{x}, \mathfrak{x}, \cdot \cdot \cdot) = \mathfrak{x}^r B\_n^{(c,r)}(\mathfrak{x}).$$

### **4. Conclusions**

In recent years, studies on various old and new special numbers and polynomials have received attention from many mathematicians. They have been carried out by several means, including generating functions, combinatorial methods, umbral calculus, *p*-adic analysis, differential equations, probability and so on.

In this paper, by making use of generating functions we introduced and studied the extended *r*-central incomplete and complete Bell polynomials, as multivariate versions of the recently studied the extended *r*-central factorial numbers of the second and the extended *r*-central Bell polynomials (see [1]), and also as multivariate versions of the *r*- Stirling numbers of the second kind and the extended *r*-Bell polynomials (see Section 2). Then we studied several properties, some identities and various explicit formulas related to these polynomials and also their connections.

In Section 1 we briefly recalled, in more detail, definitions and basic properties about the central factorial numbers of the second kind, the central Bell polynomials, the extended *r*-central factorial numbers of the second kind and the extended *r*-central Bell polynomials. In Section 2 we introduced the incomplete and complete *r*-Bell polynomials as multivariate versions of the *r*- Stirling numbers of the second kind and the extended *r*-Bell polynomials and give some of their simple properties. In Section 3, we introduced the extended *r*-central incomplete and complete Bell polynomials, and provided several properties, some identities and various explicit formulas for them.

As our immediate next project, we would like to find some further results about the extended *r*-central incomplete and complete Bell polynomials by virtue of umbral calculus and also some of their applications associated with partition polynomials.

**Author Contributions:** Conceptualization, D.S.K. and T.K.; Formal analysis, D.S.K., D.K. and T.K.; Funding acquisition, D.K.; Investigation, D.S.K., H.Y.K. and T.K.; Methodology, D.S.K. and T.K.; Project administration, T.K.; Software, D.K.; Supervision, D.S.K. and T.K.; Validation, D.S.K., H.Y.K., D.K. and T.K.; Visualization, H.Y.K. and D.K.; Writing—original draft, T.K.; Writing— review & editing, D.S.K., H.Y.K. and D.K.

**Funding:** This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2019R1C1C1003869).

**Conflicts of Interest:** The authors declare no conflict of interest.

**Acknowledgments:** We would like to thank the referees for their comments and suggestions which improved the original manuscript greatly.

### **References**



© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **Some Identities of Fully Degenerate Bernoulli Polynomials Associated with Degenerate Bernstein Polynomials**

### **Jeong Gon Lee 1, Wonjoo Kim 2,\* and Lee-Chae Jang <sup>3</sup>**


Received: 26 April 2019; Accepted: 20 May 2019; Published: 24 May 2019

**Abstract:** In this paper, we investigate some properties and identities for fully degenerate Bernoulli polynomials in connection with degenerate Bernstein polynomials by means of bosonic *p*-adic integrals on Z*<sup>p</sup>* and generating functions. Furthermore, we study two variable degenerate Bernstein polynomials and the degenerate Bernstein operators.

**Keywords:** degenerate Bernoulli polynomials; degenerate Bernstein operators

### **1. Introduction**

Let *p* be a fixed prime number. Throughout this paper, Z, Z*p*, Q*<sup>p</sup>* and C*p*, will denote the ring of rational integers, the ring of *p*-adic integers, the field of *p*-adic rational numbers and the completion of algebraic closure of <sup>Q</sup>*p*, respectively. The *<sup>p</sup>*-adic norm <sup>|</sup>*q*|*<sup>p</sup>* is normalized as <sup>|</sup>*p*|*<sup>p</sup>* <sup>=</sup> <sup>1</sup> *p* .

For *λ*, *t* ∈ C*<sup>p</sup>* with |*λt*|*<sup>p</sup>* < *p* <sup>−</sup> <sup>1</sup> *<sup>p</sup>*−<sup>1</sup> and |*t*|*<sup>p</sup>* < 1, the degenerate Bernoulli polynomials are defined by the generating function to be

$$\frac{t}{(1+\lambda t)^{\frac{1}{\lambda}}-1}(1+\lambda t)^{\frac{x}{\lambda}}=\sum\_{n=0}^{\infty}\beta\_n(x|\lambda)\frac{t^n}{n!},\tag{1}$$

(See [1–3]). When *x* = 0, *βn*(*λ*) = *βn*(0|*λ*) are called the degenerate Bernoulli numbers. Note that lim*λ*→<sup>0</sup> *βn*(*x*|*λ*) = *Bn*(*x*), where *Bn*(*x*) are the ordinary Bernoulli polynomials defined by

$$\frac{t}{e^t - 1} e^{\mathbf{x}t} = \sum\_{n=0}^{\infty} B\_n(\mathbf{x}) \frac{t^n}{n!} \,\tag{2}$$

and *Bn* = *Bn*(0) are called the Bernoulli numbers. The degenerate exponential function is defined by

$$
\sigma\_{\lambda}^{x}(t) = (1 + \lambda t)^{\frac{x}{\lambda}} = \sum\_{n=0}^{\infty} (x)\_{n,\lambda} \frac{t^n}{n!} \tag{3}
$$

where (*x*)0,*<sup>λ</sup>* = 1, (*x*)*n*,*<sup>λ</sup>* = *x*(*x* − *λ*)(*x* − 2*λ*)···(*x* − (*n* − 1)*λ*), for *n* ≥ 1. From (1), we get

$$\beta\_n(\mathbf{x}|\boldsymbol{\lambda}) = \sum\_{l=0}^n \binom{n}{l} \beta\_l(\boldsymbol{\lambda}) (\mathbf{x})\_{n-l,\boldsymbol{\lambda}}.\tag{4}$$

Recentely, Kim-Kim introduced the degenerate Bernstein polynomials given by

$$\frac{\lambda(x)\_{k,\lambda}}{k!}t^k(1+\lambda t)^{\frac{1-x}{\lambda}}=\sum\_{n=k}^{\infty}B\_{k,n}(x|\lambda)\frac{t^n}{n!},\tag{5}$$

(See [4–6]). Thus, by (5), we note that

$$B\_{k,n}(\mathbf{x}|\lambda) = \begin{cases} \binom{n}{k} (\mathbf{x})\_{k,\lambda} (1-\mathbf{x})\_{n-k,\lambda'} & \text{if } n \ge k, \\\ 0, & \text{if } n < k. \end{cases} \tag{6}$$

where *n*, *k* are non-negative integers. Let *UD*(Z*p*) be the space of uniformly differentiable functions on Z*p*. For *f* ∈ *UD*(Z*p*), the degenerate Bernstein operator of order *n* is given by

$$\begin{split} \mathbb{B}\_{\mathfrak{n},\lambda}(f|\lambda) &= \sum\_{k=0}^{\infty} f\left(\frac{k}{n}\right) \binom{n}{k} (\mathbf{x})\_{k,\lambda} (1-\mathbf{x})\_{\mathfrak{n}-k,\lambda} \\ &= \sum\_{k=0}^{\infty} f\left(\frac{k}{n}\right) B\_{k,\mathfrak{n}}(\mathbf{x}|\lambda)\_{\mathfrak{n}} \end{split} \tag{7}$$

(See [4–6]). The bosonic *p*-adic integral on Z*<sup>p</sup>* is defined by Volkenborn as

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x})d\mu\_0(\mathbf{x}) = \lim\_{N \to \infty} \frac{1}{p^N} \sum\_{\mathbf{x}=\mathbf{0}}^{p^N-1} f(\mathbf{x}),\tag{8}$$

(see [7]). By (8), we get

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x} + 1) d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} f(\mathbf{x}) d\mu\_0(\mathbf{x}) = f'(0),\tag{9}$$

where *<sup>d</sup> dx f*(*x*) *<sup>x</sup>*=<sup>0</sup> = *f* (0).

From (8), Kim-Seo [8] proposed fully degenerate Bernoulli polynomials which are reformulated in terms of bosonic *p*-adic integral on Z*<sup>p</sup>* as

$$\int\_{\mathbb{Z}\_p} (1 + \lambda t)^{\frac{x+y}{\lambda}} d\mu\_0(y) = \frac{\frac{1}{\lambda} \log(1 + \lambda t)}{(1 + \lambda t)^{\frac{1}{\lambda}} - 1} (1 + \lambda t)^{\frac{x}{\lambda}} = \sum\_{n=0}^{\infty} B\_n(x|\lambda) \frac{t^n}{n!} \tag{10}$$

and for *x* = 0, *Bn*(*λ*) = *Bn*(0|*λ*) are called fully degenerate Bernoulli numbers.

Note that the fully degenerate Bernoulli polynomial was named Daehee polynomials with *α*-parameter in [9]. On the other hand,

$$\int\_{\mathbb{Z}\_p} (1 + \lambda t)^{\frac{x+y}{\lambda}} d\mu\_0(y) = \sum\_{n=0}^{\infty} \int\_{\mathbb{Z}\_p} (x + y)\_{n\lambda} d\mu\_0(y) \frac{t^n}{n!}.\tag{11}$$

By (10) and (11), we get

$$\int\_{\mathbb{Z}\_p} (\mathbf{x} + \mathbf{y})\_{n,\lambda} d\mu\_0(\mathbf{y}) = B\_n(\mathbf{x}|\lambda), \ (n \ge 0). \tag{12}$$

Recall that the Daehee polynomials are defined by the generating function to be

$$\frac{\log\left(1+t\right)}{t}\left(1+t\right)^{x}=\sum\_{n=0}^{\infty}D\_{n}(x)\frac{t^{n}}{n!}\tag{13}$$

and for *x* = 0, *Dn* = *Dn*(0) are called the Daehee numbers (see [10,11]).

Also, the higher order Daehee polynomials are defined by the generating function to be

$$\left(\frac{\log(1+t)}{t}\right)^k (1+t)^x = \sum\_{n=0}^{\infty} D\_n^{(k)}(x) \frac{t^n}{n!} \tag{14}$$

and for *x* = 0, *D*(*k*) *<sup>n</sup>* <sup>=</sup> *<sup>D</sup>*(*k*) *<sup>n</sup>* (0) are called the higher order Daehee numbers. From (10), we observe

$$\begin{split} \frac{\frac{1}{\lambda}\log(1+\lambda t)}{(1+\lambda t)^{\frac{1}{\lambda}}-1}(1+\lambda t)^{\frac{1}{\lambda}} &= \frac{t}{(1+\lambda t)^{\frac{1}{\lambda}}-1}(1+\lambda t)^{\frac{1}{\lambda}}\frac{\log(1+\lambda t)}{\lambda t} \\ &= \left(\sum\_{m=0}^{\infty}\beta\_{m}(\mathbf{x}|\lambda)\frac{t^{m}}{m!}\right)\left(\sum\_{l=0}^{\infty}D\_{l}\frac{(\lambda t)^{l}}{l!}\right) \\ &= \sum\_{n=0}^{\infty}\left(\sum\_{m=0}^{n}\binom{n}{m}\beta\_{m}(\mathbf{x}|\lambda)D\_{n-m}\lambda^{n-m}\right)\frac{t^{n}}{n!}. \end{split} \tag{15}$$

By (10) and (14), we get

$$B\_{\boldsymbol{n}}(\mathbf{x}|\boldsymbol{\lambda}) = \sum\_{m=0}^{n} \binom{n}{m} \beta\_{\boldsymbol{m}}(\mathbf{x}|\boldsymbol{\lambda}) D\_{\boldsymbol{n}-\boldsymbol{m}} \boldsymbol{\lambda}^{n-m}, \quad (n \ge 0). \tag{16}$$

From (3) and (10), we observe that

$$\begin{split} \sum\_{n=0}^{\infty} B\_n(\mathbf{x}|\lambda) \frac{t^n}{n!} &= \frac{\frac{1}{\lambda} \log(1+\lambda t)}{(1+\lambda t)^{\frac{1}{\lambda}} - 1} (1+\lambda t)^{\frac{x}{\lambda}} \\ &= \left( \sum\_{m=0}^{\infty} B\_m(\lambda) \frac{t^m}{m!} \right) \left( \sum\_{l=0}^{\infty} (\mathbf{x})\_{l,\lambda} \frac{t^l}{l!} \right) \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^n \binom{n}{m} B\_m(\lambda) (\mathbf{x})\_{n-m,\lambda} \right) \frac{t^n}{n!} . \end{split} \tag{17}$$

By (17), we get

$$B\_{\rm ll}(\mathbf{x}|\lambda) = \sum\_{m=0}^{n} \binom{n}{m} B\_{\rm m}(\lambda) (\mathbf{x})\_{n-m,\lambda\prime} \quad (n \ge 0). \tag{18}$$

From (1) and (3), we note that

$$\begin{split} t &= \left( (1 + \lambda t)^{\frac{1}{\lambda}} - 1 \right) \sum\_{m=0}^{\infty} \beta\_m(\lambda) \frac{t^m}{m!} \\ &= \left( \sum\_{l=0}^{\infty} (1)\_{l,\lambda} \frac{t^l}{l!} \right) \left( \sum\_{m=0}^{\infty} \beta\_m(\lambda) \frac{t^m}{m!} \right) - \sum\_{m=0}^{\infty} \beta\_m(\lambda) \frac{t^m}{m!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^{n} \binom{n}{m} (1)\_{n-m,\lambda} \beta\_m(\lambda) \right) \frac{t^n}{n!} - \sum\_{n=0}^{\infty} \beta\_n(\lambda) \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^{n} \binom{n}{m} (1)\_{n-m,\lambda} \beta\_m(\lambda) - \beta\_n(\lambda) \right) \frac{t^n}{n!} .\end{split} \tag{19}$$

Comparing the cofficients on both sides of (19), we get

$$\sum\_{m=0}^{n} \binom{n}{m} (1)\_{n-m,\lambda} \beta\_m(\lambda) - \beta\_n(\lambda) = \delta\_{1,n\nu} \quad (n \ge 0),\tag{20}$$

where *δk*,*<sup>n</sup>* is the Kronecker's symbol.

By (4) and (20), we have

$$
\beta\_{\mathfrak{n}}(1|\lambda) - \beta\_{\mathfrak{n}}(\lambda) = \delta\_{1,\mathfrak{n}}.\tag{21}
$$

The generating function of fully degenerate Bernoulli polynomials introduced in (5) can be expressed as bosonic *p*-adic integral but the generating function of degenerate Bernoulli polynomials introduced in (1) is not expressed as a bosonic *p*-adic integral. This is why we considered the fully degenerate Bernoulli polynomials, and the motivation of this paper is to investigate some identities of them associated with degenerate Bernstein polynomials.

In this paper, we consider the fully degenerate Bernoulli polynomials and investigate some properties and identities for these polynomials in connection with degenerate Bernstein polynomials by means of bosonic *p*-adic integrals on Z*<sup>p</sup>* and generating functions. Furthermore, we study two variable degenerate Bernstein polynomials and the degenerate Bernstein operators.

### **2. Fully Degenerate Bernoulli and Bernstein Polynomials**

From (10), we observe that

$$\begin{split} \sum\_{n=0}^{\infty} B\_n (1 - x | \lambda) \frac{t^n}{n!} &= \frac{\frac{1}{\lambda} \log(1 + \lambda t)}{(1 + \lambda t)^{\frac{1}{\lambda}} - 1} (1 + \lambda t)^{\frac{1 - x}{\lambda}} \\ &= \frac{\left( -\frac{1}{\lambda} \right) (1 + (-\lambda)(-t))}{(1 + (-\lambda)(-t))^{-\frac{1}{\lambda}} - 1} \left( 1 + (-\lambda)(-t) \right)^{-\frac{1}{\lambda}} \\ &= \sum\_{n=0}^{\infty} B\_n (x| - \lambda)(-1)^n \frac{t^n}{n!} .\end{split} \tag{22}$$

From (22), we obtain the following Lemma.

**Lemma 1.** *For n* ∈ N ∪ {0}*, we have*

$$B\_{\rm ll}(1-\mathbf{x}|\lambda) = (-1)^{\rm n} B\_{\rm n}(\mathbf{x}|-\lambda). \tag{23}$$

From (16) and (21), we get

$$\begin{split} B\_n(1|\lambda) - B\_n(\lambda) &= \sum\_{m=0}^n \binom{n}{m} \left( \beta\_m (1|\lambda) - \beta\_m(\lambda) \right) D\_{n-m} \lambda^{n-m} \\ &= \sum\_{m=0}^n \binom{n}{m} \delta\_{1,m} D\_{n-m} \lambda^{n-m}, \quad (n \ge 0). \end{split} \tag{24}$$

From (1), we observe that

$$\begin{split} \sum\_{n=0}^{\infty} \beta\_{\mathbb{H}} (\mathbf{x} + 1 | \boldsymbol{\lambda}) \frac{t^{n}}{n!} &= (1 + \boldsymbol{\lambda} t)^{\frac{1}{\lambda}} \left( \sum\_{m=0}^{\infty} \beta\_{\mathbb{H}} (\mathbf{x} | \boldsymbol{\lambda}) \frac{t^{m}}{m!} \right) \\ &= \left( \sum\_{l=0}^{\infty} (1)\_{l,\boldsymbol{\lambda}} \frac{t^{l}}{l!} \right) \left( \sum\_{m=0}^{\infty} \beta\_{\mathbb{H}} (\mathbf{x} | \boldsymbol{\lambda}) \frac{t^{m}}{m!} \right) \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^{n} \binom{n}{m} (1)\_{n-m,\boldsymbol{\lambda}} \beta\_{\mathbb{H}} (\mathbf{x} | \boldsymbol{\lambda}) \right) \frac{t^{n}}{n!} . \end{split} \tag{25}$$

By (25), we get

$$\beta\_{\mathbb{H}}(\mathbf{x} + \mathbf{1}|\lambda) = \sum\_{m=0}^{n} \binom{n}{m} (1)\_{n-m,\lambda} \beta\_{\mathbb{H}}(\mathbf{x}|\lambda). \tag{26}$$

By (26), with *x* = 1, we have

$$\begin{split} \beta\_{n}(2|\lambda) &= \sum\_{m=0}^{n} \binom{n}{m} \beta\_{m}(1|\lambda)(1)\_{n-m,\lambda} \\ &= (1)\_{n,\lambda} \beta\_{0}(1|\lambda) + n(1)\_{n-1,\lambda} \beta\_{1}(1|\lambda) + \sum\_{m=2}^{n} \binom{n}{m} \beta\_{m}(1|\lambda)(1)\_{n-m,\lambda} \\ &= (1)\_{n,\lambda} + n(1)\_{n-1,\lambda} \left(\beta\_{1}(\lambda) - 1\right) + \sum\_{m=2}^{n} \binom{n}{m} \beta\_{m}(\lambda)(1)\_{n-m,\lambda} \\ &= (1)\_{n,\lambda} + n(1)\_{n-1,\lambda} \beta\_{1}(\lambda) - n(1)\_{n-1,\lambda} + \sum\_{m=2}^{n} \binom{n}{m} \beta\_{m}(\lambda)(1)\_{n-m,\lambda} \\ &= -n(1)\_{n-1,\lambda} + \sum\_{m=0}^{n} \binom{n}{m} \beta\_{m}(\lambda)(1)\_{n-m,\lambda} \\ &= -n(1)\_{n-1,\lambda} + \beta\_{1}(1|\lambda). \end{split} \tag{27}$$

Therefore, by (27), we obtain the following theorem.

**Theorem 1.** *For n* ∈ N*, we have*

$$
\beta\_n(2|\lambda) = -n(1)\_{n-1,\lambda} + \beta\_n(1|\lambda). \tag{28}
$$

Note that

$$(1 - \mathbf{x})\_{n, \lambda} = (-1)^n (\mathbf{x} - 1)\_{n, -\lambda}, \quad (n \ge 0). \tag{29}$$

Therefore by (12), (23), and (29), we get

$$\int\_{\mathbb{Z}\_p} (1-\mathbf{x})\_{n,\lambda} d\mu\_0(\mathbf{x}) = (-1)^n \int\_{\mathbb{Z}\_p} (\mathbf{x}-1)\_{n,-\lambda} d\mu\_0(\mathbf{x}) = \int\_{\mathbb{Z}\_p} (\mathbf{x}+2)\_{n,\lambda} d\mu\_0(\mathbf{x}).\tag{30}$$

Therefore, by (30) and Theorem 1, we obtain the following theorem.

**Theorem 2.** *For n* ∈ N*, we have*

$$\int\_{\mathbb{Z}\_p} (1-\mathbf{x})\_{\boldsymbol{n},\lambda} d\mu\_0(\mathbf{x}) = \int\_{\mathbb{Z}\_p} (\mathbf{x}+\mathbf{2})\_{\boldsymbol{n},\lambda} d\mu\_0(\mathbf{x}) = n(1)\_{\boldsymbol{n}-1,\lambda} (\lambda-1) B\_1(\lambda) + \int\_{\mathbb{Z}\_p} (\mathbf{x})\_{\boldsymbol{n},\lambda} d\mu\_0(\mathbf{x}).\tag{31}$$

**Corollary 1.** *For n* ∈ N*, we have*

$$n(-1)^{\text{ll}}B\_{\text{ll}}(-1|-\lambda) = (1)\_{n-1,\lambda} \left(1 - nB\_1(\lambda)\right) + B\_{\text{ll}}(\lambda) = B\_{\text{ll}}(2|\lambda).\tag{32}$$

By (17), we get

$$\begin{split} B\_{\mathfrak{m}}(1-\mathfrak{x}|\lambda) &= \sum\_{m=0}^{n} \binom{n}{m} B\_{\mathfrak{m}}(\lambda)(1-\mathfrak{x})\_{n-m,\lambda} \\ &= \sum\_{m=0}^{n} \binom{n}{m} (\mathfrak{x})\_{m,\lambda} (1-\mathfrak{x})\_{n-m,\lambda} \frac{B\_{\mathfrak{m}}(\lambda)}{(\mathfrak{x})\_{m,\lambda}} \\ &= \sum\_{m=0}^{n} B\_{\mathfrak{m},\mathfrak{m}}(\mathfrak{x}|\lambda) B\_{\mathfrak{m}}(\lambda) \frac{1}{(\mathfrak{x})\_{m,\lambda}}. \end{split} \tag{33}$$

In [8], we note that

$$\begin{split} \frac{1}{(\mathbf{x})\_{m,\lambda}} &= \frac{1}{\mathbf{x}(\mathbf{x}-\boldsymbol{\lambda})(\mathbf{x}-2\boldsymbol{\lambda})\cdots(\mathbf{x}-(m-1)\boldsymbol{\lambda})} \\ &= \sum\_{k=0}^{m-1} \frac{(-1)^{k}}{(m-1)!} \binom{m-1}{k} \frac{(-\boldsymbol{\lambda})^{1-m}}{\mathbf{x}-k\boldsymbol{\lambda}}, \quad (m \in \mathbb{N}). \end{split} \tag{34}$$

By (33) and (34) we get

$$\begin{split} B\_n(1-\mathbf{x}|\lambda) &= \sum\_{m=0}^n B\_{m,n}(\mathbf{x}|\lambda) B\_m(\lambda) \frac{1}{(\mathbf{x})\_{m,\lambda}} \\ &= (1-\mathbf{x})\_{n,\lambda} + \sum\_{m=1}^n B\_{m,n}(\mathbf{x}|\lambda) B\_m(\lambda) \frac{1}{(\mathbf{x})\_{m,\lambda}} \\ &= (1-\mathbf{x})\_{n,\lambda} + \sum\_{m=1}^n B\_{m,n}(\mathbf{x}|\lambda) B\_m(\lambda) \frac{(-\lambda)^{1-m}}{(m-1)!} \sum\_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{\mathbf{x}-k\lambda} .\end{split} \tag{35}$$

Therefore, by (35), we obtain the following theorem.

**Theorem 3.** *For n* ∈ N ∪ {0}*, we have*

$$B\_{\rm nl}(1-\mathbf{x}|\lambda) = (1-\mathbf{x})\_{\rm n,\lambda} + \sum\_{m=1}^{n} B\_{\rm m,n}(\mathbf{x}|\lambda) B\_{\rm m}(\lambda) \frac{(-\lambda)^{1-m}}{(m-1)!} \sum\_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{\mathbf{x}-k\lambda}.\tag{36}$$

**Corollary 2.** *For n* ∈ N ∪ {0}*, we have*

$$B\_n(2|\lambda) = (2)\_{n,\lambda} + \sum\_{m=1}^n B\_{m,n}(-1|\lambda) B\_m(\lambda) \frac{(-\lambda)^{1-m}}{(m-1)!} \sum\_{k=0}^{m-1} (-1)^{k+1} \binom{m-1}{k} \frac{1}{1+k\lambda}.\tag{37}$$

For *k* ∈ N, the higher-order fully degenerate Bernoulli polynomials are given by the generating function

$$\left(\frac{\frac{1}{\lambda}\log(1+\lambda t)}{(1+\lambda t)^{\frac{1}{\lambda}}-1}\right)^k (1+\lambda t)^{\frac{\lambda}{\lambda}} = \sum\_{n=0}^{\infty} B\_n^{(k)}(x|\lambda) \frac{t^n}{n!} \tag{38}$$

(See [8,12,13]). When *x* = 0, *B*(*k*) *<sup>n</sup>* (*λ*) = *<sup>B</sup>*(*k*) *<sup>n</sup>* (*x*|0) are called the higher-order fully degenerate Bernoulli numbers. From (5) and (38), we note that

$$\begin{split} \left(\frac{\log(1+\lambda t)}{\lambda t}\right)^k \sum\_{n=k}^{\infty} B\_{k,n}(x|\lambda) \frac{t^n}{n!} &= (x)\_{k,\lambda} t^k (1+\lambda t)^{\frac{k-1}{k}} \left(\frac{\log(1+\lambda t)}{\lambda t}\right)^k \frac{1}{k!} \\ &= \frac{\left((1+\lambda t)^{\frac{k}{k}} - 1\right)^k}{\left((1+\lambda t)^{\frac{k}{k}} - 1\right)^k} (x)\_{k,\lambda} \left(\frac{1}{\lambda} \log(1+\lambda t)\right)^k (1+\lambda t)^{\frac{k-1}{k}} \frac{1}{k!} \\ &= (x)\_{k,\lambda} \sum\_{m=0}^k \binom{k}{m} (-1)^{m-k} (1+\lambda t)^{\frac{k}{k}} \left(\frac{\frac{1}{k}\log(1+\lambda t)}{(1+\lambda t)^{\frac{k}{k}} - 1}\right)^k (1+\lambda t)^{\frac{k-1}{k}} \frac{1}{k!} \\ &= (x)\_{k,\lambda} \sum\_{m=0}^k \binom{k}{m} (-1)^{m-k} \frac{\left(\frac{1}{k}\log(1+\lambda t)\right)^k}{(1+\lambda t)^{\frac{k}{k}} - 1} (1+\lambda t)^{\frac{k-1}{k}} \frac{1}{k!} \\ &= (x)\_{k,\lambda} \sum\_{m=0}^k \binom{k}{m} (-1)^{m-k} \sum\_{n=0}^\infty B\_n (1-x+m|\lambda|) \frac{t^n}{n!} \frac{1}{k!} \\ &= \sum\_{n=0}^\infty \binom{k}{0} \zeta\_{k,\lambda} \sum\_{m=0}^k \binom{k}{m} (-1)^{m-k} B\_n (1-x+m|\lambda|) \frac{1}{k!} \Big$$

and hence, we get

$$\begin{split} \left(\frac{\log(1+\lambda t)}{\lambda t}\right)^k \sum\_{m=k}^{\infty} B\_{k,m}(\mathbf{x}|\lambda) \frac{t^m}{m!} &= \left(\sum\_{l=0}^{\infty} D\_l^{(k)} \lambda^l \frac{t^l}{l!}\right) \left(\sum\_{m=k}^{\infty} B\_{k,m}(\mathbf{x}|\lambda) \frac{t^m}{m!}\right) \\ &= \sum\_{n=k}^{\infty} \left(\sum\_{l=0}^n D\_l^{(k)} \lambda^l B\_{k,n-l}(\mathbf{x}|\lambda)\right) \frac{t^n}{n!} .\end{split} \tag{40}$$

Therefore, by (39) and (40), we obtain the following theorem.

**Theorem 4.** *For k*, *n* ∈ N*, we have*

$$\frac{1}{k!} (\mathbf{x})\_{n,\lambda} \sum\_{m=0}^{k} \binom{k}{m} (-1)^{m-k} B\_n (1 - \mathbf{x} + m|\lambda) = \begin{cases} \sum\_{l=0}^{n} D\_l^{(k)} \lambda^l B\_{k,n-l} (\mathbf{x}|\lambda), & \text{if } n \ge k, \\ 0, & \text{if } n < k. \end{cases} \tag{41}$$

Let *f* ∈ *UD*(Z*p*). For *x*1, *x*<sup>2</sup> ∈ Z*p*, we consider the degenerate Bernstein operator of order *n* given by

$$\mathbb{B}\_{n,\lambda}(f|\mathbf{x}\_1,\mathbf{x}\_2) = \sum\_{k=0}^n f\left(\frac{k}{n}\right) \binom{n}{k} (\mathbf{x}\_1)\_{k,\lambda} (\mathbf{1} - \mathbf{x}\_2)\_{n-k,\lambda} = \sum\_{k=0}^n f\left(\frac{k}{n}\right) B\_{k,n}(\mathbf{x}\_1,\mathbf{x}\_2|\lambda)\_{\prime} \tag{42}$$

where *Bn*,*k*(*x*1, *x*2|*λ*) are called two variable degenerate Bernstein polynomials of degree *n* as followings (see, [2–6,9,14–27]):

$$B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) = \binom{n}{k} (\mathbf{x}\_1)\_{k,\lambda} (1 - \mathbf{x}\_2)\_{n-k,\lambda} \quad (n \ge 0). \tag{43}$$

The authors [3] obtained the following:

$$\sum\_{k=0}^{\infty} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) \frac{t^n}{n!} = \frac{(\mathbf{x}\_1)\_{k,\lambda}}{k!} t^k e\_{\lambda}^{1-\mathbf{x}\_2}(t). \tag{44}$$

The authors [8] obtained the following:

$$\begin{split} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) &= \binom{n}{k} \left( 1 - (1 - \mathbf{x}\_1) \right)\_{n - (n - k), \lambda} (1 - \mathbf{x}\_2)\_{n - k, \lambda} \\ &= B\_{n - k, n} (1 - \mathbf{x}\_2, 1 - \mathbf{x}\_1 | \lambda), \end{split} \tag{45}$$

and

$$\begin{split} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) &= (1 - \mathbf{x}\_2 - (n - k - 1)\lambda) B\_{k,n-1}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) \\ &+ (\mathbf{x}\_1 - (k - 1)\lambda) B\_{k-1,n-1}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda). \end{split} \tag{46}$$

From (42), we note that *x*1, *x*<sup>2</sup> ∈ Z*p*, if *f*(*x*) = 1, then we have

$$\begin{split} \mathbb{B}\_{n,\lambda}(1|\mathbf{x}\_1, \mathbf{x}\_2) &= \sum\_{k=0}^{n} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) \\ &= \sum\_{k=0}^{n} \binom{n}{k} (\mathbf{x}\_1)\_{k,\lambda} (1 - \mathbf{x}\_2)\_{n-k,\lambda} \\ &= (1 + \mathbf{x}\_1 - \mathbf{x}\_2)\_{n,\lambda} \end{split} \tag{47}$$

and if *f*(*t*) = *t*, then we have

$$\mathbb{B}\_{n,\lambda}(t|\mathbf{x}\_1,\mathbf{x}\_2) = (\mathbf{x}\_1)\_{1,\lambda}(\mathbf{x}\_1 + 1 - \lambda - \mathbf{x}\_2)\_{n-1,\lambda} \tag{48}$$

and if *f*(*t*) = *t* 2, then we have

$$\mathbb{B}\_{n,\lambda}(t^2|\mathbf{x}\_1,\mathbf{x}\_2) = \frac{1}{n}(\mathbf{x}\_1)\_{1,\lambda}(\mathbf{x}\_1 + 1 - \lambda - \mathbf{x}\_2)\_{n-1,\lambda} + \frac{n-1}{n}(\mathbf{x}\_1)\_{2,\lambda}(1 + \mathbf{x}\_2 - 2\lambda - \mathbf{x}\_2)\_{n-2,\lambda}.\tag{49}$$

The authors [3] obtained the following:

$$\mathbf{B}(\mathbf{x})\_{1,\lambda} = \frac{1}{(\mathbf{x}\_1 + 1 - \lambda - \mathbf{x}\_2)\_{n-1,\lambda}} \mathbb{B}\_n(t|\mathbf{x}\_1, \mathbf{x}\_2), \tag{50}$$

and

$$f\_{\mathbf{x}}(\mathbf{x})\_{2,\lambda} = \frac{1}{(\mathbf{x}\_1 + 1 - 2\lambda - \mathbf{x}\_2)\_{n-2,\lambda}} \sum\_{k=2}^n \frac{\binom{k}{2}}{\binom{n}{2}} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda), \tag{51}$$

and

$$f\_i(\mathbf{x})\_{i,\lambda} = \frac{1}{(1 + \mathbf{x}\_1 - \mathbf{x}\_2 - i\lambda)\_{n-i,\lambda}} \sum\_{k=i}^n \frac{\binom{k}{i}}{\binom{n}{i}} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda). \tag{52}$$

Taking double bosonic *p*-adic integral on Z*p*, we get the following equation:

$$\int\_{\mathbb{Z}\_p} \int\_{\mathbb{Z}\_p} B\_{k, \mathfrak{l}}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) d\mu\_0(\mathbf{x}\_1) d\mu\_0(\mathbf{x}\_2) = \binom{n}{k} \int\_{\mathbb{Z}\_p} (\mathbf{x}\_1)\_{k, \lambda} d\mu\_0(\mathbf{x}\_1) \int\_{\mathbb{Z}\_p} (1 - \mathbf{x}\_2)\_{n - k, \lambda} d\mu\_0(\mathbf{x}\_2). \tag{53}$$

Therefore, by (53) and Theorem 2, we obtain the following theorem.

**Theorem 5.** *For n*, *k* ∈ N ∪ {0}*, we have*

$$\begin{cases} \int\_{\mathbb{Z}\_p} \int\_{\mathbb{Z}\_p} B\_{k,\mathfrak{l}}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) d\mu\_0(\mathbf{x}\_1) d\mu\_0(\mathbf{x}\_2) \\\\ = \begin{cases} \binom{n}{k} B\_{\mathfrak{l}}(\lambda) \left( (1)\_{n-1,\mathfrak{l}} n(\lambda - 1) B\_{\mathfrak{n}}(\lambda) + B\_{n-k}(\lambda) \right), & \text{if } n > k, \\ B\_{\mathfrak{n}}(\lambda), & \text{if } n = k. \end{cases} \end{cases} \tag{54}$$

We get from the symmetric properties of two variable degenerate Bernstein polynomials that for *n*, *k* ∈ N with *n* > *k*,

$$\begin{split} \int\_{\mathbb{Z}\_p} \int\_{\mathbb{Z}\_p} B\_{k,n}(\mathbf{x}\_1, \mathbf{x}\_2 | \lambda) d\mu\_0(\mathbf{x}\_1) d\mu\_0(\mathbf{x}\_2) \\ = \sum\_{m=0}^k \binom{n}{k} \binom{k}{m} (-1)^{k+m} (1)\_{m,\lambda} \\ \quad \times \int\_{\mathbb{Z}\_p} \int\_{\mathbb{Z}\_p} (1-\mathbf{x}\_1)\_{k-m,-\lambda} (1-\mathbf{x}\_2)\_{n-k,\lambda} d\mu\_0(\mathbf{x}\_1) d\mu\_0(\mathbf{x}\_2) \\ = \binom{n}{k} \int\_{\mathbb{Z}\_p} (1-\mathbf{x}\_2)\_{n-k} d\mu\_0(\mathbf{x}\_2) \sum\_{m=0}^k \binom{n}{k} \binom{k}{m} (-1)^{k+m} (1)\_{m,\lambda} d\mu\_0(\mathbf{x}\_2) \\ \quad \times \left\{ (1)\_{k-m,-\lambda} (k-m)(-\lambda-1) B\_1(-\lambda) + \int\_{\mathbb{Z}\_p} (\mathbf{x}\_1)\_{k-m,-\lambda} d\mu\_0(\mathbf{x}\_1) \right\} \\ = \binom{n}{k} B\_{n-k,\lambda}(2) \sum\_{m=0}^k \binom{n}{k} \binom{k}{m} (-1)^{k+m} (1)\_{m,\lambda} \\ \quad \times \left\{ (1)\_{k-m,-\lambda} (k-m)(-\lambda-1) B\_1(-\lambda) + B\_{k-m,-\lambda}(2) \right\} \end{split} (55)$$

Therefore, by Theorem 5, we obtain the following theorem.

**Theorem 6.** *For n*, *k* ∈ N ∪ {0}*, we have the following identities:*

*1. If n* > *k, then we have*

$$\begin{split} &B\_{n}\left((1)\_{n-1,\lambda}n(\lambda-1)B\_{1}(\lambda)+B\_{n-k}(\lambda)\right) \\ &=B\_{n-k,\lambda}(2)\sum\_{m=0}^{k}\binom{n}{k}\binom{k}{m}(-1)^{k+m}(1)\_{m,\lambda} \\ &\quad \times \left((1)\_{k-m,-\lambda}(k-m)(-\lambda-1)B\_{1}(-\lambda)+B\_{k-m,-\lambda}(2)\right). \end{split} \tag{56}$$

*2. If n* = *k, then we have*

$$B\_k(\lambda) = \sum\_{m=0}^k \binom{k}{n} (1)^{k+m} (1)\_{k,\lambda} \left( (1)\_{k-m,-\lambda} (k-m)(-\lambda -1) B\_1(-\lambda) + B\_{k-m,-\lambda}(2) \right). \tag{57}$$

### **3. Remark**

Let us assume that the probability of success in an experiment is *p*. We wondered if we could say the probability of success in the 9th trial is still *p* after failing eight times in a ten trial experiment, because there is a psychological burden to be successful. It seems plausible that the probability is less than *p*. The degenerate Bernstein polynomial *Bn*(*x*|*λ*) is used in the probability of success. Thus, we give examples in our results as follows:

**Example 1.** *Let n* = 2*, we have*

$$\begin{aligned} B\_2(2|\lambda) &= 2(1)\_{1,\lambda}(\lambda - 1)B\_1(\lambda) + B\_2(\lambda) \\ &= 2(\lambda - 1)\left(-\frac{1}{2}\right) + \frac{\lambda}{2} + \frac{1}{6} \\ &= -\frac{\lambda}{2} + \frac{7}{6} .\end{aligned}$$

**Example 2.** *Let n* = 1*, we have*

$$\begin{aligned} B\_1(1-\mathbf{x}|\lambda) &= (1-\mathbf{x})\_{1,\lambda} + \sum\_{m=1}^{1} B\_{m,1}(\mathbf{x}|\lambda) B\_m(\lambda) \frac{(-1)^{1-m}}{(m-1)!} \sum\_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{\mathbf{x}-k\lambda} \\ &= (1-\mathbf{x})\_{1,\lambda} + B\_{1,1}(\mathbf{x}|\lambda) B\_1(\lambda) \frac{1}{\mathbf{x}} \\ &= -\mathbf{x} + \frac{1}{2} .\end{aligned}$$

**Example 3.** *Let n* = 1*, k* = 2*, we have*

$$\begin{aligned} \mathbf{x}(\mathbf{x})\_{1,\lambda} \sum\_{m=0}^{2} \binom{2}{m} (-1)^{m-2} B\_1 (1 - \mathbf{x} + m|\lambda) &= \mathbf{x} \left( B\_1 (1 - \mathbf{x}|\lambda) - 2B\_1 (2 - \mathbf{x}|\lambda) + B\_1 (3 - \mathbf{x}|\lambda) \right) \\ &= -\mathbf{x} \left( \left( -\mathbf{x} + \frac{1}{2} \right) - 2 \left( -\mathbf{x} + \frac{3}{2} \right) + \left( -\mathbf{x} + \frac{5}{2} \right) \right) \\ &= 0. \end{aligned}$$

#### **4. Conclusions**

In this paper, we studied the fully degenerate Bernoulli polynomials associated with degenerate Bernstein polynomials. In Section 1, Equations (12), (18), (20) and (21) are some properties of them. In Section 2, Theorems 1–3 are results of identities for fully degenerate Bernoulli polynomials in connection with degenerate Bernstein polynomials by means of bosonic *p*-adic integrals on Z*<sup>p</sup>*

and generating functions. Theorems 4–6 are results of higher-order fully Bernoulli polynomials in connection with two variable degenerate Bernstein polynomials by means of bosonic *p*-adic integrals on Z*<sup>p</sup>* and generating functions.

**Author Contributions:** Conceptualization, W.K. and L.-C.J.; Data curation, L.-C.J.; Formal analysis, L.-C.J.; Funding acquisition, J.G.L.; Investigation, J.G.L., W.K. and L.-C.J.; Methodology, W.K. and L.-C.J.; Project administration, L.-C.J.; Resources, L.-C.J.; Supervision, L.-C.J.; Visualization, L.-C.J.; Writing—original draft, W.K. and L.-C.J.; Writing—review & editing, J.G.L. and L.-C.J.

**Funding:** This paper was supported by Wonkwang University in 2018.

**Conflicts of Interest:** The authors declare that they have no competing interests.

### **References**


© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **The Power Sums Involving Fibonacci Polynomials and Their Applications**

### **Li Chen <sup>1</sup> and Xiao Wang 1,\***


Received: 28 March 2019 ; Accepted: 25 April 2019; Published: 6 May 2019

**Abstract:** The Girard and Waring formula and mathematical induction are used to study a problem involving the sums of powers of Fibonacci polynomials in this paper, and we give it interesting divisible properties. As an application of our result, we also prove a generalized conclusion proposed by R. S. Melham.

**Keywords:** Fibonacci polynomials; Lucas polynomials; sums of powers; divisible properties; R. S. Melham's conjectures

**MSC:** 11B39

### **1. Introduction**

For any integer *n* ≥ 0, the famous Fibonacci polynomials {*Fn*(*x*)} and Lucas polynomials {*Ln*(*x*)} are defined as *F*0(*x*) = 0, *F*1(*x*) = 1, *L*0(*x*) = 2, *L*1(*x*) = *x* and *Fn*+2(*x*) = *xFn*+1(*x*) + *Fn*(*x*), *Ln*+2(*x*) = *xLn*+1(*x*) + *Ln*(*x*) for all *<sup>n</sup>* <sup>≥</sup> 0. Now, if we let *<sup>α</sup>* <sup>=</sup> *<sup>x</sup>*<sup>+</sup> <sup>√</sup>*x*2+<sup>4</sup> <sup>2</sup> and *<sup>β</sup>* <sup>=</sup> *<sup>x</sup>*<sup>−</sup> <sup>√</sup>*x*2+<sup>4</sup> <sup>2</sup> , then it is easy to prove that

$$F\_n(\mathbf{x}) = \frac{1}{n - \beta} \left(\boldsymbol{\alpha}^n - \beta^n\right) \text{ and } \ L\_n(\mathbf{x}) = \boldsymbol{\alpha}^n + \beta^n \text{ for all } n \ge 0.$$

If *x* = 1, we have that {*Fn*(*x*)} turns into Fibonacci sequences {*Fn*}, and {*Ln*(*x*)} turns into Lucas sequences {*Ln*}. If *x* = 2, then *Fn*(2) = *Pn*, the *n*th Pell numbers, they are defined by *P*<sup>0</sup> = 0, *P*<sup>1</sup> = 1 and *Pn*+<sup>2</sup> = 2*Pn*+<sup>1</sup> + *Pn* for all *n* ≥ 0. In fact, {*Fn*(*x*)} is a second-order linear recursive polynomial, when *x* takes a different value *x*0, then *Fn*(*x*0) can become a different sequence.

Since the Fibonacci numbers and Lucas numbers occupy significant positions in combinatorial mathematics and elementary number theory, they are thus studied by plenty of researchers, and have gained a large number of vital conclusions; some of them can be found in References [1–15]. For example, Yi Yuan and Zhang Wenpeng [1] studied the properties of the Fibonacci polynomials, and proved some interesting identities involving Fibonacci numbers and Lucas numbers. Ma Rong and Zhang Wenpeng [2] also studied the properties of the Chebyshev polynomials, and obtained some meaningful formulas about the Chebyshev polynomials and Fibonacci numbers. Kiyota Ozeki [3] got some identity involving sums of powers of Fibonacci numbers. That is, he proved that

$$\sum\_{k=1}^{n} F\_{2k}^{2m+1} = \frac{1}{5^m} \sum\_{j=0}^{m} \frac{(-1)^j}{L\_{2m+1-2j}} \binom{2m+1}{j} \left( F\_{(2m+1-2j)(2n+1)} - F\_{2m+1-2j} \right) \dots$$

Helmut Prodinger [4] extended the result of Kiyota Ozeki [3].

In addition, regarding many orthogonal polynomials and famous sequences, Kim et al. have done a lot of important research work, obtaining a series of interesting identities. Interested readers can refer to References [16–22]; we will not list them one by one.

In this paper, our main purpose is to care about the divisibility properties of the Fibonacci polynomials. This idea originated from R. S. Melham. In fact, in [5], R. S. Melham proposed two interesting conjectures as follows:

**Conjecture 1.** *If m* ≥ 1 *is a positive integer, then the summation*

$$L\_1 L\_3 L\_5 \cdots \cdot L\_{2m+1} \sum\_{k=1}^{n} F\_{2k}^{2m+1}$$

*can be written as* (*F*2*n*+<sup>1</sup> − 1) <sup>2</sup> *<sup>P</sup>*2*m*−<sup>1</sup> (*F*2*n*+1)*, where <sup>P</sup>*2*m*−1(*x*) *is an integer coefficients polynomial with degree* 2*m* − 1*.*

**Conjecture 2.** *If m* ≥ 0 *is an integer, then the summation*

$$L\_1 L\_3 L\_5 \cdots \cdot L\_{2m+1} \sum\_{k=1}^n L\_{2k}^{2m+1}$$

*can be written as* (*L*2*n*+<sup>1</sup> − 1) *Q*2*<sup>m</sup>* (*L*2*n*+1)*, where Q*2*m*(*x*) *is an integer coefficients polynomial with degree* 2*m.*

Wang Tingting and Zhang Wenpeng [6] solved Conjecture 2 completely. They also proved a weaker conclusion for Conjecture 1. That is,

$$L\_1 L\_3 L\_5 \cdots \cdot L\_{2m+1} \sum\_{k=1}^n F\_{2k}^{2m+1}$$

can be expressed as (*F*2*n*+<sup>1</sup> − 1) *P*2*<sup>m</sup>* (*F*2*n*+1), where *P*2*m*(*x*) is a polynomial of degree 2*m* with integer coefficients.

Sun et al. [7] solved Conjecture 1 completely. In fact, Ozeki [3] and Prodinger [4] indicated that the odd power sum of the first several consecutive Fibonacci numbers of even order is equivalent to the polynomial estimated at a Fibonacci number of odd order. Sun et al. in [7] proved that this polynomial and its derivative both disappear at 1, and it can be an integer polynomial when a product of the first consecutive Lucas numbers of odd order multiplies it. This presents an affirmative answer to Conjecture 1 of Melham.

In this paper, we are going to use a new and different method to study this problem, and give a generalized conclusion. That is, we will use the Girard and Waring formula and mathematical induction to prove the conclusions in the following:

**Theorem 1.** *If n and h are positive integers, then we have the congruence*

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\cdot\cdots L\_{2n+1}(\mathbf{x})\sum\_{m=1}^h F\_{2m}^{2n+1}(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right)^2 \dots$$

Taking *x* = 1 and *x* = 2 in Theorem 1, we can instantly infer the two corollaries:

**Corollary 1.** *Let Fn and Ln be Fibonacci numbers and Lucas numbers, respectively. Then, for any positive integers n and h, we have the congruence*

$$L\_1 L\_3 L\_5 \cdots \cdot L\_{2n+1} \sum\_{m=1}^h F\_{2m}^{2n+1} \equiv 0 \bmod \left( F\_{2l+1} - 1 \right)^2 \dots$$

**Corollary 2.** *Let Pn be nth Pell numbers. Then, for any positive integers n and h, we have the congruence*

$$L\_1(2)L\_3(2)L\_5(2)\cdots L\_{2n+1}(2)\sum\_{m=1}^h P\_{2m}^{2n+1} \equiv 0 \bmod \left(P\_{2h+1} - 1\right)^2 \pmod{p}$$

*where Ln*(2) = <sup>1</sup> <sup>+</sup> <sup>√</sup><sup>2</sup> *n* + <sup>1</sup> <sup>−</sup> <sup>√</sup><sup>2</sup> *n is called nth Pell–Lucas numbers.*

It is clear that our Corollary 1 gave a new proof for Conjecture 1.

### **2. Several Lemmas**

In this part, we will give four simple lemmas, which are essential to prove our main results.

**Lemma 1.** *Let h be any positive integer; then, we have*

$$\left(\mathbf{x}^2 + 4\mathbf{\hat{z}}F\_{2h+1}(\mathbf{x}) - 1\right) = 1\mathbf{\hat{z}}$$

*where x*<sup>2</sup> + <sup>4</sup> *and F*2*h*+1(*x*) − <sup>1</sup> *are said to be relatively prime.*

**Proof.** From the definition of *Fn*(*x*) and binomial theorem, we have

$$F\_{2h+1}(\mathbf{x}) \quad = \frac{1}{2^{2h+1}\sqrt{\mathbf{x}^2 + 4}} \sum\_{k=0}^{2h+1} \binom{2h+1}{k} \mathbf{x}^k \left(\mathbf{x}^2 + 4\right)^{\frac{2h+1-k}{2}}$$

$$-\frac{1}{2^{2h+1}\sqrt{\mathbf{x}^2 + 4}} \sum\_{k=0}^{2h+1} \binom{2h+1}{k} \mathbf{x}^k (-1)^{2h+1-k} \left(\mathbf{x}^2 + 4\right)^{\frac{2h+1-k}{2}}$$

$$= \frac{1}{4^h} \sum\_{k=0}^h \binom{2h+1}{2k} \mathbf{x}^{2k} \left(\mathbf{x}^2 + 4\right)^{h-k}.\tag{1}$$

Thus, from Equation (1), we have the polynomial congruence

$$4^h F\_{2h+1}(\mathbf{x}) = \sum\_{k=0}^h \binom{2h+1}{2k} \mathbf{x}^{2k} \left(\mathbf{x}^2 + 4\right)^{h-k} \equiv (2h+1)\mathbf{x}^{2h}$$

$$\equiv \quad (2h+1)\left(\mathbf{x}^2 + 4 - 4\right)^h \equiv (2h+1)(-4)^h \bmod \left(\mathbf{x}^2 + 4\right)$$

or

$$F\_{2h+1}(\mathbf{x}) - 1 \equiv (2h+1)(-1)^h - 1 \bmod (\mathbf{x}^2 + 4). \tag{2}$$

Since *<sup>x</sup>*<sup>2</sup> + 4 is an irreducible polynomial of *<sup>x</sup>*, and (2*<sup>h</sup>* + <sup>1</sup>)(−1)*<sup>h</sup>* − 1 is not divisible by (*x*<sup>2</sup> + <sup>4</sup>) for all integer *h* ≥ 1, so, from (2), we can deduce that

$$\left(\mathfrak{x}^2 + 4, F\_{2h+1}(\mathfrak{x}) - 1\right) = 1.$$

Lemma 1 is proved.

**Lemma 2.** *Let h and n be non-negative integers with h* ≥ 1*; then, we have*

$$F(\mathbf{x}^2 + 4)F\_{(2n+1)(2n+1)}(\mathbf{x}) - L\_{2n}(\mathbf{x}) - L\_{2n+2}(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right) \dots$$

**Proof.** We use mathematical induction to calculate the polynomial congruence for *n*. Noting *L*0(*x*) = 2, *L*1(*x*) = *x*, *L*2(*x*) = *x*<sup>2</sup> + 2. Thus, if *n* = 0, then

$$\begin{aligned} & \left(\mathbf{x}^2 + 4\right) F\_{(2\mathbf{h}+1)(2\mathbf{n}+1)}(\mathbf{x}) - L\_{2\mathbf{n}}(\mathbf{x}) - L\_{2\mathbf{n}+2}(\mathbf{x}) \\ &= \quad \left(\mathbf{x}^2 + 4\right) F\_{2\mathbf{h}+1}(\mathbf{x}) - 2 - \mathbf{x}^2 - 2 \\ &= \quad \left(\mathbf{x}^2 + 4\right) \left(F\_{2\mathbf{h}+1}(\mathbf{x}) - 1\right) \equiv 0 \text{ mod } \left(F\_{2\mathbf{h}+1}(\mathbf{x}) - 1\right) .\end{aligned}$$

If *n* = 1, then *L*2(*x*) + *L*4(*x*) = *x*<sup>2</sup> + 2 + *x*<sup>4</sup> + 4*x*<sup>2</sup> + 2 = *x*<sup>4</sup> + 5*x*<sup>2</sup> + 4. Note that the identity *F*<sup>3</sup> <sup>2</sup>*h*+1(*x*) = 1 *x*2+4 *F*3(2*h*+1)(*x*) + 3*F*2*h*+1(*x*) , so we obtain the congruence

$$\begin{aligned} &(\mathbf{x}^2 + 4)F\_{(2h+1)(2n+1)}(\mathbf{x}) - L\_{2h}(\mathbf{x}) - L\_{2n+2}(\mathbf{x}) \\ &= \quad \quad (\mathbf{x}^2 + 4)F\_{3(2h+1)}(\mathbf{x}) - \mathbf{x}^4 - 5\mathbf{x}^2 - 4 \\ &= \quad \quad \quad (\mathbf{x}^2 + 4)\left[ (\mathbf{x}^2 + 4)F\_{2h+1}^3(\mathbf{x}) - 3F\_{2h+1}(\mathbf{x}) \right] - \mathbf{x}^4 - 5\mathbf{x}^2 - 4 \\ &= \quad \quad \quad (\mathbf{x}^2 + 4)^2 \left[ F\_{2h+1}^3(\mathbf{x}) - F\_{2h+1}(\mathbf{x}) \right] + (\mathbf{x}^2 + 4)(\mathbf{x}^2 + 1)F\_{2h+1}(\mathbf{x}) - \mathbf{x}^4 - 5\mathbf{x}^2 - 4 \\ &\equiv \quad \quad (\mathbf{x}^2 + 4)^2 \left( F\_{2h+1}^2(\mathbf{x}) + F\_{2h+1}(\mathbf{x}) \right) \left( F\_{2h+1}(\mathbf{x}) - 1 \right) \equiv 0 \text{ mod } \left( F\_{2h+1}(\mathbf{x}) - 1 \right), \end{aligned}$$

which means that Lemma 2 is correct for *n* = 0 and 1.

Assume Lemma 2 is right for all integers *n* = 0, 1, 2, ··· , *k*. Namely,

$$F(\mathbf{x}^2 + 4)F\_{(2h+1)(2n+1)}(\mathbf{x}) - L\_{2n}(\mathbf{x}) - L\_{2n+2}(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right),\tag{3}$$

where 0 ≤ *n* ≤ *k*.

Thus, *n* = *k* + 1 ≥ 2, and we notice that

$$L\_{2(2l+1)}(\mathbf{x})F\_{(2l+1)(2k+1)}(\mathbf{x}) = F\_{(2l+1)(2k+3)}(\mathbf{x}) + F\_{(2l+1)(2k-1)}(\mathbf{x}),$$

$$L\_{2k+2}(\mathbf{x}) + L\_{2k+4}(\mathbf{x}) = (\mathbf{x}^2 + 2)L\_{2k}(\mathbf{x}) + (\mathbf{x}^2 + 2)L\_{2k+2}(\mathbf{x}) - (L\_{2k-2}(\mathbf{x}) + L\_{2k}(\mathbf{x})) $$

and

$$L\_{2(2h+1)}(\mathbf{x}) = (\mathbf{x}^2 + 4)F\_{2h+1}^2(\mathbf{x}) - 2 \equiv \mathbf{x}^2 + 2 \bmod \left( F\_{2h+1}(\mathbf{x}) - 1 \right) \dots$$

From inductive assumption (3), we have

$$\begin{split} & \quad (\mathbf{x}^2 + 4) F\_{(2k+1)(2n+1)}(\mathbf{x}) - L\_{2k}(\mathbf{x}) - L\_{2n+2}(\mathbf{x}) \\ &= \quad (\mathbf{x}^2 + 4) F\_{(2k+1)(2k+3)}(\mathbf{x}) - L\_{2k+2}(\mathbf{x}) - L\_{2k+4}(\mathbf{x}) \\ &= \quad (\mathbf{x}^2 + 4) L\_{2(2k+1)}(\mathbf{x}) F\_{(2k+1)(2k+1)}(\mathbf{x}) - (\mathbf{x}^2 + 4) F\_{(2k+1)(2k-1)} - L\_{2k+2}(\mathbf{x}) - L\_{2k+4}(\mathbf{x}) \\ &\equiv \quad (\mathbf{x}^2 + 4)(\mathbf{x}^2 + 2) F\_{(2k+1)(2k+1)}(\mathbf{x}) - (\mathbf{x}^2 + 2) L\_{2k}(\mathbf{x}) - (\mathbf{x}^2 + 2) L\_{2k+2}(\mathbf{x}) \\ &\quad - (\mathbf{x}^2 + 4) F\_{(2k+1)(2k-1)}(\mathbf{x}) + L\_{2k-2}(\mathbf{x}) + L\_{2k}(\mathbf{x}) \\ &\equiv \quad (\mathbf{x}^2 + 2) \left[ (\mathbf{x}^2 + 4) F\_{(2k+1)(2k+1)}(\mathbf{x}) - L\_{2k}(\mathbf{x}) - L\_{2k+2}(\mathbf{x}) \right] \\ &\quad - \left[ (\mathbf{x}^2 + 4) F\_{(2k+1)(2k-1)}(\mathbf{x}) - L\_{2k-2}(\mathbf{x}) - L\_{2k}(\mathbf{x}) \right] \\ &\equiv \quad 0 \bmod \left( F\_{2k+1}(\mathbf{x}) - 1 \right) . \end{split}$$

Now, we have achieved the results of Lemma 2.

**Lemma 3.** *Let h and n be non-negative integers with h* ≥ 1*; then, we have the polynomial congruence*

$$\begin{aligned} &L\_1(\mathbf{x})L\_3(\mathbf{x})\cdots L\_{2n+1}(\mathbf{x})\sum\_{m=1}^h \left[F\_{2m(2n+1)}(\mathbf{x}) - (2n+1)F\_{2m}(\mathbf{x})\right], \\ &\equiv \quad 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right)^2. \end{aligned}$$

**Proof.** For positive integer *<sup>n</sup>*, first note that *αβ* = −1, *Ln*(*x*) = *<sup>α</sup><sup>n</sup>* + *<sup>β</sup>n*,

$$\sum\_{m=1}^{h} F\_{2m(2n+1)}(\mathbf{x}) = \frac{1}{\sqrt{\mathbf{x}^2 + 4}} \sum\_{m=1}^{h} \left[ \mathfrak{a}^{2m(2n+1)} - \mathfrak{f}^{2m(2n+1)} \right]$$

$$= \frac{1}{\sqrt{\mathbf{x}^2 + 4}} \left[ \frac{\mathfrak{a}^{2(2n+1)} \left( \mathfrak{a}^{2h(2n+1)} - 1 \right)}{\mathfrak{a}^{2(2n+1)} - 1} - \frac{\mathfrak{f}^{2(2n+1)} \left( \mathfrak{f}^{2h(2n+1)} - 1 \right)}{\mathfrak{f}^{2(2n+1)} - 1} \right]$$

$$= -\frac{1}{L\_{2n+1}(\mathbf{x})} \left[ F\_{(2h+1)(2n+1)}(\mathbf{x}) - F\_{2n+1}(\mathbf{x}) \right] \tag{4}$$

and

$$\sum\_{m=1}^{h} F\_{2m}(\mathbf{x}) = \frac{1}{\sqrt{\mathbf{x}^2 + 4}} \sum\_{m=1}^{h} \left[ a^{2m} - \beta^{2m} \right] = \frac{1}{L\_1(\mathbf{x})} \left[ F\_{(2h+1)}(\mathbf{x}) - 1 \right]. \tag{5}$$

Thus, from Labels (4) and (5), we know that, to prove Lemma 3, now we need to obtain the polynomial congruence

$$L\_1(\mathbf{x}) \left( F\_{(2h+1)(2n+1)}(\mathbf{x}) - F\_{2n+1}(\mathbf{x}) \right) - (2n+1)L\_{2n+1}(\mathbf{x}) \left( F\_{2h+1}(\mathbf{x}) - 1 \right)$$

$$\equiv 0 \bmod \left( F\_{2h+1}(\mathbf{x}) - 1 \right)^2. \tag{6}$$

Now, we prove (6) by mathematical induction. If *n* = 0, then it is obvious that (6) is correct. If *n* = 1, we notice that *L*1(*x*) = *x*, *F*3(2*h*+1)(*x*)=(*x*<sup>2</sup> + 4)*F*<sup>3</sup> <sup>2</sup>*h*+1(*x*) − <sup>3</sup>*F*2*h*+1(*x*) and *<sup>F</sup>*<sup>3</sup> <sup>2</sup>*h*+1(*x*) ≡ (*F*2*h*+1(*x*) − 1 + 1) <sup>3</sup> <sup>≡</sup> <sup>3</sup>*F*2*h*+1(*x*) <sup>−</sup> 2 mod (*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) <sup>2</sup> we have

*L*1(*x*)*F*(2*h*+1)(2*n*+1)(*x*) − *L*1(*x*)*F*2*n*+1(*x*) − (2*n* + 1)*L*2*n*+1(*x*)(*F*2*h*+1(*x*) − 1) = *xF*3(2*h*+1)(*x*) − *xF*3(*x*) − 3*L*3(*x*)(*F*2*h*+1(*x*) − 1) = *x*(*x*<sup>2</sup> + 4)*F*<sup>3</sup> <sup>2</sup>*h*+1(*x*) <sup>−</sup> <sup>3</sup>*xF*2*h*+1(*x*) <sup>−</sup> *<sup>x</sup>*(*x*<sup>2</sup> <sup>+</sup> <sup>1</sup>) <sup>−</sup> <sup>3</sup>(*x*<sup>3</sup> <sup>+</sup> <sup>3</sup>*x*)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) <sup>≡</sup> (*x*<sup>3</sup> <sup>+</sup> <sup>4</sup>*x*)(3*F*2*h*+1(*x*) <sup>−</sup> <sup>2</sup>) <sup>−</sup> <sup>3</sup>*xF*2*h*+1(*x*) <sup>−</sup> (*x*<sup>3</sup> <sup>+</sup> *<sup>x</sup>*) <sup>−</sup> <sup>3</sup>(*x*<sup>3</sup> <sup>+</sup> <sup>3</sup>*x*)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) <sup>≡</sup> <sup>3</sup>(*x*<sup>3</sup> <sup>+</sup> <sup>3</sup>*x*)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) <sup>−</sup> <sup>3</sup>(*x*<sup>3</sup> <sup>+</sup> <sup>3</sup>*x*)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) ≡ 0 mod (*F*2*h*+1(*x*) − 1) 2 .

Thus, *n* = 1 is fit for (6). Assume that (6) is correct for all integers *n* = 0, 1, 2, ··· , *k*. Namely,

$$\begin{aligned} &L\_1(\mathbf{x}) \left( F\_{(2n+1)(2n+1)}(\mathbf{x}) - F\_{2n+1}(\mathbf{x}) \right) - (2n+1)L\_{2n+1}(\mathbf{x}) \left( F\_{2n+1}(\mathbf{x}) - 1 \right) \\ &\equiv \ 0 \text{ mod } \left( F\_{2n+1}(\mathbf{x}) - 1 \right)^2 \end{aligned} \tag{7}$$

for all *n* = 0, 1, ··· , *k*.

Where *n* = *k* + 1 ≥ 2, we notice

$$L\_{2(2h+1)}(\mathbf{x})F\_{(2h+1)(2k+1)}(\mathbf{x}) = F\_{(2h+1)(2k+3)}(\mathbf{x}) + F\_{(2h+1)(2k-1)}(\mathbf{x})$$

and

$$\begin{aligned} L\_{2(2h+1)}(\mathbf{x}) &= (\mathbf{x}^2 + 4)F\_{2h+1}^2(\mathbf{x}) - 2 = (\mathbf{x}^2 + 4) \left( F\_{2h+1}(\mathbf{x}) - 1 + 1 \right)^2 - 2 \\ &= -(\mathbf{x}^2 + 4) \left[ \left( F\_{2h+1}(\mathbf{x}) - 1 \right)^2 + 2 \left( F\_{2h+1}(\mathbf{x}) - 1 \right) \right] + \mathbf{x}^2 + 2 \\ &\equiv -2(\mathbf{x}^2 + 4)(F\_{2h+1}(\mathbf{x}) - 1) + \mathbf{x}^2 + 2 \bmod \left( F\_{2h+1}(\mathbf{x}) - 1 \right)^2. \end{aligned}$$

From inductive assumption (7) and Lemma 2, we have

*xF*(2*h*+1)(2*n*+1)(*x*) − *xF*2*n*+1(*x*) − (2*n* + 1)*L*2*n*+1(*x*)(*F*2*h*+1(*x*) − 1) = *xF*(2*h*+1)(2*k*+3)(*x*) − *xF*2*k*+3(*x*) − (2*k* + 3)*L*2*k*+3(*x*)(*F*2*h*+1(*x*) − 1) = *xL*2(2*h*+1)(*x*)*F*(2*h*+1)(2*k*+1)(*x*) − *xF*(2*h*+1)(2*k*−<sup>1</sup>)(*x*) − *xF*2*k*+3(*x*) −(2*k* + 3)*L*2*k*+3(*x*)(*F*2*h*+1(*x*) − 1) <sup>≡</sup> <sup>2</sup>*x*(*x*<sup>2</sup> <sup>+</sup> <sup>4</sup>)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>)*F*(2*h*+1)(2*k*+1)(*x*) + *<sup>x</sup>*(*x*<sup>2</sup> <sup>+</sup> <sup>2</sup>)*F*(2*h*+1)(2*k*+1)(*x*) <sup>−</sup>*xF*(2*h*+1)(2*k*−<sup>1</sup>)(*x*) <sup>−</sup> *<sup>x</sup>*(*x*<sup>2</sup> <sup>+</sup> <sup>2</sup>)*F*2*k*+1(*x*) + *xF*2*k*−1(*x*) <sup>−</sup>(*x*<sup>2</sup> <sup>+</sup> <sup>2</sup>)(2*<sup>k</sup>* <sup>+</sup> <sup>1</sup>)*L*2*k*+1(*x*)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) + (2*<sup>k</sup>* <sup>−</sup> <sup>1</sup>)*L*2*k*−1(*x*)(*F*2*h*+1(*x*) <sup>−</sup> <sup>1</sup>) −2*x* (*L*2*k*(*x*) + *L*2*k*+2(*x*)) (*F*2*h*+1(*x*) − 1) ≡ 2*x*(*F*2*h*+1(*x*) − 1) (*x*<sup>2</sup> <sup>+</sup> <sup>4</sup>)*F*(2*h*+1)(2*k*+1)(*x*) <sup>−</sup> *<sup>L</sup>*2*k*(*x*) <sup>−</sup> *<sup>L</sup>*2*k*+2(*x*) +(*x*<sup>2</sup> + 2) *xF*(2*h*+1)(2*k*+1)(*x*) − *xF*2*k*+1(*x*) − (2*k* + 1)*L*2*k*+1(*x*)(*F*2*h*+1(*x*) − 1) − *xF*(2*h*+1)(2*k*−<sup>1</sup>)(*x*) − *xF*2*k*−1(*x*) − (2*<sup>k</sup>* − <sup>1</sup>)*L*2*k*−1(*x*)(*F*2*h*+1(*x*) − <sup>1</sup>) ≡ 2*x*(*F*2*h*+1(*x*) − 1) (*x*<sup>2</sup> <sup>+</sup> <sup>4</sup>)*F*(2*h*+1)(2*k*+1)(*x*) <sup>−</sup> *<sup>L</sup>*2*k*(*x*) <sup>−</sup> *<sup>L</sup>*2*k*+2(*x*) ≡ 0 mod (*F*2*h*+1(*x*) − 1) 2 .

Now, we attain Lemma 3 by mathematical induction.

**Lemma 4.** *For all non-negative integers u and real numbers X, Y, we have the identity*

$$X^{\mu} + Y^{\mu} = \sum\_{k=0}^{\left[\frac{\mu}{2}\right]} (-1)^k \frac{\mu}{\mu - k} \binom{\mu - k}{k} \left(X + Y\right)^{\mu - 2k} \left(XY\right)^k$$

*in which* [*x*] *denotes the greatest integer* ≤ *x.*

**Proof.** This formula due to Waring [15]. It can also be found in Girard [14].

### **3. Proof of the Theorem**

We will achieve the theorem by these lemmas. Taking *<sup>X</sup>* = *<sup>α</sup>*2*m*, *<sup>Y</sup>* = −*β*2*<sup>m</sup>* and *<sup>U</sup>* = <sup>2</sup>*<sup>n</sup>* + 1 in Lemma 4, we notice that *XY* = −1, from the expression of *Fn*(*x*)

$$\begin{split} F\_{2m(2n+1)}(\mathbf{x}) &= \sum\_{k=0}^{n} (-1)^{k} \frac{2n+1}{2n+1-k} \binom{2n+1-k}{k} (\mathbf{x}^{2}+4)^{n-k} F\_{2m}^{2n+1-2k}(\mathbf{x}) (-1)^{k} \\ &= \sum\_{k=0}^{n} \frac{2n+1}{2n+1-k} \binom{2n+1-k}{k} (\mathbf{x}^{2}+4)^{n-k} F\_{2m}^{2n+1-2k}(\mathbf{x}). \end{split} \tag{8}$$

For any integer *h* ≥ 1, from (8), we get

$$\sum\_{m=1}^{h} \left[ F\_{2m(2n+1)}(\mathbf{x}) - (2n+1)F\_{2m}(\mathbf{x}) \right]$$

$$= \sum\_{k=0}^{n-1} \frac{2n+1}{2n+1-k} \binom{2n+1-k}{k} (\mathbf{x}^2 + 4)^{n-k} \sum\_{m=1}^{h} F\_{2m}^{2n+1-2k}(\mathbf{x}).\tag{9}$$

If *n* = 1, then, from (9), we can get

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\sum\_{m=1}^h \left(F\_{6m}(\mathbf{x}) - 3F\_{2m}(\mathbf{x})\right) = L\_1(\mathbf{x})L\_3(\mathbf{x})(\mathbf{x}^2 + 4)\sum\_{m=1}^h F\_{2m}^3(\mathbf{x}).\tag{10}$$

From Lemma 1, we know that (*x*<sup>2</sup> + 4, *<sup>F</sup>*2*h*+1(*x*) − <sup>1</sup>) = 1, so, applying Lemma <sup>3</sup> and (10), we deduce that

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\sum\_{m=1}^h F\_{2m}^3(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right)^2. \tag{11}$$

This means that Theorem 1 is suitable for *n* = 1.

Assume that Theorem 1 is correct for all integers *n* = 1, 2, ··· , *s*. Then,

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\cdots L\_{2n+1}(\mathbf{x})\sum\_{m=1}^h F\_{2m}^{2n+1}(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right)^2\tag{12}$$

for all integers 1 ≤ *n* ≤ *s*.

When *n* = *s* + 1, from (9), we obtain

$$\sum\_{m=1}^{h} \left( F\_{2m(2s+3)}(\mathbf{x}) - (2s+3)F\_{2m}(\mathbf{x}) \right)$$

$$\begin{split} &= \sum\_{k=0}^{s} \frac{2s+3}{2s+3-k} \binom{2s+3-k}{k} (\mathbf{x}^2 + 4)^{s+1-k} \sum\_{m=1}^{h} F\_{2m}^{2s+3-2k}(\mathbf{x}) \\ &= \sum\_{k=1}^{s} \frac{2s+3}{2s+3-k} \binom{2s+3-k}{k} (\mathbf{x}^2 + 4)^{s+1-k} \sum\_{m=1}^{h} F\_{2m}^{2s+3-2k}(\mathbf{x}) \\ &\quad + (\mathbf{x}^2 + 4)^{s+1} \sum\_{m=1}^{h} F\_{2m}^{2s+3}(\mathbf{x}). \end{split} \tag{13}$$

From Lemma 3, we have

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\cdots L\_{2s+3}(\mathbf{x})\sum\_{m=1}^h \left[F\_{2m(2s+3)}(\mathbf{x}) - (2s+3)F\_{2m}(\mathbf{x})\right]$$

$$\equiv \ 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right)^2. \tag{14}$$

Applying inductive hypothesis (12), we obtain

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\cdots L\_{2s+1}(\mathbf{x})\sum\_{k=1}^s \frac{2s+3}{2s+3-k} \binom{2s+3-k}{k}$$

$$\mathbf{x}\times(\mathbf{x}^2+4)^{s+1-k}\sum\_{m=1}^h F\_{2m}^{2s+3-2k}(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x})-1\right)^2. \tag{15}$$

Combining (13), (14), (15) and Lemma 3, we have the conclusion

$$L\_1(\mathbf{x})L\_3(\mathbf{x})\cdots L\_{2s+3}(\mathbf{x})\cdot(\mathbf{x}^2+4)^{s+1}\sum\_{m=1}^h F\_{2m}^{2s+3}(\mathbf{x})$$

$$\equiv \quad 0 \text{ mod } \left(F\_{2h+1}(\mathbf{x})-1\right)^2. \tag{16}$$

Note that *<sup>x</sup>*<sup>2</sup> + 4, *<sup>F</sup>*2*h*+1(*x*) − <sup>1</sup> = 1, so (16) indicates the conclusion

$$L\_1(\mathbf{x})L\_3(\mathbf{x}) \cdot \cdots \cdot L\_{2s+3}(\mathbf{x}) \cdot \sum\_{m=1}^{h} F\_{2m}^{2s+3}(\mathbf{x}) \equiv 0 \bmod \left(F\_{2h+1}(\mathbf{x}) - 1\right)^2 \dots$$

Now, we apply mathematical induction to achieve Theorem 1.

**Author Contributions:** Conceptualization, L.C.; methodology, L.C and X.W.; validation, L.C. and X.W.; formal analysis, L.C.; investigation, X.W.; resources, L.C.; writing—original draft preparation, L.C.; writing—review and editing, X.W.; visualization,L.C.; supervision, L.C.; project administration, X.W.; all authors have read and approved the final manuscript.

**Funding:** This work is supported by the N. S. F. (11771351) and (11826205) of P. R. China.

**Acknowledgments:** The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.

**Conflicts of Interest:** The authors state that there are no conflicts of interest concerning the publication of this paper.

### **References**


© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **Identities of Symmetry for Type 2 Bernoulli and Euler Polynomials**

### **Dae San Kim 1, Han Young Kim 2, Dojin Kim 3,\* and Taekyun Kim <sup>2</sup>**


Received: 13 April 2019; Accepted: 28 April 2019; Published: 2 May 2019

**Abstract:** The main purpose of this paper is to give several identities of symmetry for type 2 Bernoulli and Euler polynomials by considering certain quotients of bosonic *p*-adic and fermionic *p*-adic integrals on Z*p*, where *p* is an odd prime number. Indeed, they are symmetric identities involving type 2 Bernoulli polynomials and power sums of consecutive odd positive integers, and the ones involving type 2 Euler polynomials and alternating power sums of odd positive integers. Furthermore, we consider two random variables created from random variables having Laplace distributions and show their moments are given in terms of the type 2 Bernoulli and Euler numbers.

**Keywords:** type 2 Bernoulli polynomials; type 2 Euler polynomials; identities of symmetry; Laplace distribution

### **1. Introduction**

In this section, we are going to review some known results. We first recall the definitions of Bernoulli and Euler polynomials together with their type 2 polynomials. Then, we introduce the bosonic *p*-adic integrals and the fermionic *p*-adic integrals on Z*<sup>p</sup>* that we need for the derivation of an identity of symmetry. As is well known, the Bernoulli polynomials are defined by

$$\frac{t}{e^t - 1} e^{\mathbf{x}t} = \sum\_{n=0}^{\infty} B\_n(\mathbf{x}) \frac{t^n}{n!} \,\tag{1}$$

(see [1,2]).

In particular, the Bernoulli numbers are the constant terms *Bn* = *Bn*(0) of the Bernoulli polynomials. By making use of (1), we can deduce that

$$\sum\_{l=0}^{n-1} l^k = \frac{1}{k+1} (B\_{k+1}(n) - B\_{k+1})\_\prime \text{ for } k = 0, 1, 2, \cdots \text{ } \tag{2}$$

The type 2 Bernoulli polynomials are defined by generating function

$$\frac{t}{e^t - e^{-t}} e^{\ge t} = \sum\_{n=0}^{\infty} b\_n(x) \frac{t^n}{n!} \,\tag{3}$$

*Symmetry* **2019**, *11*, 613; doi:10.3390/sym11050613 www.mdpi.com/journal/symmetry

(see [3,4]).

In particular, *bn* = *bn*(0) are called type 2 Bernoulli numbers. From (3), it can be seen that

$$b\_{\mathbb{H}}(\mathbf{x}) = \sum\_{k=0}^{n} \binom{n}{k} b\_{k} \mathbf{x}^{n-k},\tag{4}$$

(see [3,4]).

Analogously to (2), we observe that

$$\begin{split} \sum\_{l=0}^{n-1} e^{(2l+1)t} &= \frac{1}{e^t - e^{-t}} (e^{2nt} - 1) \\ &= \sum\_{k=0}^{\infty} \left( \frac{b\_{k+1}(2n) - b\_{k+1}}{k+1} \right) \frac{t^k}{k!} . \end{split} \tag{5}$$

Thus, by (5), we get

$$\sum\_{l=0}^{n-1} (2l+1)^k = \frac{1}{k+1} (b\_{k+1}(2n) - b\_{k+1}), \quad k = 0, 1, 2, \cdots \\ \dots \tag{6}$$

Let *p* be a fixed odd prime number. Throughout this paper, we will use the notations Z*p*, Q*p*, C*p*, and C to denote the ring of *p*-adic rational integers, the field of *p*-adic rational numbers, the completion of an algebraic closure of Q*p*, and the field of complex numbers, respectively. The normalized valuation in <sup>C</sup>*<sup>p</sup>* is denoted by |·|*p*, with <sup>|</sup>*p*|*<sup>p</sup>* <sup>=</sup> <sup>1</sup> *<sup>p</sup>* . For a uniformly differentiable function *f* on Z*p*, the bosonic *p*-adic integral on Z*<sup>p</sup>* (or *p*-adic invariant integral on Z*p*) is defined by

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x})d\mu\_0(\mathbf{x}) = \lim\_{N \to \infty} \sum\_{\mathbf{x}=0}^{p^N - 1} f(\mathbf{x})\mu\_0(\mathbf{x} + p^N \mathbb{Z}\_p) = \lim\_{N \to \infty} \frac{1}{p^N} \sum\_{\mathbf{x}=0}^{p^N - 1} f(\mathbf{x}). \tag{7}$$

Then, by (7), we easily get

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x} + \mathbf{1}) d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} f(\mathbf{x}) d\mu\_0(\mathbf{x}) = f'(0),\tag{8}$$

(see [5,6]).

The fermionic integral on Z*<sup>p</sup>* is defined by Kim [6] as

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x})d\mu\_{-1}(\mathbf{x}) = \lim\_{N \to \infty} \sum\_{\mathbf{x}=0}^{p^N - 1} f(\mathbf{x})\mu\_{-1}(\mathbf{x} + p^N \mathbb{Z}\_p) = \lim\_{N \to \infty} \sum\_{\mathbf{x}=0}^{p^N - 1} f(\mathbf{x})(-1)^{\mathbf{x}}.\tag{9}$$

From (9), we can show that

$$\int\_{\mathbb{Z}\_p} f(\mathbf{x} + 1) d\mu\_{-1}(\mathbf{x}) + \int\_{\mathbb{Z}\_p} f(\mathbf{x}) d\mu\_{-1}(\mathbf{x}) = 2f(0),\tag{10}$$

(see [4,7–10]).

It is well known that the Euler polynomials are defined by

$$\frac{2}{e^t + 1} e^{\chi t} = \sum\_{n=0}^{\infty} E\_n^\*(\chi) \frac{t^n}{n!}. \tag{11}$$

We denote the Euler numbers by *E*∗ *<sup>n</sup>* = *E*<sup>∗</sup> *<sup>n</sup>*(0), (*n* ≥ 0). Clearly, we have

$$2\sum\_{l=0}^{n-1}(-1)^{l}e^{lt} = \frac{2}{e^{t}+1}(e^{nt}+1), \quad \text{where } n \equiv 1 \pmod{2}.\tag{12}$$

From (11) and (12), we obtain that

$$2\sum\_{l=0}^{n-1}(-1)^{l}l^{k} = E\_{k}^{\*}(n) + E\_{k'}^{\*} \tag{13}$$

where *n* is a positive odd integer.

Now, we consider the type 2 Euler polynomials which are given by

$$\frac{2}{e^t + e^{-t}} e^{xt} = \text{sech}(t) e^{\ge t} = \sum\_{n=0}^{\infty} E\_n(\mathbf{x}) \frac{t^n}{n!}. \tag{14}$$

In particular, when *x* = 0, *En* = *En*(0) are called the type 2 Euler numbers.

In this paper, we obtain some identities of symmetry involving the type 2 Bernoulli polynomials, the type 2 Euler polynomials, power sums of odd positive integers and alternating power sums of odd positive integers which are derived from certain quotients of bosonic *p*-adic and fermionic *p*-adic integrals on Z*p*. In the following section, we will construct two random variables from random variables having Laplace distributions whose moments are closely related to the type 2 Bernoulli and Euler numbers. All the results in Sections 2 and 3 are newly developed. Finally, we note that the results here have applications in such diverse areas as combinatorics, probability, algebra and analysis (see [11–13]).

### **2. Some Identities of Symmetry for Type** 2 **Bernoulli and Euler Polynomials**

In virtue of (8), we readily see that

$$\frac{1}{2} \int\_{\mathbb{Z}\_p} e^{(2\chi+1)t} d\mu\_0(\mathbf{x}) = \frac{t}{e^t - e^{-t}}.\tag{15}$$

Hence, by (15), we get

$$\frac{1}{2} \int\_{\mathbb{Z}\_p} (2\mathbf{x} + 1)^n d\mu\_0(\mathbf{x}) = b\_{n\prime} \quad (n \ge 0). \tag{16}$$

In addition, it follows from (15) that

$$\frac{1}{2} \int\_{\mathbb{Z}\_p} e^{(2y+x+1)t} d\mu\_0(y) = \frac{t}{e^t - e^{-t}} e^{xt} = \sum\_{n=0}^{\infty} b\_n(x) \frac{t^n}{n!}.\tag{17}$$

Hence, by (17), we get

$$\frac{1}{2} \int\_{\mathbb{Z}\_p} (2y + x + 1)^n d\mu\_0(y) = b\_n(x), \quad (n \ge 0). \tag{18}$$

Using (15) and (17), one can easily check that

$$\frac{1}{2} \left( \int\_{\mathbb{Z}\_p} e^{(2x+2u+1)t} d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} e^{(2x+1)t} d\mu\_0(\mathbf{x}) \right) = t \sum\_{l=0}^{n-1} e^{(2l+1)t}.\tag{19}$$

Next, we let *Tk*(*n*) = *<sup>n</sup>* ∑ *l*=0 (2*<sup>l</sup>* + <sup>1</sup>)*k*, (*<sup>k</sup>* ∈ N ∪ {0}). Note that *Tk*(*n*) represents the *<sup>k</sup>*th power sums of consecutive positive odd integers. By (19), we easily get

$$\int\_{\mathbb{Z}\_p} e^{(2x+1+2n)t} d\mu\_0(\mathbf{x}) - \int\_{\mathbb{Z}\_p} e^{(2x+1)t} d\mu\_0(\mathbf{x}) = \frac{2nt \int\_{\mathbb{Z}\_p} e^{(2x+1)t} d\mu\_0(\mathbf{x})}{\int\_{\mathbb{Z}\_p} e^{2nxt} d\mu\_0(\mathbf{x})}. \tag{20}$$

Let *w*1, *w*<sup>2</sup> be positive integers. Then, we observe that

$$\begin{split} \frac{w\_1 \int\_{Z\_p} e^{(2x+1)t} d\mu\_0(\mathbf{x})}{\int\_{Z\_p} e^{2w\_1 \mathbf{x} t} d\mu(\mathbf{x})} &= \sum\_{l=0}^{w\_1 - 1} e^{(2l+1)t} \\ &= \sum\_{k=0}^{\infty} \sum\_{l=0}^{w\_1 - 1} (2l+1)^k \frac{t^k}{k!} \\ &= \sum\_{k=0}^{\infty} T\_k (w\_1 - 1) \frac{t^k}{k!} .\end{split} \tag{21}$$

Now, we consider the next quotient of bosonic *p*-adic integrals on Z*<sup>p</sup>* from which the identities of symmetry for the type 2 Bernoulli polynomials follow:

$$I(w\_1, w\_2) = \frac{w\_1 w\_2}{2} \frac{\int\_{\mathbb{Z}\_p} \int\_{\mathbb{Z}\_p} e^{(2w\_1 \mathbf{x}\_1 + w\_1 + 2w\_2 \mathbf{x}\_2 + w\_2 + w\_1 w\_2 \mathbf{x})t} d\mu\_0(\mathbf{x}\_1) d\mu\_0(\mathbf{x}\_2)}{\int\_{\mathbb{Z}\_p} e^{2w\_1 w\_2 \mathbf{x}} d\mu\_0(\mathbf{x})}. \tag{22}$$

From (22), we have

$$\begin{split} I(w\_1, w\_2) &= \frac{w\_2}{2} \int\_{\mathbb{Z}\_p} e^{(2w\_1 + w\_2 x + 1)w\_1 t} d\mu\_0(\mathbf{x}) \, \frac{w\_1 \int\_{\mathbb{Z}\_p} e^{(2w\_2 x\_2 + w\_2)t} d\mu\_0(\mathbf{x}\_2)}{\int\_{\mathbb{Z}\_p} e^{2w\_1 w\_2 x t} d\mu\_0(\mathbf{x})} \\ &= w\_2 \sum\_{k=0}^{\infty} b\_k(w\_2 \mathbf{x}) \frac{w\_1^k}{k!} t^k \sum\_{l=0}^{\infty} T\_l(w\_1 - 1) \frac{w\_2^l}{l!} t^l \\ &= \sum\_{n=0}^{\infty} \sum\_{k=0}^n \binom{n}{k} b\_k(w\_2 \mathbf{x}) T\_{n-k}(w\_1 - 1) w\_1^k w\_2^{n-k+1} \frac{t^n}{n!} .\end{split} \tag{23}$$

We note from (22) that *I*(*w*1, *w*2) = *I*(*w*2, *w*1). Interchanging *w*<sup>1</sup> and *w*2, we get

$$I(w\_2, w\_1) = \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} \binom{n}{k} b\_k(w\_1 \mathbf{x}) T\_{n-k}(w\_2 - 1) w\_2^k w\_1^{n-k+1} \frac{t^n}{n!}. \tag{24}$$

Therefore, by (23) and (24), we obtain the following theorem.

**Theorem 1.** *For w*1, *w*<sup>2</sup> ∈ N *and n* ∈ N ∪ {0}*, we have*

$$\sum\_{k=0}^{n} \binom{n}{k} b\_k(w\_2 \ge) T\_{n-k}(w\_1 - 1) w\_1^k w\_2^{n-k+1} = \sum\_{k=0}^{n} \binom{n}{k} b\_k(w\_1 \ge) T\_{n-k}(w\_2 - 1) w\_2^k w\_1^{n-k+1}.$$

Setting *x* = 0 in Theorem 1, we obtain the following corollary.

**Corollary 1.** *For w*1, *w*<sup>2</sup> ∈ N *and n* ∈ N ∪ {0}*, we have*

$$\sum\_{k=0}^{n} \binom{n}{k} b\_k T\_{n-k} (\mathfrak{w}\_1 - 1) \mathfrak{w}\_1^k \mathfrak{w}\_2^{n-k+1} = \sum\_{k=0}^{n} \binom{n}{k} b\_k T\_{n-k} (\mathfrak{w}\_2 - 1) \mathfrak{w}\_2^k \mathfrak{w}\_1^{n-k+1}.$$

Furthermore, let us take *w*<sup>2</sup> = 1 in Corollary 1. Then, we have

$$\sum\_{k=0}^{n} \binom{n}{k} b\_k w\_1^{n-k+1} = \sum\_{k=0}^{n} \binom{n}{k} b\_k T\_{n-k}(w\_1 - 1) w\_1^k. \tag{25}$$

Therefore, by (4) and (25), we obtain the following corollary.

**Corollary 2.** *For w*<sup>1</sup> ∈ N *and n* ∈ N ∪ {0}*, we have*

$$b\_n(w\_1) = \sum\_{k=0}^n \binom{n}{k} b\_k T\_{n-k}(w\_1 - 1) w\_1^{k-1} = \sum\_{k=0}^n \binom{n}{k} b\_k w\_1^{k-1} \sum\_{l=0}^{w\_1 - 1} (2l + 1)^{n-k}.$$

From (22), we observe that

$$\begin{split} I(w\_1, w\_2) &= \frac{w\_2}{2} \epsilon^{w\_1 w\_2 \ge t} \int\_{\mathbb{Z}\_p} \epsilon^{2w\_1 x\_1 t + w\_1 t} d\mu\_0(\mathbf{x}\_1) \frac{w\_1 \int\_{\mathbb{Z}\_p} \epsilon^{(2w\_2 x\_2 + w\_2)t} d\mu\_0(\mathbf{x}\_1)}{\int\_{\mathbb{Z}\_p} \epsilon^{2w\_1 w\_2 \ge t} d\mu\_0(\mathbf{x})} \\ &= \frac{w\_2}{2} \epsilon^{w\_1 w\_2 x t} \int\_{\mathbb{Z}\_p} \epsilon^{(2w\_1 x\_1 + w\_1)t} d\mu\_0(\mathbf{x}\_1) \sum\_{l=0}^{w\_1 - 1} \epsilon^{(2l+1)w\_2 t} \\ &= \frac{w\_2}{2} \sum\_{l=0}^{w\_1 - 1} \int\_{\mathbb{Z}\_p} \epsilon^{\left(2x\_1 + 1 + w\_2 x + (2l+1)\frac{w\_2}{w\_1}\right) w\_1 t} d\mu\_0(\mathbf{x}\_1) \\ &= \sum\_{n=0}^{\infty} w\_2 \sum\_{l=0}^{w\_1 - 1} b\_n \left( w\_2 x + (2l+1) \frac{w\_2}{w\_1} \right) \frac{w\_1^n t^n}{n!} .\end{split} \tag{26}$$

By interchanging *w*<sup>1</sup> and *w*2, we obtain the following equation:

$$I(w\_2, w\_1) = \sum\_{n=0}^{\infty} w\_1 \sum\_{l=0}^{w\_2 - 1} b\_{ll} \left( w\_1 x + (2l + 1) \frac{w\_1}{w\_2} \right) \frac{w\_2^n t^n}{n!}. \tag{27}$$

As *I*(*w*1, *w*2) = *I*(*w*2, *w*1), the following theorem is immediate from (26) and (27).

**Theorem 2.** *For w*1, *w*<sup>2</sup> ∈ N *and n* ∈ N ∪ {0}*, we have*

$$w\_1^n w\_2 \sum\_{l=0}^{w\_1 - 1} b\_n \left( w\_2 \mathbf{x} + (2l + 1) \frac{w\_2}{w\_1} \right) = w\_2^n w\_1 \sum\_{l=0}^{w\_2 - 1} b\_n \left( w\_1 \mathbf{x} + (2l + 1) \frac{w\_1}{w\_2} \right).$$

**Example 1.** *We check the result in Theorem 2 in the case of n* = 2, *w*<sup>1</sup> = 3, *and w*<sup>2</sup> = 7. *We first note that b*2(*x*) = <sup>1</sup> <sup>2</sup> (*x*<sup>2</sup> <sup>−</sup> <sup>1</sup> <sup>3</sup> )*. This can be obtained from <sup>B</sup>*2(*x*) = *<sup>x</sup>*<sup>2</sup> <sup>−</sup> *<sup>x</sup>* <sup>+</sup> <sup>1</sup> <sup>6</sup> *and the relation bn*(*x*) = <sup>2</sup>*n*−1*Bn*( *<sup>x</sup>*+<sup>1</sup> <sup>2</sup> ) *which follows from* (1) *and* (3)*. Thus, we have to see that*

$$\sum\_{l=0}^{2} \left\{ \left( \mathbf{7x} + \frac{7}{3} (2l+1) \right)^2 - \frac{1}{3} \right\} = \frac{7}{3} \sum\_{l=0}^{6} \left\{ \left( \mathbf{3x} + \frac{3}{7} (2l+1) \right)^2 - \frac{1}{3} \right\}.\tag{28}$$

*Now, we can easily show that both the left and the right side of* (28) *are equal to* 147*x*<sup>2</sup> + 294*x* + <sup>1706</sup> <sup>9</sup> *.* Let us take *w*<sup>1</sup> = 1. Then, by Theorem 2, we get

$$w\_2 b\_{\mathbb{H}}(w\_2 \mathbf{x} + w\_2) = w\_2^{\eta} \sum\_{l=0}^{w\_2 - 1} b\_{\mathbb{H}} \left( \mathbf{x} + (2l + 1) \frac{1}{w\_2} \right) . \tag{29}$$

Equivalently, by (29), we have

$$b\_n(w\_2\mathbf{x} + w\_2) = w\_2^{n-1} \sum\_{l=0}^{w\_2 - 1} b\_n \left(\mathbf{x} + (2l + 1)\frac{1}{w\_2}\right). \tag{30}$$

Similarly to (13), we observe that

$$2\sum\_{l=0}^{n-1}(-1)^{l}e^{(2l+1)t} = \sum\_{k=0}^{\infty}\left(E\_{k} + E\_{k}(2n)\right)\frac{t^{k}}{k!} \tag{31}$$

where *n* ∈ N with *n* ≡ 1 (mod 2). Thus, by (31), we get

$$2\sum\_{l=0}^{n-1}(-1)^{l}(2l+1)^{k} = E\_{k} + E\_{K}(2n),\tag{32}$$

where *k* ∈ N ∪ {0} and *n* ∈ N with *n* ≡ 1 (mod 2).

From (14), we easily note that

$$E\_{\mathbb{R}}(\mathbf{x}) = \sum\_{k=0}^{n} \binom{n}{k} E\_k \mathbf{x}^{n-k}, \quad (n \ge 0). \tag{33}$$

By (10), we get

$$\int\_{\mathbb{Z}\_p} e^{(2y+x+1)t} d\mu\_{-1}(y) = \frac{2}{e^t + e^{-t}} e^{\ge t} = \sum\_{n=0}^{\infty} E\_n(\mathbf{x}) \frac{t^n}{n!}.\tag{34}$$

Thus, we have

$$\int\_{\mathbb{Z}\_p} (2y + x + 1)^n d\mu\_{-1}(y) = E\_n(x), \quad (n \ge 0).$$

The next equation follows immediately from (10):

$$\int\_{\mathbb{Z}\_p} e^{(2y+2n+1)t} d\mu\_{-1}(y) + \int\_{\mathbb{Z}\_p} e^{(2x+1)t} d\mu\_{-1}(x) = 2 \sum\_{l=0}^{n-1} e^{(2l+1)t} (-1)^l,\tag{35}$$

where *n* ∈ N with *n* ≡ 1 (mod 2).

Now, we let *Ak*(*n*) = *<sup>n</sup>* ∑ *l*=0 (−1)*<sup>l</sup>* (2*<sup>l</sup>* + <sup>1</sup>)*k*, (*<sup>k</sup>* ∈ N ∪ {0}). Here we note that *Ak*(*n*) is the alternating *<sup>k</sup>*th power sums of consecutive odd positive integers. From (35), we have

$$\int\_{\mathbb{Z}\_p} \mathfrak{e}^{(2\mathbf{x} + 2\mathbf{x} + 1)t} d\mu\_{-1}(\mathbf{x}) + \int\_{\mathbb{Z}\_p} \mathfrak{e}^{(2\mathbf{x} + 1)t} d\mu\_{-1}(\mathbf{x}) = \frac{2 \int\_{\mathbb{Z}\_p} \mathfrak{e}^{(2\mathbf{x} + 1)t} d\mu\_{-1}(\mathbf{x})}{\int\_{\mathbb{Z}\_p} \mathfrak{e}^{2\mathbf{x} \mathbf{x}t} d\mu\_{-1}(\mathbf{x})}. \tag{36}$$

Let *a*, *b* be positive integers with *a* ≡ 1 (mod 2) and *b* ≡ 1 (mod 2). Then, by using the fermionic *p*-adic integral on Z*p*, we get

$$\begin{split} \frac{2 \int\_{\mathbb{Z}\_p} e^{(2\mathbf{x} + 1)t} d\mu\_{-1}(\mathbf{x})}{\int\_{\mathbb{Z}\_p} e^{2\mathbf{x}t} d\mu\_{-1}(\mathbf{x})} &= 2 \sum\_{l=0}^{a-1} e^{(2l+1)t} (-1)^l \\ &= \sum\_{k=0}^{\infty} 2 \sum\_{l=0}^{a-1} (2l+1)^k (-1)^l \frac{t^k}{k!} \\ &= \sum\_{k=0}^{\infty} 2A\_k (a-1) \frac{t^k}{k!} .\end{split} \tag{37}$$

We now consider the next quotient of the fermionic *p*-adic integrals on Z*<sup>p</sup>* from which the identities of symmetry for the type 2 Euler polynomials follow:

$$f(a,b) = \frac{\int\_{\mathbb{Z}\_p} \int\_{\mathbb{Z}\_p} e^{(2a\mathbf{x}\_1 + a + 2b\mathbf{x}\_2 + b + ab\mathbf{x})t} d\mu\_{-1}(\mathbf{x}\_1) d\mu\_{-1}(\mathbf{x}\_2)}{\int\_{\mathbb{Z}\_p} e^{2ab\mathbf{x}t} d\mu\_{-1}(\mathbf{x})}. \tag{38}$$

From (38), we can derive the following equation given by

$$\begin{split} J(a,b) &= \frac{1}{2} \int\_{\mathbb{Z}\_p} e^{a(2\chi\_1+1+bx)t} d\mu\_{-1}(\mathbf{x}\_1) \; \frac{2 \int\_{\mathbb{Z}\_p} e^{(2bx\_2+b)t} d\mu\_{-1}(\mathbf{x}\_2)}{\int\_{\mathbb{Z}\_p} e^{2ab\mathbf{x}t} d\mu\_{-1}(\mathbf{x})} \\ &= \frac{1}{2} \sum\_{k=0}^{\infty} E\_k(b\mathbf{x}) \frac{a^k t^k}{k!} \; \mathbf{2} \sum\_{l=0}^{\infty} A\_l(a-1) \frac{b^l t^l}{l!} \\ &= \sum\_{n=0}^{\infty} \sum\_{k=0}^n \binom{n}{k} E\_k(b\mathbf{x}) A\_{n-k}(a-1) a^k b^{n-k} \frac{t^n}{n!} .\end{split} \tag{39}$$

We note from (38) that *J*(*a*, *b*) = *J*(*b*, *a*). Interchanging *a* and *b*, we get

$$f(b,a) = \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} \binom{n}{k} E\_k(ax) A\_{n-k}(b-1) b^k a^{n-k} \frac{t^n}{n!}.\tag{40}$$

The following theorem is an immediate consequence of (39) and (40).

**Theorem 3.** *For n* ≥ 0*, a*, *b* ∈ N *with a* ≡ 1 *(mod* 2*) and b* ≡ 1 *(mod* 2*), we have*

$$\sum\_{k=0}^{n} \binom{n}{k} E\_k(bx) A\_{n-k}(a-1) a^k b^{n-k} = \sum\_{k=0}^{n} \binom{n}{k} E\_k(ax) A\_{n-k}(b-1) b^k a^{n-k}.$$

The next corollary is now obtained by setting *x* = 0 in Theorem 3.

**Corollary 3.** *For n* ≥ 0*, a*, *b* ∈ N*, with a* ≡ 1 *(mod* 2*) and b* ≡ 1 *(mod* 2*), we have*

$$\sum\_{k=0}^{n} \binom{n}{k} E\_k A\_{n-k} (a-1) a^k b^{n-k} = \sum\_{k=0}^{n} \binom{n}{k} E\_k A\_{n-k} (b-1) b^k a^{n-k}.$$

Taking *b* = 1 in Corollary 3 gives us the following identities.

**Corollary 4.** *For n* ≥ 0*, a* ∈ N *with a* ≡ 1 *(mod* 2*), we have*

$$\begin{aligned} E\_n(a) &= \sum\_{k=0}^n \binom{n}{k} E\_k A\_{n-k} (a-1) a^k \\ &= \sum\_{k=0}^n \binom{n}{k} E\_k a^k \sum\_{l=0}^{a-1} (-1)^l (2l+1)^{n-k} \end{aligned}$$

From (38), we have

$$\begin{split} f(a,b) &= \frac{e^{ab\mathbf{x}t}}{2} \int\_{\mathbb{Z}\_p} e^{(2ax\_1+a)t} d\mu\_{-1}(\mathbf{x}\_1) \frac{2 \int\_{\mathbb{Z}\_p} e^{(2bx\_2+b)t} d\mu\_{-1}(\mathbf{x}\_2)}{\int\_{\mathbb{Z}\_p} e^{2ab\mathbf{x}t} d\mu\_{-1}(\mathbf{x})} \\ &= \frac{e^{ab\mathbf{x}t}}{2} \int\_{\mathbb{Z}\_p} e^{(2ax\_1+a)t} d\mu\_{-1}(\mathbf{x}\_1) \ 2 \sum\_{l=0}^{a-1} (-1)^l e^{(2l+1)bt} \\ &= \sum\_{l=0}^{a-1} (-1)^l \int\_{\mathbb{Z}\_p} e^{(2x\_1+1+b\mathbf{x}+(2l+1)\frac{b}{a})at} d\mu\_{-1}(\mathbf{x}\_1) \\ &= \sum\_{n=0}^{\infty} a^n \sum\_{l=0}^{a-1} (-1)^l E\_n(b\mathbf{x}+(2l+1)\frac{b}{a}) \frac{t^n}{n!}, \end{split} \tag{41}$$

.

where *a*, *b* ∈ N with *a* ≡ 1 (mod 2) and *b* ≡ 1 (mod 2). Interchanging *a* and *b*, we get

$$f(b,a) = \sum\_{n=0}^{\infty} b^n \sum\_{l=0}^{b-1} (-1)^l E\_n(ax + (2l+1)\frac{a}{b}) \frac{t^n}{n!}.\tag{42}$$

As *J*(*a*, *b*) = *J*(*b*, *a*), by (41) and (42), we obtain the following theorem.

**Theorem 4.** *For n* ≥ 0*, a*, *b* ∈ N *with a* ≡ 1 *(mod* 2*) and b* ≡ 1 *(mod* 2*), we have*

$$a^n \sum\_{l=0}^{a-1} (-1)^l E\_n(b\mathbf{x} + (2l+1)\frac{b}{a}) = b^n \sum\_{l=0}^{b-1} (-1)^l E\_n(a\mathbf{x} + (2l+1)\frac{a}{b}).$$

Let us take *a* = 1 in Theorem 4. Then, we have

$$E\_{\hbar}(bx+b) = b^{\hbar} \sum\_{l=0}^{b-1} (-1)^{l} E\_{\hbar}(\chi + (2l+1)\frac{1}{b}).$$

**Example 2.** *Here, we illustrate Theorem 2 in the case of n* = 2, *a* = 7, *and b* = 3*. First, we note that E*2(*x*) = *<sup>x</sup>*<sup>2</sup> − <sup>1</sup>*. This follows from <sup>E</sup>*<sup>∗</sup> <sup>2</sup> (*x*) = *<sup>x</sup>*<sup>2</sup> − *<sup>x</sup> and the relation En*(*x*) = <sup>2</sup>*nE*<sup>∗</sup> *<sup>n</sup>*( *<sup>x</sup>*+<sup>1</sup> <sup>2</sup> ) *that can be deduced from* (11) *and* (14)*. Here, we need to show that*

$$\sum\_{l=0}^{6}(-1)^{l}\left\{\left(3\mathbf{x} + \frac{3}{7}(2l+1)\right)^{2} - 1\right\} = \left(\frac{3}{7}\right)^{2}\sum\_{l=0}^{2}(-1)^{l}\left\{\left(7\mathbf{x} + \frac{7}{3}(2l+1)\right)^{2} - 1\right\}.\tag{43}$$

*Indeed, we can easily check that both the left- and right-hand side of* (43) *are equal to* 9*x*<sup>2</sup> + 18*x* + <sup>824</sup> <sup>49</sup> *.*

### **3. Further Remarks**

For *s* ∈ C, the Riemann zeta function is defined by

$$\zeta(s) = \sum\_{n=1}^{\infty} \frac{1}{n^s}, \quad (\text{Re}(s) > 1)\_s$$

(see [14–16]).

It is well known that

$$
\zeta(2n) = (-1)^{n-1} \frac{2^{2n-1}}{(2n)!} \pi^{2n} B\_{2n}, \quad (n \ge 0), \tag{44}
$$

(see [14,16]).

By (44), we get

$$\begin{split} z \cot(z) &= z \frac{\cos(z)}{\sin(z)} \\ &= \frac{z \frac{\sin(z)}{2}}{\frac{z^2 - z^{-1}}{2}}, \quad (i = \sqrt{-1}) \\ &= iz \left( 1 + \frac{2}{e^{2iz} - 1} \right) \\ &= iz + \sum\_{k=0}^{\infty} B\_k \frac{(2iz)^k}{k!} \\ &= 1 + \sum\_{k=1}^{\infty} \frac{B\_{2k}}{(2k)!} z^{2k} i^{2k} \\ &= 1 - 2 \sum\_{k=1}^{\infty} \frac{\xi(2k)}{\pi^{2k}} z^{2k} \\ &= 1 - 2 \sum\_{n=1}^{\infty} \left( \sum\_{k=1}^{\infty} \frac{z^{2k}}{(n\pi)^{2k}} \right) \\ &= 1 - 2 \sum\_{n=1}^{\infty} \left( \frac{z}{n\pi} \right)^2 \left( 1 - \left( \frac{z}{n\pi} \right)^2 \right)^{-1} . \end{split} \tag{45}$$

Thus, by (45), we get

$$\cot(z) - \frac{1}{z} = -\sum\_{n=1}^{\infty} \frac{2z}{(n\pi)^2} \left(1 - \left(\frac{z}{n\pi}\right)^2\right)^{-1}.\tag{46}$$

From (39), we easily note that

$$\frac{d}{dz}\left(\log(\sin(z)) - \log(z)\right) = \sum\_{n=1}^{\infty} \frac{d}{dz}\left(\log\left(1 - \left(\frac{z}{n\pi}\right)^2\right)\right).\tag{47}$$

By (47), we easily get

$$\frac{\sin(z)}{z} = \prod\_{n=1}^{\infty} \left( 1 - \left( \frac{z}{n\pi} \right)^2 \right). \tag{48}$$

It is not difficult to show that

$$z\cot(z) - 2z\cot(2z) = z\tan(z). \tag{49}$$

From (45) and (49), we have

$$\begin{split} z \tan(z) &= z \cot(z) - 2z \cot(2z) \\ &= 2 \sum\_{n=1}^{\infty} \left( \frac{2z}{n\pi} \right)^2 \left( 1 - \left( \frac{2z}{n\pi} \right)^2 \right)^{-1} - 2 \sum\_{n=1}^{\infty} \left( \frac{z}{n\pi} \right)^2 \left( 1 - \left( \frac{z}{n\pi} \right)^2 \right)^{-1} . \end{split} \tag{50}$$

By (50), we get

$$\frac{d}{dz}\left(-\log(\cos(z))\right) = -\sum\_{n=1}^{\infty} \frac{d}{dz}\left(\log\left(1-\frac{4z^2}{(n\pi)^2}\right)\right) + \sum\_{n=1}^{\infty} \frac{d}{dz}\left(\log\left(1-\left(\frac{z}{n\pi}\right)^2\right)\right).\tag{51}$$

Thus, from (51), we have

$$\sec(z) = \prod\_{n=1}^{\infty} \left( \frac{1 - \left(\frac{z}{n\pi}\right)^2}{1 - \left(\frac{2z}{n\pi}\right)^2} \right) = \prod\_{n=1}^{\infty} \left( 1 - \left(\frac{2z}{(2n-1)\pi}\right)^2 \right)^{-1},\tag{52}$$

which is equivalent to

$$\cos(z) = \prod\_{n=1}^{\infty} \left( 1 - \left( \frac{2z}{(2n-1)\pi} \right)^2 \right) \,. \tag{53}$$

A random variable has the Laplace distribution with positive parameter *μ* and *b* if its probability density function is

$$f(\mathbf{x}|\mu, b) = \frac{1}{2b} \exp\left(-\frac{|\mathbf{x} - \mu|}{b}\right),\tag{54}$$

(see [17]).

The shorthand notation *X* ∼ Laplace(*μ*, *b*) is used to indicate that the random variable *X* has the Laplace distribution with positive parameters *μ* and *b*. If *μ* = 0 and *b* = 1, the positive half-time is exactly an exponential scaled by <sup>1</sup> 2 .

We assume that the independent random variables *X*1, *X*2, *X*3, ··· have the Laplace distribution with parameters 0 and 1, (i.e., *Xk* ∼ Laplace(0, 1), *k* ∈ N). Let us put

$$Y = \sum\_{k=1}^{\infty} \frac{X\_k}{(2k-1)\pi}. \tag{55}$$

Then, the characteristic function of *Y* is given by

$$\begin{split} \sum\_{n=0}^{\infty} \mathbb{E}\left[Y^{n}\right] \frac{(2it)^{n}}{n!} &= \mathbb{E}\left[\sum\_{n=0}^{\infty} Y^{n} \frac{(2it)^{n}}{n!}\right] \\ &= \mathbb{E}\left[e^{2itYt}\right] \\ &= \mathbb{E}\left[e^{\left(\sum\_{k=1}^{\infty} \frac{X\_{k}}{(2k-1)\pi}\right)2it}\right] \\ &= \prod\_{k=1}^{\infty} \mathbb{E}\left[e^{\left(\frac{X\_{k}}{(2k-1)\pi} 2it\right)}\right]. \end{split} \tag{56}$$

Now, we observe that

$$\begin{split} E\left[e^{\frac{\lambda\_k}{(2k-1)\pi}2it}\right] &= \int\_{-\infty}^{\infty} \frac{1}{2} \varepsilon^{\varepsilon} \frac{\left(\frac{2it}{(2k-1)\pi}\right)x}{\varepsilon^{\varepsilon}} e^{-|x|} dx \\ &= \frac{1}{2} \int\_{-\infty}^{0} \varepsilon^{\varepsilon} \frac{\left(\frac{2it}{(2k-1)\pi}\right)x}{\varepsilon^{\varepsilon}} x^{\varepsilon} dx + \frac{1}{2} \int\_{0}^{\infty} \varepsilon^{\varepsilon} \frac{\left(\frac{2it}{(2k-1)\pi}\right)x}{\varepsilon^{\varepsilon}} x^{-x} dx \\ &= \frac{1}{2} \frac{1}{1 + \frac{2it}{(2k-1)\pi}} + \frac{1}{2} \frac{1}{1 - \frac{2it}{(2k-1)\pi}} \\ &= \left(1 + \left(\frac{2t}{(2k-1)\pi}\right)^{2}\right)^{-1} .\end{split} \tag{57}$$

By (53), (56) and (57), we get

$$\begin{split} \sum\_{n=0}^{\infty} \mathbb{E} \left[ Y^n \right] \frac{(2it)^n}{n!} &= \prod\_{k=1}^{\infty} \mathbb{E} \left[ \varepsilon \frac{\chi\_k}{(2k-1)\pi} \right] 2it \\ &= \prod\_{k=1}^{\infty} \left( 1 + \left( \frac{2t}{(2k-1)\pi} \right)^2 \right)^{-1} \\ &= \frac{2}{\varepsilon^t + \varepsilon^{-t}} \\ &= \sum\_{n=0}^{\infty} E\_n \frac{t^n}{n!} . \end{split} \tag{58}$$

Therefore, by comparing the coefficients on both sides of (58), we get

$$\mathcal{Z}^n i^n \mathbb{E}[Y^n] = E\_n, \quad (n \ge 0). \tag{59}$$

Now, we assume that

$$Z = \sum\_{k=1}^{\infty} \frac{X\_k}{2k\pi}.\tag{60}$$

Then, the characteristic function of *Z* is given by

$$\begin{split} \sum\_{n=0}^{\infty} E[Z^n] \frac{(it)^n}{n!} &= E\left[\sum\_{n=0}^{\infty} Z^n \frac{(it)^n}{n!} \right] \\ &= E[e^{Zit}] \\ &= E\left[e^{\sum\_{k=1}^{\infty} \left(\frac{X\_k}{2k\pi}\right)it} \right] \\ &= \prod\_{k=1}^{\infty} E\left[e^{\left(\frac{X\_k}{2k\pi}\right)it} \right]. \end{split} \tag{61}$$

Now, we note that

$$\begin{split} E\left[e^{\left(\frac{\lambda\_{\ell}}{2\pi\pi}\right)it}\right] &= \frac{1}{2} \int\_{-\infty}^{\infty} e^{\left(\frac{\hat{\mu}}{2\pi\pi}\right)x} e^{-|x|} dx \\ &= \frac{1}{2} \int\_{-\infty}^{0} e^{\left(\frac{\hat{\mu}}{2\pi\pi}\right)x} e^{x} dx + \frac{1}{2} \int\_{0}^{\infty} e^{\left(\frac{\hat{\mu}}{2\pi\pi}\right)x} e^{-x} dx \\ &= \frac{1}{2} \left(\frac{1}{1 + \frac{it}{2k\pi}}\right) + \frac{1}{2} \left(\frac{1}{1 - \frac{it}{2k\pi}}\right) \\ &= \frac{1}{1 + \left(\frac{1}{k\pi}\right)^{2}}. \end{split} \tag{62}$$

From (61) and (62), we have

$$\begin{split} \sum\_{n=0}^{\infty} E[Z^n] \frac{(it)^n}{n!} &= \prod\_{k=1}^{\infty} E\left[e^{\left(\frac{X\_k}{2k\pi}\right)it}\right] \\ &= \prod\_{k=1}^{\infty} \left(1 + \left(\frac{t}{2k\pi}\right)^2\right)^{-1} . \end{split} \tag{63}$$

On the other hand, by (48), we get

$$\prod\_{n=1}^{\infty} \left( 1 + \left( \frac{t}{n\pi} \right)^2 \right)^{-1} = \frac{it}{\sin(it)} = \frac{2t}{e^t - e^{-t}}.\tag{64}$$

By replacing *t* by *<sup>t</sup>* <sup>2</sup> , we have

$$\begin{split} \prod\_{n=1}^{\infty} \left( 1 + \left( \frac{t}{2n\pi} \right)^2 \right)^{-1} &= \frac{t}{\varepsilon^{\frac{t}{2}} - \varepsilon^{-\frac{t}{2}}} \\ &= \sum\_{n=0}^{\infty} \left( \frac{1}{2} \right)^{n-1} b\_n \frac{t^n}{n!} . \end{split} \tag{65}$$

Therefore, by (63) and (65), we obtain the following equation

$$\mathrm{d}^{\mathfrak{n}}\mathbb{E}[Z^{\mathfrak{n}}] = \left(\frac{1}{2}\right)^{n-1} b\_{\mathfrak{n}\_{\ell}} \quad (n \ge 0). \tag{66}$$

#### **4. Conclusions**

In this paper, we obtained several identities of symmetry for the type 2 Bernoulli and Euler polynomials (see Theorems 1–4). Indeed, they are symmetric identities involving type 2 Bernoulli polynomials and power sums of consecutive odd positive integers, and the ones involving type 2 Euler polynomials and alternating power sums of odd positive integers. For the derivation of those identities, we introduced certain quotients of bosonic *p*-adic and fermionic *p*-adic integrals on Z*p*, which have built-in symmetries. We note that this idea of using certain quotients of *p*-adic integrals has produced abundant symmetric identities (see [5,7,8,18–21] and references therein).

We emphasize here that, even though there have been many results on symmetric identities relating to some special numbers and polynomials, this paper is the first one that deals with symmetric identities

involving type 2 Bernoulli polynomials, type 2 Euler polynomials, power sums of odd positive integers and alternating power sums of odd positive integers.

In [22,23], we derived some identities involving special numbers and moments of random variables by using the generating functions of the moments of certain random variables. The related special numbers are Stirling numbers of the first and second kinds, degenerate Stirling numbers of the first and second kinds, derangement numbers, higher-order Bernoulli numbers and Bernoulli numbers of the second kind.

In this paper, we considered two random variables created from random variables having Laplace distributions and showed that their moments are closely connected with the type 2 Bernoulli and Euler numbers. Again, this is the first paper that interprets the type 2 Bernoulli and Euler numbers as the moments of certain random variables.

**Author Contributions:** Conceptualization, T.K.; Formal analysis, D.S.K. and T.K.; Funding acquisition, D.K.; Investigation, D.S.K., H.Y.K., D.K. and T.K.; Methodology, D.S.K. and T.K.; Project administration, D.K. and T.K.; Supervision, D.S.K. and T.K.; Validation, H.Y.K. and D.K.; Writing—original draft, T.K.; Writing—review and editing, D.S.K., H.Y.K. and D.K.

**Funding:** This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2019R1C1C1003869).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**



© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **Extended Degenerate** *r***-Central Factorial Numbers of the Second Kind and Extended Degenerate** *r***-Central Bell Polynomials**

**Dae San Kim 1, Dmitry V. Dolgy 2, Taekyun Kim <sup>3</sup> and Dojin Kim 4,\***


Received: 2 March 2019; Accepted: 22 April 2019; Published: 24 April 2019

**Abstract:** In this paper, we introduce the extended degenerate *r*-central factorial numbers of the second kind and the extended degenerate *r*-central Bell polynomials. They are extended versions of the degenerate central factorial numbers of the second kind and the degenerate central Bell polynomials, and also degenerate versions of the extended *r*-central factorial numbers of the second kind and the extended *r*-central Bell polynomials, all of which have been studied by Kim and Kim. We study various properties and identities concerning those numbers and polynomials and also their connections.

**Keywords:** extended degenerate *r*-central factorial numbers of the second kind; extended degenerate *r*-central bell polynomials

### **1. Introduction**

For *<sup>λ</sup>* ∈ R, we recall that the degenerate exponential function *<sup>e</sup><sup>x</sup> <sup>λ</sup>*(*t*) is defined by (see [1–7])

$$
\sigma\_{\lambda}^{\chi}(t) = (1 + \lambda t)^{\frac{x}{\chi}} \tag{1}
$$

When *x* = 1, we let *eλ*(*t*) = *e*<sup>1</sup> *<sup>λ</sup>*(*t*). Note that lim*λ*→<sup>0</sup> *ex <sup>λ</sup>*(*t*) = *<sup>e</sup>xt*.

We use the notation (*x*)*<sup>n</sup>* to denote the falling factorial sequence (*x*)*n*, which is defined by (see [8–14])

$$\mathbf{x}^\*(\mathbf{x})\_0 = 1, \quad (\mathbf{x})\_n = \mathbf{x}(\mathbf{x} - 1) \cdots (\mathbf{x} - n + 1), \quad (n \ge 1) \tag{2}$$

More generally, for *λ* ∈ R, the *λ*-falling factorial sequence (*x*)*n*,*<sup>λ</sup>* is given by (see [4])

$$(\mathbf{x})\_{0,\lambda} = 1, \quad (\mathbf{x})\_{n,\lambda} = \mathbf{x}(\mathbf{x} - \lambda)(\mathbf{x} - 2\lambda) \cdots (\mathbf{x} - (n - 1)\lambda), \quad (n \ge 1) \tag{3}$$

Obviously, it is noted that lim*λ*→<sup>1</sup> (*x*)*n*,*<sup>λ</sup>* = (*x*)*n*, lim*λ*→<sup>0</sup> (*x*)*n*,*<sup>λ</sup>* = *<sup>x</sup>n*, (*<sup>n</sup>* ≥ <sup>0</sup>).

In Reference [4], the *λ*- binomial expansion is defined by

$$(1+\lambda t)^{\frac{x}{\lambda}} = \sum\_{l=0}^{\infty} \binom{x}{l}\_{\lambda} t^l = \sum\_{l=0}^{\infty} (x)\_{l,\lambda} \frac{t^l}{l!} \tag{4}$$

*Symmetry* **2019**, *11*, 595; doi:10.3390/sym11040595 www.mdpi.com/journal/symmetry

where

$$
\binom{\mathbf{x}}{l}\_{\lambda} = \frac{(\mathbf{x})\_{l,\lambda}}{l!} = \frac{\mathbf{x}(\mathbf{x}-\lambda)(\mathbf{x}-2\lambda)\cdots(\mathbf{x}-(l-1)\lambda)}{l!}.
$$

The central factorial sequence is given by

$$\mathbf{x}^{[0]} = \mathbf{1}, \quad \mathbf{x}^{[n]} = \mathbf{x}(\mathbf{x} + \frac{n}{2} - 1)(\mathbf{x} + \frac{n}{2} - 2)\cdots(\mathbf{x} - \frac{n}{2} + 1), \quad (n \ge 1).$$

One can then easily show that the generating function of central factorial *x*[*n*] , (*n* ≥ 0), is given by (see [3,15–20])

$$\left(\frac{t}{2} + \sqrt{1 + \frac{t^2}{4}}\right)^{2x} = \sum\_{n=0}^{\infty} x^{[n]} \frac{t^n}{n!} \tag{5}$$

As is defined in [18], for any non-negative integer *n*, the central factorial numbers of the first kind are given by

$$\mathbf{x}^{[n]} = \sum\_{k=0}^{n} t(n,k)\mathbf{x}^{k}.\tag{6}$$

Then, from (5) and (6), we can show that the generating function of *t*(*n*, *k*) satisfies the following equation:

$$\frac{1}{k!} \left( 2 \log \left( \frac{t}{2} + \sqrt{1 + \frac{t^2}{4}} \right) \right)^k = \sum\_{n=k}^{\infty} t \left( n, k \right) \frac{t^n}{n!}.$$

As the inverse to the central factorial numbers of the first kind, the central factorial numbers of the second kind are defined by (see [18,20–22])

$$\mathbf{x}^n = \sum\_{k=0}^n T\_2(n,k)\mathbf{x}^{[k]}, \quad (n \ge 0) \tag{7}$$

The generating function of *T*2(*n*, *k*) can be easily derived from (7), which is given by (see [18])

$$\frac{1}{k!} \left( e^{\frac{k}{2}} - e^{-\frac{k}{2}} \right)^k = \sum\_{n=k}^{\infty} T\_2(n, k) \frac{t^n}{n!}, \quad (k \ge 0) \tag{8}$$

It can immediately be seen from (8) that

$$k!T\_2(n,k) = \sum\_{j=0}^{k} \binom{k}{j} (-1)^j (\frac{1}{2}k - j)^n. \tag{9}$$

In Reference [22] were introduced the central Bell polynomials defined by

$$e^{\frac{\chi}{\hbar}\left(\frac{\hbar}{e^{\frac{\hbar}{2}}-e^{-\frac{\hbar}{2}}\right)}}=\sum\_{n=0}^{\infty}B\_{n}^{(c)}(\chi)\frac{t^{n}}{n!}.\tag{10}$$

The Dobinski-like formula for *B*(*c*) *<sup>n</sup>* (*x*) is given by (see [22])

$$B\_n^{(c)}(\mathbf{x}) = \sum\_{l=0}^{\infty} \sum\_{k=0}^{\infty} \binom{l+k}{k} (-1)^k \frac{1}{(l+k)!} \left(\frac{l}{2} - \frac{k}{2}\right)^{l+1} \tag{11}$$

In Reference [3], the degenerate central factorial polynomials of the second kind are defined by

$$\frac{1}{k!} \left( \varepsilon\_{\lambda}^{\frac{1}{2}}(t) - \varepsilon\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} e\_{\lambda}^{\mathbf{x}}(t) = \sum\_{n=k}^{\infty} T\_{2,\lambda}(n, k|\mathbf{x}) \frac{t^{n}}{n!}, \quad (k \ge 0). \tag{12}$$

When *x* = 0, *T*2,*λ*(*n*, *k*) = *T*2,*λ*(*n*, *k*|0), these are called degenerate central factorial numbers of the second kind.

Let us recall that the degenerate central Bell polynomials are defined by (see [3])

$$\left(\epsilon^{x}\right)^{x}\Big(\epsilon^{\frac{1}{\lambda}\left(t\right)-\left(x\right)^{-1}\left(t\right)}\Big) = \sum\_{n=0}^{\infty} B\_{n,\lambda}^{\left(c\right)}\left(x\right)\frac{t^{n}}{n!}\Big.\tag{13}$$

In particular, *B*(*c*) *<sup>n</sup>*,*<sup>λ</sup>* <sup>=</sup> *<sup>B</sup>*(*c*) *<sup>n</sup>*,*λ*(1) are called the degenerate central Bell numbers.

Note that lim*λ*→<sup>0</sup> *B*(*c*) *<sup>n</sup>*,*λ*(*x*) = *<sup>B</sup>*(*c*) *<sup>n</sup>* (*x*), (*n* ≥ 0).

Carlitz [1] introduced the degenerate Stirling, Bernoulli, and Eulerian numbers as the first degenerate special numbers. Broder [23] investigated the *r*-Stirling numbers of the first and second kind as the numbers counting restricted permutations and restricted partitions, respectively. We recall here that the *r*-Stirling numbers of the second kind are given by (see [23])

$$\frac{1}{k!} \mathbf{e}^{rt} (\mathbf{e}^t - 1)^k = \sum\_{n=k}^{\infty} \mathcal{S}\_2^{(r)} (n+r, k+r) \frac{t^n}{n!} \tag{14}$$

In this paper, we will introduce the extended degenerate *r*-central factorial numbers of the second kind and the extended degenerate *r*-central Bell polynomials. Central analogues of Stirling numbers of the second kind and Bell polynomials are, respectively, the central factorial numbers of the second kind and the central Bell polynomials. Degenerate versions of the central factorial numbers of the second kind and the central Bell polynomials are, respectively, the degenerate central factorial numbers of the second kind and the degenerate central Bell polynomials. Extended versions of the degenerate central factorial numbers of the second kind and the degenerate central Bell polynomials are, respectively, the extended degenerate *r*-central factorial numbers of the second kind and the extended degenerate *r*-central Bell polynomials. The central factorial numbers of the second kind have many applications in such diverse areas as approximation theory [21], finite difference calculus, spline theory, spectral theory of differential operators [24,25], and algebraic geometry [26,27]. For broad applications of the related complete and incomplete Bell polynomials, we let the reader consult the introduction in [11]. Here, we will study various properties and identities relating to those numbers and polynomials, and also their connections. Finally, we note that the present paper can be useful in the area of non-integer systems and let the reader refer to [28] for more research in this direction.

### **2. Extended Degenerate** *r***-Central Factorial Numbers of the Second Kind and Extended Degenerate** *r***-Central Bell Polynomials**

From (12) and (13), we note that

$$\begin{split} \sum\_{n=0}^{\infty} B\_{n,\lambda}^{(c)}(\mathbf{x}) \frac{t^n}{n!} &= \sum\_{k=0}^{\infty} \mathbf{x}^k \sum\_{n=k}^{\infty} T\_{2,\lambda}(n,k) \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \sum\_{k=0}^n \mathbf{x}^k T\_{2,\lambda}(n,k) \frac{t^n}{n!} . \end{split} \tag{15}$$

One can compare the coefficients on both sides of (15) to obtain

$$B\_{n,\lambda}^{(\varepsilon)}(\mathbf{x}) = \sum\_{k=0}^{n} T\_{2,\lambda}(n,k)\mathbf{x}^{k}, \quad (n \ge 0). \tag{16}$$

Throughout this paper, we assume that *r* is a nonnegative integer. The following definition is motivated by (14).

**Definition 1.** *The extended degenerate r-central factorial numbers of the second kind <sup>T</sup>*(*r*) *<sup>λ</sup>* (*n* + *r*, *k* + *r*) *are defined as*

$$\frac{1}{k!} \boldsymbol{\varepsilon}\_{\lambda}^{r}(t) \left( \boldsymbol{\varepsilon}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{\varepsilon}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} = \sum\_{n=k}^{\infty} T\_{\lambda}^{(r)}(n+r, k+r) \frac{t^{n}}{n!}. \tag{17}$$

Note that lim*λ*→<sup>0</sup> *T*(*r*) *<sup>λ</sup>* (*<sup>n</sup>* <sup>+</sup> *<sup>r</sup>*, *<sup>k</sup>* <sup>+</sup> *<sup>r</sup>*) = *<sup>T</sup>*(*r*)(*<sup>n</sup>* <sup>+</sup> *<sup>r</sup>*, *<sup>k</sup>* <sup>+</sup> *<sup>r</sup>*), (*n*, *<sup>k</sup>* <sup>≥</sup> <sup>0</sup>),

where *T*(*r*)(*n* + *r*, *k* + *r*) is the extended *r*-central factorial numbers of the second kind given by

$$\frac{1}{k!}e^{r^t}\left(e^{\frac{t}{2}}-e^{-\frac{t}{2}}\right)^k = \sum\_{n=k}^{\infty} T^{(r)}(n+r,k+r)\frac{t^n}{n!}.\tag{18}$$

**Theorem 1.** *For n*, *k* ∈ N ∪ {0}*, with n* ≥ *k, we have*

$$T\_{\lambda}^{(r)}(n+r,k+r) = \sum\_{l=k}^{n} \binom{n}{l} T\_{2,\lambda}(l,k)(r)\_{n-l,\lambda}.$$

**Proof.** By (17), we get

$$\begin{split} \frac{1}{k!} \left( e\_{\lambda}^{\frac{1}{2}}(t) - e\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} e\_{\lambda}^{r}(t) &= \sum\_{l=k}^{\infty} T\_{2,\lambda}(l,k) \frac{t^{l}}{l!} \sum\_{m=0}^{\infty} (r)\_{m,\lambda} \frac{t^{m}}{m!} \\ &= \sum\_{n=k}^{\infty} \sum\_{l=k}^{n} \binom{n}{l} T\_{2,\lambda}(l,k) (r)\_{n-l,\lambda} \frac{t^{n}}{n!} . \end{split} \tag{19}$$

Therefore, by (17) and (19), we obtain the result.

We note that by taking the limit as *λ* tends to 0, we get

$$T^{(r)}(n+r,k+r) = \sum\_{l=k}^{n} \binom{n}{l} r^{n-l} T\_2(l,k). \tag{20}$$

**Theorem 2.** *For n*, *k* ≥ 0*, with n* ≥ *k, we have*

$$T\_{\lambda}^{(r)}(n+r,k+r) = \sum\_{m=k}^{n} \sum\_{l=k}^{m} \binom{m}{l} \mathcal{S}\_1(n,m) T\_2(l,k) \lambda^{n-m} r^{m-l},\tag{21}$$

*where S*1(*n*, *m*) *are the signed Stirling numbers of the first kind.*

**Proof.** Replacing *t* by <sup>1</sup> *<sup>λ</sup>* log(1 + *λt*) in (18), we obtain

$$\begin{split} \frac{1}{k!} \left( \boldsymbol{e}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{e}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} \boldsymbol{e}\_{\lambda}^{r}(t) &= \sum\_{m=k}^{\infty} \lambda^{-m} T^{(r)}(m+r, k+r) \frac{1}{m!} \Big( \log(1 + \lambda t) \Big)^{m} \\ &= \sum\_{m=k}^{\infty} \lambda^{-m} T^{(r)}(m+r, k+r) \sum\_{n=m}^{\infty} \mathcal{S}\_{1}(n, m) \frac{\lambda^{n} t^{n}}{n!} \\ &= \sum\_{n=k}^{\infty} \sum\_{m=k}^{n} \lambda^{n-m} \mathcal{S}\_{1}(n, m) T^{(r)}(m+r, k+r) \frac{t^{n}}{n!} . \end{split} \tag{22}$$

Now, by substituting the expression of *T*(*r*)(*m* + *r*, *k* + *r*) in (20) into (22), we finally get

$$\frac{1}{k!} \left( \varepsilon\_{\lambda}^{\frac{1}{2}}(t) - \varepsilon\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} \varepsilon\_{\lambda}^{r}(t) = \sum\_{n=k}^{\infty} \sum\_{m=k}^{n} \binom{m}{l} S\_{1}(n,m) T\_{2}(l,k) \lambda^{n-m} r^{m-l} \frac{t^{n}}{n!},$$

from which the result follows.

**Example 1.** *Here, we will illustrate the formula* (21) *for small values of n. The following values of T*2(*n*, *k*) *can be determined, for example, from the formula in* (9)*:*

$$T\_2(n,n) = 1,\ T\_2(n,0) = \delta\_{n,0},\ T\_2(2,1) = T\_2(3,2) = T\_2(4,1) = T\_2(4,3) = 0,\\ T\_2(3,1) = \frac{1}{4},\ T(4,2) = 1.\tag{23}$$

*In addition, we recall the following values of S*1(*n*, *k*)*:*

$$\begin{aligned} S\_1(n, n) &= 1, \ S\_1(n, 0) = \delta\_{n, 0}, \ S\_1(2, 1) = -1, \ S\_1(3, 1) = 2, \\ S\_1(3, 2) &= -3, \ S\_1(4, 1) = S\_1(4, 3) = -6, \ S\_1(4, 2) = 11. \end{aligned} \tag{24}$$

*Now, from* (21)*,* (23)*, and* (24)*, we easily have*

$$\begin{split} T\_{\lambda}^{(r)}(\mathbf{1}+r,\mathbf{r}+\mathbf{r})&=1,\ T\_{\lambda}^{(r)}(\mathbf{1}+r,\mathbf{r})=r,\ T\_{\lambda}^{(r)}(2+r,\mathbf{r})=-\lambda r+r^{2},\\ T\_{\lambda}^{(r)}(\mathbf{3}+r,\mathbf{r})&=2\lambda^{2}r-3\lambda r^{2}+r^{3},\ T\_{\lambda}^{(r)}(4+r,\mathbf{r})=-6\lambda^{3}r+11\lambda^{2}r^{2}-6\lambda r^{3}+r^{4},\\ T\_{\lambda}^{(r)}(2+r,\mathbf{1}+r)&=-\lambda+2r,\ T\_{\lambda}^{(r)}(3+r,\mathbf{1}+r)=2\lambda^{2}-6\lambda r+3r^{2}+\frac{1}{4},\\ T\_{\lambda}^{(r)}(3+r,\mathbf{2}+r)&=-3\lambda+3r,\ T\_{\lambda}^{(r)}(4+r,\mathbf{1}+r)=-6\lambda^{3}+22\lambda^{2}r-18\lambda r^{2}-\frac{3}{2}\lambda+4r^{3}+r,\\ T\_{\lambda}^{(r)}(4+r,\mathbf{2}+r)&=11\lambda^{2}-18\lambda r+6r^{2}+1,\ T\_{\lambda}^{(r)}(4+r,\mathbf{3}+r)=-6\lambda+4r. \end{split}$$

**Theorem 3.** *For n*, *k* ≥ 0*, with n* ≥ *k, we have*

$$T\_{\lambda}^{(r)}(n+r,k+r) = \sum\_{m=0}^{n-k} \binom{m+k}{m} m! \binom{r}{m} T\_{2,\lambda}(n,m+k|\frac{m}{2})... $$

**Proof.** Now, we observe that

1 *k*! *e r <sup>λ</sup>*(*t*) - *e* 1 2 *<sup>λ</sup>* (*t*) − *e* − 1 2 *<sup>λ</sup>* (*t*) *k* = 1 *k*! *e r* 2 *<sup>λ</sup>* (*t*) - *e* 1 2 *<sup>λ</sup>* (*t*) − *e* − 1 2 *<sup>λ</sup>* (*t*) + *e* − 1 2 *<sup>λ</sup>* (*t*) *<sup>r</sup>* - *e* 1 2 *<sup>λ</sup>* (*t*) − *e* − 1 2 *<sup>λ</sup>* (*t*) *k* = 1 *k*! ∞ ∑ *m*=0 - *r m e* 1 2 *<sup>λ</sup>* (*t*) − *e* − 1 2 *<sup>λ</sup>* (*t*) *m*+*<sup>k</sup> e m* 2 *<sup>λ</sup>* (*t*) = ∞ ∑ *m*=0 - *r m* (*m* + *k*)! *k*! 1 (*m* + *k*)! - *e* 1 2 *<sup>λ</sup>* (*t*) − *e* − 1 2 *<sup>λ</sup>* (*t*) *m*+*<sup>k</sup> e m* 2 *<sup>λ</sup>* (*t*) = ∞ ∑ *m*=0 - *r m m*! *m* + *k m* <sup>∞</sup> ∑ *n*=*m*+*k T*2,*λ*(*n*, *m* + *k*| *m* 2 ) *t n n*! = ∞ ∑ *n*=*k n*−*k* ∑ *m*=0 - *r m m*! *m* + *k m T*2,*λ*(*n*, *m* + *k*| *m* 2 ) *t n n*! . (25)

Therefore, by (17) and (25), we obtain the theorem.

One can easily show that the inverse function of *eλ*(*t*) is given by

$$
\log\_{\lambda}(t) = \frac{t^{\lambda} - 1}{\lambda}, \quad (t > 0),
$$

so that *eλ*(log*λ*(*t*)) = log*λ*(*eλ*(*t*)) = *t*, lim *λ*→0 log*λ*(*t*) = log(*t*). If *g*(*t*) = *e* 1 2 *<sup>λ</sup>* (*t*) − *e* − 1 2 *<sup>λ</sup>* (*t*), then one can see that

$$\log^{-1}(t) = \log\_{\lambda} \left( \frac{t}{2} + \sqrt{1 + \frac{t^2}{4}} \right)^2. \tag{26}$$

where *<sup>g</sup>* ◦ *<sup>g</sup>*−1(*t*) = *<sup>g</sup>*−<sup>1</sup> ◦ *<sup>g</sup>*(*t*) = *<sup>t</sup>*.

**Theorem 4.** *For n* ≥ 0*, we have*

$$\begin{aligned} (\mathbf{x} + r)\_{n,\lambda} &= \sum\_{k=0}^{n} T\_{\lambda}^{(r)} (n + r, k + r) \mathbf{x}^{[k]} \\ &= \sum\_{k=0}^{n} T\_{2,\lambda} (n, k | \frac{k}{2} + r) (\mathbf{x})\_k \end{aligned}$$

### **Proof.** By (1) and (4), we get

$$\begin{split} \epsilon\_{\lambda}^{x+r}(t) &= \epsilon\_{\lambda}^{t}(t) \left( \epsilon\_{\lambda}(t) - 1 + 1 \right)^{x} \\ &= \epsilon\_{\lambda}^{t}(t) \sum\_{k=0}^{\infty} (x)\_{k} \frac{1}{k!} \left( e\_{\lambda}(t) - 1 \right)^{k} \\ &= \sum\_{k=0}^{\infty} (x)\_{k} \frac{1}{k!} e\_{\lambda}^{\frac{k}{2} + r} \left( t \right) \left( e\_{\lambda}^{\frac{1}{2}}(t) - e\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} \\ &= \sum\_{k=0}^{\infty} (x)\_{k} \sum\_{n=k}^{\infty} T\_{2,\lambda}(n, k|\frac{k}{2} + r) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} (x)\_{k} T\_{2,\lambda}(n, k|\frac{k}{2} + r) \frac{t^{n}}{n!} . \end{split} \tag{27}$$

Now, from the observations in (26) and (5), we have

$$\begin{split} \epsilon^{++}\_{\lambda}(t) &= \epsilon^{\prime}\_{\lambda}(t) \epsilon^{\lambda}\_{\lambda}(t) \\ &= \epsilon^{\prime}\_{\lambda}(t) \left( \epsilon\_{\lambda} \left( \log\_{\lambda} \left( \frac{\mathbf{g}(t)}{2} + \sqrt{1 + \frac{\mathbf{g}(t)^{2}}{4}} \right)^{2} \right) \right)^{\cdot} \\ &= \epsilon^{\prime}\_{\lambda}(t) \left( \frac{\mathbf{g}(t)}{2} + \sqrt{1 + \frac{\mathbf{g}(t)^{2}}{4}} \right)^{2\times} \\ &= \sum\_{k=0}^{\infty} \mathbf{x}^{[k]} \frac{1}{k!} \epsilon^{\prime}\_{\lambda}(t) \left( \epsilon^{\frac{1}{\lambda}}\_{\lambda}(t) - \epsilon^{-\frac{1}{\lambda}}\_{\lambda}(t) \right)^{k} \\ &= \sum\_{k=0}^{\infty} \mathbf{x}^{[k]} \sum\_{n=-k}^{\infty} T^{(k)}\_{\lambda}(n+r, k+r) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} \mathbf{x}^{[k]} T^{(k)}\_{\lambda}(n+r, k+r) \frac{t^{n}}{n!} . \end{split} \tag{28}$$

From (4), we note also that

$$e\_{\lambda}^{x+r}(t) = \sum\_{n=0}^{\infty} (x+r)\_{n,\lambda} \frac{t^n}{n!}. \tag{29}$$

Therefore, by (27), (28), and (29), we have the desired result.

Note that, taking the limit as *λ* tends to 0, we have

$$(x+r)^n = \sum\_{k=0}^n T^{(r)}(n+r,k+r)x^{[k]} = \sum\_{k=0}^n T\_2(n,k|\frac{k}{2}+r)(x)\_k \dots$$

**Definition 2.** *The extended degenerate r-central Bell polynomials B*(*c*,*r*) *<sup>n</sup>*,*<sup>λ</sup>* (*x*) *are defined by*

$$e\_{\lambda}^{r}(t)e^{\mathbf{x}\left(\frac{\mathbf{b}}{\kappa^{2}}(t) - \varepsilon\_{\lambda}^{-\frac{1}{2}}(t)\right)} = \sum\_{n=0}^{\infty} B\_{n,\lambda}^{(c,r)}(\mathbf{x}) \frac{t^{n}}{n!}.\tag{30}$$

*Specifically, B*(*c*,*r*) *<sup>n</sup>*,*<sup>λ</sup>* (1) = *<sup>B</sup>*(*c*,*r*) *<sup>n</sup>*,*<sup>λ</sup> are called the extended degenerate r-central Bell numbers.*

**Theorem 5.** *For n* ≥ 0*, we have*

$$B\_{n,\lambda}^{(c,r)}(\mathbf{x}) = \sum\_{k=0}^{n} \mathbf{x}^k T\_{\lambda}^{(r)}(n+r, k+r).$$

**Proof.** From (30), we note that

$$\begin{split} \sigma\_{\lambda}^{r}(t)\varepsilon^{x}\Big(\varepsilon^{\frac{1}{\lambda}(t)-\varepsilon^{-\frac{1}{\lambda}}(t)}\Big) &= \sum\_{k=0}^{\infty}x^{k}\frac{1}{k!}\left(\varepsilon^{\frac{1}{\lambda}}\_{\lambda}(t)-\varepsilon^{-\frac{1}{\lambda}}(t)\right)^{k}\varepsilon\_{\lambda}^{r}(t) \\ &= \sum\_{k=0}^{\infty}x^{k}\sum\_{n=k}^{\infty}T\_{\lambda}^{(r)}(n+r,k+r)\frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty}\sum\_{k=0}^{n}x^{k}T\_{\lambda}^{(r)}(n+r,k+r)\frac{t^{n}}{n!}.\end{split} \tag{31}$$

Therefore, from (30) and (31), the theorem follows.

The central difference operator *δ* for a given function *f* is given by

$$
\delta f(\mathbf{x}) = f(\mathbf{x} + \frac{1}{2}) - f(\mathbf{x} - \frac{1}{2}),
$$

and by induction we can show

$$\delta^k f(\mathbf{x}) = \sum\_{l=0}^k \binom{k}{l} (-1)^{k-l} f(\mathbf{x} + l - \frac{k}{2}), \quad (k \ge 0). \tag{32}$$

**Theorem 6.** *Let n*, *k be nonnegative integers. Then, we have*

$$\frac{1}{k!} \delta^k(r)\_{n,\lambda} = \begin{cases} 0, & \text{if } n < k, \\ T\_{\lambda}^{(r)}(n + r, k + r), & \text{if } n \ge k. \end{cases}$$

**Proof.** By the binomial theorem, we have

$$\begin{split} \frac{1}{k!} e\_{\lambda}^{r}(t) \left( e\_{\lambda}^{\frac{1}{2}}(t) - e^{-\frac{1}{2}}(t) \right)^{k} &= \frac{1}{k!} e\_{\lambda}^{r-\frac{k}{2}}(t) \sum\_{l=0}^{k} \binom{k}{l} (-1)^{k-l} e\_{\lambda}^{l}(t) \\ &= \frac{1}{k!} \sum\_{l=0}^{k} \binom{k}{l} (-1)^{k-l} e\_{\lambda}^{r-\frac{k}{2}+l}(t) \\ &= \sum\_{n=0}^{\infty} \frac{1}{k!} \sum\_{l=0}^{k} \binom{k}{l} (-1)^{k-l} (r - \frac{k}{2} + l)\_{n,\lambda} \frac{t^{n}}{n!} . \end{split} \tag{33}$$

If we choose *f*(*x*)=(*x*)*n*,*λ*, (*n* ≥ 0) in (32), then we have

$$\delta^k(r)\_{n,\lambda} = \sum\_{l=0}^k \binom{k}{l} (r+l-\frac{k}{2})\_{n,\lambda} (-1)^{k-l}.\tag{34}$$

From (33) and (34), the following equation is obtained.

$$\frac{1}{k!}e\_{\lambda}^{r}(t)\left(e\_{\lambda}^{\frac{1}{2}}(t) - e\_{\lambda}^{-\frac{1}{2}}(t)\right)^{k} = \sum\_{n=0}^{\infty} \frac{1}{k!} \delta^{k}(r)\_{n,\lambda} \frac{t^{n}}{n!}.\tag{35}$$

Therefore, by (17) and (35), we have the result.

From Theorem 4 and Theorem 5, we have

$$\begin{split} B\_{n,\lambda}^{(c,r)}(\mathbf{x}) &= \sum\_{k=0}^{n} T\_{\lambda}^{(r)}(n+r,k+r) \mathbf{x}^{k} \\ &= \sum\_{k=0}^{n} \mathbf{x}^{k} \frac{1}{k!} \boldsymbol{\delta}^{k}(r)\_{n,\lambda} \quad (n \ge 0). \end{split} \tag{36}$$

**Theorem 7.** *For n* ≥ 0*, we have*

$$B\_{n,\lambda}^{(c,r)}(\mathbf{x}) = \sum\_{m=0}^{n} \binom{n}{m} (r)\_{n-m,\lambda} B\_{m,\lambda}^{(c)}(\mathbf{x}) .$$

**Proof.** From (30), we note that

$$\begin{split} \sum\_{n=0}^{\infty} B\_{n,\lambda}^{(c,r)}(x) \frac{t^n}{n!} &= \varepsilon\_{\lambda}'(t) e^{\frac{t}{\lambda} \left(\frac{1}{\lambda}(t) - \varepsilon\_{\lambda}^{-\frac{1}{\lambda}}(t)\right)} \\ &= \sum\_{l=0}^{\infty} (r)\_{l,\lambda} \frac{t^l}{l!} \sum\_{m=0}^{\infty} B\_{m,\lambda}^{(c)} \frac{t^m}{m!} \\ &= \sum\_{n=0}^{\infty} \sum\_{m=0}^{n} \binom{n}{m} (r)\_{n-m,\lambda} B\_{m,\lambda}^{(c)} \left(x\right) \frac{t^n}{n!} .\end{split} \tag{37}$$

Therefore, by comparing the coefficients on both sides of (37), the desired result is achieved.

**Theorem 8.** *For m*, *n*, *k* ≥ 0*, with n* ≥ *m* + *k, we have*

$$
\binom{m+k}{m} T\_{\lambda}^{(r)}(n+r,m+k+r) = \sum\_{l=m}^{n-k} \binom{n}{l} T\_{\lambda}^{(r)}(l+r,m+r) T\_{2,\lambda}(n-l,k).
$$

**Proof.** We further observe that

$$\begin{split} \frac{1}{m!} \boldsymbol{\varepsilon}\_{\lambda}^{r}(t) \left( \boldsymbol{\varepsilon}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{\varepsilon}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{m} \frac{1}{k!} \left( \boldsymbol{\varepsilon}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{\varepsilon}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} &= \frac{(m+k)!}{m!k!} \frac{1}{(m+k)!} \boldsymbol{\varepsilon}\_{\lambda}^{r}(t) \left( \boldsymbol{\varepsilon}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{\varepsilon}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{m+k} \\ &= \binom{m+k}{m} \sum\_{n=m+k}^{\infty} T\_{\lambda}^{(r)}(n+r, m+k+r) \frac{t^{n}}{n!}, \end{split} \tag{38}$$

where *m*, *k* are nonnegative integers. Alternatively, the left-hand side of (38) can be expressed by

$$\begin{split} \frac{1}{m!} \boldsymbol{\varepsilon}\_{\lambda}^{r}(t) \left( \boldsymbol{\varepsilon}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{\varepsilon}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{m} \frac{1}{k!} \left( \boldsymbol{\varepsilon}\_{\lambda}^{\frac{1}{2}}(t) - \boldsymbol{\varepsilon}\_{\lambda}^{-\frac{1}{2}}(t) \right)^{k} &= \sum\_{l=m}^{\infty} \boldsymbol{T}\_{\lambda}^{(r)}(l+r, m+r) \frac{t^{l}}{l!} \sum\_{j=k}^{\infty} \boldsymbol{T}\_{2, \lambda}(j, k) \frac{t^{j}}{j!} \\ &= \sum\_{n=m+l}^{\infty} \sum\_{l=m}^{n-k} \binom{n}{l} \boldsymbol{T}\_{\lambda}^{(r)}(l+r, m+r) \boldsymbol{T}\_{2, \lambda}(n-l, k) \frac{t^{n}}{n!}. \end{split} \tag{39}$$

Therefore, by (38) and (39), the desired identity is obtained.

### **3. Conclusions**

In recent years, many researchers have studied a lot of old and new special numbers and polynomials by means of generating functions, through combinatorial methods, umbral calculus, differential equations, *p*-adic integrals, *p*-adic *q*-integrals, special functions, complex analyses, and so on.

The study of degenerate versions of special numbers and polynomials began with Carlitz [1]. Kim and his colleagues have been studying degenerate versions of various special numbers and polynomials by making use of the same methods. Studying degenerate versions of known special numbers and polynomials can be very a fruitful research and is highly rewarding. For example, this line of study led even to the introduction of degenerate Laplace transforms and degenerate gamma functions (see [4]).

In this paper, we introduced the extended degenerate *r*-central factorial numbers of the second kind and the extended degenerate *r*-central Bell polynomials. We studied various properties and identities relating to those numbers and polynomials and also their connections. This study was done by using generating function techniques.

Central analogues of Stirling numbers of the second kind and Bell polynomials are, respectively, the central factorial numbers of the second kind and the central Bell polynomials. Degenerate versions of the central factorial numbers of the second kind and the central Bell polynomials are, respectively, the degenerate central factorial numbers of the second kind and the degenerate central Bell polynomials. Extended versions of the degenerate central factorial numbers of the second kind and the degenerate central Bell polynomials are, respectively, the extended degenerate *r*-central factorial numbers of the second kind and the extended degenerate *r*-central Bell polynomials. The central factorial numbers of the second kind have many applications in diverse areas such as approximation theory [21], finite difference calculus, spline theory, spectral theory of differential operators [24,25], and algebraic geometry [26,27].

For future research projects, we would like to continue to work on some special numbers and polynomials and their degenerate versions, as well as try to explore their applications not only in mathematics but also in the sciences and engineering [29].

**Author Contributions:** Conceptualization, D.S.K., T.K. and D.K.; Formal analysis, D.S.K., D.V.D., T.K. and D.K.; Investigation, D.S.K., D.V.D., T.K. and D.K.; Methodology, D.S.K., T.K. and D.K.; Supervision, D.S.K.; Writing—original draft, T.K.; Writing—review & editing, D.S.K., D.V.D., T.K. and D.K.

**Funding:** This work was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT) (No. 2019R1C1C1003869).

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**



© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **The Extended Minimax Disparity RIM Quantifier Problem**

### **Dug Hun Hong**

Department of Mathematics, Myongji University, Yongin Kyunggido 449-728, Korea; dhhong@mju.ac.kr

Received: 8 March 2019; Accepted: 1 April 2019; Published: 3 April 2019

**Abstract:** An interesting regular increasing monotone (RIM) quantifier problem is investigated. Amin and Emrouznejad [Computers & Industrial Engineering 50(2006) 312–316] have introduced the extended minimax disparity OWA operator problem to determine the OWA operator weights. In this paper, we propose a corresponding continuous extension of an extended minimax disparity OWA model, which is the extended minimax disparity RIM quantifier problem, under the given orness level and prove it analytically.

**Keywords:** fuzzy sets; RIM quantifier; extended minimax disparity; OWA model; RIM quantifier problem

### **1. Introduction**

One of the important topic in the theory of ordered weighted averaging (OWA) operators is the determination of the associated weights. Several authors have suggested a number of methods for obtaining associated weights in various areas such as decision making, approximate reasoning, expert systems, data mining, fuzzy systems and control [1–18]. Researchers can easily see most of OWA papers in the recent bibliography published in Emrouznejad and Marra [5]. Yager [16] proposed RIM quantifiers as a method for finding OWA weight vectors through fuzzy linguistic quantifiers. Liu [19] and Liu and Da [20] gave solutions to the maximum-entropy RIM quantifier model when the generating functions are differentiable. Liu and Lou [21] studied the equivalence of solutions to the minimax ratio and maximum-entropy RIM quantifier models, and the equivalence of solutions to the minimax disparity and minimum-variance RIM quantifier problems. Hong [22,23] gave the proof of the minimax ratio RIM quantifier problem and the minimax disparity RIM quantifier model when the generating functions are absolutely continuous. He also gave solutions to the maximum-entropy RIM quantifier model and the minimum-variance RIM quantifier model when the generating functions are Lebesgue integrable. Liu [24] proposed a general RIM quantifier determination model, proved it analytically using the optimal control method and investigated the solution equivalence to the minimax problem for the RIM quantifier. However, Hong [11] recently provided a modified model for the general RIM quantifier model and the correct formulation of Liu's result.

Amin and Emrouznejad [1] have introduced the following the extended minimax disparity OWA operator model to determine the OWA operator weights:

$$\begin{aligned} \text{Minimize } & \max\_{i \in \{1, \dots, n-1\}, \ j \in \{i+1, \dots, n\}} |w\_i - w\_j| \\ \text{subject to } & \operatorname{cross}(W) = \sum\_{i=1}^n \frac{n-i}{n-1} w\_i = n, \ 0 \le n \le 1, \\ & w\_1 + \dots + w\_n = 1, 0 \le w\_i, i = 1, \dots, n. \end{aligned}$$

In this paper, we propose a corresponding extended minimax disparity model for RIM quantifier determination under given orness level and prove it analytically. This paper is organized as follows: Section 2 presents the preliminaries and Section 3 reviews some models for the RIM quantifier problems and propose the extended minimax disparity model for the RIM quantifier problem. In Section 4, we prove the extended minimax disparity model problem mathematically for the case in which the generating functions are Lesbegue integrable functions.

### **2. Preliminaries**

Yager [15] introduced a new aggregation technique based on the OWA operators. An OWA operator of dimension *<sup>n</sup>* is a function *<sup>F</sup>* : *<sup>R</sup><sup>n</sup>* → *<sup>R</sup>* that has an associated weighting vector *<sup>W</sup>* = (*w*1, ··· , *wn*)*<sup>T</sup>* of having the properties 0 ≤ *wi* ≤ 1, *<sup>i</sup>* = 1, ··· , *<sup>n</sup>*, *<sup>w</sup>*<sup>1</sup> + ··· + *wn* = 1, and such that

$$F(a\_1, \dots, a\_n) = \sum\_{i=1}^n w\_i b\_{i\prime}$$

where *bj* is the *j*th largest element of the collection of the aggregated objects {*a*1, ··· , *an*}. In [15], Yager defined a measure of "orness" associated with the vector *W* of an OWA operator as

$$orness(\mathcal{W}) = \sum\_{i=1}^{n} \frac{n-i}{n-1} w\_{i\prime}$$

and it characterizes the degree to which the aggregation is like an *or* operation.

The RIM quantifiers was introduced by Yager [16] as a method for obtaining the OWA weight vectors via fuzzy linguistic quantifiers. The RIM quantifiers can provide information aggregation procedures guided by a dimension independent description and verbally expressed concepts of the desired aggregation.

**Definition 1** ([14])**.** *A fuzzy subset <sup>Q</sup> is called a RIM quantifier if <sup>Q</sup>*(0) = 0, *<sup>Q</sup>*(1) = <sup>1</sup> *and <sup>Q</sup>*(*x*) <sup>≥</sup> *<sup>Q</sup>*(*y*) *for x* > *y*.

The quantifier *f or all* is represented by the fuzzy set

$$Q\_\*(r) = \begin{cases} 1, & \text{x = 1, \\ 0, & \text{x \neq 1.} \end{cases}$$

The quantifier *there exist*, not none, is defined as

$$Q^\*(r) = \begin{cases} 0, & \text{x = 0,} \\ 1, & \text{x \neq 0.} \end{cases}$$

Both of these are examples of RIM quantifier. To analyze the relationship between OWA and RIM quantifier, a generating function representation of RIM quantifier was proposed.

**Definition 2.** *For <sup>f</sup>*(*t*) *on [0, 1] and a RIM quantifier <sup>Q</sup>*(*x*), *<sup>f</sup>*(*t*) *is called generating function of <sup>Q</sup>*(*x*), *if it satisfies*

$$Q(x) = \int\_0^x f(t)dt$$

*where f*(*t*) <sup>≥</sup> <sup>0</sup> *and* " <sup>1</sup> <sup>0</sup> *f*(*t*)*dt* = 1.

If *Q*(*x*) is an absolutely continuous function, then *f*(*x*) is a Lesbegue integrable function; moreover, *f*(*x*) is unique in the sense of "almost everywhere" in abbreviated form, *a.e.*

Yager extended the *orness* measure of OWA operator, and defined the *orness* of a RIM quantifier [16].

$$
ormals(Q) = \int\_0^1 Q(x)dx = \int\_0^1 (1-t)f(t)dt.
$$

As the RIM quantifier can be seen as the continuous form of OWA operator with generating function, OWA optimization problem is extended to the RIM quantifier case.

The definitions of *essential supremum* and *essential infimum* [21] of *f* are as follows:

$$\operatorname{ess}\sup f = \inf \left\{ t : \left| \{ \mathbf{x} \in [0,1] : f(\mathbf{x}) > t \} \right| = 0 \right\},$$

$$\operatorname{ess}\inf f = \sup \left\{ t : \left| \{ \mathbf{x} \in [0,1] : f(\mathbf{x}) < t \} \right| = 0 \right\},$$

where |*E*| is the Lebesgue measure of the Lebesgue measurable set *E*.

### **3. Models for the RIM Quantifier Problems**

Fullér and Majlender [8] proposed the minimum variance model, which minimizes the variance of OWA operator weights under a given level of orness. Their method requires the proof of the following mathematical programming problem:

$$\begin{aligned} \text{Minimize} \quad & D(W) = \frac{1}{n} \sum\_{i=1}^{n-1} \left( w\_i - \frac{1}{n} \right)^2 \\ \text{subject to} \quad & \text{or} \quad \text{cross}(W) = \sum\_{i=1}^n \frac{n-i}{n-1} w\_i = n, \quad 0 \le n \le 1, \\ & w\_1 + \dots + w\_n = 1, 0 \le w\_{i'} i = 1, \dots, n. \end{aligned}$$

Liu [19,24] extended the minimum variance problem for OWA operator to the RIM quantifier problem case:

$$\begin{aligned} \text{Minimize } \quad & D\_f = \int\_0^1 f^2(r) dr - 1\\ \text{subject to } \quad & \int\_0^1 r f(r) dr = 1 - a, \ 0 < a < 1/a\\ & \int\_0^1 f(r) dr = 1, \ f(r) \ge 0. \end{aligned}$$

Wang and Parkan [13] proposed the minimax disparity problem as follows:

$$\begin{aligned} \text{Minimize } & \max\_{i \in \{1, \dots, n-1\}} |w\_i - w\_{i+1}| \\ \text{subject to } & \operatorname{argmax}(W) = \sum\_{i=1}^n \frac{n-i}{n-1} w\_i = n, \ 0 \le n \le 1, \\ & w\_1 + \dots + w\_n = 1, 0 \le w\_i, i = 1, \dots, n. \end{aligned}$$

Similar to the minimax disparity OWA operator problem, Hong [11] proposed the minimax disparity RIM quantifier problem as follows:

$$\begin{aligned} \text{Minimize } & \quad \operatorname{ess\,sup}\_{t \in [0,1]} \left| f'(t) \right| \\ \text{subject to } & \quad \int\_0^1 r f(r) dr = 1 - a, \ 0 < a < 1, \\ & \quad \int\_0^1 f(r) dr = 1, \text{ absolutely continuous } f(r) \ge 0. \end{aligned}$$

Wang et al. [14] have introduced the following least squares deviation (LSD) method as an alternative approach to determine the OWA operator weights.

$$\begin{aligned} \text{Minimize } \quad & \sum\_{i=1}^{n-1} \left( w\_i - w\_{i-1} \right)^2 \\ \text{subject to } \quad & \operatorname{or} \operatorname{ess}(W) = \sum\_{i=1}^n \frac{n-i}{n-1} w\_i = n, \ 0 \le n \le 1, \\ & w\_1 + \dots + w\_n = 1, 0 \le w\_{i\prime} i = 1, \dots, n. \end{aligned}$$

Hong [25] proposed the following corresponding least squares disparity RIM quantifier problem under a given orness level:

$$\begin{aligned} \text{Minimize } \quad & D\_f = \int\_0^1 (f')^2(r) dr\\ \text{subject to } \quad & \int\_0^1 (1-r)f(r) dr = a, \ 0 < a < 1, \\ & \int\_0^1 f(r) dr = 1, \\ & f(r) > 0. \end{aligned}$$

Recently, Amin and Emrouznejad [1] proposed a problem of minimizing the maximum disparity of any distinct pairs of weights instead of adjacent weights. that is:

$$\begin{aligned} \text{Minimize } & \max\_{i \in \{1, \dots, n-1\}, j \in \{i+1, \dots, n\}} |w\_i - w\_j| \\ \text{subject to } & \operatorname{cross}(\mathcal{W}) = \sum\_{i=1}^n \frac{n-i}{n-1} w\_i = \alpha, \ 0 \le \alpha \le 1, \\ & w\_1 + \dots + w\_n = 1, 0 \le w\_i, i = 1, \dots, n. \end{aligned}$$

We consider the following easy important fact. **Note**

$$
\max\_{i \in \{1, \dots, n-1\}, j \in \{i+1, \dots, n\}} |w\_i - w\_j| = \max w\_i - \min w\_i.
$$

For this, first it is trivial that

$$\max\_{i \in \{1, \dots, n-1\}, j \in \{i+1, \dots, n\}} |w\_i - w\_j| \le \max |w\_i - \min |w\_i|.$$

Next, suppose that *max wi* = *wi*<sup>0</sup> , *min wi* = *wj*<sup>0</sup> . If *i*<sup>0</sup> < *j*0, then

$$\begin{aligned} \max w\_i - \min w\_i &=& w\_{i\_0} - w\_{j\_0} \\ &=& |w\_{i\_0} - w\_{j\_0}| \\ &\le& \max\_{i \in \{1, \dots, n-1\}, \, j \in \{i\_0+1, \dots, n\}} |w\_i - w\_j| \end{aligned}$$

If *i*<sup>0</sup> > *j*0, then

$$\begin{aligned} \max w\_i - \min w\_i &=& w\_{i0} - w\_{j0} \\ &=& |w\_{j0} - w\_{i0}| \\ &\leq& \max\_{i \in \{1, \dots, \mathfrak{n} - 1\}, j \in \{j + 1, \dots, \mathfrak{n}\}} |w\_i - w\_j|. \end{aligned}$$

and hence the equality holds.

Then the corresponding extended minimax disparity model for RIM quantifier problem with given orness level can be proposed as follows:

$$\begin{aligned} \text{Minimize } & \quad \operatorname{ess}\sup f - \operatorname{ess}\inf f \\ \text{subject to } & \quad \int\_0^1 r f(r) dr = 1 - a, \ 0 < a < 1, \\ & \quad \int\_0^1 f(r) dr = 1, \ f(r) \ge 0. \end{aligned} $$

### **4. Relation of Solutions between OWA Operator Model and RIM Quantifier Model**

The following result is the solution of the extended minimax OWA operator problem given by Hong [26].

**Theorem 1** (*n* = 2*k*:even)**.** *An optimal weight for the constrained optimization problem (2) for a given level of α* = *orness*(*W*) *should satisfy the following equation:*

$$H(a) = \text{Minimize}\left\{ \max\_{i \in \{1, \dots, n-1\}, \ j \in \{i+1, \dots, n\}} |w\_i - w\_j| \right\} = \left| \frac{(1 - 2\alpha)(n - 1)}{(n - m)m} \right|^2$$

$$w\_1^\* = w\_2^\* = \dots = w\_{m'}^\* \ w\_{k+1}^\* = w\_{k+2}^\* = \dots = w\_{n'}^\*$$

*where*

$$w\_1^\* = \frac{m - (1 - 2a)(n - 1)}{nm}$$

*and*

$$w\_{m+1}^\* = \frac{n - m - (2\alpha - 1)(n - 1)}{n(n - m)}.$$

*Here m satisfies the following:*

$$m = \begin{cases} \lceil (1 - 2\alpha)(n - 1) \rceil, & \text{if} \quad 0 \le \alpha \le \frac{n - 2}{4(n - 1)},\\ k, & \text{if} \quad \frac{n - 2}{4(n - 1)} \le \alpha \le \frac{3n - 2}{4(n - 1)},\\ n - \lceil (2\alpha - 1)(n - 1) \rceil, & \text{if} \quad \frac{3n - 2}{4(n - 1)} \le \alpha \le 1. \end{cases}$$

*where x* = *m* + 1 ⇐⇒ *m* < *x* ≤ *m* + 1 *for any integer m*.

Can we get a hint about the solution of the extended minimax Rim quantifier problem? Here, we suggest an idea.

For a given associated weighting vector *Wn* = (*w*1, ··· , *wn*) of having the property *w*<sup>1</sup> + ··· + *wn* = 1, 0 ≤ *wi* ≤ 1, *i* = 1, ··· , *n*, we define a generating function *f*(*t*)

$$f\_{W\_n}(\mathbf{x}) = nw\_{i\prime} \quad \mathbf{x} \in \left[\frac{i}{n}, \frac{i+1}{n}\right) , \ i = 0, 1, \cdots, n-1, \cdots$$

having the property " <sup>1</sup> <sup>0</sup> *<sup>f</sup> <sup>n</sup> <sup>W</sup>*(*x*)*dx* = 1 and let

$$f^\*(\mathbf{x}) = \lim\_{n \to \infty} = f\_{\mathbb{W}\_n}(\mathbf{x}).$$

Can this function *f* ∗(*x*) be a solution of the corresponding extended minimax Rim quantifier problem? Maybe, yes! Let's try to follow this idea.

For given *W*∗ *<sup>n</sup>* = (*w*<sup>∗</sup> <sup>1</sup>, ··· , *w*<sup>∗</sup> *<sup>n</sup>*) from above Theorem 1, we have for 0 <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup> 4 ,

$$f\_{W\_n^\pm}(\mathbf{x}) = \begin{cases} \frac{\left[ \left( 1 - 2a \right) (n - 1) \right] - (1 - 2a)(n - 1)}{\left[ (1 - 2a)(n - 1) \right]}, & \text{if} \quad \mathbf{x} \in \left[ 0, \frac{\left[ \left( 1 - 2a \right) (n - 1) \right]}{n} \right] \\\frac{n - \left[ \left( 1 - 2a \right) (n - 1) \right] - (2a - 1)(n - 1)}{n - \left[ \left( 1 - 2a \right) (n - 1) \right]}, & \text{if} \quad \mathbf{x} \in \left[ \frac{\left[ \left( 1 - 2a \right) (n - 1) \right]}{n}, 1 \right]. \end{cases}$$

for <sup>1</sup> <sup>4</sup> <sup>≤</sup> *<sup>α</sup>* <sup>≤</sup> <sup>3</sup> 4 ,

$$f\_{W\_n^\*} (\mathbf{x}) = \begin{cases} \frac{n/2 - (1 - 2\alpha)(n - 1)}{n/2}, & \text{if} \quad \mathbf{x} \in \left[0, \frac{1}{2}\right) \\\frac{n/2 - (2\alpha - 1)(n - 1)}{(n/2)}, & \text{if} \quad \mathbf{x} \in \left[\frac{1}{2}, 1\right]. \end{cases}$$

for 3/4 ≤ *α* ≤ 1,

$$f\_{\mathcal{W}\_n^\pi}(\mathbf{x}) = \begin{cases} \frac{n - \lceil (2a-1)(n-1) \rceil - (1-2a)(n-1)}{n - \lceil (2a-1)(n-1) \rceil}, & \text{if} \quad \mathbf{x} \in \left[0, \mathbf{1} - \frac{\lceil (1-2a)(n-1) \rceil}{n} \right] \\\frac{\lceil (2a-1)(n-1) \rceil - (2a-1)(n-1)}{\lceil (2a-1)(n-1) \rceil} & \text{if} \quad \mathbf{x} \in \left[1 - \frac{\lceil (1-2a)(n-1) \rceil}{n}, 1 \right]. \end{cases}$$

Let lim*n*→<sup>∞</sup> *fW*<sup>∗</sup> *<sup>n</sup>* (*x*) = *f* <sup>∗</sup>(*x*), then

1. for 0 <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup> 4 ,

$$f^\*(r) = \begin{cases} 0, & \text{if} \quad r \in [0, 1 - 2\alpha), \\ \frac{1}{2\alpha'} & \text{if} \quad r \in [1 - 2\alpha, 1]. \end{cases}$$

2. for <sup>1</sup> <sup>4</sup> <sup>≤</sup> *<sup>α</sup>* <sup>≤</sup> <sup>3</sup> 4 ,

$$f^\*(r) = \begin{cases} 4\alpha - 1, & \text{if } \quad r \in \left[0, \frac{1}{2}\right), \\ 3 - 4\alpha, & \text{if } \quad r \in \left[\frac{1}{2}, 1\right]. \end{cases}$$

3. for <sup>3</sup> <sup>4</sup> < *α* ≤ 1,

$$f^\*(r) = \begin{cases} \frac{1}{2(1-\kappa)'} & \text{if} \quad r \in [0, 2\kappa],\\ 0, & \text{elsewhere.} \end{cases}$$

In the following section, we will show that *f* ∗ can be the solution of the extended minimax RIM quantifier problem.

### **5. Proof of the Extended Minimax RIM Quantifier Problem**

In this section, we prove the following main result.

**Theorem 2.** *The optimal solution for problem (2) for given orness level α is the weighting function f* ∗ *such that*

*1. for* <sup>0</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup> 4 ,

$$f^\*(r) = \begin{cases} 0 \ a.e.\_\prime & \text{if} \quad r \in [0, 1 - 2\alpha)\_\prime\\ \frac{1}{2\alpha} \ a.e.\_\prime & \text{if} \quad r \in [1 - 2\alpha, 1]\_\prime \end{cases}$$

*2. for* <sup>1</sup> <sup>4</sup> <sup>≤</sup> *<sup>α</sup>* <sup>≤</sup> <sup>3</sup> 4 ,

$$f^\*(r) = \begin{cases} 4\alpha - 1 & a.c., \qquad \text{if} \quad r \in \left[0, \frac{1}{2}\right), \\ 3 - 4\alpha & a.c., \qquad \text{if} \quad r \in \left[\frac{1}{2}, 1\right]. \end{cases}$$

*3. for* <sup>3</sup> <sup>4</sup> < *α* ≤ 1,

$$f^\*(r) = \begin{cases} \frac{1}{2(1-a)} & a.e.\_ {\prime} \\ 0 & a.e.\_ {\prime} \end{cases} \quad \text{if} \quad r \in [0, 2a]\_{\prime}$$

*and*

$$H(a) = \text{Minimize } \left| \operatorname{ess} \sup f - \operatorname{ess} \inf f \right| = \begin{cases} \frac{1}{2a} & \text{if } \quad 0 < a \le \frac{1}{4}, \\ 4|(1 - 2a)| & \text{if } \quad \frac{1}{4} \le a \le \frac{3}{4}, \\ \frac{1}{2a} & \text{if } \quad \frac{3}{4} < a \le 1. \end{cases}$$

We need the following two lemma's to prove the main result. We denote *Df*(*x*) = " *<sup>x</sup>* <sup>0</sup> *f*(*t*)*dt*, <sup>0</sup> <sup>≤</sup> *<sup>x</sup>* <sup>≤</sup> 1 and *<sup>E</sup>*(*f*) = " <sup>1</sup> <sup>0</sup> *r f*(*r*)*dr*.

The following result is known.

**Lemma 1.** *E*(*f*) = " <sup>1</sup> <sup>0</sup> (1 − *Df*(*t*))*dt*.

**Lemma 2.** *Let ess inf f* <sup>=</sup> *<sup>β</sup>*<sup>0</sup> <sup>≥</sup> <sup>0</sup> *and ess sup f* <sup>=</sup> *<sup>β</sup>*<sup>1</sup> <sup>&</sup>gt; <sup>0</sup> *such that* " <sup>1</sup> <sup>0</sup> *f*(*r*)*dr* = 1 *and define a function f*<sup>0</sup> *as*

$$f\_0(r) = \begin{cases} \beta\_0 \ a.e., & \text{if} \quad r \in [0, c\_0)\_{\prime\prime} \\ \beta\_1 \ a.e., & \text{if} \quad r \in [c\_0, 1]. \end{cases}$$

*for some <sup>c</sup>*<sup>0</sup> <sup>∈</sup> (0, 1) *such that* " <sup>1</sup> <sup>0</sup> *f*0(*r*)*dr* = 1. *Then we have E*(*f*) ≤ *E*(*f*0) *and the equality holds iff f* = *f*<sup>0</sup> *a*.*e*.

**Proof.** The result follows immediately from Lemma 1 if we show that *Df*<sup>0</sup> (*x*) ≤ *Df*(*x*), *x* ∈ [0, 1]. It is clear that *Df*<sup>0</sup> (*x*) ≤ *Df*(*x*), *x* ∈ [0, *c*0]. Suppose that there exists a point *t*<sup>0</sup> ∈ (*c*0, 1) such that *Df*<sup>0</sup> (*t*0) > *Df*(*t*0). Then

$$\int\_{t\_0}^1 \beta\_1 dr = \int\_{t\_0}^1 f\_0(r) dr = 1 - D\_{f\_0}(t\_0) < 1 - D\_f(t\_0) = \int\_{t\_0}^1 f(r) dr$$

which implies *ess sup*(*t*0,1) *f* > *β*1. It is a contradiction.

**Proof of Theorem 2.** If *α* = <sup>1</sup> <sup>2</sup> , we clearly have the optimal solution is *f* <sup>∗</sup>(*r*) = 1 *a*.*e*. for *r* ∈ [0, 1]. Note that *ess inf f* <sup>∗</sup> < 1 < *ess sup f* <sup>∗</sup> for *α* ∈ 0, <sup>1</sup> 2 . Without loss of generality, we can assume that *α* ∈ 0, <sup>1</sup> 2 , since if a weighting function *f* ∗(*r*) is optimal to problem (2) for some given level of preference *α* ∈ 0, <sup>1</sup> 2 , then *f* <sup>∗</sup>(1 − *r*) is optimal to the problem (2) for a given level of preference 1 − *α*. Indeed, since *Df* <sup>=</sup> *Df <sup>R</sup>* , " <sup>1</sup> <sup>0</sup> *<sup>f</sup>*(*r*)*dr* <sup>=</sup> " <sup>1</sup> <sup>0</sup> *<sup>f</sup> <sup>R</sup>*(*r*)*dr* and *<sup>E</sup>*(*<sup>f</sup> <sup>R</sup>*) = <sup>1</sup> − *<sup>E</sup>*(*f*), where *<sup>f</sup> <sup>R</sup>*(*r*) = *<sup>f</sup>*(<sup>1</sup> − *<sup>r</sup>*) hence for *α* > <sup>1</sup> <sup>2</sup> , we can consider problem (2) for the level of preference with index 1 − *α*, and then take the reverse of that optimal solution. We can easily check that the weighting functions, *f* ∗, given above are feasible for problem (2). We show that *f* ∗ is the unique optimal solution for a given *α*. Let nonnegative function *f* satisfy 1 = " <sup>1</sup> <sup>0</sup> *<sup>f</sup>*(*r*)*dr* and *<sup>E</sup>*(*f*) = " <sup>1</sup> <sup>0</sup> *r f*(*r*)*dr* = 1 − *α*. Let *ess inf f* = *β*<sup>0</sup> and *ess sup f* = *β*1.

Case (A): *α* ∈ 0, <sup>1</sup> 4 

.

We note that *ess inf f* <sup>∗</sup> <sup>−</sup> *ess inf f* <sup>∗</sup> <sup>=</sup> <sup>1</sup> <sup>2</sup>*<sup>α</sup>* . We will show that *<sup>β</sup>*<sup>1</sup> <sup>−</sup> *<sup>β</sup>*<sup>0</sup> <sup>≥</sup> <sup>1</sup> <sup>2</sup>*<sup>α</sup>* . To show this, we define a function *f*<sup>0</sup> as

$$f\_0(r) = \begin{cases} \beta\_0 & \text{if } \quad r \in [0, \infty\_0)\_{\prime \prime} \\ \beta\_1 & \text{if } \quad r \in [\infty\_0, 1]\_{\prime \prime} \end{cases}$$

for some *<sup>x</sup>*<sup>0</sup> <sup>∈</sup> (0, 1) such that " <sup>1</sup> <sup>0</sup> *f*0(*r*)*dr* = 1. Then by Lemma 2, *E*(*f*) ≤ *E*(*f*0). Suppose that *<sup>β</sup>*<sup>1</sup> <sup>−</sup> *<sup>β</sup>*<sup>0</sup> <sup>&</sup>lt; <sup>1</sup> <sup>2</sup>*<sup>α</sup>* and define another function *f* <sup>∗</sup> <sup>0</sup> as

$$f\_0^\*(r) = \begin{cases} \beta\_0 & \text{if } \quad r \in [0, \mathfrak{x}\_0^\*)\_{\prime\prime} \\ \beta\_0 + \frac{1}{2\kappa} & \text{if } \quad r \in [\mathfrak{x}\_0^\*, 1]\_{\prime\prime} \end{cases}$$

for some *x*∗ <sup>0</sup> <sup>∈</sup> (0, 1) such that " <sup>1</sup> <sup>0</sup> *f* <sup>∗</sup> <sup>0</sup> (*r*)*dr* = 1. Then *E*(*f*0) < *E*(*f* <sup>∗</sup> <sup>0</sup> ). We note that 1 = *β*0*x*<sup>∗</sup> <sup>0</sup> + (1 − *x*∗ <sup>0</sup> )(*β*<sup>0</sup> + <sup>1</sup> <sup>2</sup>*<sup>α</sup>* ). Then

$$
\Delta x\_0^\* = 2\alpha\beta\_0 + 1 - 2\alpha. \tag{3}
$$

We know that

$$\begin{aligned} E(f\_0^\*) &= -\beta\_0 \int\_0^{\mathbf{x}\_0^\*} \mathbf{x} d\mathbf{x} + \left(\beta\_0 + \frac{1}{2\alpha}\right) \int\_{\mathbf{x}\_0^\*}^1 \mathbf{x} d\mathbf{x} \\ &= -\frac{\beta\_0}{2} + \frac{1}{4\alpha} - \frac{\mathbf{x}\_0^{\*2}}{4\alpha} \end{aligned}$$

and

$$E(f^\*) \quad = \quad \frac{1}{2\alpha} \int\_{1-2\alpha}^1 \mathfrak{x} d\mathfrak{x} = 1 - \alpha.$$

And we have

$$\begin{aligned} E(f^\*) - E(f\_0^\*) &= \quad \frac{1}{2} \frac{\mathbf{x}\_0^{\*2}}{2\alpha} - \frac{1}{2} \frac{(1 - 2\alpha)^2}{2\alpha} - \frac{\beta\_0}{2} \\ &= \quad \frac{1}{2} \left[ \frac{1}{2\alpha} \mathbf{x}\_0^{\*2} - \frac{(1 - 2\alpha)^2}{2\alpha} - \beta\_0 \right] \\ &= \quad \frac{1}{2} \left[ \frac{1}{2\alpha} (2\alpha\beta\_0 + 1 - 2\alpha)^2 - \frac{(1 - 2\alpha)^2}{2\alpha} - \beta\alpha \right] \\ &= \quad \frac{\beta\_0}{2} \left[ 2\alpha\beta\_0 + 2(1 - 2\alpha) - 1 \right] \\ &\ge \quad 0 \end{aligned}$$

where the third equality comes from (3) and the last inequality comes from the facts that 1 <sup>−</sup> <sup>2</sup>*<sup>α</sup>* <sup>≥</sup> <sup>1</sup> 2 , *β*<sup>0</sup> ≥ 0 and *α* > 0. This proves *E*(*f*) < *E*(*f* <sup>∗</sup> <sup>0</sup> ) ≤ *E*(*f* <sup>∗</sup>) = 1 − *α*, which is a contradiction. Hence *f* <sup>∗</sup> is an optimal solution for the case of *α* ∈ 0, <sup>1</sup> 4 .

Case (B): *α* ∈ 1 4 , 1 2 .

We note that *ess inf f* <sup>∗</sup> − *ess inf f* <sup>∗</sup> = 4(1 − 2*α*). We will show that *β*<sup>1</sup> − *β*<sup>0</sup> ≥ 4(1 − 2*α*). As in the Case (A), we define a function *f*<sup>0</sup> as

$$f\_0(r) = \begin{cases} \beta\_0 & \text{if } \quad r \in [0, \infty\_0)\_{\prime \prime} \\ \beta\_1 & \text{if } \quad r \in [\infty\_0, 1]\_{\prime \prime} \end{cases}$$

for some *<sup>x</sup>*<sup>0</sup> <sup>∈</sup> (0, 1) such that " <sup>1</sup> <sup>0</sup> *f*0(*r*)*dr* = 1. Then by lemma 2, *E*(*f*) ≤ *E*(*f*0). Suppose that *<sup>β</sup>*<sup>1</sup> <sup>−</sup> *<sup>β</sup>*<sup>0</sup> <sup>&</sup>lt; <sup>1</sup> <sup>2</sup>*<sup>α</sup>* and define another function *f* <sup>∗</sup> <sup>1</sup> as

$$f\_1^\*(r) = \begin{cases} \beta\_0 & \text{if } \quad r \in [0, \mathfrak{x}\_1^\*)\_{\prime} \\ \beta\_0 + 4(1 - 2\mathfrak{a}) & \text{if } \quad r \in [\mathfrak{x}\_1^\*, 1]\_{\prime} \end{cases}$$

for some *x*∗ <sup>1</sup> <sup>∈</sup> (0, 1) such that " <sup>1</sup> <sup>0</sup> *f* <sup>∗</sup> <sup>1</sup> (*r*)*dr* = 1. Then, since *x*<sup>0</sup> < *x*<sup>∗</sup> <sup>1</sup> , by lemma 2 *E*(*f*0) < *E*(*f* <sup>∗</sup> <sup>1</sup> ). We note that 1 = *β*0*x*<sup>∗</sup> <sup>1</sup> + (1 − *x*<sup>∗</sup> <sup>1</sup> )(*β*<sup>0</sup> + 4(1 − 2*α*)). Then

$$x\_1^\* = 1 + \frac{\beta\_0 - 1}{4(1 - 2\alpha)}$$

and

$$\mathbf{x}\_{1}^{\*2} = 1 + \frac{\beta\_0 - 1}{2(1 - 2a)} + \frac{(\beta\_0 - 1)^2}{16(1 - 2a)^2} \tag{4}$$

We know that

$$E(f\_0^\*) = \beta\_0 \int\_0^{x\_1^\*} x dx + (\beta\_0 + 4(1 - 2\alpha)) \int\_{x\_1^\*}^1 x dx$$

$$= \frac{1}{2} [\beta\_0 + 4(1 - 2\alpha)] - 2(1 - 2\alpha)x\_1^{\*2}$$

and

$$\begin{aligned} E(f^\*) &= -(4\alpha - 1) \int\_0^{\frac{1}{2}} \mathbf{x} d\mathbf{x} + (3 - 4\alpha) \int\_{\frac{1}{2}}^1 \mathbf{x} d\mathbf{x} \\ &= -1 - \alpha. \end{aligned}$$

Then we have that

$$\begin{aligned} E(f^\*) - E(f\_1^\*) &= -3\alpha - 1 - \frac{\beta\_0}{2} + 2(1 - 2\alpha)x\_1^{\*2} \\ &= \frac{(\beta\_0 - 1)^2}{8(1 - 2\alpha)} + \frac{\beta\_0}{2} - \alpha \\ &= \frac{[\beta\_0 - (4\alpha - 1)]^2}{8(1 - 2\alpha)} \\ &\ge 0 \end{aligned}$$

where the second equality comes from (4) and hence *E*(*f*) < *E*(*f* ∗ <sup>1</sup> ) ≤ *E*(*f* <sup>∗</sup>) = 1 − *α*, which is a contradiction. This completes the proof.

### **6. Conclusions**

Previous studies have suggested a number of methods for obtaining optimal solution of the RIM quantifier problem. This paper proposes the extended minimax disparity RIM quantifier problem under a given orness level. We completely prove it analytically.

**Funding:** This research was supported by Basic Science Research Program through the National 247 Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03027869).

**Conflicts of Interest:** The authors declare no conflict of interest.

### **References**


© 2019 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

### *Article* **The Solution Equivalence to General Models for the RIM Quantifier Problem**

### **Dug Hun Hong**

Department of Mathematics, Myongji University, Yongin 449-728, Kyunggido, Korea; dhhong@mju.ac.kr

Received: 3 March 2019; Accepted: 28 March 2019; Published: 1 April 2019

**Abstract:** Hong investigated the relationship between the minimax disparity minimum variance regular increasing monotone (RIM) quantifier problems. He also proved the equivalence of their solutions to minimum variance and minimax disparity RIM quantifier problems. Hong investigated the relationship between the minimax ratio and maximum entropy RIM quantifier problems and proved the equivalence of their solutions to the maximum entropy and minimax ratio RIM quantifier problems. Liu proposed a general RIM quantifier determination model and proved it analytically by using the optimal control technique. He also gave the equivalence of solutions to the minimax problem for the RIM quantifier. Recently, Hong proposed a modified model for the general minimax RIM quantifier problem and provided correct formulation of the result of Liu. Thus, we examine the general minimum model for the RIM quantifier problem when the generating functions are Lebesgue integrable under the more general assumption of the RIM quantifier operator. We also provide a solution equivalent relationship between the general maximum model and the general minimax model for RIM quantifier problems, which is the corrected and generalized version of the equivalence of solutions to the general maximum model and the general minimax model for RIM quantifier problems of Liu's result.

**Keywords:** OWA operator; RIM quantifier; maximum entropy; minimax ratio; generating function; minimal variability; minimax disparity; solution equivalence

### **1. Introduction**

One of the important topics in the theory of ordered weighted averaging (OWA) operators is the determination of the associated weights. Several authors have suggested a number of methods for obtaining associated weights in various areas such as decision-making, approximate reasoning, expert systems, data mining, fuzzy systems and control [1–22]. Yager [12] proposed RIM quantifiers as a method for finding OWA weight vectors through fuzzy linguistic quantifiers. Liu [15] and Liu and Da [16] gave solutions to the maximum-entropy RIM quantifier model when the generating functions are differentiable. Liu and Lou [9] studied the equivalence of solutions to the minimax ratio and maximum-entropy RIM quantifier models, and the equivalence of solutions to the minimax disparity and minimum-variance RIM quantifier problems. Hong [17,18] gave the proof of the minimax ratio RIM quantifier problem and the minimax disparity RIM quantifier model when the generating functions are absolutely continuous. He also gave solutions to the maximum-entropy RIM quantifier model and the minimum-variance RIM quantifier model when the generating functions are Lebesgue integrable.

Based on these results, Hong [17,18] provided a relationship between the minimax disparity and minimum-variance RIM quantifier problems. He also provided a correct relationship between the minimax ratio and maximum-entropy RIM quantifier models. Liu [19] suggested a general RIM quantifier determination model and proved it analytically using the optimal control methods. He also studied the solution equivalence to the minimax problem for the RIM quantifier.

This paper investigates the general minimax model for the RIM quantifier problem for the case in which the generating functions are absolutely continuous and a generalized solution to the general minimum model for the RIM quantifier problem for the case in which the generating functions are Lebesgue integrable. Moreover, this paper provides a solution equivalent relationship between the general maximum model and the general minimax model for RIM quantifier problems and generalizes the results of Hong [17,18]. In this paper, we improve and extend Liu's theorems to be suitable for absolutely continuous generating functions. We have corrected and improved Theorem 13 [19] by using the absolutely continuous condition of generating functions and the absolute continuity condition of *F* for the general minimax model for the RIM quantifier problem. Theorem 9 [19] has been improved using the Lebesgue integrability condition of generating functions and the continuity condition of *F* for the general maximum model for the RIM quantifier problem.

Based on these results, we give a correct relationship between the general minimum model and the general minimax model for RIM quantifier problems.

### **2. Preliminaries**

Yager [11] proposed a new aggregation technique based on OWA operators. An OWA operator of dimension *<sup>n</sup>* is a mapping *<sup>F</sup>* : *<sup>R</sup><sup>n</sup>* → *<sup>R</sup>* that has an associated weight vector *<sup>W</sup>* = (*w*1, ··· , *wn*)*<sup>T</sup>* with the properties *w*<sup>1</sup> + ··· + *wn* = 1, 0 ≤ *wi* ≤ 1, *i* = 1, ··· , *n*, such that

$$F(a\_{1\prime}\cdots\cdot, a\_n) = \sum\_{i=1}^n w\_i b\_{i\prime}$$

where *bj* is the *j*th largest element of the collection of the aggregated objects {*a*1, ··· , *an*}. In [11], Yager introduced a measure of "orness" associated with the weight vector *W* of an OWA operator:

$$orness(\mathcal{W}) = \sum\_{i=1}^{n} \frac{n-i}{n-1} w\_i.$$

This measure characterizes the degree to which the aggregation is like an *OR* operation.

Here, *min*, *max*, and *average* correspond to *W*∗, *W*<sup>∗</sup> and *WA* respectively, where *W*<sup>∗</sup> = (1, 0, ··· , 0), *W*<sup>∗</sup> = (0, 0, ··· , 1) and *WA* = (1/*n*, 1/*n*, ··· , 1/*n*). Clearly, *orness*(*W*∗) = 1, *orness*(*W*∗) = 0 and *orness*(*WA*) = 1/2.

Yager [12] introduced RIM quantifiers as a method for obtaining OWA weight vectors through fuzzy linguistic quantifiers.

**Definition 1** ([12])**.** *A fuzzy subset Q on the real line is called a RIM quantifier if Q*(0) = 0, *Q*(1) = 1 *and Q*(*x*) ≥ *Q*(*y*) *for x* > *y.*

The quantifier *f or all* is represented by the fuzzy set

$$Q\_\*(r) = \begin{cases} 1, & \text{if } x = 1, \\ 0, & \text{if } x \neq 1. \end{cases}$$

The quantifier *there exists* is defined as

$$Q^\*(r) = \begin{cases} 0, & \text{if } x = 0, \\ 1, & \text{if } x \neq 0. \end{cases}$$

Both of these are examples of the RIM quantifier. A generating function representation of RIM quantifiers has been proposed for analyzing the relationship between OWA operators and RIM quantifiers.

**Definition 2.** *For f*(*t*) *on [0, 1] and the RIM quantifier Q*(*x*), *f*(*t*) *is called the generating function of Q*(*x*), *if it satisfies*

$$Q(x) = \int\_0^x f(t)dt\_\prime$$

*where f*(*t*) <sup>≥</sup> <sup>0</sup> *and* " <sup>1</sup> <sup>0</sup> *f*(*t*)*dt* = 1.

If the RIM quantifier *Q*(*x*) is smooth, then *f*(*x*) should be continuous; if *Q*(*x*) is a piecewise linear function, then *f*(*x*) is a jump piecewise function of some constants; and if *Q*(*x*) is an absolutely continuous function, then *f*(*x*) is a Lesbegue integrable function and unique in the sense of being "almost everywhere" [23].

Yager extended the *orness* measure of OWA operators, and defined the *orness* of RIM quantifiers [10] as:

$$ormes(Q) = \int\_0^1 Q(\mathfrak{x})d\mathfrak{x} = \int\_0^1 (1-t)f(t)dt.$$

We see that *Q*<sup>∗</sup> leads to the weight vector *W*∗, *Q*<sup>∗</sup> leads to the weight vector *W*∗, and the ordinary *average* RIM quantifier *QA*(*x*) = *x* leads to the weight vector *WA*. We also have *orness*(*Q*∗) = 1, *rness*(*Q*∗) = 0, and *orness*(*QA*) = 1/2.

As the RIM quantifier can be seen as a continuous form of OWA, an operator with a generating function, the OWA optimization problem can be extended to the case of the RIM quantifier.

### **3. The General Model for the Minimax RIM Quantifier Problem**

In this section, we consider the general model for the minimax RIM quantifier problem and generalize some results of Hong [17,18]. Hong [7] provided a modified model for the minimax RIM quantifier problem and the correct formulation of a result of Liu [19]. We summarize briefly.

### ∗ **The minimax disparity RIM quantifier problem [15,17].**

The minimax disparity RIM quantifier problem with a given orness level 0 < *α* < 1 consists of finding a solution *f* : [0, 1] → [0, 1] to the following optimization problem:

$$\begin{aligned} \text{Minimize } & \quad \max\_{t \in (0,1)} |f'(t)|, \\ \text{subject to } & \quad \int\_0^1 (1-r)f(r) dr = a, \ 0 < a < 1, \\ & \quad \int\_0^1 f(r) dr = 1, \\ & \quad f(r) \ge 0. \end{aligned}$$

### ∗ **The minimax ratio RIM quantifier problem [9,18].**

The minimax ratio RIM quantifier problem with a given orness level 0 < *α* < 1 consists of finding a solution *f* : [0, 1] → [0, 1] to the following optimization problem:

$$\begin{aligned} \text{Minimize } & \quad \max\_{t \in (0,1)} \left| \frac{f'(t)}{f(t)} \right|, \\ \text{subject to } & \quad \int\_0^1 (1-r)f(r) dr = a, \ 0 < a < 1, \\ & \quad \int\_0^1 f(r) dr = 1, \\ & \quad f(r) > 0. \end{aligned}$$

In regard to the above optimization problem, Liu [19] considered a general model for the minimax RIM quantifier problem:

$$\begin{aligned} \text{Minimize } \quad & M\_f = \max\_{r \in (0,1)} |F''(f(r))f'(r)|, \\ \text{subject to } \quad & \int\_0^1 r f(r) dr = a, \ 0 < a < 1, \\ & \int\_0^1 f(r) dr = 1, \\ & f(r) > 0, \end{aligned} \tag{1}$$

where the generating functions are continuous and *F* is a strictly convex function on [0, ∞), which is differentiable to at least the 2nd order.

The above two cases are special cases of this model with *F*(*x*) = *x*<sup>2</sup> and *F*(*x*) = *x* ln *x*. Hong [7] gave a corrected and modified general model for the minimax RIM quantifier problem as follows:

### ∗ **The general model for the minimax RIM quantifier problem.**

$$\begin{aligned} \text{Minimize } \qquad &M\_f = \operatorname{ess}\sup\_{r \in (0,1)} |F'(f(x))f'(x)|, \\ \text{subject to } \qquad &\int\_0^1 r f(r) dr = a, \ 0 < a < 1, \\ &\int\_0^1 f(r) dr = 1, \\ &f(r) > 0. \end{aligned} \tag{2}$$

**Theorem 1.** *Supposing that the generating functions are absolutely continuous, F is a strictly convex function on* [0, ∞), *and F is absolutely continuous, then there is a unique optimal solution for problem (2), and that the optimal solution has the form*

$$f^\*(r) = \max\left\{ (F')^{-1} \left( a^\*r + b^\* \right), 0 \right\},$$

*where a*∗ *and b*∗ *are determined by the constraints:*

$$\begin{cases} \int\_0^1 r f^\*(r) dr = a, \\ \int\_0^1 f^\*(r) dr = 1, \\ f^\*(r) \ge 0. \end{cases}$$

The next example shows that the condition of *F* being *absolutely continuous* on [0, ∞) in Theorem 1 is essential.

**Example 1.** *Letting <sup>F</sup>*1(*x*) = " *<sup>x</sup>* <sup>0</sup> (*C*(*r*) + *r*)*dr where C*(*x*) *is a Cantor function, then F* (*x*) = *C*(*x*) + *x and F* <sup>1</sup> (*x*) = 1 *a*.*e*. *but F* <sup>1</sup>(*x*) <sup>=</sup> " *<sup>x</sup>* <sup>0</sup> *F* <sup>1</sup> (*r*)*dr*, *that is, F* <sup>1</sup> *is not absolutely continuous on* [0, ∞). *Let F*2(*x*) = (1/2)*x*2, *then F* <sup>2</sup> (*x*) = 1*. Since*

$$\text{ess}\,\text{ess}\,p\_{r\in(0,1)}|F\_1^{\prime\prime}(f(\mathbf{x}))f^{\prime}(\mathbf{x})| = \text{ess}\,sup\_{r\in(0,1)}|f^{\prime}(\mathbf{x})| = \text{ess}\,sup\_{r\in(0,1)}|F\_2^{\prime\prime}(f(\mathbf{x}))f^{\prime}(\mathbf{x})|\_{\prime}$$

*the optimal solution of problem (2) with respect to F*<sup>1</sup> *and F*<sup>1</sup> *are the same. However, since F* <sup>1</sup>(*x*) = *F* <sup>2</sup>(*x*), *the optimal solution of problem (2) with respect to F*<sup>1</sup> *and F*<sup>1</sup> *cannot be the same by Theorem 2, which is a contradiction. This example shows the Theorem 2 is incorrect if F is not absolutely continuous on* [0, ∞).

### **4. The General Model for the Minimum RIM Quantifier Problem**

In this section, we consider the general model for the minimum RIM quantifier problem. We improve the results of Liu [19] and generalize Theorem 4 of Hong [17] and Theorem 5 of Hong [18]. Liu [19] obtained solutions to the general minimum RIM quantifier problem for the case in which the generating functions are continuous and *F* is differentiable to at least the 2nd order by considering a variational optimization problem using the Lagrangian multiplier method ([24], Chapter 2). In this section, we consider a generalized result for this problem.

### ∗ **The minimum variance RIM quantifier problem [17,18].**

The minimum variance RIM quantifier problem under a given orness level is

$$\begin{aligned} \text{Minimize } \quad & D\_f = \int\_0^1 f^2(r) dr, \\ \text{subject to } \quad & \int\_0^1 r f(r) dr = a, \ 0 < a < 1, \\ & \int\_0^1 f(r) dr = 1, \\ & f(r) > 0. \end{aligned}$$

### ∗ **The maximum entropy RIM quantifier problem [9,18].**

The maximum entropy RIM quantifier problem with a given orness level 0 < *α* < 1 consists of finding a solution *f* : [0, 1] → [0, 1] to the following optimization problem:

$$\begin{aligned} \text{Maximize} \quad & -\int\_0^1 f(r) \ln f(r) dr, \\ \text{subject to} \quad & \int\_0^1 r f(r) dr = a, \ 0 < a < 1, \\ & \int\_0^1 f(r) dr = 1, \\ & f(r) > 0. \end{aligned}$$

Recently, Liu [19] considered the general model for the minimum variance and maximum entropy RIM quantifier problems, under a given orness level formulated as follows:

### ∗ **The general model for the minimum RIM quantifier problem.**

$$\begin{aligned} \text{Minimize } \qquad &V\_f = \int\_0^1 F(f(r)) dr, \\ \text{subject to } \qquad &\int\_0^1 r f(r) dr = a, \ 0 < a < 1, \\ &\int\_0^1 f(r) dr = 1, \\ &f(r) > 0, \end{aligned} \tag{3}$$

where *F* is a strictly convex function on [0, ∞), and differentiable to at least the 2nd order.

The above two cases are special cases of the model where *F*(*x*) = *x*<sup>2</sup> and *F*(*x*) = *xlnx*.

Liu (Theorem 9, [19]) proved the following problem for the case in which generating functions are continuous and *F* is differentiable to at least the 2nd order:

**Theorem 2** (Theorem 9, [19])**.** *There is a unique optimal solution for (3), and the optimal solution has the form*

$$f^\*(r) = \begin{cases} (F')^{-1}(a^\*r + b^\*), & \text{if } (F')^{-1}(a^\*r + b^\*) \ge 0, \\ 0, & \text{elsewhere,} \end{cases}$$

*where a*∗, *b*∗ *are determined by the constraints:*

$$\begin{cases} \int\_0^1 r f^\*(r) dr = a, \\ \int\_0^1 f^\*(r) dr = 1, \\ f^\*(r) \ge 0. \end{cases}$$

Here, we consider a generalized result for Theorem 3 when *f*(*x*) is Lebesgue integrable and *F* is continuous.

**Theorem 3.** *Suppose that the generating functions are Lebesgue integrable, F is a strictly convex function on* [0, ∞)*, and F is continuous. Then, there is a unique optimal solution for problem (3), and that optimal solution has the form*

$$f^\*(r) = \begin{cases} (F')^{-1}(a^\*r + b^\*) & a.e., \quad \text{if } (F')^{-1}(a^\*r + b^\*) > 0, \\ 0 & \text{elsewhere,} \end{cases}$$

*where a*∗ *and b*∗ *are determined by the constraints:*

$$\begin{cases} \int\_0^1 r f^\*\left(r\right) dr = a, \\ \int\_0^1 f^\*\left(r\right) dr = 1, \\ f^\*\left(r\right) \ge 0. \end{cases}$$

**Proof.** As shown in Theorem 2, we consider the case where *α* ∈ (0, 1/2] and assume that {*r* < 1 : *f* <sup>∗</sup>(*r*) > 0} = [0, *t*) for some *t* ∈ (0, 1) and {*r* < 1 : *f* <sup>∗</sup>(*r*) = 0} = [*t*, 1). We also note that for *r* ∈ [0, *t*],

$$F'(f^\*(r)) = a^\*r + b^\*$$

and for *r* ∈ (*t*, 1),

$$a^\*r + b^\* < F'(0).$$

if *F* (0) exists. Let the nonnegative function *f* satisfy 1 = " <sup>1</sup> <sup>0</sup> *<sup>f</sup>*(*r*)*dr* and " <sup>1</sup> <sup>0</sup> *r f*(*r*)*dr* = *α*. We set *f*(*r*) = *f* <sup>∗</sup>(*r*) + *g*(*r*), *r* ∈ [0, 1]. Then, noting that *f*(*r*) = *g*(*r*), *r* ∈ [*t*, 1], we have

$$\int\_{0}^{t} \mathbf{g}(r) dr + \int\_{t}^{1} f(r) dr = \int\_{0}^{1} \mathbf{g}(r) dr = 0,\tag{4}$$

since 1 = " <sup>1</sup> *<sup>f</sup>*(*r*)*dr* <sup>=</sup> " <sup>1</sup> *<sup>f</sup>* <sup>∗</sup>(*r*)*dr* <sup>+</sup> " <sup>1</sup> *<sup>g</sup>*(*r*)*dr* <sup>=</sup> <sup>1</sup> <sup>+</sup> " <sup>1</sup> *g*(*r*)*dr*. We also have

$$\int\_{0}^{t} r\mathbf{g}(r)dr + \int\_{t}^{1} r f(r)dr = \int\_{0}^{1} r\mathbf{g}(r)dr = 0,\tag{5}$$

since *α* = " <sup>1</sup> *r f*(*r*)*dr* <sup>=</sup> " <sup>1</sup> *r f* <sup>∗</sup>(*r*)*dr* <sup>+</sup> " <sup>1</sup> *rg*(*r*)*dr* <sup>=</sup> *<sup>α</sup>* <sup>+</sup> " <sup>1</sup> *rg*(*r*)*dr*. We now show that

$$\int\_0^1 F\left(f\left(r\right)\right) dr \ge \int\_0^1 F\left(f^\*\left(r\right)\right) dr.$$

Since *F*(*x*) − *F*(*x*0) ≥ *F* (*x*0)(*x* − *x*0) (the equality holds if and only if *x* = *x*0), we have that

$$\begin{split} &\quad \int\_{0}^{1} F(f(r)) dr - \int\_{0}^{1} F(f^\*(r)) dr \\ &= \quad \int\_{0}^{1} F((f^\*(r) + \operatorname{g}(r))) dr - \int\_{0}^{1} F(f^\*(r)) dr \\ &\ge \quad \int\_{0}^{1} F'(f^\*(r)) \operatorname{g}(r) dr \\ &= \quad \int\_{0}^{t} (a^\* \tau + b^\*) \operatorname{g}(r) dr + \int\_{t}^{1} F'(0) \operatorname{g}(r) dr \\ &= \quad a^\* \int\_{0}^{t} r \operatorname{g}(r) dr + b^\* \int\_{0}^{t} \operatorname{g}(r) dr + \int\_{t}^{1} F'(0) \operatorname{g}(r) dr \\ &= \quad a^\* (-\int\_{t}^{1} r f(r) dr) + b^\*(-\int\_{t}^{1} f(r) dr) + \int\_{t}^{1} F'(0) \operatorname{g}(r) dr \\ &= \quad \int\_{t}^{1} (F'(0) - a^\* \tau - b^\*) f(r) dr \\ &\ge \quad 0, \end{split}$$

where the fourth equality comes from (4) and (5) and the second inequality comes from the fact that *a*∗*r* + *b*<sup>∗</sup> ≤ *F* (0) *a*.*e*. for *r* ∈ [*t*, 1]. The equalities hold if and only if *f* <sup>∗</sup> = *f a*.*e*. This completes the proof.

Combining Theorems 2 and 4, we now have a solution equivalent relationship between the general minimum RIM quantifier problem and the general minimax RIM quantifier problem. This result generalizes Theorem 6 of Hong [17] and Theorem 5 of Hong [18] and provides a corrected version of Theorem 13 [19].

**Theorem 4.** *Suppose that the generating functions are absolutely continuous and F is increasing and absolutely continuous. Then, the general minimum RIM quantifier problem has the same solution as the general minimax RIM quantifier problem.*

### **5. Numerical Example**

We consider a RIM quantifier operator *F* which is not differentiable to at least the second order, but *F* is absolutely continuous, and find an optimal solution of two RIM quantifier problems.

Let a RIM quantifier operator *F* be

$$F(\mathbf{x}) = \begin{cases} \frac{\mathbf{x}^2}{2} & \text{if } 0 \le \mathbf{x} < \frac{1}{2},\\ \mathbf{x}^2 - \frac{1}{2}\mathbf{x} + \frac{1}{8\prime} & \text{if } \frac{1}{2} \le \mathbf{x} \le 1. \end{cases}$$

Then,

$$F'(\mathbf{x}) = \begin{cases} \mathbf{x}, & \text{if } 0 \le \mathbf{x} < \frac{1}{2}, \\ 2\mathbf{x} - \frac{1}{2}, & \text{if } \frac{1}{2} \le \mathbf{x} \le 1. \end{cases}$$

Hence, *F*(*x*) is strictly convex and *F* (*x*) is absolutely continuous, but *F*(*x*) is not the second order differentiable. Let

$$f^\*(r) = \begin{cases} \ (F')^{-1}(a^\*r + b^\*), & \text{if } (F')^{-1}\ (a^\*r + b^\*) > 0, \\ 0, & \text{elsewhere } r \end{cases}$$

where *a*∗ and *b*∗ are determined by the constraints:

$$\begin{cases} \int\_{0}^{1} rf^\*(r) dr = a, \\ \int\_{0}^{1} f^\*(r) dr = 1, \\ f^\*(r) \ge 0. \end{cases} \tag{6}$$

We consider the case for 0 < *α* ≤ 1/2. Then, *a*<sup>∗</sup> ≤ 0 and *b*<sup>∗</sup> > 0.

Case (1) (See Figure 1) There exists *m*, *d* ∈ [0, 1] such that *m* < *d* and

$$f^\*(r) = \begin{cases} \frac{1}{2}(a^\*r + b^\*) + \frac{1}{4}, & \text{if } 0 \le r \le m\_{\text{tot}}\\\ a^\*r + b^\*, & m < r \le d\_{\text{tot}}\\\ 0, & d < r \le 1. \end{cases}$$

Since *a*∗*m* + *b*<sup>∗</sup> = <sup>1</sup> and *<sup>a</sup>*∗*<sup>d</sup>* <sup>+</sup> *<sup>b</sup>*<sup>∗</sup> <sup>=</sup> 0, *<sup>b</sup>*<sup>∗</sup> <sup>=</sup> <sup>−</sup>*a*∗*<sup>m</sup>* <sup>+</sup> <sup>1</sup> and *<sup>d</sup>* <sup>=</sup> *<sup>m</sup>* <sup>−</sup> <sup>1</sup> *a*<sup>∗</sup> . Hence,

$$f^\*(r) = \begin{cases} \frac{1}{2}a^\*(r-m) + \frac{1}{2}, & \text{if } 0 \le r \le m, \\\ a^\*(r-m) + \frac{1}{2}, & m < r \le m - \frac{1}{2a^\*}, \\\ 0, & m - \frac{1}{2a^\*} < r \le 1. \end{cases}$$

From (6),

$$\begin{array}{rcl} a^\* &=& \frac{2m - 4 - \sqrt{2m^2 - 16m + 16}}{2m^2}, \\ \kappa &=& -\frac{4m^3 a^{\*3} - 12m^2 a^{\*2} + 6ma^\* - 1}{48a^{\*2}} \end{array}$$

hold. In addition, since *a*∗ < 0 and *f* ∗(1) < 0,

$$0 < m < 4 - \sqrt{10}, \quad 0 < a < \frac{17 - 4\sqrt{10}}{12}.$$

**Figure 1.** The graph of *f* <sup>∗</sup> (0 < *α* < <sup>17</sup>−<sup>4</sup> <sup>√</sup><sup>10</sup> <sup>12</sup> ).

Case (2) (See Figure 2) There exists *m* ∈ [0, 1] such that

$$f^\*(r) = \begin{cases} \frac{1}{2}(a^\*r + b^\*) + \frac{1}{4}, & \text{if } 0 \le r \le m, \\\ a^\*r + b^\*, & m < r \le 1. \end{cases}$$

Since *a*∗*m* + *b*<sup>∗</sup> = <sup>1</sup> 2 ,

$$f^\*(r) = \begin{cases} \frac{1}{2}a^\*(r-m) + \frac{1}{2}, & \text{if } 0 \le r \le m, \\\ a^\*(r-m) + \frac{1}{2}, & m < r \le 1. \end{cases}$$

From (6),

$$\begin{array}{rcl} a^\* &=& \frac{2}{m^2 - 4m + 2} \\ a &=& \frac{2m^3 + 3m^2 - 24m + 14}{12(m^2 - 4m + 2)} \end{array}$$

hold. In addition, since *a*<sup>∗</sup> < 0 and *f* <sup>∗</sup>(1) ≥ 0,

$$4 - \sqrt{10} \le m \le 1, \quad \frac{17 - 4\sqrt{10}}{12} \le \alpha \le \frac{5}{12}.$$

**Figure 2.** The graph *f* <sup>∗</sup> ( <sup>17</sup>−<sup>4</sup> <sup>√</sup><sup>10</sup> <sup>12</sup> <sup>≤</sup> *<sup>α</sup>* <sup>≤</sup> <sup>5</sup> <sup>12</sup> ).

Case (3) (See Figure 3) For all 0 ≤ *r* ≤ 1,

$$f^\*(r) = \frac{1}{2}(a^\*r + b^\*) + \frac{1}{4}.$$

From (6),

$$\begin{array}{rcl} a^\* &=& -12 + 24\alpha\_{\prime} \\ b^\* &=& \frac{15}{2} - 12\alpha\_{\prime} \end{array}$$

hold. In addition, since *<sup>a</sup>*<sup>∗</sup> <sup>≤</sup> 0 and *<sup>f</sup>* <sup>∗</sup>(1) <sup>&</sup>gt; <sup>1</sup> <sup>2</sup> , <sup>5</sup> <sup>12</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup> 2 .

**Figure 3.** The graph *f* <sup>∗</sup> ( <sup>5</sup> <sup>12</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>≤</sup> <sup>1</sup> 2 ).

### **6. Conclusions**

In this paper, we examined the general minimax model for the RIM quantifier problem for the case in which the generating functions are absolutely continuous and a generalized solution to the general minimum model for the RIM quantifier problem for the case in which the generating functions are Lebesgue integrable. In addition, we provided a solution equivalent relationship between the general maximum model and the general minimax model for RIM quantifier problems and generalizes results of Hong based on these results. We also corrected Liu's theorems from a mathematical perspective as their theorems are not suitable for absolutely continuous generating functions.

**Funding:** This research was funded by the Basic Science Research Program through the National Research Foundation of Korea (NRF) grant number 2017R1D1A1B03027869.

**Conflicts of Interest:** The author declares no conflict of interest.

### **References**


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*Correction*
