**Current Trends in Symmetric Polynomials with Their Applications II**

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**Taekyun Kim**

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*Editor* Taekyun Kim Department of Mathematics, Kwangwoon University Korea

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This is a reprint of articles from the Special Issue published online in the open access journal *Symmetry* (ISSN 2073-8994) (available at: https://www.mdpi.com/journal/symmetry/special issues/Symmetric Polynomials).

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## **Contents**



### **About the Editor**

**Taekyun Kim** received a PhD in the Department of Mathematics, Kyushu University in Japan(1994). He worked as a lecturer in Kyungpook National University in 1994–1996, a research professor in the Institute of Science Education, Kongju National University in 2001–2006, a professor(BK) in the Department of Electrical and Computer Engineering, Kyungpook National University in 2006–2008, and a chair professor in Tianjin Polytechnic University in 2015–2019. He has been working as a professor at the Department of Mathematics in Kwangwoon University since 2008 and has also been serving as the editor-in -chief in Advanced Studies in Contemporary Mathematics (http://www.jangjeonopen.or.kr/) since 1999.

### **Preface to "Current Trends in Symmetric Polynomials with Their Applications II"**

The special numbers and polynomials play an extremely important role in various applications in such diverse areas as mathemaics, probability and statistics, mathematical physics, and engineering. Due to their powerful expressions, the combinations of special numbers and polynomials can be seen almost ubiquitously as the solutions of differential equations in the diverse fields by orthogonality condition, generating functions, recurrence relations, bosonic and fermionic p-adic integrals and etc.

Further, their importance can be also found in the developments of classical analysis, number theory, mathematical analysis, mathematical physics, symmetric functions, combinatorics, and other parts of the natural sciences.

In many years, a great amount of effort has been paid by many researchers to find new representations of families of special functions and polynomials with its practical applications.

This special issue will be contributed to the fields of special functions and orthogonal polynomials (or q-special functions and orthogonal polynomials) along the modern trends.

> **Taekyun Kim** *Editor*

### *Article* **Some Identities on the Poly-Genocchi Polynomials and Numbers**

**Dmitry V. Dolgy <sup>1</sup> and Lee-Chae Jang 2,\***


Received: 28 May 2020; Accepted: 9 June 2020; Published: 14 June 2020

**Abstract:** Recently, Kim-Kim (2019) introduced polyexponential and unipoly functions. By using these functions, they defined type 2 poly-Bernoulli and type 2 unipoly-Bernoulli polynomials and obtained some interesting properties of them. Motivated by the latter, in this paper, we construct the poly-Genocchi polynomials and derive various properties of them. Furthermore, we define unipoly Genocchi polynomials attached to an arithmetic function and investigate some identities of them.

**Keywords:** polylogarithm functions; poly-Genocchi polynomials; unipoly functions; unipoly Genocchi polynomials

**MSC:** 11B83; 11S80

### **1. Introduction**

The study of the generalized versions of Bernoulli and Euler polynomials and numbers was carried out in [1,2]. In recent years, various special polynomials and numbers regained the interest of mathematicians and quite a few results have been discovered. They include the Stirling numbers of the first and the second kind, central factorial numbers of the second kind, Bernoulli numbers of the second kind, Bernstein polynomials, Bell numbers and polynomials, central Bell numbers and polynomials, degenerate complete Bell polynomials and numbers, Cauchy numbers, and others (see [3–8] and the references therein). We mention that the study of a generalized version of the special polynomials and numbers can be done also for the transcendental functions like hypergeometric ones. For this, we let the reader refer to the papers [3,5,6,8,9]. The poly-Bernoulli numbers are defined by means of the polylogarithm functions and represent the usual Bernoulli numbers (more precisely, the values of Bernoulli polynomials at 1) when *k* = 1. At the same time, the degenerate poly-Bernoulli polynomials are defined by using the polyexponential functions (see [8]) and they are reduced to the degenerate Bernoulli polynomials if *k* = 1. The polyexponential functions were first studied by Hardy [10] and reconsidered by Kim [6,9,11,12] in view of an inverse to the polylogarithm functions which were studied by Zagier [13], Lewin [14], and Jaonqui*e*`re [15]. In 1997, Kaneko [16] introduced poly-Bernoulli numbers which are defined by the polylogaritm function.

Recently, Kim-Kim introduced polyexponential and unipoly functions [9]. By using these functions, they defined type 2 poly-Bernoulli and type 2 unipoly-Bernoulli polynomials and obtained several interesting properties of them.

In this paper, we consider poly-Genocchi polynomials which are derived from polyexponential functions. Similarly motivated, in the final section, we define unipoly Genocchi polynomials attached to an arithmetic function and investigate some identities for them. In addition, we give explicit expressions and identities involving those polynomials.

It is well known, the Bernoulli polynomials of order *α* are defined by their generating function as follows (see [1–3,17,18]):

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

We note that for *<sup>α</sup>* <sup>=</sup> 1, *Bn*(*x*) = *<sup>B</sup>*(1) *<sup>n</sup>* (*x*) are the ordinary Bernoulli polynomials. When *<sup>x</sup>* <sup>=</sup> 0, *Bα <sup>n</sup>* = *B<sup>α</sup> <sup>n</sup>*(0) are called the Bernoulli numbers of order *α*. The Genocchi polynomials *Gn*(*x*) are defined by (see [19–24]).

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

When *x* = 0, *Gn* = *Gn*(0) are called the Genocchi numbers.

As is well-known, the Euler polynomials are defined by the generating function to be (see [1,4]).

$$\frac{2}{\varepsilon^t + 1} \varepsilon^{xt} = \sum\_{n=0}^{\infty} E\_n(x) \frac{t^n}{n!} \tag{3}$$

For *n* ≥ 0, the Stirling numbers of the first kind are defined by (see [5,7,25]),

$$\mathbf{y}(\mathbf{x})\_{\mathbb{N}} = \sum\_{l=0}^{n} S\_1(n, l) \mathbf{x}^l,\tag{4}$$

where (*x*)<sup>0</sup> = 1, (*x*)*<sup>n</sup>* = *x*(*x* − 1)...(*x* − *n* + 1), (*n* ≥ 1). From (4), it is easy to see that

$$\frac{1}{k!} (\log(1+t))^k = \sum\_{n=k}^{\infty} S\_1(n,k) \frac{t^n}{n!}.\tag{5}$$

In the inverse expression to (4), for *n* ≥ 0, the Stirling numbers of the second kind are defined by

$$\mathbf{x}^n = \sum\_{l=0}^n \mathcal{S}\_2(n, l)(\mathbf{x})\_l. \tag{6}$$

From (6), it is easy to see that

$$\frac{1}{k!}(e^t - 1)^k = \sum\_{n=k}^{\infty} S\_2(n, k) \frac{t^n}{n!}. \tag{7}$$

### **2. The Poly-Genocchi Polynomials**

For *k* ∈ Z, by (2) and (14), we define the poly-Genocchi polynomials which are given by

$$\frac{2\varepsilon\_k(\log(1+t))}{c^t+1}e^{xt} = \sum\_{n=0}^{\infty} G\_n^{(k)}(x)\frac{t^n}{n!}.\tag{8}$$

When *x* = 0, *G*(*k*) *<sup>n</sup>* <sup>=</sup> *<sup>G</sup>*(*k*) *<sup>n</sup>* (0) are called the poly-Genocchi numbers. From (8), we see that

$$\mathcal{G}\_n^{(1)}(\mathbf{x}) = \mathcal{G}\_n(\mathbf{x}), \ (n \in \mathbb{N} \cup \{0\}) \tag{9}$$

are the ordinary Genocchi polynomials. From (2), (4) and (8) , we observe that

$$\begin{split} &\sum\_{n=0}^{\infty} G\_n^{(l)} \frac{t^n}{m!} \\ &= \frac{2t\_k(\log(1+t))}{\epsilon^t + 1} \\ &= \frac{2}{\epsilon^t + 1} \sum\_{m=1}^{\infty} \frac{(\log(1+t))^m}{(m-1)! m^k} \\ &= \frac{2}{\epsilon^t + 1} \sum\_{m=0}^{\infty} \frac{(\log(1+t))^{m+1}}{m!(m+1)^k} \\ &= \frac{2}{\epsilon^t - 1} \sum\_{m=0}^{\infty} \frac{1}{(m+1)^{k-1}} \sum\_{l=m+1}^{\infty} S\_1(l, m+1) \frac{t^l}{l!} \\ &= \frac{2t}{\epsilon^t + 1} \sum\_{m=0}^{\infty} \frac{1}{(m+1)^{k-1}} \sum\_{l=m}^{\infty} \frac{S\_1(l+1, m+1)}{l+1} \frac{t^l}{l!} \\ &= \left(\sum\_{j=0}^{\infty} G\_j^{(j)} \frac{t^j}{j!} \right) \sum\_{l=0}^{\infty} \left(\sum\_{m=0}^l \frac{1}{(m+1)^{k-1}} \frac{S\_1(l+1, m+1)}{l+1} \right) \frac{t^l}{l!} \\ &= \sum\_{n=0}^{\infty} \left(\sum\_{l=0}^n \sum\_{m=0}^l \binom{n}{l} \frac{1}{(m+1)^{k-1}} \frac{S\_1(l+1, m+1)}{l+1} G\_{n-l} \right) \frac{t^n}{n!}. \end{split} \tag{10}$$

Therefore, by (10), we obtain the following theorem.

**Theorem 1.** *For k* ∈ Z *and n* ∈ N ∪ {0}*, we have*

$$G\_n^{(k)} = \sum\_{l=0}^n \sum\_{m=0}^l \binom{n}{l} \frac{1}{(m+1)^{k-1}} \frac{S\_1(l+1, m+1)}{l+1} G\_{n-l}.\tag{11}$$

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

$$\mathcal{G}\_{\mathfrak{n}}^{(1)} = \mathcal{G}\_{\mathfrak{n}} = \sum\_{l=0}^{n} \sum\_{m=0}^{l} \binom{n}{l} \frac{\mathcal{S}\_1(l+1, m+1)}{l+1} \mathcal{G}\_{n-l}.\tag{12}$$

*Moreover,*

$$\sum\_{l=1}^{n} \sum\_{m=0}^{l} \binom{n}{l} \frac{S\_1(l+1, m+1)}{l+1} G\_{n-l} = 0, \quad (n \in \mathbb{N}).\tag{13}$$

Kim-Kim ([9]) defined the polyexponential function by (see [6,9–12,26]).

$$c\_k(\mathbf{x}) = \sum\_{n=1}^{\infty} \frac{\mathbf{x}^n}{(n-1)! n^{k'}} \tag{14}$$

In [18], it is well known that for *k* ≥ 2,

$$\frac{d}{dx}e\_k(\mathbf{x}) = \frac{1}{\mathbf{x}}e\_{k-1}(\mathbf{x}).\tag{15}$$

Thus, by (15), for *k* ≥ 2, we get

$$e\_k(\mathbf{x}) = \int\_0^\mathbf{x} \underbrace{\frac{1}{t\_1} \int\_0^{t\_1} \frac{1}{t\_1} \cdots \int\_0^{t\_{k-2}}}\_{(k-2)\text{times}} \frac{1}{t\_{k-1}} (e^{t\_{k-1}} - 1) d\_{k-1} t dt\_{k-1} \cdots dt\_1. \tag{16}$$

From (16), we obtain the following equation.

$$\begin{aligned} &\sum\_{n=0}^{\infty} G\_{n}^{(k)} \frac{x^{n}}{n!} \\ &= \frac{2}{e^{x} + 1} e\_{k} (\log(1+x)) \\ &= \frac{2}{e^{x} + 1} \int\_{0}^{x} \frac{1}{(1+t)\log(1+t)} t\_{k-1} (\log(1+t)) dt \\ &= \frac{2}{e^{x} + 1} \int\_{0}^{x} \frac{1}{(1+t\_{1})\log(1+t\_{1})} \\ &\underbrace{\int\_{0}^{t\_{1}} \frac{1}{(1+t\_{2})\log(1+t\_{2})} \cdots \int\_{0}^{t\_{k-2}} \frac{t\_{k-1}}{(1+t\_{k-1})\log(1+t\_{k-1})} dt\_{k-1} dt\_{k-2} \cdots dt\_{1}, (k \ge 2). \end{aligned} \tag{17}$$

Let us take *k* = 2. Then, by (2) and (16), we get

$$\sum\_{n=0}^{\infty} G\_n^{(2)} \frac{\mathbf{x}^n}{n!} = \frac{2}{\epsilon^{\mathbf{x}} + 1} \int\_0^{\infty} \frac{t}{(1+t)\log(1+t)} dt$$

$$= \frac{2}{\epsilon^{\mathbf{x}} + 1} \sum\_{l=0}^{\infty} \frac{B\_l^{(l)}}{l!} \int\_0^{\infty} t^l dt$$

$$= \frac{2}{\epsilon^{\mathbf{x}} + 1} \sum\_{l=0}^{\infty} \frac{B\_l^{(l)}}{l+1} \frac{\mathbf{x}^{l+1}}{l!}$$

$$= \frac{2\mathbf{x}}{\epsilon^{\mathbf{x}} + 1} \sum\_{l=0}^{\infty} \frac{B\_l^{(l)}}{l+1} \frac{\mathbf{x}^l}{l!}$$

$$= \left(\sum\_{m=0}^{\infty} G\_m \frac{\mathbf{x}^m}{m!}\right) \left(\sum\_{l=0}^{\infty} \frac{B\_l^{(l)}}{l+1} \frac{\mathbf{x}^l}{l!}\right)$$

$$= \sum\_{n=0}^{\infty} \left(\sum\_{l=0}^n \binom{n}{l} \frac{B\_l^{(l)}}{l+1} G\_{n-l}\right) \frac{\mathbf{x}^n}{n!}.\tag{18}$$

Therefore, by (18), we obtain the following theorem.

**Theorem 2.** *Let n* ∈ N ∪ {0}*, we have*

$$\mathbf{G}\_n^{(2)} = \sum\_{l=0}^n \binom{n}{l} \frac{B\_l^{(l)}}{l+1} \mathbf{G}\_{n-l}. \tag{19}$$

From (3) and (16), we also get

$$\sum\_{n=0}^{\infty} G\_n^{(2)} \frac{\mathbf{x}^n}{n!} = \frac{2}{c^\mathbf{x} + 1} \int\_0^\mathbf{x} \frac{t}{(1+t)\log(1+t)} dt$$

$$= \frac{2}{c^\mathbf{x} + 1} \sum\_{l=0}^\infty B\_l^{(l)} \frac{\mathbf{x}^{l+1}}{(l+1)!}$$

$$= \frac{2}{c^\mathbf{x} + 1} \sum\_{l=1}^\infty B\_{l-1}^{(l-1)} \frac{\mathbf{x}^l}{l!}$$

$$= \left(\sum\_{m=0}^\infty E\_m \frac{\mathbf{x}^m}{m!}\right) \left(\sum\_{l=1}^\infty B\_{l-1}^{(l-1)} \frac{\mathbf{x}^l}{l!}\right)$$

$$= \sum\_{n=1}^\infty \left(\sum\_{l=1}^n \binom{n}{l} B\_{l-1}^{(l-1)} E\_{n-l}\right) \frac{\mathbf{x}^n}{n!}.\tag{20}$$

Therefore, by (20), we obtain the following theorem.

**Theorem 3.** *Let n* ≥ 1*, we have*

$$G\_n^{(2)} = \sum\_{l=1}^n \binom{n}{l} B\_{l-1}^{(l-1)} E\_{n-l}.\tag{21}$$

From (8), we observe that

$$\sum\_{n=0}^{\infty} G\_n^{(k)}(x) \frac{t^n}{n!} = \frac{2e\_k(\log(1+t))}{e^t + 1} e^{xt}$$

$$\begin{split} &= \left(\sum\_{l=0}^{\infty} G\_l^{(k)} \frac{t^l}{l!} \right) \left(\sum\_{m=0}^{\infty} x^m \frac{t^m}{m!} \right) \\ &= \sum\_{n=0}^{\infty} \left(\sum\_{l=0}^n \binom{n}{l} G\_l^{(k)} x^{n-l} \right) \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \left(\sum\_{l=0}^n \binom{n}{l} G\_{n-l}^{(k)} x^l \right) \frac{t^n}{n!} . \end{split} \tag{22}$$

From (22), we obtain the following theorem.

**Theorem 4.** *Let n* ∈ N*, we have*

$$G\_n^{(k)}(\mathbf{x}) = \sum\_{l=0}^n \binom{n}{l} G\_{n-l}^{(k)} \mathbf{x}^l. \tag{23}$$

From (23), we observe that

$$\begin{split} \frac{d}{dx}G\_n^{(k)}(x) &= \sum\_{l=1}^n \binom{n}{l} \mathbf{G}\_{n-l}^{(k)} l x^{l-1} \\ &= \sum\_{l=0}^{n-1} \binom{n}{l+1} \mathbf{G}\_{n-l-1}^{(k)} (l+1) x^l \\ &= \sum\_{l=0}^{n-1} \frac{n!}{(l+1)!(n-l-1)!} \mathbf{G}\_{n-1-l}^{(k)} (l+1) x^l \\ &= n \sum\_{l=0}^{n-1} \frac{(n-1)!}{l!(n-1-l)!} \mathbf{G}\_{n-1-l}^{(k)} x^l \\ &= n \mathbf{G}\_{n-1}^{(k)}(x). \end{split} \tag{24}$$

From (24), we obtain the following theorem.

**Theorem 5.** *Let n* ∈ N ∪ {0} *and k* ∈ Z*, we have*

$$\frac{d}{d\mathbf{x}}\mathcal{G}\_{n}^{(k)}(\mathbf{x}) = n\mathcal{G}\_{n-1}^{(k)}(\mathbf{x}).\tag{25}$$

### **3. The Unipoly Genocchi Polynomials and Numbers**

Let *p* be any arithmetic function which is real or complex valued function defined on the set of positive integers N. Then, Kim-Kim ([9]) defined the unipoly function attached to polynomials by

$$\mu\_k(\mathbf{x}|p) = \sum\_{n=1}^{\infty} \frac{p(n)\mathbf{x}^n}{n^k}, \ (k \in \mathbb{Z}). \tag{26}$$

It is well known that

$$\mu\_k(\mathbf{x}|1) = \sum\_{n=1}^{\infty} \frac{\mathbf{x}^n}{n^k} = L i\_k(\mathbf{x}) \tag{27}$$

is the ordinary polylogarithm function, and for *k* ≥ 2,

$$\frac{d}{dx}u\_k(x|p) = \frac{1}{x}u\_{k-1}(x|p),\tag{28}$$

and

$$u\_k(x|p) = \int\_0^x \frac{1}{t} \underbrace{\int\_0^t \frac{1}{t} \cdots}\_{(k-2)\text{times}} \frac{1}{t} u\_1(t|p) dt dt \cdot \cdots dt \tag{29}$$

By using (26), we define the unipoly Genocchi polynomials as follows:

$$\frac{2}{e^t + 1} \mu\_k(\log(1+t)|p)e^{\mathbf{x}t} = \sum\_{n=0}^{\infty} G\_{n,p}^{(k)}(\mathbf{x}) \frac{t^n}{n!}.\tag{30}$$

Let us take *p*(*n*) = <sup>1</sup> (*n*−1)! . Then we have

$$\sum\_{n=0}^{\infty} G\_{n,p}^{(k)}(x) \frac{t^n}{n!} = \frac{2}{\epsilon^t + 1} \mu\_k \left( \log(1+t) \left| \frac{1}{(n-1)!} \right. \right) e^{xt}$$

$$= \frac{2}{\epsilon^t + 1} \sum\_{m=1}^{\infty} \frac{(\log(1+t))^m}{m^k (m-1)!} e^{xt}$$

$$= \frac{2\epsilon\_k (\log(1+t))}{\epsilon^t + 1} e^{xt}$$

$$= \sum\_{n=0}^{\infty} G\_n^{(k)}(x) \frac{t^n}{n!}. \tag{31}$$

Thus, by (31), we have the following theorem.

**Theorem 6.** *If we take p*(*n*) = <sup>1</sup> (*n*−1)! *for n* <sup>∈</sup> <sup>N</sup> ∪ {0} *and k* <sup>∈</sup> <sup>Z</sup>*, then we have*

$$\mathcal{G}\_{\mathfrak{n},p}^{(k)}(\mathbf{x}) = \mathcal{G}\_{\mathfrak{n}}^{(k)}(\mathbf{x}).\tag{32}$$

From (4) and (30) with *x* = 0, we have

$$\begin{split} &\sum\_{n=0}^{\infty} G\_{n,p}^{(l)} \frac{l^n}{n!} \\ &= \frac{2}{\epsilon^l + 1} \sum\_{m=1}^{\infty} \frac{p(m)}{m^k} (\log(1+t))^m \\ &= \frac{2}{\epsilon^l + 1} \sum\_{m=0}^{\infty} \frac{p(m+1)(m+1)!}{(m+1)^k} \sum\_{l=m+1}^{\infty} S\_1(l, m+1) \frac{l^l}{l!} \\ &= \frac{2}{\epsilon^l + 1} \sum\_{m=0}^{\infty} \frac{p(m+1)(m+1)!}{(m+1)^k} \sum\_{l=m}^{\infty} S\_1(l+1, m+1) \frac{l^{l+1}}{(l+1)!} \\ &= \frac{2t}{\epsilon^l + 1} \sum\_{m=0}^{\infty} \frac{p(m+1)(m+1)!}{(m+1)^k} \sum\_{l=m}^{\infty} S\_1(l+1, m+1) \frac{l^l}{(l+1)!} \\ &= \left(\sum\_{j=0}^{\infty} G\_j \frac{l^j}{j!}\right) \sum\_{l=0}^{\infty} \left(\sum\_{m=0}^l \frac{p(m+1)(m+1)!}{(m+1)^k} \frac{S\_1(l+1, m+1)}{l+1}\right) \frac{l^l}{l!} \\ &= \sum\_{n=0}^{\infty} \left(\sum\_{l=0}^n \sum\_{m=0}^l \binom{n}{l} \frac{p(m+1)(m+1)!}{(m+1)^k} \frac{S\_1(l+1, m+1)}{l+1} G\_{n-l}\right) \frac{l^n}{n!}. \end{split}$$

Therefore, by comparing the coefficients on both sides of (33), we obtain the following theorem.

**Remark 1.** *Let n* ∈ N *and k* ∈ Z*. Then, we have*

$$G\_{n,p}^{(k)} = \sum\_{l=0}^{n} \sum\_{m=0}^{l} \binom{n}{l} \frac{p(m+1)(m+1)!}{(m+1)^k} \frac{S\_1(l+1, m+1)}{l+1} G\_{n-l}. \tag{34}$$

*In particular,*

$$G\_{n, \frac{1}{(n-1)!}}^{(k)} = \sum\_{l=0}^{n} \sum\_{m=0}^{l} \binom{n}{l} \frac{G\_{n-l}}{(m+1)^{k-1}} \frac{S\_1(l+1, m+1)}{l+1} \tag{35}$$

*arrives at* (11)*.*

From (30), we easily obtain the following theorem.

**Theorem 7.** *Let n* ∈ N ∪ {0} *and k* ∈ Z*. Then, we have*

$$\mathbf{G}\_{n,p}^{(k)}(\mathbf{x}) = \sum\_{l=0}^{n} \binom{n}{l} \mathbf{G}\_{n-l,p}^{(k)} \mathbf{x}^{l}. \tag{36}$$

From (36), we easily obtain the following theorem.

**Theorem 8.** *Let n* ∈ N ∪ {0} *and k* ∈ Z*. Then, we have*

$$\frac{d}{dx}G\_{n,p}^{(k)}(x) = nG\_{n-1,p}^{(k)}(x). \tag{37}$$

Finally, by (4) and (30), we observe that

$$\begin{split} &\sum\_{n=0}^{\infty} G\_{n,p}^{(k)} \frac{t^n}{n!} \\ &= \frac{2}{\epsilon^t + 1} \sum\_{m=1}^{\infty} \frac{p(m)}{m^k} \frac{m!}{m!} (\log(1+t))^m \\ &= \frac{2}{\epsilon^t + 1} \sum\_{m=1}^{\infty} \frac{p(m+1)}{(m+1)^k} \frac{(m+1)!}{(m+1)!} (\log(1+t))^{m+1} \\ &= \sum\_{j=0}^{\infty} E\_j \frac{t^j}{j!} \sum\_{m=0}^{\infty} \frac{p(m+1)(m+1)!}{(m+1)^k} \sum\_{l=m+1}^{\infty} S\_1(l, m+1) \frac{t^l}{l!} \\ &= \sum\_{j=0}^{\infty} E\_j \frac{t^j}{j!} \sum\_{l=0}^{\infty} \sum\_{m=0}^l \frac{p(m+1)(m+1)!}{(m+1)^k} \sum\_{l=m}^{\infty} S\_1(l+1, m+1) \frac{t^{l+1}}{(l+1)!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{l=0}^n \sum\_{m=0}^l \binom{n}{l} \frac{p(m+1)(m+1)}{(m+1)^k} \frac{S\_1(l+1, m+1)}{l+1} E\_{n-l} \right) \frac{t^n}{n!} . \end{split} \tag{38}$$

From (37) , we obtain the following theorem.

**Theorem 9.** *Let n* ∈ N *and k* ∈ Z*, we have*

$$G\_{n,p}^{(k)} = \sum\_{l=0}^{n} \sum\_{m=0}^{l} \binom{n}{l} \frac{p(m+1)(m+1)!}{(m+1)^k} \frac{S\_1(l+1, m+1)}{l+1} E\_{n-l}.\tag{39}$$

### **4. Conclusions**

In 2019, Kim-Kim considered the polyexponential functions and poly-Bernoulli polynomials. In the same view as these functions and polynomials, we defined the poly-Genocchi polynomials (Equation (8)) and obtained some identities (Theorem 1 and Corollary 1). In particular, we observed explicit poly-Genocchi numbers for *k* = 2 (Theorems 2, 3 and 4). Furthermore, by using the unipoly functions, we defined the unipoly Genocchi polynomials (Equation (30)) and obtained some their properties (Theorems 6 and 7). Finally, we obtained the derivative of the unipoly Genocchi polynomials (Theorem 8) and gave the identity indicating the relationship of unipoly Genocchi polynomials and Euler polynomials (Theorem 9). It is recommended that our readers look at references [27–31] if they want to know the applications related to this paper.

**Author Contributions:** L.-C.J. and D.V.D. conceived the framework and structured the whole paper; D.V.D. and L.-C.J. checked the results of the paper and completed the revision of the article. All authors have read and agreed to the published version of the manuscript.

**Funding:** The present research has been conducted by the Research Grant of Kwangwoon University in 2020.

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

### **References**


© 2020 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* **An Erd ˝os-Ko-Rado Type Theorem via the Polynomial Method**

### **Kyung-Won Hwang 1,†, Younjin Kim 2,\*,† and Naeem N. Sheikh 3,†**


Received: 29 March 2020; Accepted: 16 April 2020; Published: 17 April 2020

**Abstract:** A family F is an intersecting family if any two members have a nonempty intersection. Erd˝os, Ko, and Rado showed that |F| ≤ ( *n*−1 *<sup>k</sup>*−1) holds for a *<sup>k</sup>*-uniform intersecting family <sup>F</sup> of subsets of [*n*]. The Erd˝os-Ko-Rado theorem for non-uniform intersecting families of subsets of [*n*] of size at most *k* can be easily proved by applying the above result to each uniform subfamily of a given family. It establishes that |F| ≤ ( *n*−1 *<sup>k</sup>*−1) <sup>+</sup> ( *n*−1 *<sup>k</sup>*−2) <sup>+</sup> ··· <sup>+</sup> ( *n*−1 <sup>0</sup> ) holds for non-uniform intersecting families of subsets of [*n*] of size at most *k*. In this paper, we prove that the same upper bound of the Erd˝os-Ko-Rado Theorem for *k*-uniform intersecting families of subsets of [*n*] holds also in the non-uniform family of subsets of [*n*] of size at least *k* and at most *n* − *k* with one more additional intersection condition. Our proof is based on the method of linearly independent polynomials.

**Keywords:** Erd˝os-Ko-Rado theorem; intersecting families; polynomial method

### **1. Introduction**

Let [*n*] be the set {1, 2, ··· , *n*}. A family F of subsets of [*n*] is *intersecting* if *F* ∩ *F* is non-empty for all *F*, *F* ∈ F. A family F of subsets of [*n*] is *t-intersecting* if |*F* ∩ *F* | ≥ *t* holds for any *F*, *F* ∈ F. A family F is *k*-*uniform* if it is a collection of *k*-subsets of [*n*]. In 1961, Erd˝os, Ko, and Rado [1] were interested in obtaining an upper bound on the maximum size that an intersecting *k*-uniform family can have and proved the following theorem which bounds the cardinality of an intersecting *k*-uniform family.

**Theorem 1** (Erd˝os-Ko-Rado Theorem [1])**.** *If n* ≥ 2*k and* F *is an intersecting k-uniform family of subsets of* [*n*]*, then*

$$|\mathcal{F}| \le \binom{n-1}{k-1}.$$

Erd˝os-Ko-Rado Theorem is an important result of extremal set theory and has been an inspiration for various generalizations by many authors for over 50 years. Erd˝os, Ko, and Rado [1] also proved that there exists an integer *n*0(*k*, *t*) such that if *n* ≥ *n*0(*k*, *t*), then the maximum size of a *t*-intersecting *k*-uniform family of subsets of [*n*] is ( *n*−*t k*−*t* ). The following generalization of the Erd˝os-Ko-Rado Theorem was proved by Frankl [2] for *t* ≥ 15, and was completed by Wilson [3] for all *t*. It establishes that the generalized EKR theorem is true if *n* ≥ (*k* − *t* + 1)(*t* + 1).

**Theorem 2** (Generalized Erd˝os-Ko-Rado Theorem [2,3])**.** *If n* ≥ (*k* − *t* + 1)(*t* + 1) *and* F *is a t-intersecting k-uniform family of subsets of* [*n*]*, then we have*

$$|\mathcal{F}| \le \binom{n-t}{k-t}.$$

The Erd˝os-Ko-Rado Theorem can be restated as follows.

**Theorem 3** (Erd˝os-Ko-Rado Theorem [1])**.** *If* F *is a family of subsets Fi of* [*n*] *with* |*Fi*| = *k and* |*Fi*| ≤ *n* − *k that satisfies the following two conditions, for i* = *j*

*(a)* 1 ≤ |*Fi* ∩ *Fj*| ≤ *k* − 1 *(b)* <sup>1</sup> ≤ |*Fi* ∩ *<sup>F</sup><sup>c</sup> <sup>j</sup>* | ≤ *k* − 1

*then we have*

$$|\mathcal{F}| \le \binom{n-1}{k-1}.$$

#### **2. Results**

The following EKR-type theorem for non-uniform intersecting families of subsets of [*n*] of size at most *k* can be easily proved by applying Theorem 3 to each uniform subfamily of the given non-uniform family.

**Theorem 4.** *If* F *is a family of subsets Fi of* [*n*]*, with* |*Fi*| ≤ *k and n* ≥ 2*k, that satisfies the following two conditions, for i* = *j*

*(a)* 1 ≤ |*Fi* ∩ *Fj*| ≤ *k* − 1 *(b)* <sup>1</sup> ≤ |*Fi* ∩ *<sup>F</sup><sup>c</sup> <sup>j</sup>* | ≤ *k* − 1

*then we have*

$$|\mathcal{F}| \le \binom{n-1}{k-1} + \binom{n-1}{k-2} + \dots + \binom{n-1}{0}.$$

In 2014, Alon, Aydinian, and Huang [4] gave the following strengthening of the bounded rank Erd˝os-Ko-Rado theorem by obtaining the same upper bound under a weaker condition as follows.

**Theorem 5** (Alon, Aydinian, and Huang [4])**.** *Let* F *be a family of subsets of* [*n*] *of size at most k,* 1 ≤ *k* ≤ *n* − 1*. Suppose that for every two subsets A*, *B* ∈ F*, if A* ∩ *B* = ∅*, then* |*A*| + |*B*| ≤ *k. Then we have*

$$|\mathcal{F}| \le \binom{n-1}{k-1} + \binom{n-1}{k-2} + \dots + \binom{n-1}{0} \cdot \frac{1}{k}$$

Since the bound ( *n*−1 *<sup>k</sup>*−1) <sup>+</sup> ( *n*−1 *<sup>k</sup>*−2) <sup>+</sup> ··· <sup>+</sup> ( *n*−1 <sup>0</sup> ) is much larger than ( *n*−1 *<sup>k</sup>*−1), this leads to the following interesting question: when is it possible to get the same bound as in the Erd˝os-Ko-Rado theorem for uniform intersecting families for the non-uniform intersecting families? We answer this question in the main result of this paper, where we prove that the same upper bound of the EKR Theorem for *k*-uniform intersecting families of subsets of [*n*] also holds in the non-uniform family of subsets of [*n*] of size at least *k* and at most *n* − *k* with one more additional intersection condition, as follows.

**Theorem 6.** *If* F *is a family of subsets Fi of* [*n*] *with k* ≤ |*Fi*| ≤ *n* − *k that satisfies the following three conditions, for i* = *j*

*(a)* 1 ≤ |*Fi* ∩ *Fj*| ≤ *k* − 1 *(b)* <sup>1</sup> ≤ |*Fi* ∩ *<sup>F</sup><sup>c</sup> <sup>j</sup>* | ≤ *k* − 1 *(c)* <sup>1</sup> ≤ |*F<sup>c</sup> <sup>i</sup>* ∩ *<sup>F</sup><sup>c</sup> <sup>j</sup>* | ≤ *k* − 1

*then we have*

$$|\mathcal{F}| \le \binom{n-1}{k-1}.$$

Please note that if we remove the third condition in Theorem 6, we get the same bound of the Erd˝os-Ko-Rado theorem for *k*-uniform intersecting families under the same condition for subsets of [*n*] that are of size at least *k* and at most *n* − *k*.

Erd˝os-Ko-Rado Theorem is a seminal result in extremal combinatorics and has been proved by various methods (see a survey in [5]). There have been many results that have generalized EKR in various ways over the decades. The aim of this paper is to give a generalization of the EKR Theorem to non-uniform families with some extra conditions. Our proof is based on the method of linearly independent multilinear polynomials.

Our paper is organized as follows. In Section 3, we will introduce our main tool, the method of linearly independent multilinear polynomials. In Section 4, we will give the proof of our main result, Theorem 6.

### **3. Polynomial Method**

The method of linearly independent polynomials is one of the most powerful methods for counting the number of sets in various combinatorial settings. In this method, we correspond multilinear polynomials to the sets and then prove that these polynomials are linearly independent in some space. In 1975, Ray-Chaudhuri and Wilson [6] obtained the following result by using the method of linearly independent polynomials.

**Theorem 7** (Ray-Chaudhuri and Wilson [6])**.** *Let l*1, *l*2, ··· , *ls* < *n be nonnegative integers. If* F *is a k-uniform family of subsets of* [*n*] *such that* |*A* ∩ *B*| ∈ *L* = {*l*1, *l*2, ··· , *ls*} *holds for every pair of distinct subsets A*, *B* ∈ F*, then* |F| ≤ ( *n <sup>s</sup>*) *holds.*

In 1981, Frankl and Wilson [7] obtained the following nonuniform version of the Ray-Chaudhuri-Wilson Theorem using the polynomial method. Their proof is given underneath.

**Theorem 8** (Frankl and Wilson [7])**.** *Let l*1, *l*2, ··· , *ls* < *n be nonnegative integers. If* F *is a family of subsets of* [*n*] *such that* |*A* ∩ *B*| ∈ *L* = {*l*1, *l*2, ··· , *ls*} *holds for every pair of distinct subsets A*, *B* ∈ F*, then* |F| ≤ <sup>∑</sup>*<sup>s</sup> <sup>k</sup>*=<sup>0</sup> ( *n <sup>k</sup>*) *holds.*

**Proof.** Let *x* be the *n*-tuple of variables *x*1, *x*2, ··· , *xn*, where *xi* takes the values only 0 and 1. Then all the polynomials we will work with have the relation *x*<sup>2</sup> *<sup>i</sup>* = *xi* in their domain. Let *F*1, *F*2, ··· , *Fm* be the distinct sets in F, listed in non-decreasing order according to their sizes. We define the characteristic vector *vi* = (*vi*<sup>1</sup> , *vi*<sup>2</sup> , ··· , *vin* ) of *Fi* such that *vij* = 1 if *j* ∈ *Fi* and *vij* = 0 if *j* ∈ *Fi*. We consider the following multilinear polynomial

$$f\_l(\mathbf{x}) = \prod\_{\substack{l \in L\_r \ l < |F\_l|}} (v\_i \cdot \mathbf{x} - l)^l$$

where *x* = (*x*1, *x*2, ··· , *xn*).

Then we obtain that *fi*(*vi*) = 0 and *fi*(*vj*) = 0 for *j* < *i*. As the vectors *vi* are 0 − 1 vectors, we have an another multilinear polynomial *gi*(*x*) such that *fi*(*x*) = *gi*(*x*) holds for all *<sup>x</sup>* ∈ {0, 1}*<sup>n</sup>* by substituting *xk* for the powers of *xk*, where *k* = 1, 2, ··· , *n*. Then it is easy to see that the polynomials *g*1, *g*2, ··· , *gm* are linearly independent over R. Since the dimension of *n*-variable multilinear polynomials of degree at most *s* is ∑*<sup>s</sup> <sup>k</sup>*=<sup>0</sup> ( *n <sup>k</sup>*), we have

$$|\mathcal{F}| \le \sum\_{k=0}^{s} \binom{n}{k}$$

finishing the proof of Theorem 8.

In the same paper, Frankl and Wilson [7] obtained the following modular version of Theorem 7.

**Theorem 9** (Frankl and Wilson [7])**.** *If* F *is a family of subsets of* [*n*] *such that* |*A* ∩ *B*| ≡ *l* ∈ *L (mod p) holds for every pair of distinct subsets A*, *<sup>B</sup>* ∈ F*, then* |F| ≤ ( *<sup>n</sup>* |*L*| ) *holds.*

In 1983, Deza, Frankl and Singhi [8] obtained the following modular version of Theorem 8.

**Theorem 10** (Deza, Frankl and Singhi [8])**.** *If* F *is a family of subsets of* [*n*] *such that* |*A* ∩ *B*| ≡ *l* ∈ *L (mod p) holds for every pair of distinct subsets <sup>A</sup>*, *<sup>B</sup>* ∈ F *and* <sup>|</sup>*A*<sup>|</sup> ≡ *<sup>l</sup> (mod p) for every <sup>A</sup>* ∈ F*, then* |F| ≤ <sup>∑</sup>|*L*<sup>|</sup> *<sup>i</sup>*=<sup>0</sup> ( *n <sup>i</sup>*) *holds.*

In 1991, Alon, Babai, and Suzuki [9] gave another modular version of Theorem 8 by replacing the condition of nonuniformity with the condition that the members of F have *r* different sizes as follows. Their proof was also based on the polynomial method.

**Theorem 11** (Alon-Babai-Suzuki [9])**.** *Let K* = {*k*1, *k*2, ··· , *kr*} *and L* = {*l*1, *l*2, ··· , *ls*} *be two disjoint subsets of* {0, 1, ··· , *p* −1}*, where p is a prime, and let* F *be a family of subsets of* [*n*] *whose sizes modulo p are in the set K, and* |*A* ∩ *B*| (mod *p*) ∈ *L holds for every distinct two subsets A*, *B in* F*, then the largest size of such a family* F *is* ( *n <sup>s</sup>*) <sup>+</sup> ( *<sup>n</sup> <sup>s</sup>*−1) <sup>+</sup> ··· <sup>+</sup> ( *<sup>n</sup> <sup>s</sup>*−*r*+1) *under the conditions <sup>r</sup>*(*<sup>s</sup>* <sup>−</sup> *<sup>r</sup>* <sup>+</sup> <sup>1</sup>) <sup>≤</sup> *<sup>p</sup>* <sup>−</sup> <sup>1</sup> *and <sup>n</sup>* <sup>≥</sup> *<sup>s</sup>* <sup>+</sup> max1≤*i*≤*<sup>r</sup> ki.*

In the same paper, Alon, Babai, and Suzuki [9] also conjectured that the statement of Theorem 11 remains true if the condition *r*(*s* − *r* + 1) ≤ *p* − 1 is dropped. Recently Hwang and Kim [10] proved this conjecture of Alon, Babai and Suzuki (1991), using the method of linearly independent polynomials. This result is as follows.

**Theorem 12** (Hwang and Kim [10])**.** *Let K* = {*k*1, *k*2, ··· , *kr*} *and L* = {*l*1, *l*2, ··· , *ls*} *be two disjoint subsets of* {0, 1, ··· , *p* − 1}*, where p is a prime, and let* F *be a family of subsets of* [*n*] *whose sizes modulo p are in the set K, and* |*A* ∩ *B*| (mod *p*) ∈ *L for every distinct two subsets A*, *B in* F*, then the largest size of such a family* F *is* ( *n <sup>s</sup>*) <sup>+</sup> ( *<sup>n</sup> <sup>s</sup>*−1) <sup>+</sup> ··· <sup>+</sup> ( *<sup>n</sup> <sup>s</sup>*−*r*+1) *under the only condition that n* <sup>≥</sup> *<sup>s</sup>* <sup>+</sup> max1≤*i*≤*<sup>r</sup> ki.*

The method of linearly independent polynomials has also been used to prove many intersection theorems about set families by Blokhuis [11], Chen and Liu [12], Furedi, Hwang, and Weichsel [13], Liu and Yang [14], Qian and Ray-Chaudhuri [15], Ramanan [16], Snevily [17,18], Wang, Wei, and Ge [19], and others.

### **4. Proof of the Main Result**

In this section, we prove Theorem 6. As we have mentioned before, our proof is based on the polynomial method. Let *x* be the *n*-tuple of variables *x*1, *x*2, ··· , *xn*, where *xi* takes the values only 0 and 1. Then all the polynomials we will work with have the relation *x*<sup>2</sup> *<sup>i</sup>* = *xi* in their domain.

**Proof of Theorem 6.** The result is immediate if |F| = 1. Suppose |F| > 1. Let *F*1, *F*2, ··· , *Ff* be the distinct sets in F, listed in non-decreasing order of size. We define the characteristic vector *vi* = (*vi*<sup>1</sup> , *vi*<sup>2</sup> , ··· , *vin* ) of *Fi* such that *vij* = 1 if *j* ∈ *Fi* and *vij* = 0 if *j* ∈ *Fi*.

We consider the following family of multilinear polynomials

$$f\_i(\mathbf{x}) = \prod\_{j=1}^{k-1} (v\_i \cdot \mathbf{x} - j)$$

where *x* = (*x*1, *x*2, ··· , *xn*).

Since |*F*1|≤|*F*2|, there exists some *p* ∈ *F*<sup>2</sup> such that *p* ∈ *F*1. Let G = {*G*1, *G*2, ··· , *Gg*} be the family of subsets of [*n*] with the size at most *k* − 2, which is listed in non-decreasing order of size, and not containing *p*. Next, we consider the second family of multilinear polynomials

$$\mathfrak{g}\_i(\mathfrak{x}) = (\mathfrak{x}\_p - 1) \prod\_{j \in G\_i} \mathfrak{x}\_j$$

where 1 ≤ *i* ≤ *g*. Let H = {*H*1, *H*2, ··· , *Hh*} be the family of subsets of [*n*] with the size at most *k* − 1, which is listed in non-decreasing order of size, and containing *p*. Then, we consider our third and last family of multilinear polynomials

$$\begin{split} h\_{i}(\mathbf{x}) = \prod\_{j=0}^{|H\_{i}|-1} (w\_{i} \cdot \mathbf{x} - j) - \quad \sum\_{l: p \notin F\_{l}} \frac{\prod\_{j=0}^{|H\_{i}|-1} (w\_{i} \cdot v\_{l}^{c} - j)}{\prod\_{j=1}^{k-1} (v\_{l}^{c} \cdot v\_{l}^{c} - j)} \prod\_{j=1}^{k-1} (v\_{l}^{c} \cdot \mathbf{x} - j) \\ - \quad \sum\_{l: p \in F\_{l}} \frac{\prod\_{j=0}^{|H\_{i}|-1} (w\_{l} \cdot v\_{l} - j)}{\prod\_{j=1}^{k-1} (v\_{l'} \cdot v\_{l} - j)} \prod\_{j=1}^{k-1} (v\_{l} \cdot \mathbf{x} - j) \end{split}$$

where *wi* is the characteristic vector of *Hi*.

We claim that the functions *fi*(*x*), *gi*(*x*), and *hi*(*x*) taken together are linearly independent. Assume that

$$\sum\_{i=1}^{f} a\_i f\_i(\mathbf{x}) + \sum\_{i=1}^{\mathcal{S}} \beta\_i g\_i(\mathbf{x}) + \sum\_{i=1}^{h} \gamma\_i h\_i(\mathbf{x}) = 0 \tag{1}$$

We substitute the characteristic vector *vs* of *Fs* containing *p* into Equation (1). Because of the (*xp* − 1) factor, we have

$$
\lg\_i(v\_s) = 0 \quad \text{for all} \ 1 \le i \le \emptyset \cdot 1
$$

Let *v<sup>c</sup> <sup>l</sup>* be the characteristic vector of *<sup>F</sup><sup>c</sup> <sup>l</sup>* . Next, let us consider *hi*(*vs*) :

$$\begin{split} h\_{i}(\boldsymbol{\upsilon}\_{\boldsymbol{s}}) = \prod\_{j=0}^{|H\_{i}|-1} (\boldsymbol{w}\_{l} \cdot \boldsymbol{\upsilon}\_{\boldsymbol{s}} - j) - \sum\_{l:\, p \notin F\_{l}} \frac{\prod\_{j=0}^{|H\_{i}|-1} (\boldsymbol{w}\_{l} \cdot \boldsymbol{\upsilon}\_{l}^{\boldsymbol{c}} - j)}{\prod\_{j=1}^{k-1} (\boldsymbol{\upsilon}\_{l}^{\boldsymbol{c}} \cdot \boldsymbol{\upsilon}\_{l}^{\boldsymbol{c}} - j)} \prod\_{j=1}^{k-1} (\boldsymbol{\upsilon}\_{l}^{\boldsymbol{c}} \cdot \boldsymbol{\upsilon}\_{\boldsymbol{s}} - j) \\ - \sum\_{l:\, p \in F\_{l}} \frac{\prod\_{j=0}^{|H\_{i}|-1} (\boldsymbol{w}\_{l} \cdot \boldsymbol{\upsilon}\_{l} - j)}{\prod\_{j=1}^{k-1} (\boldsymbol{\upsilon}\_{l} \cdot \boldsymbol{\upsilon}\_{l} - j)} \prod\_{j=1}^{k-1} (\boldsymbol{\upsilon}\_{l} \cdot \boldsymbol{\upsilon}\_{\boldsymbol{s}} - j) . \end{split}$$

Since 1 ≤ |*Fl* <sup>∩</sup> *Fs*| ≤ *<sup>k</sup>* <sup>−</sup> 1, we have <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (*vl* · *vs* − *j*) = 0 except when *s* = *l*. Since |*Fi*| ≥ *k* for all *i*, we have

$$\begin{aligned} -\sum\_{l:\,p\in F\_{l}} \frac{\prod\_{j=0}^{|H\_{l}|-1} (w\_{l}\cdot v\_{l} - j)}{\prod\_{j=1}^{k} (v\_{l}\cdot v\_{l} - j)} \prod\_{j=1}^{k-1} (v\_{l}\cdot v\_{s} - j) &= -\frac{\prod\_{j=0}^{|H\_{l}|-1} (w\_{l}\cdot v\_{s} - j)}{\prod\_{j=1}^{k-1} (v\_{s}\cdot v\_{s} - j)} \prod\_{j=1}^{k-1} (v\_{s}\cdot v\_{s} - j) \\ &= -\prod\_{j=0}^{|H\_{l}|-1} (w\_{l}\cdot v\_{s} - j) .\end{aligned}$$

Since 1 ≤ |*F<sup>c</sup> <sup>l</sup>* <sup>∩</sup> *Fs*| ≤ *<sup>k</sup>* <sup>−</sup> 1 for *<sup>s</sup>* <sup>=</sup> *<sup>l</sup>*, we have <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (*v<sup>c</sup> <sup>l</sup>* · *vs* <sup>−</sup> *<sup>j</sup>*) = <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (|*F<sup>c</sup> <sup>l</sup>* ∩ *Fs*| − *j*) = 0. Thus, we have

$$h\_i(v\_s) = \prod\_{j=0}^{|H\_i|-1} (w\_i \cdot v\_s - j) - \prod\_{j=0}^{|H\_i|-1} (w\_i \cdot v\_s - j) = 0 \quad \text{for all } 1 \le i \le h.$$

Finally, we consider *fi*(*vs*). Since *fs*(*vs*) = 0 and 1 ≤ |*Fi* ∩ *Fs*| ≤ *k* − 1 for *i* = *s*, we get *α<sup>s</sup>* = 0 whenever *p* ∈ *Fs*.

Next, we substitute the characteristic vector *v<sup>c</sup> <sup>s</sup>* of *F<sup>c</sup> <sup>s</sup>* into Equation (1), where *p* ∈ *Fs*. Because of the (*xp* − 1) factor, we have

$$
\mathcal{g}\_i(v\_s^c) = 0 \quad \text{for all } 1 \le i \le \mathcal{g}\_i
$$

Next, let us consider *hi*(*v<sup>c</sup> <sup>s</sup>*). Since 1 ≤ |*F<sup>c</sup> <sup>l</sup>* ∩ *<sup>F</sup><sup>c</sup> <sup>s</sup>* | ≤ *<sup>k</sup>* <sup>−</sup> 1, we have <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (*v<sup>c</sup> <sup>l</sup>* · *<sup>v</sup><sup>c</sup> <sup>s</sup>* − *j*) = 0 except when *s* = *l*. Since *n* − |*Fi*| ≥ *k*, we have

$$-\sum\_{l:\,p\notin F\_l} \frac{\prod\_{j=0}^{|H\_i|-1} (w\_l \cdot v\_l^\varepsilon - j)}{\prod\_{j=1}^{k-1} (v\_l^\varepsilon \cdot v\_l^\varepsilon - j)} \prod\_{j=1}^{k-1} (v\_l^\varepsilon \cdot v\_s^\varepsilon - j) = -\prod\_{j=0}^{|H\_i|-1} (w\_l \cdot v\_s^\varepsilon - j).$$

Since 1 ≤ |*Fl* ∩ *<sup>F</sup><sup>c</sup> <sup>s</sup>* | ≤ *<sup>k</sup>* <sup>−</sup> 1 for *<sup>s</sup>* <sup>=</sup> *<sup>l</sup>*, we have <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (*vl* · *<sup>v</sup><sup>c</sup> <sup>s</sup>* <sup>−</sup> *<sup>j</sup>*) = <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (|*Fl* ∩ *<sup>F</sup><sup>c</sup> <sup>s</sup>* | − *j*) = 0. Thus, we have

$$h\_i(\boldsymbol{v}\_s^\varepsilon) = \prod\_{j=0}^{|H\_i|-1} \left( w\_i \cdot \boldsymbol{v}\_s^\varepsilon - j \right) - \prod\_{j=0}^{|H\_i|-1} \left( w\_i \cdot \boldsymbol{v}\_s^\varepsilon - j \right) = 0 \quad \text{for all} \ 1 \le i \le h.$$

Finally we consider *fi*(*v<sup>c</sup> <sup>s</sup>*). Since 1 ≤ |*Fi* ∩ *<sup>F</sup><sup>c</sup> <sup>s</sup>* | ≤ *<sup>k</sup>* − 1, by the hypothesis *fi*(*v<sup>c</sup> <sup>s</sup>*) is also 0 except for *fs*(*v<sup>c</sup> <sup>s</sup>*). Since *fs*(*v<sup>c</sup> <sup>s</sup>*) = 0, we get *α<sup>s</sup>* = 0 whenever *p* ∈ *Fs*.

So Equation (1) is reduced to :

$$\sum\_{i=1}^{\mathcal{S}} \beta\_i g\_i(\mathbf{x}) + \sum\_{i=1}^h \gamma\_i h\_i(\mathbf{x}) = 0 \tag{2}$$

Next, we substitute the characteristic vector *ws* of *Hs* in order of increasing size into Equation (2). Now we note that *p* ∈ *Hs*. Because of the (*xp* − 1) factor, we have *gi*(*ws*) = 0 for all 1 ≤ *i* ≤ *g*. Since the size of *Hi* is at most *<sup>k</sup>* − 1 for all *<sup>i</sup>*, we have 1 ≤ |*F<sup>c</sup> <sup>l</sup>* ∩ *Hs*| ≤ *<sup>k</sup>* − 1 for *<sup>p</sup>* ∈ *<sup>F</sup><sup>c</sup> <sup>l</sup>* . Thus, the factor ∏*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (*v<sup>c</sup> <sup>l</sup>* · *ws* <sup>−</sup> *<sup>j</sup>*) is 0. Similarly, the factor <sup>∏</sup>*k*−<sup>1</sup> *<sup>j</sup>*=<sup>1</sup> (*vl* · *ws* − *j*) is 0 for *p* ∈ *Fl*. Thus, we have *hi*(*ws*) = <sup>∏</sup>|*Hi*|−<sup>1</sup> *<sup>j</sup>*=<sup>0</sup> (*wi* · *ws* <sup>−</sup> *<sup>j</sup>*). Since *hs*(*ws*) <sup>=</sup> 0, and *hi*(*ws*) = 0 for *<sup>i</sup>* <sup>&</sup>gt; *<sup>s</sup>*, we have <sup>∑</sup>*<sup>h</sup> <sup>i</sup>*=<sup>1</sup> *γihi*(*ws*) = ∑*s <sup>i</sup>*=<sup>1</sup> *γihi*(*ws*).

Recall that we substitute the vector *ws* in order of increasing size. When we first plug *w*<sup>1</sup> into Equation (2), we have *γ*1*h*1(*w*1) = 0, and thus *γ*<sup>1</sup> = 0. Next, we plug *w*<sup>2</sup> into (2) after dropping *γ*1*h*1(*w*1) term from (2). Then we have *γ*2*h*2(*w*2) = 0, and thus *γ*<sup>2</sup> = 0. Similarly, we have *γ<sup>i</sup>* = 0 for all *i*.

Thus, Equation (1) becomes

$$\sum\_{i} \beta\_{i} \mathbf{g}\_{i}(\mathbf{x}) = 0. \tag{3}$$

Next, we substitute the characteristic vector *ys* of *Gs* in order of increasing size into Equation (3). Thus, we have

$$g\_i(y\_s) = (y\_{s\_p} - 1) \prod\_{j \in G\_i} y\_{s\_j} = - \prod\_{j \in G\_i} y\_{s\_j} \quad \text{for all} \ 1 \le i \le g.$$

Recall that we substitute the vector *ys* in order of increasing size. Please note that *gi*(0) is the empty product, which is taken to be 1. When we first plug *y*<sup>1</sup> into Equation (3), we have *g*1(*y*1) = 0 and *gi*(*y*1) = 0 for all *i* > 1, and thus *β*<sup>1</sup> = 0. Next, we plug *y*<sup>2</sup> into (3) after dropping *β*1*g*1(*x*) term from (3). Then we have *g*2(*y*2) = 0 and *gi*(*y*2) = 0 for all *i* > 2, and thus *β*<sup>2</sup> = 0. Similarly, we have *β<sup>i</sup>* = 0 for all *i*.

This concludes that all the polynomials *fi*(*x*), *gi*(*x*), and *hi*(*x*) are linearly independent. We found |F| + |G| + |H| linearly independent polynomials. All these polynomials are of degree less than or equal to *<sup>k</sup>* <sup>−</sup> 1. The space of these multilinear polynomials has dimension <sup>∑</sup>*k*−<sup>1</sup> *<sup>i</sup>*=<sup>0</sup> ( *n <sup>i</sup>*). We have

$$|\mathcal{F}| + |\mathcal{G}| + |\mathcal{H}| \le \sum\_{i=0}^{k-1} \binom{n}{i} \dots$$

Since |G| <sup>=</sup> <sup>∑</sup>*k*−<sup>2</sup> *<sup>i</sup>*=<sup>0</sup> ( *n*−1 *<sup>i</sup>* ) and |H| <sup>=</sup> <sup>∑</sup>*k*−<sup>2</sup> *<sup>i</sup>*=<sup>0</sup> ( *n*−1 *<sup>i</sup>* ), we have |F| <sup>+</sup> <sup>2</sup> <sup>∑</sup>*k*−<sup>2</sup> *<sup>i</sup>*=<sup>0</sup> ( *n*−1 *<sup>i</sup>* ) <sup>≤</sup> <sup>∑</sup>*k*−<sup>1</sup> *<sup>i</sup>*=<sup>0</sup> ( *n <sup>i</sup>*). This gives us

$$|\mathcal{F}| \le \binom{n-1}{k-1}.$$

finishing the proof of Theorem 6.

### **5. Conclusions**

We have answered the following question: when is it possible to get the same bound of the Erd˝os-Ko-Rado theorem for uniform intersecting families in the non-uniform intersecting families? Since the EKR-type bound for the non-uniform family of subsets of [*n*], which is ( *n*−1 *<sup>k</sup>*−1) <sup>+</sup> ( *n*−1 *<sup>k</sup>*−2) <sup>+</sup> ··· <sup>+</sup> ( *n*−1 <sup>0</sup> ), is much larger than ( *n*−1 *<sup>k</sup>*−1), this question is interesting and deserves further study.

Please note that if we can delete the condition (*c*) in Theorem 6, we can get the same bound of the Erd˝os-Ko-Rado theorem for *k*-uniform intersecting families under the same condition for non-uniform intersecting families of size at least *k* and at most *n* − *k*. Another intriguing question motivated by our result is the problem of getting the same bound of Theorem 6 without the condition (*c*) or finding a better bound for the non-uniform intersecting families than the previous results by the others.

**Author Contributions:** All authors have contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

**Funding:** The first author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2011-0025252). The second author was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education (2017R1A6A3A04005963).

**Acknowledgments:** All authors sincerely appreciate the reviewers for their valuable comments and suggestions to improve the paper.

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

### **References**


© 2020 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* **A Note on Parametric Kinds of the Degenerate Poly-Bernoulli and Poly-Genocchi Polynomials**

### **Taekyun Kim 1, Waseem A. Khan 2, Sunil Kumar Sharma 3,\* and Mohd Ghayasuddin <sup>4</sup>**


Received: 20 March 2020; Accepted: 2 April 2020; Published: 13 April 2020

**Abstract:** Recently, the parametric kind of some well known polynomials have been presented by many authors. In a sequel of such type of works, in this paper, we introduce the two parametric kinds of degenerate poly-Bernoulli and poly-Genocchi polynomials. Some analytical properties of these parametric polynomials are also derived in a systematic manner. We will be able to find some identities of symmetry for those polynomials and numbers.

**Keywords:** degenerate poly-Bernoulli polynomials; degenerate poly-Genocchi polynomials; stirling numbers

### **1. Introduction**

Special functions, polynomials and numbers play a prominent role in the study of many areas of mathematics, physics and engineering. In particular, the Appell polynomials and numbers are frequently used in the development of pure and applied mathematics related to functional equations in differential equations, approximation theories, interpolation problems, summation methods, quadrature rules and their multidimensional extensions (see [1] ).The sequence of Appell polynomials *Aj*(*z*) can be signified as follows:

$$\frac{d}{dz}A\_j(z) = jA\_{j-1}(z), \quad A\_0(z) \neq 0,\\ z = \eta + i\xi \in \mathbb{C}, \quad j \in \mathbb{N}.\tag{1}$$

or equivalently

$$A(z)e^{\eta z} = \sum\_{j=0}^{\infty} A\_j(\eta) \frac{z^j}{j!} \, ^{\prime} \, ^{\prime} \, ^{\prime} \, ^{\prime} \tag{2}$$

where

$$A(z) = A\_0 + A\_1 \frac{z}{1!} + A\_2 \frac{z^2}{2!} + \dots + A\_j \frac{z^j}{j!} + \dots , \ A\_0 \neq 0,$$

is a formal power series with coefficients *Aj* known as Appell numbers.

The well known degenerate exponential function is defined by (see [2])

$$
\varepsilon^{\eta}\_{\mu}(z) = (1 + \mu z)^{\frac{\eta}{\mu}}, \qquad \varepsilon\_{\mu}(z) = \varepsilon^{1}\_{\mu}(z), (\mu \in \mathbb{R}).\tag{3}
$$

In 1956 and 1979, Carlitz [3,4] introduced and investigated the following degenerate Bernoulli and Euler polynomials:

$$\frac{z}{e\_{\mu}(z) - 1} e\_{\mu}^{\eta}(z) = \frac{z}{(1 + \mu z)^{\frac{1}{\mu}} - 1} (1 + \mu z)^{\frac{\eta}{\mu}} = \sum\_{s=0}^{\infty} \beta\_s(\eta; \mu) \frac{z^s}{s!} \tag{4}$$

and

$$\frac{2}{\varepsilon\_{\mu}(z) + 1} \varepsilon\_{\mu}^{\eta}(z) = \frac{2}{(1 + \mu z)^{\frac{1}{\mu}} - 1} (1 + \mu z)^{\frac{\eta}{\mu}} = \sum\_{s=0}^{\infty} \mathfrak{E}\_{s}(\eta; \mu) \frac{z^{s}}{s!}. \tag{5}$$

Note that

$$\lim\_{\mu \longrightarrow 0} \beta\_s(\eta; \mu) = B\_s(\eta), \quad \lim\_{\mu \longrightarrow 0} \mathfrak{E}\_s(\eta; \mu) = E\_s(\eta),$$

where *Bs*(*η*) and *Es*(*η*) are the classical Bernoulli and Euler polynomials (see [5,6]).

Lim [7] introduced the degenerate Genocchi polynomials *<sup>G</sup>*(*p*) *<sup>j</sup>* (*η*; *μ*) of order *p* by means of the undermentioned generating function:

$$\left(\frac{2z}{\varepsilon\_{\mu}(z)+1}\right)^{p}\varepsilon\_{\mu}^{\eta}(z) = \left(\frac{2z}{(1+\mu z)^{\frac{1}{p}}-1}\right)^{p}(1+\mu z)^{\frac{\eta}{p}} = \sum\_{j=0}^{\infty}G\_{j}^{(p)}(\eta;\mu)\frac{z^{j}}{j!},\tag{6}$$

so that

$$\mathcal{G}\_{\dot{\boldsymbol{\eta}}}^{(p)}(\boldsymbol{\eta};\boldsymbol{\mu}) = \sum\_{s=0}^{\dot{\boldsymbol{\eta}}} \binom{\dot{\boldsymbol{\eta}}}{s} \mathcal{G}\_{s}^{(p)}(\boldsymbol{\mu}) \left(\frac{\boldsymbol{\eta}}{\mu}\right)\_{\dot{\boldsymbol{\eta}}-\boldsymbol{s}}.\tag{7}$$

From Equation (6), we note that

$$\lim\_{\mu \longrightarrow 0} \sum\_{s=0}^{\infty} G\_j^{(p)}(\eta; \mu) \frac{z^j}{j!} = \lim\_{\mu \longrightarrow 0} \left( \frac{2z}{(1 + \mu z)^{\frac{1}{p}} - 1} \right)^p (1 + \mu z)^{\frac{p}{p}}$$

$$= \left( \frac{2z}{e^z + 1} \right)^p e^{\eta z} = \sum\_{j=0}^{\infty} G\_j^{(p)}(\eta) \frac{z^j}{j!}$$

where *<sup>G</sup>*(*p*) *<sup>j</sup>* (*η*) are the generalized Genocchi polynomials of order *p* (see [8–11]).

The degenerate poly-Bernoulli and poly-Genocchi polynomials are defined by (see [12–14])

$$\frac{\operatorname{Li}\_k(1-e^{-z})}{e\_\mu(z)-1}e\_\mu^\eta(z) = \frac{\operatorname{Li}\_k(1-e^{-z})}{(1+\mu z)^{\frac{1}{p}}-1}(1+\mu z)^{\frac{\eta}{p}} = \sum\_{s=0}^\infty B\_s^{(k)}(\eta;\mu)\frac{z^s}{s!},(k\in\mathbb{Z}),\tag{8}$$

and

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{e\_\mu(z)+1}e\_\mu^\eta(z) = \frac{2\operatorname{Li}\_k(1-e^{-z})}{(1+\mu z)^{\frac{1}{\lambda}}+1}(1+\mu z)^{\frac{\eta}{\mu}} = \sum\_{s=0}^\infty G\_s^{(k)}(\eta;\mu)\frac{z^s}{s!},(k\in\mathbb{Z}).\tag{9}$$

Here, we note that (see [5,15]).

$$\lim\_{\mu \longrightarrow 0} B\_s^{(k)}(\eta;\mu) = B\_s^{(k)}(\eta), \quad \lim\_{\mu \stackrel{-}{\longrightarrow} 0} G\_s^{(k)}(\eta;\mu) = G\_s^{(k)}(\eta), \mu$$

The Stirling numbers of the first kind are given by (see, [16–18])

$$s(a)\_s = a(a-1)\cdots(a-s+1) = \sum\_{k=0}^{s} S^{(1)}(s,k)a^k, (k \ge 0), \tag{10}$$

and the Stirling numbers of the second kind are defined by (see [19,20])

$$a^s = \sum\_{k=0}^s S^{(2)}(k, s)(a)\_k. \tag{11}$$

The degenerate Stirling numbers of the of the second kind are defined by (see [10,21,22])

$$\frac{1}{k!} (\varepsilon\_{\mu}(t) - 1)^k = \sum\_{k=s}^{\infty} S\_{\mu}^{(2)}(k, s) \frac{t^k}{k!}, (k \ge 0). \tag{12}$$

Note that lim*μ*−→<sup>0</sup> *<sup>S</sup>*(2) *<sup>μ</sup>* (*k*,*s*) = *<sup>S</sup>*(2)(*k*,*s*),(*s*, *<sup>k</sup>* <sup>≥</sup> <sup>0</sup>).

In the year (2017, 2018), Jamei et al. [23,24] introduced the two parametric kinds of exponential functions as follows (see also [6,23–25]):

$$\epsilon^{\eta z}\cos\xi z = \sum\_{k=0}^{\infty} \mathcal{C}\_k(\eta\_\prime \xi) \frac{z^k}{k!} \tag{13}$$

and

$$
\epsilon^{\eta z} \sin \zeta z = \sum\_{k=0}^{\infty} S\_k(\eta, \zeta) \frac{z^k}{k!} \tag{14}
$$

where

$$\mathbb{C}\_{k}(\eta,\xi) = \sum\_{j=0}^{\left[\frac{k}{2}\right]} \binom{k}{2j} (-1)^{j} \eta^{k-2j} \xi^{2j} \,\_{\prime} \tag{15}$$

and

$$S\_k(\eta, \xi) = \sum\_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} (-1)^j \eta^{k-2j-1} \xi^{2j+1}.\tag{16}$$

Recently, Kim et al. [2] introduced the following degenerate type parametric exponential functions:

$$e^{\eta}\_{\mu}(z)\cos^{\overline{\varsigma}}\_{\mu}(z) = \sum\_{k=0}^{\infty} \mathbb{C}\_{k,\mu}(\eta,\xi)\frac{z^k}{k!},\tag{17}$$

and

$$e^{\eta}\_{\mu}(z)\sin^{\overline{\xi}}\_{\mu}(z) = \sum\_{k=0}^{\infty} \mathcal{S}\_{k,\mu}(\eta\_{\prime}\overline{\xi})\frac{z^{k}}{k!},\tag{18}$$

where

$$\mathbb{C}\_{r,\mu}(\eta,\xi) = \sum\_{k=0}^{\left[\frac{r}{2}\right]} \sum\_{q=2k}^{r} \binom{r}{q} (-1)^{k} \mu^{q-2k} \xi^{2k} S^{1}(q, 2k) (\eta)\_{r-q,\mu} \tag{19}$$

and

$$S\_{r, \mu}(\eta, \xi) = \sum\_{k=0}^{\left[\frac{r-1}{2}\right]} \sum\_{q=2k+1}^{r} \binom{r}{q} (-1)^k \mu^{q-2k-1} \xi^{2k+1} S^1(q, 2k+1)(\eta)\_{r-q, \mu}. \tag{20}$$

Motivated by the importance and potential applications in certain problems in number theory, combinatorics, classical and numerical analysis and physics, several families of degenerate Bernoulli and Euler polynomials and degenerate versions of special polynomials have been recently studied

by many authors, (see [3–5,11–13,16]). Recently, Kim and Kim [2] have introduced the degenerate Bernoulli and degenerate Euler polynomials of a complex variable. By separating the real and imaginary parts, they introduced the parametric kinds of these degenerate polynomials.

The main object of this article is to present the parametric kinds of degenerate poly-Bernoulli and poly-Genocchi polynomials in terms of the degenerate type parametric exponential functions. We also investigate some fundamental properties of our introduced parametric polynomials.

### **2. Parametric Kinds of the Degenerate Poly-Bernoulli Polynomials**

In this section, we define the two parametric kinds of degenerate poly-Bernoulli polynomials by means of the two special generating functions involving the degenerate exponential as well as trigonometric functions.

It is well known that (see [2])

$$\epsilon^{(\eta+i\vec{z})z} = \epsilon^{\eta z}\epsilon^{i\vec{\xi}z} = \epsilon^{\eta z}(\cos\vec{\zeta}z + i\sin\vec{\zeta}z),\tag{21}$$

The degenerate trigonometric functions are defined by (see [19])

$$\cos\_{\mu} z = \frac{e^i\_{\mu}(z) + e^{-i}\_{\mu}(z)}{2}, \quad \sin\_{\mu} z = \frac{e^i\_{\mu}(z) - e^{-i}\_{\mu}(z)}{2i}. \tag{22}$$

Note that, we have

$$\lim\_{\mu \to 0} \cos\_{\mu} z = \cos z, \quad \lim\_{\mu \to 0} \sin\_{\mu} z = \sin z.$$

In view of Equation (8), we have

$$\frac{\operatorname{Li}\_k(1-e^{-z})}{\operatorname{c}\_{\mu}(z)-1} \epsilon\_{\mu}^{\eta+i\frac{x}{\eta}}(z) = \sum\_{j=0}^{\infty} B\_{j,\mu}^{(k)}(\eta+i\xi) \frac{z^j}{j!} \,. \tag{23}$$

and

$$\frac{\operatorname{Li}\_k(1-\epsilon^{-z})}{\epsilon\_\mu(z)-1}\epsilon\_\mu^{\eta-i\frac{x}{z}}(z) = \sum\_{j=0}^\infty B\_{j,\mu}^{(k)}(\eta-i\xi)\frac{z^j}{j!}.\tag{24}$$

From Equations (23) and (24), we note that

$$\frac{\operatorname{Li}\_k(1-e^{-z})}{\varepsilon\_{\mu}(z)-1}e^{\eta}\_{\mu}(z)\cos^{\overline{\zeta}}\_{\mu}(z) = \sum\_{j=0}^{\infty} \left( \frac{B^{(k)}\_{j,\mu}(\eta+i\xi) + B^{(k)}\_{j,\mu}(\eta-i\xi)}{2} \right) \frac{z^j}{j!} \tag{25}$$

and

$$\frac{\operatorname{Li}\_k(1-e^{-z})}{\varepsilon\_{\mu}(z)-1}e^{\eta}\_{\mu}(z)\sin^{\frac{\pi}{\mu}}\_{\mu}(z) = \sum\_{j=0}^{\infty} \left( \frac{B^{(k)}\_{j,\mu}(\eta+i\xi)-B^{(k)}\_{j,\mu}(\eta-i\xi)}{2i} \right) \frac{z^j}{j!}.\tag{26}$$

**Definition 1.** *The degenerate cosine-poly-Bernoulli polynomials <sup>B</sup>*(*k*,*c*) *<sup>p</sup>*,*<sup>μ</sup>* (*η*, *ξ*) *and degenerate sine-poly-Bernoulli polynomials B*(*k*,*s*) *<sup>p</sup>*,*<sup>μ</sup>* (*η*, *ξ*) *for nonnegative integer p are defined, respectively, by*

$$\frac{\operatorname{Li}\_k(1-\varepsilon^{-z})}{e\_\mu(z)-1} \varepsilon^{\eta}\_{\mu}(z) \cos^{\overline{\xi}}\_{\mu}(z) = \sum\_{p=0}^{\infty} B^{(k,c)}\_{p,\mu}(\eta,\xi) \frac{z^p}{p!},\tag{27}$$

*and*

$$\frac{\operatorname{Li}\_k(1-\varepsilon^{-z})}{\varepsilon\_{\mu}(z)-1} \varepsilon\_{\mu}^{\eta}(z) \operatorname{sin}\_{\mu}^{\overline{z}}(z) = \sum\_{p=0}^{\infty} B\_{p,\mu}^{(k,s)}(\eta,\xi) \frac{z^p}{p!}.\tag{28}$$

*For η* = *ξ* = 0 *in Equations (27) and (28), we get*

$$B\_{p,\mu}^{(k,\varepsilon)}(0,0) = B\_{p,\mu\prime}^{(k)} B\_{p,\mu}^{(k,\varepsilon)}(0,0) = 0, (p \ge 0).$$

*Note that* lim*μ*−→<sup>0</sup> *<sup>B</sup>*(*k*,*c*) *<sup>p</sup>*,*<sup>μ</sup>* (*η*, *<sup>ξ</sup>*) = *<sup>B</sup>*(*k*,*c*) *<sup>p</sup>* (*η*, *<sup>ξ</sup>*)*,* lim*μ*−→<sup>0</sup> *<sup>B</sup>*(*k*,*s*) *<sup>p</sup>*,*<sup>μ</sup>* (*η*, *<sup>ξ</sup>*) = *<sup>B</sup>*(*k*,*s*) *<sup>p</sup>* (*η*, *ξ*)*,* (*p* ≥ 0)*, where B*(*k*,*c*) *<sup>p</sup>* (*η*, *<sup>ξ</sup>*) *and B*(*k*,*s*) *<sup>p</sup>* (*η*, *ξ*) *are the new type of poly-Bernoulli polynomials.*

*Based on Equations (25)–(28), we determine*

$$B\_{p,\mu}^{(k,\varepsilon)}(\eta,\xi) = \frac{B\_{p,\mu}^{(k)}(\eta + i\xi) + B\_{p,\mu}^{(k)}(\eta - i\xi)}{2},\tag{29}$$

*and*

$$B\_{p,\mu}^{(k,\mathfrak{s})}(\eta,\mathfrak{z}) = \frac{B\_{p,\mu}^{(k)}(\eta + i\mathfrak{z}^{\mathfrak{x}}) - B\_{p,\mu}^{(k)}(\eta - i\mathfrak{z}^{\mathfrak{x}}\_{\mathfrak{z}})}{2i}.\tag{30}$$

**Theorem 1.** *Let k* ∈ Z *and j* ≥ 0*. Then*

$$B\_{j,\mu}^{(k)}(\eta+i\xi) = \sum\_{q=0}^{j} \binom{j}{q} B\_{j-q,\mu}^{(k)}(\eta) (i\xi)\_{q,\mu}$$

$$= \sum\_{q=0}^{j} \binom{j}{q} B\_{j-q,\mu}^{(k)}(\eta+i\xi)\_{q,\mu} \tag{31}$$

*and*

$$B\_{j,\mu}^{(k)}(\eta - i\xi^x\_\circ) = \sum\_{q=0}^j \binom{j}{q} B\_{j-q,\mu}^{(k)}(\eta)(-1)^q (i\xi)\_{q,\mu}$$

$$= \sum\_{q=0}^j \binom{j}{q} B\_{j-q,\mu}^{(k)}(\eta - i\xi^x\_\circ)\_{q,\mu}.\tag{32}$$

**Proof.** From Equation (23), we have

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k)}(\eta + i\xi) \frac{z^j}{j!} = \frac{\operatorname{Li}\_k(1 - e^{-z})}{e\_\mu(z) - 1} e\_\mu^\eta(z) e\_{\mu}^{i\xi}(z)$$

$$= \left( \sum\_{j=0}^{\infty} B\_{j,\mu}^{(k)}(\eta) \frac{z^j}{j!} \right) \left( \sum\_{q=0}^{\infty} (i\xi)\_{q,\mu} \frac{z^q}{q!} \right)$$

$$= \sum\_{j=0}^{\infty} \left( \sum\_{q=0}^{j} \binom{j}{q} B\_{j-q,\mu}^{(k)}(\eta) (i\xi)\_{q,\mu} \right) \frac{z^j}{j!} . \tag{33}$$

Similarly, we find

$$\frac{\operatorname{Li}\_k(1-e^{-z})}{\epsilon\_\mu(z)-1} \epsilon\_\mu^\eta(z) \epsilon\_\mu^{j\overline{j}}(z) = \left(\sum\_{j=0}^\infty B\_{j,\mu}^{(k)} \frac{z^j}{j!} \right) \left(\sum\_{q=0}^\infty (\eta+i\xi^q)\_{q,\mu} \frac{z^q}{q!} \right)$$

$$=\sum\_{j=0}^\infty \left(\sum\_{q=0}^j \binom{j}{q} B\_{j-q,\mu}^{(k)} (\eta+i\xi^q)\_{q,\mu} \right) \frac{z^j}{j!}.\tag{34}$$

In view of Equations (33) and (34), we obtain our first claimed result shown in Equation (31). Similarly, we can establish our second result shown in Equation (32).

**Theorem 2.** *The following results hold true:*

$$B\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} B\_{r,\mu}^{(k)} \, \mathbb{C}\_{j-r,\mu}(\eta,\xi)$$

$$= \sum\_{r=0}^{\lfloor \frac{q}{2} \rfloor} \sum\_{q=2r}^{j} \binom{j}{q} \, \mu^{q-2r}(-1)^r \xi^{2r} S^{(1)}(q,2r) B\_{j-q,\mu}^{(k)}(\eta),\tag{35}$$

*and*

$$B\_{j,\mu}^{(k,\mathfrak{s})}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} B\_{r,\mu}^{(k)} S\_{j-r,\mu}(\eta,\xi)$$

$$\xi = \sum\_{r=0}^{\lfloor \frac{q-1}{2} \rfloor} \sum\_{q=2r+1}^{j} \binom{j}{q} \mu^{q-2r-1}(-1)^r \xi^{2r+1} S^{(1)}(q, 2r+1) B\_{j-q,\mu}^{(k)}(\eta). \tag{36}$$

**Proof.** From Equations (27) and (17), we see

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \frac{\text{Li}\_k(1-e^{-z})}{e\_\mu(z)-1} e\_\mu^\eta(z) \cos\_\mu^\xi(z)$$

$$= \left(\sum\_{r=0}^{\infty} B\_{r,\mu}^{(k)} \frac{z^r}{r!} \right) \left(\sum\_{j=0}^{\infty} C\_{j,\mu}(\eta,\xi) \frac{z^j}{j!} \right)$$

$$= \sum\_{j=0}^{\infty} \left(\sum\_{r=0}^j \binom{j}{r} B\_{r,\mu}^{(k)} C\_{j-r,\mu}(\eta,\xi) \right) \frac{z^j}{j!} . \tag{37}$$

Now, by using Equations (27) and (10), we find

$$\frac{\operatorname{Li}\_{k}(1-e^{-z})}{e\_{\mu}(z)-1}e\_{\mu}^{\eta}(z)\cos\_{\mu}^{\xi}(z) = \sum\_{j=0}^{\infty}B\_{j,\mu}^{(k)}(\eta)\frac{z^{j}}{j!}\sum\_{p=0}^{\infty}\sum\_{r=0}^{\left[\frac{q}{2}\right]}\mu^{l-2r}(-1)^{r}\mathcal{J}^{2r}S^{(1)}(q,2r)\frac{z^{r}}{r!}$$

$$=\sum\_{j=0}^{\infty}\left(\sum\_{q=0}^{j}\sum\_{r=0}^{\left[\frac{q}{2}\right]}\binom{j}{q}\mu^{q-2r}(-1)^{r}\xi^{2r}S^{(1)}(q,2r)B\_{j-r,\mu}^{(k)}(\eta)\right)\frac{z^{j}}{j!}$$

$$=\sum\_{j=0}^{\infty}\left(\sum\_{r=0}^{\left[\frac{q}{2}\right]}\sum\_{q=-2r}^{j}\binom{j}{q}\mu^{q-2r}(-1)^{r}\xi^{2r}S^{(1)}(q,2r)B\_{j-q,\mu}^{(k)}(\eta)\right)\frac{z^{j}}{j!}.\tag{38}$$

Therefore, from Equations (37) and (38), we attain our needed result, Equation (35). Similarly, we can obtain Equation (36).

**Theorem 3.** *Each of the following identities holds true:*

$$B\_{r,\mu}^{(2,c)}(\eta,\xi) = \sum\_{q=0}^{r} \binom{r}{q} \frac{q! B\_q}{q+1} B\_{r-q,\mu}^{(c)}(\eta,\xi),\tag{39}$$

*and*

$$B\_{r,\mu}^{(2,s)}\left(\eta,\xi\right) = \sum\_{q=0}^{r} \binom{r}{q} \frac{q! B\_q}{q+1} B\_{r-q,\mu}^{(s)}(\eta,\xi) \,. \tag{40}$$

**Proof.** In view of Equation (27), we have

$$\sum\_{r=0}^{\infty} B\_{r, \mu}^{(k, c)}(\eta, \xi) \frac{z^r}{r!} = \frac{\text{Li}\_k(1 - e^{-z})}{\text{e}\_{\mu}(z) - 1} e^{\eta}\_{\mu}(z) \cos\_{\mu}^{\overline{z}}(z)$$

$$= \frac{e^{\eta}\_{\mu}(z) \cos\_{\mu}^{\overline{z}}(z)}{e\_{\mu}(z) - 1} \int\_{0}^{z} \underbrace{\frac{1}{e^{\mu} - 1} \int\_{0}^{u} \frac{1}{e^{u} - 1} \cdot \dots \cdot \frac{1}{e^{u} - 1} \int\_{0}^{u} \frac{u}{e^{u} - 1} du}\_{(k - 1) \cdots \text{times}} du \cdot \dots \, du. \tag{41}$$

Upon setting *k* = 2, we obtain

$$\sum\_{r=0}^{\infty} B\_{r,\mu}^{(2,c)}(\eta,\xi) \frac{z^r}{r!} = \frac{e\_{\mu}^{\eta}(z) \cos\_{\mu}^{\xi}(z)}{e\_{\mu}(z) - 1} \int\_0^z \frac{u}{e^u - 1} du$$

$$= \left( \sum\_{q=0}^{\infty} \frac{B\_q z^q}{(q+1)} \right) \frac{e\_{\mu}^{\eta}(z) \cos\_{\mu}^{\xi}(z)}{e\_{\mu}(z) - 1}$$

$$= \left( \sum\_{q=0}^{\infty} \frac{q! B\_q z^q}{(q+1) q!} \right) \left( \sum\_{r=0}^{\infty} B\_{r,\mu}^{(c)}(\eta, \xi) \frac{z^r}{r!} \right).$$

$$= \sum\_{r=0}^{\infty} \sum\_{q=0}^{r} \left( \begin{array}{c} r \\ q \end{array} \right) \frac{q! B\_q}{q+1} B\_{r-q,\mu}^{(c)}(\eta, \xi) \frac{z^r}{r!}.$$

which gives our required result, Equation (39). The proof of Equation (40) is similar; therefore, we omit the proof.

**Theorem 4.** *Let k* ∈ Z*, then*

$$B\_{j,\mu}^{(k,\varepsilon)}(\eta\_{\prime}\xi) = \sum\_{r=0}^{j} \binom{j}{r} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1} l! S\_2(r+1,q)}{q^k (r+1)} \right) B\_{j-r,\mu}^{(\varepsilon)}(\eta\_{\prime}\xi),\tag{42}$$

*and*

$$B\_{j,\mu}^{(k,s)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1} q! S\_2(r+1,q)}{q^k (r+1)} \right) B\_{j-r,\mu}^{(s)}(\eta,\xi). \tag{43}$$

**Proof.** From Equations (27) and (11), we see

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \left(\frac{\operatorname{Li}\_k(1-e^{-z})}{z}\right) \left(\frac{z e\_{\mu}^{\eta}(z) \cos\_{\mu}^{\frac{\pi}{2}}(z)}{e\_{\mu}(z) - 1}\right). \tag{44}$$

Now

$$\frac{1}{z} \text{Li}\_k(1 - e^{-z}) = \frac{1}{z} \sum\_{q=1}^{\infty} \frac{(1 - e^{-z})^q}{q^k}$$

$$= \frac{1}{z} \sum\_{q=1}^{\infty} \frac{(-1)^q}{q^k} q! \sum\_{r=l}^{\infty} (-1)^r S\_2(r, q) \frac{z^r}{r!}$$

$$= \frac{1}{z} \sum\_{r=q}^{\infty} \sum\_{q=1}^{r} \frac{(-1)^{q+r}}{q^k} q! S\_2(r, q) \frac{z^r}{r!}$$

$$\hat{\lambda} = \sum\_{r=0}^{\infty} \left( \sum\_{q=1}^{q+1} \frac{(-1)^{q+r+1}}{q^k} l! \frac{\mathcal{S}\_2(r+1, q)}{r+1} \right) \frac{z^r}{r!}. \tag{45}$$

On using Equation (45) in (44), we find

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,c)}(\eta,\xi) \frac{z^j}{j!} = \sum\_{r=0}^{\infty} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1}}{q^k} l! \frac{S\_2(r+1,q)}{r+1} \right) \frac{z^r}{r!} \left( \sum\_{j=0}^{\infty} B\_{j,\mu}^{(c)}(\eta,\xi) \frac{z^j}{j!} \right).$$

Replacing *<sup>j</sup>* by *<sup>j</sup>* − *<sup>r</sup>* in the right side of above expression and after equating the coefficients of *<sup>z</sup><sup>j</sup>* , we obtain our needed result, Equation (42). Similarly, we can derive our second result, Equation (43).

**Theorem 5.** *The following recurrence relation holds true:*

$$B\_{j,\mu}^{(k,c)}\left(\eta+1,\xi\right)-B\_{j,\mu}^{(k,c)}\left(\eta,\xi\right)$$

$$\xi = \sum\_{r=1}^{j} \binom{j}{r} \left(\sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1)\right) \mathbb{C}\_{j-r,\mu}(\eta,\xi),\tag{46}$$

*and*

$$B\_{j,\mu}^{(k,s)}\left(\eta+1,\xi\right)-B\_{j,\mu}^{(k,s)}\left(\eta,\xi\right)$$

$$\xi = \sum\_{r=1}^{j} \binom{j}{r} \left(\sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1)\right) S\_{j-r,\mu}(\eta,\xi). \tag{47}$$

**Proof.** In view of Equation (27), we have

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,\varepsilon)}(\eta+1,\varepsilon) \frac{z^j}{j!} - \sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,\varepsilon)}(\eta,\varepsilon) \frac{z^j}{j!}$$

$$= \frac{\text{Li}\_k(1-e^{-z})}{e\_\mu(z)-1} e\_\mu^{(q+1)}(z) \cos\_\mu^\xi(z) - \frac{\text{Li}\_k(1-e^{-z})}{e\_\mu(z)-1} e\_\mu^{(q)}(z) \cos\_\mu^\xi(z)$$

$$= \text{Li}\_k(1-e^{-z}) e\_\mu^{(q)}(z) \cos\_\mu^\xi(z)$$

$$= \sum\_{q=0}^{\infty} \frac{(1-e^{-z})^{q+1}}{(q+1)^k} e\_\mu^{(q)}(z) \cos\_\mu^\xi(z)$$

$$= \sum\_{r=1}^{\infty} \left( \sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1) \right) \frac{z^r}{r!} e\_\mu^{(q)}(z) \cos\_\mu^\xi(z)$$

$$= \left( \sum\_{r=1}^{\infty} \left( \sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1) \right) \frac{z^r}{r!} \right) \left( \sum\_{j=0}^{\infty} C\_{j,\mu}(q, \xi) \frac{z^j}{j!} \right)\_r$$

which upon replacing *j* by *j* − *r* in the right side of above expression and after equating the coefficients of *z<sup>j</sup>* , yields our first claimed result, Equation (46). Similarly, we can establish our second result, Equation (47).

**Theorem 6.** *Let k* ∈ Z *and j* ≥ 0*, then we have*

$$B\_{j,\mu}^{(k,c)}(\eta+\gamma,\xi) = \sum\_{r=0}^{j} \binom{j}{r} B\_{j-r,\mu}^{(k,c)}(\eta,\xi) (\gamma)\_{r,\mu} \tag{48}$$

*and*

$$B\_{j,\mu}^{(k,s)}\left(\eta+\gamma,\xi\right) = \sum\_{r=0}^{j} \binom{j}{r} B\_{j-r,\mu}^{(k,s)}(\eta,\xi) \left(\gamma\right)\_{r,\mu}.\tag{49}$$

**Proof.** On using Equation (27), we find

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,c)}(\eta+\gamma,\xi) \frac{z^j}{j!} = \frac{\operatorname{Li}\_k(1-e^{-z})}{\operatorname{c}\_{\mu}(z)-1} \operatorname{e}\_{\mu}^{(\eta+\gamma)}(z) \cos\_{\mu}^{\xi}(z)$$

$$= \left(\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,c)}(\eta,\xi) \frac{z^j}{j!} \right) \left(\sum\_{r=0}^{\infty} (\gamma)\_{r,\mu} \frac{z^r}{r!} \right)$$

$$= \sum\_{j=0}^{\infty} \left(\sum\_{r=0}^{j} \binom{j}{r}\_{r} B\_{j-r,\mu}^{(k,c)}(\eta,\xi) (\gamma)\_{r,\mu} \right) \frac{z^j}{j!}.$$

By comparing the coefficients of *z<sup>j</sup>* on both sides, we obtain the result, Equation (48). The proof of Equation (49) is similar to Equation (48).

**Theorem 7.** *If k* ∈ Z *and j* ≥ 0*, then*

$$B\_{j,\mu}^{(k,\varepsilon)}\left(\eta,\xi\right) = \sum\_{r=0}^{j} \sum\_{q=0}^{r} \binom{j}{r} \left(\eta\right) {}\_{q}\xi\_{\mu}^{(2)}\left(r,\eta\right)B\_{j-r,\mu}^{(k,\varepsilon)}\left(0,\xi\right),\tag{50}$$

*and*

$$B\_{j,\mu}^{(k,\varepsilon)}\left(\eta,\xi\right) = \sum\_{r=0}^{j} \sum\_{q=0}^{r} \binom{j}{r} \left(\eta\right)\_{q} S\_{\mu}^{(2)}\left(r,q\right) B\_{j-r,\mu}^{(k,\varepsilon)}\left(0,\xi\right). \tag{51}$$

**Proof.** From Equations (27) and (12), we find

$$\sum\_{j=0}^{\infty} B\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \frac{\text{Li}\_k(1-e^{-z})}{e\_\mu(z)-1} (\epsilon\_\mu(z) - 1+1)^\eta \cos\_\mu^\xi(z)$$

$$= \frac{\text{Li}\_k(1-e^{-z})}{e\_\mu(z)-1} \sum\_{q=0}^\infty \binom{\eta}{q} (\epsilon\_\mu(z) - 1)^q \cos\_\mu^\xi(z)$$

$$= \frac{\text{Li}\_k(1-e^{-z})}{e\_\mu(z)-1} \cos\_\mu^\xi(z) \sum\_{q=0}^\infty (\eta)\_q \sum\_{r=q}^\infty S\_\mu^{(2)}(r,q) \frac{z^r}{r!}$$

$$= \sum\_{j=0}^\infty B\_{j,\mu}^{(k,\varepsilon)}(0,\xi) \frac{z^j}{j!} \sum\_{r=0}^\infty \left(\sum\_{q=0}^r (\eta)\_q S\_\mu^{(2)}(r,q)\right) \frac{z^r}{r!}$$

$$= \sum\_{j=0}^\infty \left(\sum\_{r=0}^j \sum\_{q=0}^r \left(\frac{j}{r}\right)\_q (\eta)\_q S\_\mu^{(2)}(r,q B\_{j-r,\mu}^{(k,\varepsilon)}(0,\xi)) \right) \frac{z^j}{j!}$$

On comparing the coefficients of *z<sup>j</sup>* on both sides, we obtain our required result, Equation (50). The proof of Equation (51) is similar to Equation (50).

#### **3. Parametric Kinds of Degenerate Poly-Genocchi Polynomials**

In this section, we introduce the two parametric kinds of degenerate poly-Genocchi polynomials by defining the two special generating functions involving the degenerate exponential as well as trigonometric functions.

In view of Equation (9), we have

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{\varepsilon\_{\mu}(z)+1}e^{\eta+i\xi}\_{\mu}(z) = \sum\_{j=0}^{\infty} G^{(k)}\_{j,\mu}(\eta+i\xi)\frac{z^j}{j!} \tag{52}$$

and

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{e\_\mu(z)+1}e^{\eta-i\frac{x}{\mu}}(z) = \sum\_{j=0}^\infty G\_{j,\mu}^{(k)}(\eta-i\xi)\frac{z^j}{j!}.\tag{53}$$

From Equations (52) and (53), we can easily get

$$\frac{2\operatorname{Li}\_k(1-\varepsilon^{-z})}{\varepsilon\_{\mu}(z)+1}e\_{\mu}^{\eta}(z)\cos\_{\mu}^{\overline{z}}(z)=\sum\_{j=0}^{\infty}\left(\frac{G\_{j,\mu}^{(k)}(\eta+i\xi)+G\_{j,\mu}^{(k)}(\eta-i\xi)}{2}\right)\frac{\tau^j}{j!},\tag{54}$$

and

$$\frac{2\operatorname{Li}\_{\mathrm{k}}(1-e^{-z})}{\varepsilon\_{\mu}(z)+1}e^{\eta}\_{\mu}(z)\sin^{\mathbb{E}}\_{\mu}(z)=\sum\_{j=0}^{\infty}\left(\frac{G^{(k)}\_{j,\mu}(\eta+i\xi)-G^{(k)}\_{j,\mu}(\eta-i\xi)}{2i}\right)\frac{\mathbb{Z}^{j}}{j!}.\tag{55}$$

**Definition 2.** *The degenerate cosine-poly-Genocchi polynomials <sup>G</sup>*(*k*,*c*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *ξ*) *and degenerate sine-poly-Genocchi polynomials G*(*k*,*s*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *ξ*) *for nonnegative integer j are defined, respectively, by*

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{e\_\mu(z)+1}e\_\mu^\eta(z)\cos\_\mu^\xi(z) = \sum\_{j=0}^\infty G\_{j,\mu}^{(k,\epsilon)}(\eta,\xi)\frac{z^j}{j!},\tag{56}$$

*and*

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{\varepsilon\_{\mu}(z)+1}\varepsilon\_{\mu}^{\eta}(z)\sin\_{\mu}^{\overline{z}}(z)=\sum\_{j=0}^{\infty}G\_{j,\mu}^{(k,s)}(\eta,\xi)\frac{z^j}{j!}.\tag{57}$$

*On setting η* = *ξ* = 0 *in Equations (56) and (57), we get*

$$G\_{j,\mu}^{(k,\iota)}(0,0) = G\_{j,\mu}^{(k)} \quad G\_{j,\mu}^{(k,\iota)}(0,0) = 0, (j \ge 0).$$

*Note that* lim*μ*−→<sup>0</sup> *<sup>G</sup>*(*k*,*c*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *<sup>ξ</sup>*) = *<sup>G</sup>*(*k*,*c*) *<sup>j</sup>* (*η*, *<sup>ξ</sup>*)*,* lim*μ*−→<sup>0</sup> *<sup>G</sup>*(*k*,*s*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *<sup>ξ</sup>*) = *<sup>G</sup>*(*k*,*s*) *<sup>j</sup>* (*η*, *ξ*)*,* (*j* ≥ 0)*, where <sup>G</sup>*(*k*,*c*) *<sup>n</sup>* (*η*, *<sup>ξ</sup>*) *and G*(*k*,*s*) *<sup>j</sup>* (*η*, *ξ*) *are the new type of poly-Genocchi polynomials.*

*From Equations (54)–(57), we determine*

$$G\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) = \frac{G\_{j,\mu}^{(k)}(\eta+i\xi) + G\_{j,\mu}^{(k)}(\eta-i\xi)}{2} \tag{58}$$

*and*

$$G\_{j,\mu}^{(k,\flat)}(\eta,\xi) = \frac{G\_{j,\mu}^{(k)}(\eta+i\xi) - G\_{j,\mu}^{(k)}(\eta-i\xi)}{2i}.\tag{59}$$

**Theorem 8.** *For k* ∈ Z *and j* ≥ 0*, we have*

$$G\_{j,\mu}^{(k)}(\eta+i\xi) = \sum\_{q=0}^{j} \binom{j}{q} \, G\_{j-q,\mu}^{(k)}(\eta) (i\xi)\_{q,\mu}$$

$$=\sum\_{q=0}^{j} \binom{j}{q} \, \mathrm{G}\_{j-q,\mu}^{(k)} \, (\eta + i\xi^{\mu})\_{q,\mu} \,. \tag{60}$$

*and*

$$G\_{j, \mu}^{(k)}(\eta - i\xi) = \sum\_{q=0}^{j} \binom{j}{q} G\_{j-q, \mu}^{(k)}(\eta) (-1)^q (i\xi)\_{q, \mu}$$

$$= \sum\_{q=0}^{j} \binom{j}{q} G\_{j-q, \mu}^{(k)}(\eta - i\xi)\_{q, \mu}. \tag{61}$$

**Proof.** On using Equation (52), we see

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k)}(\eta + i\xi) \frac{z^j}{j!} = \frac{2\operatorname{Li}\_k(1-\epsilon^{-z})}{e\_{\mu}(z)+1} c\_{\mu}^{\eta}(z) c\_{\mu}^{i\xi}(z)$$

$$= \left(\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k)}(\eta) \frac{z^j}{j!} \right) \left(\sum\_{q=0}^{\infty} (i\xi)\_{q,\mu} \frac{z^q}{q!} \right)$$

$$= \sum\_{j=0}^{\infty} \left(\sum\_{q=0}^{j} \binom{j}{q} G\_{j-q,\mu}^{(k)}(\eta) (i\xi)\_{q,\mu} \right) \frac{z^j}{j!} . \tag{62}$$

Similarly, we find

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{e\_\mu(z)+1}e^\eta\_\mu(z)e^{i\bar{j}}\_\mu(z) = \left(\sum\_{j=0}^\infty G^{(k)}\_{j,\mu}\frac{z^j}{j!}\right)\left(\sum\_{q=0}^\infty (\eta+i\xi)\_{q,\mu}\frac{z^q}{q!}\right)$$

$$=\sum\_{j=0}^\infty \left(\sum\_{q=0}^j \binom{j}{q}G^{(k)}\_{j-q,\mu}(\eta+i\xi)\_{q,\mu}\right)\frac{z^j}{j!}.\tag{63}$$

By comparing the coefficients of *z<sup>j</sup>* on both sides in Equations (62) and (63), we obtain our desired result, Equation (60). The proof of Equation (61) is similar to Equation (60).

**Theorem 9.** *If k* ∈ Z *and j* ≥ 0*, then*

$$\mathcal{G}\_{j,\mu}^{(k,\ell)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} \mathcal{G}\_{r,\mu}^{(k)} \mathcal{C}\_{j-r,\mu}(\eta,\xi)$$

$$= \sum\_{r=0}^{\lfloor \frac{3}{2} \rfloor} \sum\_{q=2r}^{j} \binom{j}{q} \mu^{q-2r}(-1)^r \xi^{2r} S^{(1)}(q,2r) G\_{j-q,\mu}^{(k)}(\xi),\tag{64}$$

*and*

$$\mathcal{G}\_{j,\mu}^{(k,\succ)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} \mathcal{B}\_{r,\mu}^{(k)} \mathcal{S}\_{j-r,\mu}(\eta,\xi)$$

$$= \sum\_{r=0}^{\lfloor \frac{q-1}{2} \rfloor} \sum\_{q-2r+1}^{j} \binom{j}{q} \mu^{q-2r-1}(-1)^r \xi^{2r+1} S^{(1)}(q, 2r+1) \mathcal{G}\_{j-q,\mu}^{(k)}(\eta). \tag{65}$$

**Proof.** From Equations (56) and (10), we see

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,c)}(\eta,\xi) \frac{z^j}{j!} = \frac{2\operatorname{Li}\_k(1-e^{-z})}{e\_{\mu}(z)+1} e\_{\mu}^{\eta}(t) \cos\_{\mu}^{\frac{\pi}{2}}(z)$$

$$= \left(\sum\_{r=0}^{\infty} G\_{r,\mu}^{(k)} \frac{z^r}{r!} \right) \left(\sum\_{j=0}^{\infty} \mathbb{C}\_{j,\mu}(\eta\_r, \xi) \frac{z^j}{j!} \right)$$

$$= \sum\_{j=0}^{\infty} \left(\sum\_{r=0}^{j} \binom{j}{r} \mathcal{G}\_{r,\mu}^{(k)} \mathbb{C}\_{j-r,\mu}(\eta\_r, \xi) \right) \frac{z^j}{j!}.\tag{66}$$

Similarly, we find

$$\frac{2\operatorname{Li}\_k(1-e^{-z})}{e\_\mu(z)+1}e\_\mu^\eta(z)\cos\_\mu^\xi(z) = \sum\_{j=0}^\infty G\_{j,\mu}^{(k)}(\eta)\frac{z^j}{j!}\sum\_{q=0}^\infty \sum\_{r=0}^{\left[\frac{q}{2}\right]} \mu^{q-2r}(-1)^r \xi^{2r} S^{(1)}(q,2r) \frac{z^r}{r!}$$

$$=\sum\_{j=0}^\infty \left(\sum\_{l=0}^j \sum\_{m=0}^{\left[\frac{l}{2}\right]} \binom{j}{l} \mu^{l-2m}(-1)^m \xi^{2m} S^{(1)}(q,2r) G\_{j-q,\mu}^{(k)}(\eta)\right) \frac{z^j}{j!}$$

$$=\sum\_{j=0}^\infty \left(\sum\_{r=0}^{\left[\frac{q}{2}\right]} \sum\_{q=2r}^j \binom{j}{q} \mu^{q-2r}(-1)^r \xi^{2r} S^{(1)}(q,2r) G\_{j-q,\mu}^{(k)}(\eta)\right) \frac{z^j}{j!}.\tag{67}$$

By comparing the coefficients of *z<sup>j</sup>* on both sides of Equations (66) and (67), we easily get our first claimed result, Equation (64). Similarly, we can establish our second needed result, Equation (65).

**Theorem 10.** *Let j* ≥ 0*. Then, we have*

$$G\_{j,\mu}^{(2,c)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} \frac{r! B\_r}{r+1} G\_{j-r,\mu}^{(c)}(\eta,\xi),\tag{68}$$

*and*

$$G\_{j,\mu}^{(2,\succ)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} \frac{r!B\_r}{r+1} G\_{j-r,\mu}^{(s)}(\eta,\xi) \,. \tag{69}$$

**Proof.** By using Equation (56), we determine

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \frac{2\operatorname{Li}\_k(1-\varepsilon^{-z})}{\varepsilon\_{\mu}(z)+1} c\_{\mu}^{\eta}(z) \cos\_{\mu}^{\overline{z}}(z)$$

$$= \frac{2\varepsilon\_{\mu}^{\eta}(z)\cos\_{\mu}^{\overline{z}}(z)}{\varepsilon\_{\mu}(z)+1} \int\_{0}^{z} \underbrace{\frac{1}{\varepsilon^{\mu}-1} \int\_{0}^{u} \frac{1}{\varepsilon^{u}-1} \cdots \frac{1}{\varepsilon^{u}-1} \int\_{0}^{u} \frac{u}{\varepsilon^{u}-1} du}\_{(k-1)\text{-times}} \cdot \operatorname{du}\left(\begin{array}{c} \\ \end{array}\right) \frac{1}{\varepsilon^{u}-1}$$

On setting *k* = 2 in Equation (70), we find

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(2,c)}(\eta,\xi) \frac{z^j}{j!} = \frac{2\varepsilon\_{\mu}^{\eta}(z)\cos\_{\mu}^{\frac{\pi}{\mu}}(z)}{\varepsilon\_{\mu}(z)+1} \int\_{0}^{z} \frac{u}{e^{u}-1} dz$$

$$= \left(\sum\_{r=0}^{\infty} \frac{r!B\_{r}z^{r}}{(r+1)r!} \right) \frac{2ze\_{\mu}^{\eta}(z)\cos\_{\mu}^{\frac{\pi}{\mu}}(z)}{\varepsilon\_{\mu}(z)+1}$$

$$= \left(\sum\_{r=0}^{\infty} \frac{r!B\_{r}z^{r}}{(r+1)r!} \right) \left(\sum\_{j=0}^{\infty} G\_{j,\mu}^{(c)}(\eta,\xi) \frac{z^{j}}{j!} \right).$$

On replacing *j* by *j* − *r* in the above equation, we obtain

$$=\sum\_{j=0}^{\infty} \sum\_{r=0}^{j} \binom{j}{r} \frac{r! B\_r}{r+1} G\_{j-r,\mu}^{(c)}(\eta,\xi) \frac{z^j}{j!} .$$

Finally, by equating the coefficients of the like powers of *z* in the last expression, we get the result, Equation (68). The proof of Equation (69) is similar to Equation (68).

**Theorem 11.** *For k* ∈ Z *and j* ≥ 0*, we have*

$$\mathcal{G}\_{j,\mu}^{(k,c)}(\eta,\xi) = \sum\_{r=0}^{j} \binom{j}{r} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1} q! S\_2(r+1,q)}{q^k (r+1)} \right) \mathcal{G}\_{j-r,\mu}^{(c)}(\eta,\xi),\tag{71}$$

*and*

$$\mathcal{G}\_{j,\mu}^{(k,s)}(\eta\_{\prime}\circ\xi) = \sum\_{r=0}^{j} \binom{j}{r} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1} q! S\_2(r+1, q)}{q^k (r+1)} \right) \mathcal{G}\_{j-r,\mu}^{(s)}(\eta\_{\prime}\xi). \tag{72}$$

**Proof.** In view of Equations (56) and (11), we see

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \left(\frac{2\operatorname{Li}\_k(1-e^{-z})}{z}\right) \left(\frac{z e\_{\mu}^{\eta}(z) \cos\_{\mu}^{\overline{z}}(z)}{e\_{\mu}(z)+1}\right). \tag{73}$$

Now

$$\frac{1}{z} \mathrm{Li}\_k (1 - e^{-z}) = \frac{1}{z} \sum\_{q=1}^{\infty} \frac{(1 - e^{-z})^q}{q^k}$$

$$= \frac{1}{z} \sum\_{q=1}^{\infty} \frac{(-1)^q}{q^k} q! \sum\_{r=l}^{\infty} (-1)^r S\_2(r, q) \frac{z^r}{r!}$$

$$= \frac{1}{z} \sum\_{r=1}^{\infty} \sum\_{q=1}^{r} \frac{(-1)^{q+r}}{q^k} q! S\_2(r, q) \frac{t^r}{r!}$$

$$= \sum\_{r=0}^{\infty} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1}}{q^k} q! \frac{S\_2(r+1, q)}{q+1} \right) \frac{z^r}{r!}. \tag{74}$$

Using Equation (74) in (73), we find

$$\sum\_{j=0}^{\infty} \mathcal{G}\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \sum\_{r=0}^{\infty} \left( \sum\_{q=1}^{r+1} \frac{(-1)^{q+r+1}}{q^k} q! \frac{\mathcal{S}\_2(r+1,q)}{r+1} \right) \frac{z^r}{r!} \left( \sum\_{j=0}^{\infty} \mathcal{G}\_{j,\mu}^{(c)}(\eta,\xi) \frac{z^j}{j!} \right) \dots$$

which on comparing the coefficients of *z<sup>j</sup>* on both sides, yields our desired result, Equation (71). Similarly, we can derive our second result, Equation (72).

**Theorem 12.** *Let k* ∈ Z *and j* ≥ 0*, then we have*

$$
\frac{1}{2} \left[ G\_{j,\mu}^{(k,\varepsilon)}(\eta+1,\xi) + G\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \right]
$$

$$
= \sum\_{r=1}^{j} \binom{j}{r} \left( \sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1) \right) C\_{j-r,\mu}(\eta,\xi),
\tag{75}
$$

$$
\frac{1}{2} \left[ G\_{j,\mu}^{(k,\varepsilon)}(\eta+1,\xi) + G\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \right]
$$

*and*

$$\hat{\lambda} = \sum\_{r=1}^{j} \binom{j}{r} \left( \sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1) \right) S\_{j-r, \mu}(\eta, \xi) . \tag{76}$$

**Proof.** Taking

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,\varepsilon)}(\eta+1,\underline{\varepsilon}) \frac{z^j}{j!} + \sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,\varepsilon)}(\eta,\underline{\varepsilon}) \frac{z^j}{j!}$$

$$= \frac{2\text{Li}\_k(1-\varepsilon^{-z})}{\varepsilon\_{\mu}(z)+1} e\_{\mu}^{(\eta+1)}(z) \cos\_{\mu}^{\underline{\varepsilon}}(z) + \frac{2\text{Li}\_k(1-\varepsilon^{-z})}{\varepsilon\_{\mu}(z)+1} e\_{\mu}^{(\eta)}(z) \cos\_{\mu}^{\underline{\varepsilon}}(z)$$

$$= 2\text{Li}\_k(1-\varepsilon^{-z}) e\_{\mu}^{(\eta)}(z) \cos\_{\mu}^{\underline{\varepsilon}}(z)$$

$$= \sum\_{q=0}^{\infty} \frac{(1-\varepsilon^{-z})^{q+1}}{(q+1)^k} 2e\_{\mu}^{\eta}(z) \cos\_{\mu}^{(\xi)}(z)$$

$$= \sum\_{r=1}^{\infty} \left( \sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1) \right) \frac{z^r}{r!} 2e\_{\mu}^{\chi}(z) \cos\_{\mu}^{(\xi)}(z)$$

$$= 2 \left( \sum\_{r=1}^{\infty} \left( \sum\_{q=0}^{r-1} \frac{(-1)^{q+r+1}}{(q+1)^k} (q+1)! S\_2(r, q+1) \right) \frac{z^r}{r!} \right) \left( \sum\_{j=0}^{\infty} C\_{j,\mu}(\eta, \xi) \frac{z^j}{j!} \right).$$

On replacing *j* by *j* − *r* in the right side of the above equation, and after comparing the coefficients of *z<sup>j</sup>* on both sides, we acquire the desired result, Equation (75). Similarly, we can obtain the result, Equation (76).

**Theorem 13.** *For k* ∈ Z *and j* ≥ 0*, we have*

$$G\_{j,\mu}^{(k,\varepsilon)}\left(\eta+\mathfrak{a},\mathfrak{z}\right) = \sum\_{m=0}^{j} \binom{j}{m} G\_{j-m,\mu}^{(k,\varepsilon)}(\eta,\mathfrak{z}) (\mathfrak{a})\_{m,\mu},\tag{77}$$

*and*

$$G\_{j,\mu}^{(k,\varepsilon)}(\eta+\mathfrak{a},\mathfrak{z}) = \sum\_{m=0}^{j} \binom{j}{m} \, \mathrm{G}\_{j-m,\mu}^{(k,\varepsilon)}(\eta,\mathfrak{z}) (\mathfrak{a})\_{m,\mu} \,. \tag{78}$$

**Proof.** By using Equation (56), we have

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,c)}(\eta+a,\xi) \frac{z^j}{j!} = \frac{2\text{Li}\_k(1-e^{-z})}{e\_\mu(z)+1} c\_{\mu}^{(\eta+a)}(z) \cos\_{\mu}^{(\xi)}(z),$$

$$= \left(\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,c)}(\eta,\xi) \frac{z^j}{j!} \right) \left(\sum\_{m=0}^{\infty} (\alpha)\_{m,\mu} \frac{z^m}{m!} \right)$$

$$= \sum\_{j=0}^{\infty} \left(\sum\_{m=0}^{j} \binom{j}{m} \begin{pmatrix} j \\ m \end{pmatrix} G\_{j-m,\mu}^{(k,c)}(\eta,\xi) (\alpha)\_{m,\mu} \right) \frac{z^j}{j!}.$$

By comparing the coefficients of *z<sup>j</sup>* on both sides in the last expression, we acquire our desired result, Equation (77). Similarly, we can derive our second result, Equation (78).

**Theorem 14.** *If k* ∈ Z *and j* ≥ 0*, then*

$$\mathcal{G}^{(k,c)}\_{j,\mu}(\eta,\xi) = \sum\_{r=0}^{j} \sum\_{q=0}^{r} \binom{j}{r} (\eta)\_l \mathcal{S}^{(2)}\_{\mu}(r,\eta) \mathcal{G}^{(k,c)}\_{j-r,\mu}(0,\xi),\tag{79}$$

*and*

$$G\_{j,\mu}^{(k,s)}(\eta,\xi) = \sum\_{r=0}^{j} \sum\_{q=0}^{r} \binom{j}{r} (\eta)\_l S\_{\mu}^{(2)}(r,\eta) G\_{j-r,\mu}^{(k,s)}(0,\xi). \tag{80}$$

**Proof.** From Equations (56) and (12), we have

$$\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,\varepsilon)}(\eta,\xi) \frac{z^j}{j!} = \frac{2\text{Li}\_k(1-\varepsilon^{-z})}{\varepsilon\_{\mu}(z)+1} (e\_{\mu}(z)-1+1)^{\eta} \cos\_{\mu}^{\xi}(z)$$

$$=\frac{2\text{Li}\_k(1-\varepsilon^{-z})}{e\_{\mu}(z)+1} \sum\_{q=0}^{\infty} \binom{\eta}{q} (e\_{\mu}(z)-1)^{q} \cos\_{\mu}^{\xi}(z)$$

$$=\frac{2\text{Li}\_k(1-\varepsilon^{-z})}{e\_{\mu}(z)+1} \cos\_{\mu}^{\xi}(z) \sum\_{q=0}^{\infty} (\eta)\_{q} \prod\_{r=q}^{\infty} S\_{\mu}^{(2)}(r,q) \frac{z^r}{r!}$$

$$=\sum\_{j=0}^{\infty} G\_{j,\mu}^{(k,\varepsilon)}(0,\xi) \frac{z^j}{j!} \sum\_{r=0}^{\infty} \left(\sum\_{q=0}^{r} (\eta)\_{q} \, S\_{\mu}^{(2)}(r,q)\right) \frac{z^r}{r!}$$

$$=\sum\_{j=0}^{\infty} \left(\sum\_{r=0}^{j} \sum\_{q=0}^{r} \left(\begin{array}{c} j\\ r \end{array}\right) (\eta)\_{q} S\_{\mu}^{(2)}(r,q) G\_{j-r,\mu}^{(k,\varepsilon)}(0,\xi)\right) \frac{z^j}{j!}$$

Finally, by comparing the coefficients of *z<sup>j</sup>* on both sides in the last expression, we arrive at our claimed result, Equation (79). Similarly, we can establish our second result, Equation (80).

#### **4. Conclusions**

In the present article, we have considered the parametric kinds of degenerate poly-Bernoulli and poly-Genocchi polynomials by making use of the degenerate type exponential as well as trigonometric functions. We have also derived some analytical properties of our newly introduced parametric polynomials by using the series manipulation technique. Furthermore, it is noticed that, if we consider any Appell polynomials of a complex variable (as discussed in the present article), then we can easily define its parametric kinds by separating the complex variable into real and imaginary parts.

**Author Contributions:** All authors contributed equally to the manuscript and typed, read, and approved the final manuscript. All authors have read and agreed to the published version of the manuscript.

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

### **Abbreviations**

The following abbreviations are used in this manuscript:

MKdV modified Korteweg–de Vries equation

### **References**


© 2020 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 on Type 2 Degenerate Bernoulli Polynomials of the Second Kind**

**Taekyun Kim 1,2, Lee-Chae Jang 3,***∗***, Dae San Kim <sup>4</sup> and Han Young Kim <sup>2</sup>**


Received: 9 March 2020; Accepted: 23 March 2020; Published: 2 April 2020

**Abstract:** In recent years, many mathematicians studied various degenerate versions of some special polynomials for which quite a few interesting results were discovered. In this paper, we introduce the type 2 degenerate Bernoulli polynomials of the second kind and their higher-order analogues, and study some identities and expressions for these polynomials. Specifically, we obtain a relation between the type 2 degenerate Bernoulli polynomials of the second and the degenerate Bernoulli polynomials of the second, an identity involving higher-order analogues of those polynomials and the degenerate Stirling numbers of the second kind, and an expression of higher-order analogues of those polynomials in terms of the higher-order type 2 degenerate Bernoulli polynomials and the degenerate Stirling numbers of the first kind.

**Keywords:** type 2 degenerate Bernoulli polynomials of the second kind; degenerate central factorial numbers of the second kind

### **1. Introduction**

In [1,2], Carlitz initiated study of the degenerate Bernoulli and Euler polynomials and obtained some arithmetic and combinatorial results on them. In recent years, many mathematicians have drawn their attention to various degenerate versions of some old and new polynomials and numbers, namely some degenerate versions of Bernoulli numbers and polynomials of the second kind, Changhee numbers of the second kind, Daehee numbers of the second kind, Bernstein polynomials, central Bell numbers and polynomials, central factorial numbers of the second kind, Cauchy numbers, Eulerian numbers and polynomials, Fubini polynomials, Stirling numbers of the first kind, Stirling polynomials of the second kind, central complete Bell polynomials, Bell numbers and polynomials, type 2 Bernoulli numbers and polynomials, type 2 Bernoulli polynomials of the second kind, poly-Bernoulli numbers and polynomials, poly-Cauchy polynomials, and of Frobenius–Euler polynomials, to name a few [3–10] and the references therein.

They have studied those polynomials and numbers with their interest not only in combinatorial and arithmetic properties but also in differential equations and certain symmetric identities [7,9] and references therein, and found many interesting results related to them [3–6,8,10]. It is remarkable that studying degenerate versions is not only limited to polynomials but also extended to transcendental functions. Indeed, the degenerate gamma functions were introduced in connection with degenerate Laplace transforms [11,12].

The motivation for this research is to introduce the type 2 degenerate Bernoulli polynomials of the second kind defined by

$$\frac{(1+t) - (1+t)^{-1}}{\log\_{\lambda}(1+t)}(1+t)^{\chi} = \sum\_{n=0}^{\infty} b\_{n,\lambda}^\*(\mathbf{x}) \frac{t^n}{n!}.$$

and investigate its arithmetic and combinatorial properties. The facts in Section 1 are some known definitions and results that are needed throughout this paper. However, all of the results in Section 2 are new.

We will spend the rest of this section in recalling some necessary stuffs for the next section.

As is known, the type 2 Bernoulli polynomials are defined by the generating function [5,13]

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

From (1), we note that

$$B\_n^\*(\mathbf{x}) = 2^{n-1} B\_n \left( \frac{\mathbf{x} + 1}{2} \right), \ (n \ge 0), \tag{2}$$

where *Bn*(*x*) are the ordinary Bernoulli polynomials given by

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

Also, the type 2 Euler polynomials are given by [5,13]

$$
\varepsilon^{xt}\text{sech}t = \frac{2}{e^t + e^{-t}}\varepsilon^{xt} = \sum\_{n=0}^{\infty} E\_n^\*(x)\frac{t^n}{n!}.\tag{3}
$$

Note that

$$E\_n^\*(\mathbf{x}) = 2^n E\_n \left( \frac{\mathbf{x} + 1}{2} \right), \quad (n \ge 0), \tag{4}$$

where *En*(*x*) are the ordinary Euler polynomials given by [14,15]

$$\frac{2}{e^t + 1} e^{\mathbf{x}t} = \sum\_{n=0}^{\infty} E\_n(\mathbf{x}) \frac{t^n}{n!}.$$

The central factorial numbers of the second kind are defined as [5,8]

$$\mathbf{x}^n = \sum\_{k=0}^n T(n,k)\mathbf{x}^{[k]},\tag{5}$$

or equivalently as

$$\frac{1}{k!} (e^{\frac{t}{2}} - e^{-\frac{t}{2}})^k = \sum\_{n=k}^{\infty} T(n, k) \frac{t^n}{n!} \tag{6}$$

where *x*[0] = 1, *x*[*n*] = *x x* + *<sup>n</sup>* <sup>2</sup> − 1 *x* + *<sup>n</sup>* <sup>2</sup> − 2 ··· *<sup>x</sup>* <sup>−</sup> *<sup>n</sup>* <sup>2</sup> + 1 , (*n* ≥ 1).

It is well known that the Daehee polynomials are defined by [16,17]

$$\frac{\log\left(1+t\right)}{t}(1+t)^{x} = \sum\_{k=0}^{n} D\_{n}(x)\frac{t^{n}}{n!}.\tag{7}$$

When *x* = 0, *Dn* = *Dn*(0) are called the Daehee numbers.

The Bernoulli polynomials of the second kind of order *r* are defined by [15]

$$a\left(\frac{t}{\log(1+t)}\right)^r (1+t)^x = \sum\_{k=0}^n b\_n^{(r)}(x) \frac{t^n}{n!}.\tag{8}$$

Note that *b* (*r*) *<sup>n</sup>* (*x*) = *<sup>B</sup>*(*n*−*r*+1) *<sup>n</sup>* (*<sup>x</sup>* <sup>+</sup>1), (*<sup>n</sup>* <sup>≥</sup> <sup>0</sup>). Here *<sup>B</sup>*(*r*) *<sup>n</sup>* (*x*) are the ordinary Bernoulli polynomials of order *r* given by [8,15–18]

$$\left(\frac{t}{e^t - 1}\right)^r e^{xt} = \sum\_{k=0}^n B\_n^{(r)}(x) \frac{t^n}{n!}.\tag{9}$$

It is known that the Stirling numbers of the second kind are defined by [8]

$$\frac{1}{k!} \left( e^t - 1 \right)^k = \sum\_{n=k}^{\infty} S\_2(n, k) \frac{t^n}{n!} \tag{10}$$

and the Stirling numbers of the first kind by [8]

$$\frac{1}{k!} \log^k(1+t) = \sum\_{n=k}^{\infty} S\_1(n,k) \frac{t^n}{n!}. \tag{11}$$

For any nonzero *λ* ∈ R, the degenerate exponential function is defined by [11,12]

$$
\sigma\_{\lambda}^{x}(t) = (1 + \lambda t)^{\frac{x}{\lambda}} = \sum\_{n=0}^{\infty} (x)\_{n\lambda} \frac{t^n}{n!} \tag{12}
$$

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

In particular, we let

$$e\_{\lambda}(t) = e\_{\lambda}^{1}(t) = (1 + \lambda t)^{\frac{1}{\lambda}}.\tag{13}$$

In [1,2], Carlitz introduced the degenerate Bernoulli polynomials which are given by the generating function

$$\frac{t}{\varepsilon\_{\lambda}(t) - 1} \varepsilon\_{\lambda}^{x}(t) = \sum\_{n=0}^{\infty} \beta\_{n, \lambda}(x) \frac{t^{n}}{n!}. \tag{14}$$

Also, he considered the degenerate Euler polynomials given by [1,2]

$$\frac{2}{\varepsilon\_{\lambda}(t) + 1} e\_{\lambda}^{x}(t) = \sum\_{n=0}^{\infty} \mathcal{E}\_{n,\lambda}(x) \frac{t^{n}}{n!}. \tag{15}$$

Recently, Kim-Kim considered the degenerate central factorial numbers of the second kind given by [8,13]

$$\frac{1}{k!} \left( e^{\frac{1}{\lambda}}\_{\lambda}(t) - e^{-\frac{1}{\lambda}}\_{\lambda}(t) \right)^k = \sum\_{n=k}^{\infty} T\_{\lambda}(n,k) \frac{t^n}{n!}. \tag{16}$$

Note that lim*λ*→<sup>0</sup> *Tλ*(*n*, *k*) = *T*(*n*, *k*).

### **2. Type 2 Degenerate Bernoulli Polynomials of the Second Kind**

Let log*<sup>λ</sup> t* be the compositional inverse of *eλ*(*t*) in (13). Then we have

$$
\log\_{\lambda} t = \frac{1}{\lambda} \left( t^{\lambda} - 1 \right). \tag{17}
$$

Note that lim*λ*→<sup>0</sup> log*<sup>λ</sup> t* = log *t*. Now, we define the *degenerate Daehee polynomials* by

$$\frac{\log\_{\lambda}(1+t)}{t}(1+t)^{x} = \sum\_{n=0}^{\infty} D\_{n,\lambda}(x) \frac{t^{n}}{n!}. \tag{18}$$

Note that lim*λ*→<sup>0</sup> *Dn*,*λ*(*x*) = *Dn*(*x*), (*n* ≥ 0). In view of (8), we also consider the degenerate Bernoulli polynomials of the second kind of order *α* given by

$$\left(\frac{t}{\log\_{\lambda}(1+t)}\right)^{a}(1+t)^{x} = \sum\_{n=0}^{\infty} b\_{n,\lambda}^{(a)}(x) \frac{t^{n}}{n!}.\tag{19}$$

Note that lim*λ*→<sup>0</sup> *b* (*α*) *<sup>n</sup>*,*λ*(*x*) = *b* (*α*) *<sup>n</sup>* (*x*), (*<sup>n</sup>* <sup>≥</sup> <sup>0</sup>). From (19), we have

$$\left(\frac{\lambda t}{(1+t)^{\frac{\lambda}{2}} - (1+t)^{-\frac{\lambda}{2}}}\right)^{a} (1+t)^{x-\frac{\lambda a}{2}} = \sum\_{n=0}^{\infty} b\_{n,\lambda}^{(a)}(x) \frac{t^n}{n!}.\tag{20}$$

For *<sup>α</sup>* = *<sup>r</sup>* ∈ N, and replacing *<sup>t</sup>* by *<sup>e</sup>*2*<sup>t</sup>* − 1 in (20), we get

$$\begin{split} \sum\_{m=0}^{\infty} b\_{m,\lambda}^{(r)}(\mathbf{x}) \frac{1}{m!} (e^{2t} - 1)^m &= \left(\frac{\lambda t}{e^{t\lambda} - e^{-t\lambda}}\right)^r \frac{1}{t^r} (e^{2t} - 1)^r e^{(2x - \lambda r)t} \\ &= \sum\_{k=0}^{\infty} B\_k^\* \left(\frac{2\mathbf{x}}{\lambda} - r\right) \frac{\lambda^k t^k}{k!} \sum\_{m=0}^{\infty} S\_2(m + r, r) 2^{m+r} \frac{1}{\binom{m+r}{r}} \frac{t^m}{m!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^n \binom{n}{m} B\_{n-m}^\* \left(\frac{2\mathbf{x}}{\lambda} - r\right) \lambda^{n-m} \frac{S\_2(m + r, r)}{\binom{m+r}{r}} 2^{m+r} \right) \frac{t^n}{n!} . \end{split} \tag{21}$$

On the other hand,

$$\begin{split} \sum\_{m=0}^{\infty} b\_{m,\lambda}^{(r)}(\mathbf{x}) \frac{1}{m!} (e^{2t} - 1)^m &= \sum\_{m=0}^{\infty} b\_{m,\lambda}^{(r)}(\mathbf{x}) \sum\_{n=m}^{\infty} S\_2(n, m) 2^n \frac{t^n}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{m=0}^n b\_{m,\lambda}^{(r)}(\mathbf{x}) 2^n S\_2(n, m) \right) \frac{t^n}{n!} . \end{split} \tag{22}$$

From (21) and (22), we have

$$\sum\_{m=0}^{n} b\_{m,\lambda}^{(r)}(\mathbf{x}) S\_2(n,m) = \sum\_{m=0}^{n} \binom{n}{m} B\_{n-m}^\* \left(\frac{2\mathbf{x}}{\lambda} - r\right) \lambda^{n-m} \frac{S\_2(m+r,r)}{\binom{m+r}{r}} 2^{m+r-n}.\tag{23}$$

Now, we define the *type* 2 *degenerate Bernoulli polynomials of the second kind* by

$$\frac{(1+t) - (1+t)^{-1}}{\log\_{\lambda}(1+t)}(1+t)^{x} = \sum\_{n=0}^{\infty} b^{\*}\_{n,\lambda}(x) \frac{t^{n}}{n!}.\tag{24}$$

When *x* = 0, *b*∗ *<sup>n</sup>*,*<sup>λ</sup>* = *b*<sup>∗</sup> *<sup>n</sup>*,*λ*(0) are called the type 2 degenerate Bernoulli numbers of the second kind. Note that lim*λ*→<sup>0</sup> *b*<sup>∗</sup> *<sup>n</sup>*,*λ*(*x*) = *b*<sup>∗</sup> *<sup>n</sup>*(*x*), where *b*<sup>∗</sup> *<sup>n</sup>*(*x*) are the type 2 Bernoulli polynomials of the second kind given by

$$\frac{(1+t) - (1+t)^{-1}}{\log(1+t)}(1+t)^x = \sum\_{n=0}^{\infty} b\_n^\*(\mathbf{x}) \frac{t^n}{n!}.$$

From (19) and (24), we note that

$$\begin{split} \frac{(1+t) - (1+t)^{-1}}{\log\_{\lambda}(1+t)} (1+t)^{x} &= \frac{t}{\log\_{\lambda}(1+t)} (1+t)^{x} \left(1 + \frac{1}{1+t}\right) \\ &= \frac{t}{\log\_{\lambda}(1+t)} (1+t)^{x} + \frac{t}{\log\_{\lambda}(1+t)} (1+t)^{x-1} \\ &= \sum\_{n=0}^{\infty} \left( b\_{n,\lambda}^{(1)}(\mathbf{x}) + b\_{n,\lambda}^{(1)}(\mathbf{x}-1) \right) \frac{t^{n}}{n!} . \end{split} \tag{25}$$

Therefore, we obtain the following theorem.

**Theorem 1.** *For n* ≥ 0*, we have*

$$b\_{n,\lambda}^\*(\mathfrak{x}) = b\_{n,\lambda}^{(1)}(\mathfrak{x}) + b\_{n,\lambda}^{(1)}(\mathfrak{x} - 1).$$

*Moreover,*

$$\sum\_{m=0}^{n} b\_{m,\lambda}^{(r)}(\mathbf{x}) S\_2(n,m) = \sum\_{m=0}^{n} \binom{n}{m} B\_{n-m}^\* \left(\frac{2\mathbf{x}}{\lambda} - r\right) \lambda^{n-m} \frac{S\_2(m+r,r)}{\binom{m+r}{r}} 2^{m+r-n} \lambda$$

*where r is a positive integer.*

Now, we observe that

$$\begin{split} \frac{(1+t) - (1+t)^{-1}}{\log\_{\lambda}(1+t)} (1+t)^{\mathbf{x}} &= \sum\_{l=0}^{\infty} b\_{l,\lambda}^{\*} \frac{t^{l}}{l!} \sum\_{m=0}^{\infty} (\mathbf{x})\_{m} \frac{t^{m}}{m!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{l=0}^{n} \binom{n}{l} b\_{l,\lambda}^{\*} (\mathbf{x})\_{n-l} \right) \frac{t^{n}}{n!} \end{split} \tag{26}$$

where (*x*)<sup>0</sup> = 1, (*x*)*<sup>n</sup>* = *x*(*x* − 1)···(*x* − *n* + 1), (*n* ≥ 1). From (24) and (26), we get

$$b\_{n,\lambda}^\*(\mathbf{x}) = \sum\_{l=0}^n \binom{n}{l} b\_{l,\lambda}^\*(\mathbf{x})\_{n-l\prime} \quad (n \ge 0). \tag{27}$$

For *α* ∈ R, let us define the *type* 2 *degenerate Bernoulli polynomials of the second kind of order α* by

$$\left(\frac{(1+t)-(1+t)^{-1}}{\log\_{\lambda}(1+t)}\right)^{a}(1+t)^{x} = \sum\_{n=0}^{\infty} b\_{n,\lambda}^{\*(a)}(x) \frac{t^{n}}{n!} \tag{28}$$

When *x* = 0, *b* ∗(*α*) *<sup>n</sup>*,*<sup>λ</sup>* = *b* ∗(*α*) *<sup>n</sup>*,*<sup>λ</sup>* (0) are called the type 2 degenerate Bernoulli numbers of the second kind of order *α*.

Let *α* = *k* ∈ N. Then we have

$$\sum\_{n=0}^{\infty} b\_{n,\lambda}^{\*(k)}(x) \frac{t^n}{n!} = \left( \frac{(1+t) - (1+t)^{-1}}{\log\_{\lambda}(1+t)} \right)^k (1+t)^x. \tag{29}$$

By replacing *t* by *eλ*(*t*) − 1 in (29), we get

$$\begin{split} \frac{k!}{t^k} \frac{1}{k!} \left( e\_\lambda(t) - e\_\lambda^{-1}(t) \right)^k e\_\lambda^x(t) &= \sum\_{l=0}^\infty b\_{l,\lambda}^{\*(k)}(\mathbf{x}) \frac{1}{l!} (e\_\lambda(t) - 1)^l \\ &= \sum\_{l=0}^\infty b\_{l,\lambda}^{\*(k)}(\mathbf{x}) \sum\_{n=l}^\infty S\_{2,\lambda}(n,l) \frac{t^n}{n!} \\ &= \sum\_{n=0}^\infty \left( \sum\_{l=0}^n b\_{l,\lambda}^{\*(k)}(\mathbf{x}) S\_{2,\lambda}(n,l) \right) \frac{t^n}{n!} \end{split} \tag{30}$$

where *S*2,*λ*(*n*, *l*) are the degenerate Stirling numbers of the second kind given by [6]

$$\frac{1}{k!} \left( e\_{\lambda}(t) - 1 \right)^{k} = \sum\_{n=k}^{\infty} S\_{2,\lambda}(n, k) \frac{t^{n}}{n!}. \tag{31}$$

On the other hand, we also have

$$\begin{split} \frac{k!}{t^k} \frac{1}{k!} \left( e\_\lambda(t) - e\_\lambda^{-1}(t) \right)^k e\_\lambda^x(t) &= \frac{k!}{t^k} \frac{1}{k!} \left( e\_\lambda^2(t) - 1 \right)^k e\_\lambda^{x-k}(t) \\ &= \frac{k!}{t^k} \frac{1}{k!} \left( e\_\frac{1}{2}(2t) - 1 \right)^k e\_\lambda^{x-k}(t) \\ &= \sum\_{m=0}^\infty S\_{2,\frac{k}{2}}(m+k, k) \frac{2^{m+k}}{\binom{m+k}{k}} \frac{t^m}{m!} \sum\_{l=0}^\infty (x-k)\_{l,\lambda} \frac{t^l}{l!} \\ &= \sum\_{n=0}^\infty \left( \sum\_{m=0}^n \frac{\binom{n}{m} 2^{m+k}}{\binom{m+k}{k}} S\_{2,\frac{k}{2}}(m+k, k) (x-k)\_{n-m,\lambda} \right) \frac{t^n}{n!} . \end{split}$$

Therefore, by (30) and (32), we obtain the following theorem.

**Theorem 2.** *For n* ≥ 0*, we have*

$$\sum\_{l=0}^{n} b\_{l,\lambda}^{\*(k)}(x) S\_{2,\lambda}(n,l) = \sum\_{l=0}^{n} \frac{\binom{n}{l} 2^{l+k}}{\binom{l+k}{k}} S\_{2,\frac{l}{k}}(l+k,k)(x-k)\_{n-l,\lambda}...$$

In particular,

$$2^{n+k}S\_{2,\frac{k}{2}}(n+k,k) = \binom{n+k}{k} \sum\_{l=0}^{n} b\_{l,\lambda}^{\*(k)}(k)S\_{2,\lambda}(n,l).$$

For *α* ∈ R, we recall that the type 2 degenerate Bernoulli polynomials of order *α* are defined by [5,13]

$$\left(\frac{t}{\varepsilon\_{\lambda}(t) - \varepsilon\_{\lambda}^{-1}(t)}\right)^{n} \boldsymbol{e}\_{\lambda}^{\boldsymbol{x}}(t) = \sum\_{n=0}^{\infty} \beta\_{n,\lambda}^{\*(\boldsymbol{a})}(\boldsymbol{x}) \frac{t^{n}}{n!}.\tag{33}$$

For *k* ∈ N, let us take *α* = −*k* and replace *t* by log*λ*(1 + *t*) in (33). Then we have

$$\begin{split} \left( \frac{(1+t) - (1+t)^{-1}}{\log\_{\lambda}(1+t)} \right)^{k} (1+t)^{x} &= \sum\_{l=0}^{\infty} \beta\_{l,\lambda}^{\*(-k)}(x) \frac{1}{l!} \left( \log\_{\lambda}(1+t) \right)^{l} \\ &= \sum\_{l=0}^{\infty} \beta\_{l,\lambda}^{\*(-k)}(x) \sum\_{n=l}^{\infty} S\_{1,\lambda}(n.l) \frac{t^{n}}{n!} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{l=0}^{n} \beta\_{l,\lambda}^{\*(-k)} S\_{1,\lambda}(n.l) \right) \frac{t^{n}}{n!} \end{split} \tag{34}$$

where *S*1,*λ*(*n*, *l*) are the degenerate Stirling numbers of the first kind given by

$$\frac{1}{k!} (\log\_{\lambda}(1+t))^k = \sum\_{n=k}^{\infty} S\_{1,\lambda}(n,k) \frac{t^n}{n!}. \tag{35}$$

Note here that lim*λ*→<sup>0</sup> *S*1,*λ*(*n*, *l*) = *S*1(*n*, *l*). Therefore, by (26) and (34), we obtain the following theorem.

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

$$b\_{n,\lambda}^{\*(k)}(\mathfrak{x}) = \sum\_{l=0}^{n} \beta\_{l,\lambda}^{\*(-k)}(\mathfrak{x}) S\_{1,\lambda}(n,l).$$

We observe that

$$\begin{split} \frac{1}{k!}t^k &= \frac{1}{k!} \left( (1+t)^{\frac{1}{2}} - (1+t)^{-\frac{1}{2}} \right)^k (1+t)^{\frac{1}{2}} \\ &= \frac{1}{k!} \left( e^{\frac{1}{2}} (\log\_{\lambda}(1+t)) - e^{-\frac{1}{2}} \left( \log\_{\lambda}(1+t) \right) \right)^k (1+t)^{\frac{1}{2}} \\ &= \sum\_{l=k}^{\infty} T\_{\lambda}(l,k) \frac{1}{l!} (\log\_{\lambda}(1+t))^l \sum\_{r=0}^{\infty} \left( \frac{k}{2} \right)\_r \frac{t^r}{r!} \\ &= \sum\_{l=k}^{\infty} T\_{\lambda}(l,k) \sum\_{m=l}^{\infty} S\_{1,\lambda}(m,l) \frac{t^m}{m!} \sum\_{r=0}^{\infty} \left( \frac{k}{2} \right)\_r \frac{t^r}{r!} \\ &= \sum\_{m=k}^{\infty} \sum\_{l=k}^{m} T\_{\lambda}(l,k) S\_{1,\lambda}(m,l) \frac{t^m}{m!} \sum\_{r=0}^{\infty} \left( \frac{k}{2} \right)\_r \frac{t^r}{r!} \\ &= \sum\_{n=k}^{\infty} \left( \sum\_{m=k}^{n} \sum\_{l=k}^{m} T\_{\lambda}(l,k) S\_{1,\lambda}(m,l) \binom{n}{m} \left( \frac{k}{2} \right)\_{n-m} \right) \frac{t^n}{n!} . \end{split} \tag{36}$$

On the other hand,

$$\begin{split} \frac{1}{k!}t^k &= \left(\frac{t}{\log\_{\lambda}(1+t)}\right)^k \frac{1}{k!} \left(\log\_{\lambda}(1+t)\right)^k\\ &= \sum\_{l=0}^{\infty} b\_{l,\lambda}^{(k)} \frac{t^l}{l!} \sum\_{m=k}^{\infty} S\_{1,\lambda}(m,k) \frac{t^m}{m!}\\ &= \sum\_{n=k}^{\infty} \left(\sum\_{m=k}^n S\_{1,\lambda}(m,k) b\_{n-m,\lambda}^{(k)} \binom{n}{m}\right) \frac{t^n}{n!}. \end{split} \tag{37}$$

Therefore, by (36) and (37), we obtain the following theorem.

**Theorem 4.** *For n*, *k* ≥ 0*, we have*

$$\sum\_{m=k}^{n} \sum\_{l=k}^{m} T\_{\lambda}(l,k) S\_{1,\lambda}(m,l) \binom{n}{m} \left(\frac{k}{2}\right)\_{n-m} = \sum\_{m=k}^{n} S\_{1,\lambda}(m,k) b\_{n-m,\lambda}^{(k)} \binom{n}{m} \dots$$

### **3. Conclusions**

In this paper, we introduced the type 2 degenerate Bernoulli polynomials of the second kind and their higher-order analogues, and studied some identities and expressions for these polynomials. Specifically, we obtained a relation between the type 2 degenerate Bernoulli polynomials of the second and the degenerate Bernoulli polynomials of the second, an identity involving higher-order analogues of those polynomials and the degenerate Stirling numbers of second kind, and an expression of higher-order analogues of those polynomials in terms of the higher-order type 2 degenerate Bernoulli

polynomials and the degenerate Stirling numbers of the first kind.

In addition, we obtained an identity involving the higher-order degenerate Bernoulli polynomials of the second kind, the type 2 Bernoulli polynomials and Stirling numbers of the second kind, and an identity involving the degenerate central factorial numbers of the second kind, the degenerate Stirling numbers of the first kind and the higher-order degenerate Bernoulli polynomials of the second kind.

Next, we would like to mention three possible applications of our results. The first one is their applications to identities of symmetry. For instance, in [7] by using the *p*-adic fermionic integrals it was possible for us to find many symmetric identities in three variables related to degenerate Euler polynomials and alternating generalized falling factorial sums.

The second one is their applications to differential equations. Indeed, in [9] we derived an infinite family of nonlinear differential equations having the generating function of the degenerate Changhee numbers of the second kind as a solution. As a result, from those differential equations we obtained an interesting identity involving the degenerate Changhee and higher-order degenerate Changhee numbers of the second kind.

The third one is their applications to probability. For example, in [19,20] we showed that both the degenerate *λ*-Stirling polynomials of the second and the *r*-truncated degenerate *λ*-Stirling polynomials of the second kind appear in certain expressions of the probability distributions of appropriate random variables.

These possible applications of our results require a considerable amount of work and they should appear as separate papers. We have witnessed in recent years that studying various degenerate versions of some special polynomials and numbers are very fruitful and promising [21]. It is our plan to continue to do this line of research, as one of our near future projects.

**Author Contributions:** T.K. and D.S.K. conceived of the framework and structured the whole paper; D.S.K. and T.K. wrote the paper; L.-C.J. and H.Y.K. checked the results of the paper; D.S.K. and T.K. completed the revision of the article. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

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

#### **References**


© 2020 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* **Exceptional Set for Sums of Symmetric Mixed Powers of Primes**

### **Jinjiang Li 1, Chao Liu 1, Zhuo Zhang <sup>1</sup> and Min Zhang 2,\***


Received: 17 January 2020; Accepted: 25 February 2020; Published: 2 March 2020

**Abstract:** The main purpose of this paper is to use the Hardy–Littlewood method to study the solvability of mixed powers of primes. To be specific, we consider the even integers represented as the sum of one prime, one square of prime, one cube of prime, and one biquadrate of prime. However, this representation can not be realized for all even integers. In this paper, we establish the exceptional set of this kind of representation and give an upper bound estimate.

**Keywords:** Waring–Goldbach problem; circle method; exceptional set; symmetric form

**MSC:** 11P05, 11P32, 11P55

### **1. Introduction and Main Result**

Let *N*, *k*1, *k*2, ... , *ks* be natural numbers which satisfy 2 *k*<sup>1</sup> *k*<sup>2</sup> - ··· *ks*, *N* > *s*. Waring's problem of unlike powers concerns the possibility of representation of *N* in the form

$$N = \mathfrak{x}\_1^{k\_1} + \mathfrak{x}\_2^{k\_2} + \dots + \mathfrak{x}\_s^{k\_s}.\tag{1}$$

For previous literature, the reader could refer to section P12 of LeVeque's *Reviews in number theory* and the bibliography of Vaughan [1]. For the special case, *k*<sup>1</sup> = *k*<sup>2</sup> = ··· = *ks*, an interesting problem is to determine the value for *k* 2, called Waring's problem, of the function *G*(*k*), the least positive number *s* such that every *sufficiently large* number can be represented the sum of at most *s k*-th powers of natural numbers. For this problem, there are only two values of the function *G*(*k*) determined exactly. To be specific, *G*(2) = 4, by Lagrange in 1770, and *G*(4) = 16, by Davenport [2]. The majority of information for *G*(*k*) has been derived from the Hardy–Littlewood method. This method has arised from a celebrated paper of Hardy and Ramanujan [3], which focused on the partition function.

There are many authors who devoted to establish many kinds of generalisations of this classical version of Waring's problem. Among these results, it is necessary to illustrate some of the majority variants. We begin with the most famous Waring–Goldbach problem, for which one devotes to investigate the possibility of the representation of integers as sums of *k*-th powers of prime numbers. In order to explain the associated congruence conditions, we denote by *k* a natural number and *p* a prime number. We write *<sup>θ</sup>* <sup>=</sup> *<sup>θ</sup>*(*k*; *<sup>p</sup>*) as the integer with the properties *<sup>p</sup><sup>θ</sup>* <sup>|</sup>*<sup>k</sup>* and *<sup>p</sup><sup>θ</sup> k*, and then define *γ* = *γ*(*k*, *p*) by

$$\gamma(k,p) = \begin{cases} \theta + 2, & \text{when } p = 2 \text{ and } \theta > 0, \\ \theta + 1, & \text{otherwise.} \end{cases}$$

Also, we set

$$K(k) = \prod\_{(p-1)|k} p^{\gamma}.$$

Denote by *H*(*k*) the smallest integer *s*, which satisfies every sufficiently large integer congruent to *s* modulo *K*(*k*) can be represented as the sum of *s k*-th powers of primes . By noting the fact that for (*<sup>p</sup>* <sup>−</sup> <sup>1</sup>)|*k*, we have *<sup>p</sup><sup>θ</sup>* (*<sup>p</sup>* <sup>−</sup> <sup>1</sup>)|*k*, provided that *<sup>a</sup><sup>k</sup>* <sup>≡</sup> <sup>1</sup> (mod *<sup>p</sup>γ*) and (*p*, *<sup>a</sup>*) = 1. This states the seemingly awkward definition of *H*(*k*), because if *n* is the sum of *s k*-th powers of primes exceeding *k* + 1, then it must satisfy *n* ≡ *s* (mod *K*(*k*)). Trivially, further congruence conditions could arise from the primes *p* which satisfy (*p* − 1) *k*. Following the previous investigations of Vinogradov [4,5], Hua systematically considered and investigated the additive problems involving prime variables in his famous book (see Hua [6,7]).

For the nonhomogeneous case, the most optimistic conjecture suggests that, for each prime *p*, if the Equation (1) has *p*-adic solutions and satisfies

$$k\_1^{-1} + k\_2^{-1} + \dots + k\_s^{-1} > 1,\tag{2}$$

then *n* can be written as the sum of unlike powers of positive integers (1) provided that *n* is sufficiently large in terms of *k*. For *s* = 3, such an claim maybe not true in certain situations (see Jagy and Kaplansky [8], or Exercise 5 of Chapter 8 of Vaughan [1]). However, a guide of application for the Hardy–Littlewood method suggests that the condition (2) should ensure at least that *almost all* integers satisfying the expected congruence conditions can be represented. Moreover, once subject to the following condition

$$k\_1^{-1} + k\_2^{-1} + \dots + k\_s^{-1} > 2,\tag{3}$$

a standard application of the Hardy–Littlewood method suggests that all the integers, which satisfy necessary congruence conditions, could be written in the form (1). Meanwhile, a conventional argument of the circle method shows that in situations in which the condition (2) does not hold, then every sufficiently large integer can not be represented in the expected form.

Since the Hardy–Littlewood method, the investigation of Waring's problem for unlike powers has produced splendid progress in circle method, especially for the classical version of Waring's problem. Additive Waring's problems of unlike powers involving squares, cubes or biquadrates offen attract greater interest of many mathematicians than those cases with higher mixed powers, and the current circumstance is quite satisfactory. For example, the reader can refer to references [9–19].

The Waring–Goldbach problem of mixed powers concerns the representation of *N* which satisfying some necessary congruence conditions as the form

$$N = p\_1^{k\_1} + p\_2^{k\_2} + \dots + p\_s^{k\_s}.$$

where *p*1, *p*2,..., *ps* are prime variables.

In 2002, Brüdern and Kawada [20] proved that for every sufficiently large even integer *N*, the equation

$$N = \mathbf{x} + p\_2^2 + p\_3^3 + p\_4^4$$

is solvable with *x* being an almost–prime P<sup>2</sup> and the *pj* (*j* = 2, 3, 4) primes. As usual, P*<sup>r</sup>* denotes an almost–prime with at most *r* prime factors, counted according to multiplicity. On the other hand, in 2015, Zhao [21] established that, for *k* = 3 or 4, every sufficiently large even integer *N* can be represented as the form

$$N = p\_1 + p\_2^2 + p\_3^3 + p\_4^k + 2^{\nu\_1} + 2^{\nu\_2} + \dots + 2^{\nu\_{t(k)}},$$

where *p*1, ... , *p*<sup>4</sup> are primes, *ν*1, *ν*2, ... , *νt*(*k*) are natural numbers, and *t*(3) = 16, *t*(4) = 18, which is an improvement result of Liu and Lü [22]. Afterwards, Lü [23] improved the result of Zhao [21] and showed that every sufficiently large even integer *N* can be represented as a sum of one prime, one square of prime, one cube of prime, one biquadrate of prime and 16 powers of 2.

In view of the results of Brüdern and Kawada [20], Zhao [21], Liu and Lü [22] and Lü [23], it is reasonable to conjecture that, for sufficiently large integer *N* satisfying *N* ≡ 0 (mod2), the following Diophantine equation

$$N = p\_1 + p\_2^2 + p\_3^3 + p\_4^4$$

is solvable, here and below the letter *p*, with or without subscript, always denotes a prime number. However, this conjecture may be out of reach at present with the known methods and techniques.

In this paper, we shall consider the exceptional set of the problem (4) and establish the following result.

**Theorem 1.** *Let E*(*N*) *denote the number of positive integers n, which satisfy n* ≡ 0 (mod2)*, up to N, which can not be represented as*

$$m = p\_1 + p\_2^2 + p\_3^3 + p\_4^4. \tag{4}$$

*Then, for any ε* > 0*, we have*

$$E(N) \ll N^{\frac{61}{10} + \varepsilon}.$$

We will establish Theorem 1 by using a pruning process into the Hardy–Littlewood circle method. For the treatment on minor arcs, we will employ the argument developed by Wooley in [24] combined with the new estimates for exponential sum over primes developed by Zhao [25]. For the treatment on major arcs, we shall prune the major arcs further and deal with them respectively. The explicit details will be given in the related sections.

**Notation.** In this paper, let *p*, with or without subscripts, always denote a prime number; *ε* always denotes a sufficiently small positive constant, which may not be the same at different occurrences. The letter *c* always denotes a positive constant. As usual, we use *χ* mod *q* to denote a Dirichlet character modulo *q*, and *χ*<sup>0</sup> mod *q* the principal character. Moreover, we use *ϕ*(*n*) and *d*(*n*) to denote the Euler's function and Dirichlet's divisor function, respectively. *<sup>e</sup>*(*x*) = *<sup>e</sup>*2*πix*; *<sup>f</sup>*(*x*)  *<sup>g</sup>*(*x*) means that *f*(*x*) = *O*(*g*(*x*)); *f*(*x*) *g*(*x*) means that *f*(*x*)  *g*(*x*)  *f*(*x*). *N* is a sufficiently large integer and *n* ∈ (*N*/2, *N*], and hence log *N* log *n*.

### **2. Outline of the Proof of Theorem 1**

Let *N* be a sufficiently large positive integer. By a splitting argument, it is sufficient to consider the even integers *n* ∈ (*N*/2, *N*]. For the application of the Hardy–Littlewood method, it is necessary to define the Farey dissection. For this purpose, we set the parameters as follows

$$A = 100^{100}, \quad Q\_0 = \log^A N, \quad Q\_1 = N^{\frac{1}{6}}, \quad Q\_2 = N^{\frac{5}{6}}, \quad \Im\_0 = \left[ -\frac{1}{Q\_2}, 1 - \frac{1}{Q\_2} \right].$$

By Dirichlet's rational approximation lemma (for instance, see Lemma 12 on p.104 of [26], or Lemma 2.1 of [1]), each *α* ∈ (−1/*Q*2, 1 − 1/*Q*2] can be represented in the form

$$a = \frac{a}{q} + \lambda\_\prime \qquad |\lambda| \ll \frac{1}{qQ\_2\prime},$$

for some integers *a*, *q* with 1 *a q* -*Q*<sup>2</sup> and (*a*, *q*) = 1. Define

$$\mathfrak{M}(q,a) = \left[\frac{a}{q} - \frac{1}{qQ\_2}, \frac{a}{q} + \frac{1}{qQ\_2}\right], \qquad \mathfrak{M} = \bigcup\_{1 \le q \le Q\_1} \bigcup\_{\substack{1 \le a \le q \\ (a,q) = 1}} \mathfrak{M}(q,a),$$

$$\mathfrak{M}\_0(q,a) = \left[\frac{a}{q} - \frac{Q\_0^{100}}{qN}, \frac{a}{q} + \frac{Q\_0^{100}}{qN}\right], \qquad \mathfrak{M}\_0 = \bigcup\_{1 \le q \le Q\_0^{100}} \bigcup\_{\substack{1 \le a \le q \\ (a,q) = 1}} \mathfrak{M}\_0(q,a),$$

$$\mathfrak{m}\_1 = \mathfrak{I}\_0 \backslash \mathfrak{M}, \qquad \mathfrak{m}\_2 = \mathfrak{M} \backslash \mathfrak{M}\_0.$$

Then we obtain the Farey dissection

$$
\mathfrak{J}\_0 = \mathfrak{M}\_0 \cup \mathfrak{m}\_1 \cup \mathfrak{m}\_2.\tag{5}
$$

For *k* = 1, 2, 3, 4, we define

$$f\_k(a) = \sum\_{X\_k < p \in 2X\_k} e(p^k a),$$

where *Xk* = (*N*/16) 1 *<sup>k</sup>* . Let

$$\mathcal{R}(n) = \sum\_{\substack{\mathfrak{n} = p\_1 + p\_2^2 + p\_3^3 + p\_4^4\\ \mathfrak{X}\_i < p\_i \le 2\mathcal{X}\_i\\ i = 1, 2, 3, 4}} 1.$$

From (5), one has

$$\begin{split} \mathcal{R}(n) &= \int\_{0}^{1} \left( \prod\_{k=1}^{4} f\_{k}(a) \right) \mathbf{e}(-na) \mathbf{d}a = \int\_{-\frac{1}{\mathsf{Q}\_{2}}}^{1-\frac{1}{\mathsf{Q}\_{2}}} \left( \prod\_{k=1}^{4} f\_{k}(a) \right) \mathbf{e}(-na) \mathbf{d}a \\ &= \left\{ \int\_{\mathfrak{M}\_{0}} + \int\_{\mathfrak{m}\_{1}} + \int\_{\mathfrak{m}\_{2}} \right\} \left( \prod\_{k=1}^{4} f\_{k}(a) \right) \mathbf{e}(-na) \mathbf{d}a. \end{split}$$

In order to prove Theroem 1, we need the two following propositions:

**Proposition 1.** *For n* ∈ (*N*/2, *N*]*, there holds*

$$\int\_{\mathfrak{M}\_0} \left( \prod\_{k=1}^4 f\_k(a) \right) e(-na) da = \frac{\Gamma(2)\Gamma(\frac{3}{2})\Gamma(\frac{4}{3})\Gamma(\frac{5}{4})}{\Gamma(\frac{25}{12})} \mathfrak{S}(n) \frac{n^{\frac{13}{12}}}{\log^4 n} + O\left(\frac{n^{\frac{13}{12}}}{\log^5 n}\right),\tag{6}$$

*where* S(*n*) *is the singular series defined in (10), which is absolutely convergent and satisfies*

$$(\log \log n)^{-\varepsilon^\*} \ll \mathfrak{S}(n) \ll d(n) \tag{7}$$

*for any integer n satisfying n* ≡ 0 (mod2) *and some fixed constant c*<sup>∗</sup> > 0*.*

The proof of (6) in Proposition 1 follows from the well–know standard technique in the Hardy–Littlewood method. For more information, one can see pp. 90–99 of Hua [7], so we omit the details herein. For the properties (7) of singular series, we shall give the proof in Section 4.

**Proposition 2.** *Let* Z(*N*) *denote the number of integers n* ∈ (*N*/2, *N*] *satisfying n* ≡ 0 (mod2) *such that*

$$\sum\_{j=1}^{2} \left| \int\_{\mathbf{m}\_j} \left( \prod\_{k=1}^{4} f\_k(\alpha) \right) \varepsilon(-n\alpha) \mathrm{d}\alpha \right| \gg \frac{n^{\frac{13}{12}}}{\log^5 n}.$$

*Then we have*

$$Z(N) \ll N^{\frac{61}{10} + \varepsilon}.$$

The proof of Proposition 2 will be given in Section 5. The remaining part of this section is devoted to establishing Theorem 1 by using Proposition 1 and Proposition 2.

*Proof of Theorem 1.* From Proposition 2, we deduce that, with at most *<sup>O</sup> N* <sup>61</sup> <sup>144</sup> +*ε* exceptions, all even integers *n* ∈ (*N*/2, *N*] satisfy

$$\sum\_{j=1}^{2} \left| \int\_{\mathfrak{m}\_j} \left( \prod\_{k=1}^{4} f\_k(\alpha) \right) \varepsilon(-n\alpha) \mathrm{d}\alpha \right| \ll \frac{n^{\frac{13}{12}}}{\log^5 n} \lambda$$

from which and Proposition 1, we conclude that, with at most *O N* <sup>61</sup> <sup>144</sup> +*ε* exceptions, for all even integers *n* ∈ (*N*/2, *N*], R(*n*) holds the asymptotic formula

$$\mathcal{A}(n) = \frac{\Gamma(2)\Gamma(\frac{3}{2})\Gamma(\frac{4}{3})\Gamma(\frac{5}{4})}{\Gamma(\frac{25}{12})} \mathfrak{S}(n) \frac{n^{\frac{15}{12}}}{\log^4 n} + O\left(\frac{n^{\frac{15}{12}}}{\log^5 n}\right).$$

In other words, all even integers *<sup>n</sup>* ∈ (*N*/2, *<sup>N</sup>*] can be represented in the form *<sup>p</sup>*<sup>1</sup> + *<sup>p</sup>*<sup>2</sup> <sup>2</sup> + *<sup>p</sup>*<sup>3</sup> <sup>3</sup> + *<sup>p</sup>*<sup>4</sup> 4 with at most *O N* <sup>61</sup> <sup>144</sup> +*ε* exceptions, where *p*1, *p*2, *p*3, *p*<sup>4</sup> are prime numbers. By a splitting argument, we get

$$E(N) \ll \sum\_{0 \le \ell \ll \log N} \mathcal{Z}\left(\frac{N}{2^{\ell}}\right) \ll \sum\_{0 \le \ell \ll \log N} \left(\frac{N}{2^{\ell}}\right)^{\frac{61}{144} + \varepsilon} \ll N^{\frac{61}{144} + \varepsilon}.$$

This completes the proof of Theorem 1.

### **3. Some Auxiliary Lemmas**

In this section, we shall list some necessary lemmas which will be used in proving Proposition 2.

**Lemma 1.** *Suppose that α is a real number, and that* |*α* − *a*/*q*| *<sup>q</sup>*−<sup>2</sup> *with* (*a*, *<sup>q</sup>*) = <sup>1</sup>*. Let <sup>β</sup>* = *<sup>α</sup>* − *<sup>a</sup>*/*q. Then we have*

$$f\_k(\mathbf{a}) \ll d^{\delta\_k}(q) (\log \mathbf{x})^c \left( X\_k^{1/2} \sqrt{q(1+N|\beta|)} + X\_k^{4/5} + \frac{X\_k}{\sqrt{q(1+N|\beta|)}} \right),$$

*where δ<sup>k</sup>* = <sup>1</sup> <sup>2</sup> <sup>+</sup> log *<sup>k</sup>* log 2 *and c is a constant.*

**Proof.** See Theorem 1.1 of Ren [27].

**Lemma 2.** *Suppose that α is a real number, and that there exist a* ∈ Z *and q* ∈ N *with*

$$(a,q) = 1, \qquad 1 \preccurlyeq q \lessgtr X \qquad \text{and} \qquad |qa - a| \lessgtr X^{-1}.$$

*If P*2*δ*21−*<sup>k</sup>* - *X* - *Pk*−2*δ*21−*<sup>k</sup> , then one has*

$$\sum\_{P < p \le 2P} \varepsilon \left( p^k \mathfrak{a} \right) \ll P^{1 - \delta 2^{1-k} + \varepsilon} + \frac{P^{1+\varepsilon}}{q^{1/2} \left( 1 + P^k |\mathfrak{a} - \mathfrak{a}/q| \right)^{1/2}}.$$

*where δ* = 1/3 *for k* 4*.*

**Proof.** See Lemma 2.4 of Zhao [25].

**Lemma 3.** *Suppose that α is a real number, and that there are a* ∈ Z *and q* ∈ N *with*

$$(a,q) = 1, \qquad 1 \lessapprox q \lessapprox Q \qquad \text{and} \qquad |qa - a| \lessapprox Q^{-1}.$$

*If P*<sup>1</sup> <sup>2</sup> - *Q* - *P*<sup>5</sup> <sup>2</sup> *, then one has*

$$\sum\_{P$$

**Proof.** See Lemma 8.5 of Zhao [25].

**Lemma 4.** *For α* ∈ m1*, we have*

$$f\_3(\alpha) \ll N^{\frac{11}{96} + \varepsilon} \qquad \text{and} \qquad f\_4(\alpha) \ll N^{\frac{23}{96} + \varepsilon}.$$

**Proof.** For *α* ∈ m1, we have *Q*<sup>1</sup> *q* -*Q*2. By Lemma 3, we get

$$f\_3(\alpha) \ll X\_3^{\frac{11}{12} + \varepsilon} + X\_3^{1 + \varepsilon} Q\_1^{-\frac{1}{6}} \ll N^{\frac{1}{3} + \varepsilon}.$$

From Lemma 2, we obtain

$$f\_4(\alpha) \ll X\_4^{\frac{23}{74} + \varepsilon} + X\_4^{1 + \varepsilon} Q\_1^{-\frac{1}{2}} \ll N^{\frac{23}{76} + \varepsilon}.$$

This completes the proof of Lemma 4.

For 1 *a q* with (*a*, *q*) = 1, set

$$\mathcal{Z}(q,a) = \left[\frac{a}{q} - \frac{1}{q\mathbb{Q}\_0}, \frac{a}{q} + \frac{1}{q\mathbb{Q}\_0}\right], \qquad \mathcal{Z} = \bigcup\_{1 \le q \le Q\_0} \bigcup\_{\substack{a=-q\\(a,q)=1}}^{2q} \mathcal{Z}(q,a). \tag{8}$$

For *α* ∈ m2, by Lemma 1, we have

$$f\_3(a) \ll \frac{N^{\frac{1}{5}} \log^{\varepsilon} N}{q^{\frac{1}{2} - \varepsilon} \left(1 + N|\lambda|\right)^{1/2}} + N^{\frac{4}{15} + \varepsilon} = V\_3(a) + N^{\frac{4}{15} + \varepsilon},\tag{9}$$

say. Then we obtain the following Lemma.

**Lemma 5.** *We have*

$$\int\_{\mathcal{T}} |V \mathfrak{z}(\mathfrak{a})|^4 \mathrm{d}\mathfrak{a} = \sum\_{1 \le q \le Q\_0} \sum\_{\substack{\mathfrak{a} = -q \\ (\mathfrak{a}, q) = 1}}^{2q} \int\_{\mathcal{T}(q, \mathfrak{a})} |V \mathfrak{z}(\mathfrak{a})|^4 \mathrm{d}\mathfrak{a} \ll N^{\frac{1}{3}} \log^{\epsilon} N.$$

**Proof.** We have

$$\begin{split} &\sum\_{1\leq q\leq Q\_{0}}\sum\_{\begin{subarray}{c}a=-q\\(a,q)=1\end{subarray}}^{2q}\int\_{\mathbb{T}(q,a)}|V\_{3}(\alpha)|^{4}\mathbf{d}a \\ &\ll\sum\_{1\leq q\leq Q\_{0}}q^{-2+\varepsilon}\sum\_{\begin{subarray}{c}a=-q\\(a,q)=1\end{subarray}}^{2q}\int\_{|\lambda|\leq\frac{1}{\log b}}\frac{N^{\frac{\delta}{\delta}}\log^{\varepsilon}N}{(1+N|\lambda|)^{2}}\mathbf{d}\lambda \\ &\ll\sum\_{1\leq q\leq Q\_{0}}q^{-2+\varepsilon}\sum\_{\begin{subarray}{c}a=-q\\(a,q)=1\end{subarray}}^{2q}\left(\int\_{|\lambda|\leq\frac{1}{\delta}}N^{\frac{\delta}{\delta}}\log^{\varepsilon}N\mathbf{d}\lambda+\int\_{\frac{1}{\delta}}\frac{N^{\frac{\delta}{\delta}}\log^{\varepsilon}N}{N^{2}\lambda^{2}}\mathbf{d}\lambda\right) \\ &\ll N^{\frac{1}{2}}\log^{\varepsilon}N\sum\_{1\leq q\leq Q\_{0}}q^{-2+\varepsilon}\varrho(q)\ll N^{\frac{1}{2}}Q\_{0}^{\varepsilon}\log^{\varepsilon}N\ll N^{\frac{1}{2}}\log^{\varepsilon}N. \end{split}$$

This completes the proof of Lemma 5.

### **4. The Singular Series**

In this section, we shall concentrate on investigating the properties of the singular series which appear in Proposition 1. First, we illustrate some notations. For *k* ∈ {1, 2, 3, 4} and a Dirichlet character *χ* mod *q*, we define

$$\mathbb{C}\_k(\chi, a) = \sum\_{h=1}^q \overline{\chi(h)} e\left(\frac{ah^k}{q}\right), \qquad \mathbb{C}\_k(q, a) = \mathbb{C}\_k(\chi^0, a),$$

where *χ*<sup>0</sup> is the principal character modulo *q*. Let *χ*1, *χ*2, *χ*3, *χ*<sup>4</sup> be Dirichlet characters modulo *q*. Set

$$B(n,q,\chi\_1,\chi\_2,\chi\_3,\chi\_4) = \sum\_{\substack{a=1\\(a,q)=1}}^q \mathbb{C}\_1(\chi\_1,a)\mathbb{C}\_2(\chi\_2,a)\mathbb{C}\_3(\chi\_3,a)\mathbb{C}\_4(\chi\_4,a)\varepsilon \left(-\frac{an}{q}\right),$$

$$B(n,q) = B\left(n,q,\chi^0,\chi^0,\chi^0,\chi^0\right),$$

and write

$$A(n,q) = \frac{B(n,q)}{q^4(q)}, \qquad \mathfrak{S}(n) = \sum\_{q=1}^{\infty} A(n,q). \tag{10}$$

**Lemma 6.** *For* (*a*, *q*) = 1 *and any Dirichlet character χ* mod *q, there holds*

$$|\mathbb{C}\_k(\chi, a)| \ll 2q^{1/2} d^{\beta\_k}(q)$$

*with β<sup>k</sup>* = (log *k*)/ log 2*.*

**Proof.** See the Problem 14 of Chapter VI of Vinogradov [28].

**Lemma 7.** *Let p be a prime and pαk. For* (*a*, *<sup>p</sup>*) = <sup>1</sup>*, if* - *γ*(*p*)*, we have Ck*(*p*-, *a*) = 0*, where*

$$\gamma(p) = \begin{cases} \mathfrak{a} + \mathfrak{2}, & \text{if } p \neq 2 \text{ or } p = 2, \mathfrak{a} = 0; \\ \mathfrak{a} + \mathfrak{3}, & \text{if } p = 2, \mathfrak{a} > 0. \end{cases}$$

**Proof.** See Lemma 8.3 of Hua [7].

For *k* 1, we define

$$S\_k(q, a) = \sum\_{m=1}^q e\left(\frac{am^k}{q}\right).$$

**Lemma 8.** *Suppose that* (*p*, *a*) = 1*. Then*

$$S\_k(p, a) = \sum\_{\chi \in \alpha\_k^{\ell}} \overline{\chi(a)} \tau(\chi)\_{\ell}$$

*where* A*<sup>k</sup> denotes the set of non–principal characters χ modulo p for which χ<sup>k</sup> is principal, and τ*(*χ*) *denotes the Gauss sum*

$$\sum\_{m=-1}^{p} \chi(m) e\left(\frac{m}{p}\right).$$

*Also, there hold* |*τ*(*χ*)| = *<sup>p</sup>*1/2 *and* |A*k*| = (*k*, *<sup>p</sup>* − <sup>1</sup>) − <sup>1</sup>*.*

**Proof.** See Lemma 4.3 of Vaughan [1].

**Lemma 9.** *For* (*p*, *n*) = 1*, we have*

$$\left| \sum\_{a=1}^{p-1} \left( \prod\_{k=1}^{4} \frac{S\_k(p,a)}{p} \right) \mathfrak{e} \left( -\frac{an}{p} \right) \right| \leqslant 24p^{-\frac{3}{2}}.\tag{11}$$

**Proof.** We denote by S the left-hand side of (11). It follows from Lemma 8 that

$$\mathcal{S} = \frac{1}{p^4} \sum\_{a=1}^{p-1} \left( \prod\_{k=1}^4 \left( \sum\_{\chi\_k \in \mathcal{\wp}\_k'} \overline{\chi\_k(a)} \tau(\chi\_k) \right) \right) \varepsilon \left( -\frac{an}{p} \right).$$

If |A*k*| = 0 for some *k* ∈ {1, 2, 3, 4}, then S = 0. If this is not the case, then

$$\begin{split} \mathcal{S} &= \frac{1}{p^4} \sum\_{\chi\_1 \in \mathcal{A}\_1'} \sum\_{\chi\_2 \in \mathcal{A}\_2'} \sum\_{\chi\_3 \in \mathcal{A}\_3'} \sum\_{\chi\_4 \in \mathcal{A}\_4'} \tau(\chi\_1) \tau(\chi\_2) \tau(\chi\_3) \tau(\chi\_4) \\ & \times \sum\_{a=1}^{p-1} \overline{\chi\_1(a) \chi\_2(a) \chi\_3(a) \chi\_4(a)} \varepsilon \left( - \frac{an}{p} \right). \end{split}$$

From Lemma 8, the quadruple outer sums have no more than 4! = 24 terms. For each of these terms, there holds

$$|\pi(\chi\_1)\pi(\chi\_2)\pi(\chi\_3)\pi(\chi\_4)| = p^2.$$

Since in any one of these terms *χ*1(*a*)*χ*2(*a*)*χ*3(*a*)*χ*4(*a*) is a Dirichlet character *χ* (mod*p*), the inner sum is

$$\sum\_{a=1}^{p-1} \chi(a)e\left(-\frac{an}{p}\right) = \overline{\chi(-n)} \sum\_{a=1}^{p-1} \chi(-an)e\left(-\frac{an}{p}\right) = \overline{\chi(-n)}\tau(\chi).$$

By noting the fact that *<sup>τ</sup>*(*χ*0) = −1 for principal character *<sup>χ</sup>*<sup>0</sup> mod *<sup>p</sup>*, we derive that

$$|\overline{\chi(-n)}\tau(\chi)| \ll p^{\frac{1}{2}}.$$

From the above arguments, we deduce that

$$|\mathcal{S}| \leqslant \frac{1}{p^4} \cdot 24 \cdot p^2 \cdot p^{\frac{1}{2}} = 24p^{-\frac{3}{2}}\prime$$

which completes the proof of Lemma 9.

**Lemma 10.** *Let* L(*p*, *n*) *denote the number of solutions of the congruence*

$$x\_1 + x\_2^2 + x\_3^3 + x\_4^4 \equiv n \pmod{p}, \qquad 1 \le x\_1, x\_2, x\_3, x\_4 \le p - 1.$$

*Then, for n* ≡ 0 (mod2)*, we have* L(*p*, *n*) > 0*.*

**Proof.** We have

$$p \cdot \mathcal{L}(p, n) = \sum\_{a=1}^{p} \mathbb{C}\_1(p, a) \mathbb{C}\_2(p, a) \mathbb{C}\_3(p, a) \mathbb{C}\_4(p, a) e\left(-\frac{an}{p}\right) = (p - 1)^4 + E\_{p, 1}$$

where

$$E\_p = \sum\_{a=1}^{p-1} \mathbb{C}\_1(p, a) \mathbb{C}\_2(p, a) \mathbb{C}\_3(p, a) \mathbb{C}\_4(p, a) \varepsilon \left( -\frac{an}{p} \right).$$

By Lemma 8, we obtain

$$|E\_p| \lesssim (p-1)(\sqrt{p}+1)(2\sqrt{p}+1)(3\sqrt{p}+1)\dots$$

It is easy to check that |*Ep*| < (*<sup>p</sup>* − <sup>1</sup>)<sup>4</sup> for *<sup>p</sup>* 7. Therefore, we obtain L(*p*, *<sup>n</sup>*) > 0 for *<sup>p</sup>* 7. For *p* = 2, 3, 5, we can check L(*p*, *n*) > 0 one by one. This completes the proof of Lemma 10.

**Lemma 11.** *A*(*n*, *q*) *is multiplicative in q.*

**Proof.** From the definition of *A*(*n*, *q*) in (10), it is sufficient to show that *B*(*n*, *q*) is multiplicative in *q*. Suppose *q* = *q*1*q*<sup>2</sup> with (*q*1, *q*2) = 1. Then we obtain

$$\begin{aligned} B(n, q\_1 q\_2) &= \sum\_{\substack{a=1\\(a, q\_1 q\_2) = 1}}^{q\_1 q\_2} \left( \prod\_{k=1}^4 \mathbb{C}\_k(q\_1 q\_2, a) \right) \varepsilon \left( -\frac{an}{q\_1 q\_2} \right) \\ &= \sum\_{\substack{a\_1=1\\(a\_1 q\_1) = 1}}^{q\_1} \sum\_{\substack{a\_2=1\\(a\_2 q\_2) = 1}}^{q\_2} \left( \prod\_{k=1}^4 \mathbb{C}\_k(q\_1 q\_2, a\_1 q\_2 + a\_2 q\_1) \right) \varepsilon \left( -\frac{a\_1 n}{q\_1} \right) \varepsilon \left( -\frac{a\_2 n}{q\_2} \right) . \end{aligned} \tag{12}$$

For (*q*1, *q*2) = 1, there holds

$$\begin{split} \mathsf{C}\_{k}(q\_{1}q\_{2},a\_{1}q\_{2}+a\_{2}q\_{1}) &= \sum\_{\begin{subarray}{c}m=1\\(m,q\_{1}q\_{2})=1\end{subarray}}^{q\_{1}q\_{2}} \varepsilon\left(\frac{(a\_{1}q\_{2}+a\_{2}q\_{1})m^{k}}{q\_{1}q\_{2}}\right) \\ &= \sum\_{\begin{subarray}{c}m\_{1}=1\\(m\_{1},q\_{1})=1\end{subarray}}^{q\_{1}} \sum\_{\begin{subarray}{c}m\_{2}=1\\m\_{2}=1\end{subarray}}^{q\_{2}} \varepsilon\left(\frac{(a\_{1}q\_{2}+a\_{2}q\_{1})(m\_{1}q\_{2}+m\_{2}q\_{1})^{k}}{q\_{1}q\_{2}}\right) \\ &= \sum\_{\begin{subarray}{c}m\_{1}=1\\(m\_{1},q\_{1})=1\end{subarray}}^{q\_{1}} \varepsilon\left(\frac{a\_{1}(m\_{1}q\_{2})^{k}}{q\_{1}}\right) \sum\_{\begin{subarray}{c}m\_{2}=1\\(m\_{2},q\_{2})=1\end{subarray}}^{q\_{2}} \varepsilon\left(\frac{a\_{2}(m\_{2}q\_{1})^{k}}{q\_{2}}\right) \\ &= \mathsf{C}\_{k}(q\_{1},a\_{1})\mathsf{C}\_{k}(q\_{2},a\_{2}). \end{split} \tag{13}$$

Putting (13) into (12), we deduce that

$$\begin{aligned} B(n, q\_1 q\_2) &= \sum\_{\substack{a\_1=1\\(a\_1, q\_1)=1}}^{q\_1} \left( \prod\_{k=1}^4 \mathbb{C}\_k(q\_1, a\_1) \right) \varepsilon \left( -\frac{a\_1 n}{q\_1} \right) \sum\_{\substack{a\_2=1\\(a\_2, q\_2)=1}}^{q\_2} \left( \prod\_{k=1}^4 \mathbb{C}\_k(q\_2, a\_2) \right) \varepsilon \left( -\frac{a\_2 n}{q\_2} \right) \\ &= B(n, q\_1) B(n, q\_2) . \end{aligned}$$

This completes the proof of Lemma 11.

**Lemma 12.** *Let A*(*n*, *q*) *be as defined in (10). Then*

*(i) we have*

$$\sum\_{q>Z} |A(n,q)| \ll Z^{-\frac{1}{2}+\iota} d(n)\_{\iota}$$

*and thus the singular series* S(*n*) *is absolutely convergent and satisfies* S(*n*)  *d*(*n*)*.*

*(ii) there exists an absolute positive constant c*<sup>∗</sup> > 0*, such that, for n* ≡ 0 (mod 2)*,*

$$\mathfrak{S}(n) \gg (\log \log n)^{-c^\*}$$

**Proof.** From Lemma 11, we know that *B*(*n*, *q*) is multiplicative in *q*. Therefore, there holds

$$B(n,q) = \prod\_{p^t \parallel q} B(n,p^t) = \prod\_{p^t \parallel q} \sum\_{\substack{a=1 \\ (a,p)=1}}^{p^t} \left( \prod\_{k=1}^4 \mathbb{C}\_k(p^t, a) \right) e\left(-\frac{an}{p^t}\right). \tag{14}$$

.

From (14) and Lemma 7, we deduce that *B*(*n*, *q*) = ∏ *pq B*(*n*, *p*) or 0 according to *q* is square–free or not. Thus, one has <sup>∞</sup>

$$\sum\_{q=1}^{\infty} A(n, q) = \sum\_{\substack{q=1 \\ q \text{ square-free}}}^{\infty} A(n, q). \tag{15}$$

Write

$$\mathcal{R}\left(p,a\right) := \prod\_{k=1}^4 \mathbb{C}\_k(p,a) - \prod\_{k=1}^4 \mathbb{S}\_k(p,a).$$

Then

$$A(n,p) = \frac{1}{(p-1)^4} \sum\_{a=1}^{p-1} \left( \prod\_{k=1}^4 \mathcal{S}\_k(p,a) \right) \varepsilon \left( -\frac{an}{p} \right) + \frac{1}{(p-1)^4} \sum\_{a=1}^{p-1} \mathcal{R}(p,a) \varepsilon \left( -\frac{an}{p} \right). \tag{16}$$

Applying Lemma 6 and noticing that *Sk*(*p*, *a*) = *Ck*(*p*, *a*) + 1, we get *Sk*(*p*, *a*)  *p* 1 <sup>2</sup> , and thus R(*p*, *a*)  *p* 3 <sup>2</sup> . Therefore, the second term in (16) is *<sup>c</sup>*<sup>1</sup> *<sup>p</sup>*<sup>−</sup> <sup>3</sup> <sup>2</sup> . On the other hand, from Lemma 9, we can see that the first term in (16) is - <sup>2</sup><sup>4</sup> · <sup>24</sup>*p*<sup>−</sup> <sup>3</sup> <sup>2</sup> = 384*p*<sup>−</sup> <sup>3</sup> <sup>2</sup> . Let *c*<sup>2</sup> = max(*c*1, 384). Then we have proved that, for *p n*, there holds

$$|A(n,p)| \leqslant c\_2 p^{-\frac{3}{2}}.\tag{17}$$

Moreover, if we use Lemma 6 directly, it follows that

$$\begin{aligned} |B(n,p)| &= \left| \sum\_{a=1}^{p-1} \left( \prod\_{k=1}^4 \mathbb{C}\_k(p,a) \right) \mathfrak{e} \left( -\frac{an}{p} \right) \right| \lesssim \sum\_{a=1}^{p-1} \prod\_{k=1}^4 |\mathbb{C}\_k(p,a)| \\ &\lesssim (p-1) \cdot 2^4 \cdot p^2 \cdot 24 = 384p^2(p-1), \end{aligned}$$

and therefore

$$\left|A(n,p)\right| = \frac{|B(n,p)|}{\varrho^4(p)} \lessapprox \frac{384p^2}{(p-1)^3} \lessapprox \frac{2^3 \cdot 384p^2}{p^3} = \frac{3072}{p}.\tag{18}$$

Let *c*<sup>3</sup> = max(*c*2, 3072). Then, for square–free *q*, we have

$$\begin{split} |A(n,q)| &= \left(\prod\_{\substack{p|q\\p\nmid n}} |A(n,p)| \right) \left(\prod\_{\substack{p|q\\p\mid n}} |A(n,p)| \right) \leqslant \left(\prod\_{\substack{p|q\\p\nmid n}} (c\_3 p^{-\frac{3}{2}}) \right) \left(\prod\_{\substack{p|q\\p\mid n}} (c\_3 p^{-1}) \right) \\ &= c\_3^{\omega(q)} \left(\prod\_{p|q} p^{-\frac{3}{2}} \right) \left(\prod\_{p|(n,q)} p^{\frac{1}{2}} \right) \ll q^{-\frac{3}{2} + \epsilon} (n,q)^{\frac{1}{2}}. \end{split}$$

Hence, by (15), we obtain

$$\begin{split} \sum\_{q>Z} |A(n,q)| \ll \sum\_{q>Z} q^{-\frac{3}{2}+\varepsilon} (n,q)^{\frac{1}{2}} &= \sum\_{d|n} \sum\_{q>\frac{Z}{d}} (dq)^{-\frac{3}{2}+\varepsilon} d^{\frac{1}{2}} = \sum\_{d|n} d^{-1+\varepsilon} \sum\_{q>\frac{Z}{d}} q^{-\frac{3}{2}+\varepsilon} \\ &\ll \sum\_{d|n} d^{-1+\varepsilon} \left(\frac{Z}{d}\right)^{-\frac{1}{2}+\varepsilon} = Z^{-\frac{1}{2}+\varepsilon} \sum\_{d|n} d^{-\frac{1}{2}+\varepsilon} \ll Z^{-\frac{1}{2}+\varepsilon} d(n). \end{split}$$

This proves (i) of Lemma 12.

To prove (ii) of Lemma 12, by Lemma 11, we first note that

$$\begin{split} \mathfrak{S}(n) &= \prod\_{p} \left( 1 + \sum\_{t=1}^{\infty} A(n, p^t) \right) = \prod\_{p} \left( 1 + A(n, p) \right) \\ &= \left( \prod\_{\substack{p < c\_3 \\ p \nmid n}} \left( 1 + A(n, p) \right) \right) \left( \prod\_{\substack{p > c\_3 \\ p \mid n}} \left( 1 + A(n, p) \right) \right) \left( \prod\_{\substack{p > c\_3 \\ p \mid n}} \left( 1 + A(n, p) \right) \right) . \end{split} \tag{19}$$

From (17), we have

$$\prod\_{\substack{p>c\_3\\p\nmid n}} \left(1 + A(n, p)\right) \geqslant \prod\_{p>c\_3} \left(1 - \frac{c\_3}{p^{3/2}}\right) \geqslant c\_4 > 0. \tag{20}$$

By (18), we know that there are *c*<sup>5</sup> > 0 such that

$$\prod\_{\substack{p>c\_3\\p\nmid n}} \left(1 + A(n, p)\right) \geqslant \prod\_{\substack{p>c\_3\\p\mid n}} \left(1 - \frac{c\_3}{p}\right) \geqslant \prod\_{p\mid n} \left(1 - \frac{c\_3}{p}\right) \gg (\log \log n)^{-c\_3}.\tag{21}$$

On the other hand, it is easy to see that

$$1 + A(n, p) = \frac{p \cdot \mathcal{L}(p, n)}{q^4(p)}.$$

By Lemma 10, we know that L(*p*, *n*) > 0 for all *p* with *n* ≡ 0 (mod 2), and thus 1 + *A*(*n*, *p*) > 0. Therefore, there holds

$$\prod\_{p \le c\_6 \atop p \le c\_6} (1 + A(n, p)) \gg c\_6 > 0. \tag{22}$$

.

Combining the estimates (19)–(22), and taking *c*<sup>∗</sup> = *c*<sup>5</sup> > 0, we derive that

$$\circledast(n) \gg (\log \log n)^{-c^\*}$$

This completes the proof Lemma 12.

### **5. Proof of Proposition 2**

In this section, we shall give the proof of Proposition 2. We denote by Z*j*(*N*) the set of integers *n* satisfying *n* ∈ [*N*/2, *N*] and *n* ≡ 0 (mod 2) for which the following estimate

$$\left| \int\_{\mathbf{m}\_j} \left( \prod\_{k=1}^4 f\_k(\alpha) \right) e(-n\alpha) \mathrm{d}\alpha \right| \gg \frac{n^{\frac{13}{12}}}{\log^5 n} \tag{23}$$

holds. For convenience, we use Z*<sup>j</sup>* to denote the cardinality of Z*j*(*N*) for abbreviation. Also, we define the complex number *ξj*(*n*) by taking *ξj*(*n*) = 0 for *n* ∈ Z*j*(*N*), and when *n* ∈ Z*j*(*N*) by means of the equation

$$\left| \int\_{\mathfrak{m}\_j} \left( \prod\_{k=1}^4 f\_k(n) \right) \varepsilon(-na) \mathrm{d}a \right| = \mathfrak{z}\_j(n) \int\_{\mathfrak{m}\_j} \left( \prod\_{k=1}^4 f\_k(n) \right) \varepsilon(-na) \mathrm{d}a. \tag{24}$$

Plainly, one has |*ξj*(*n*)| = 1 whenever *ξj*(*n*) is nonzero. Therefore, we obtain

$$\sum\_{\mathbf{m}\in\mathbb{Z}\_{\bar{\jmath}}(N)}\xi\_{\bar{\jmath}}(n)\int\_{\mathbf{m}\_{\bar{\jmath}}}\left(\prod\_{k=1}^{4}f\_{k}(a)\right)\mathbf{e}(-na)\,\mathrm{d}a=\int\_{\mathbf{m}\_{\bar{\jmath}}}\left(\prod\_{k=1}^{4}f\_{k}(a)\right)\mathcal{K}\_{\bar{\jmath}}(a)\,\mathrm{d}a,\tag{25}$$

where the exponential sum K*j*(*α*) is defined by

$$\mathcal{K}\_{\boldsymbol{\jmath}}(\boldsymbol{\alpha}) = \sum\_{\boldsymbol{n} \in \mathcal{Z}\_{\boldsymbol{\jmath}}(\boldsymbol{N})} \xi\_{\boldsymbol{\jmath}}(\boldsymbol{n}) \boldsymbol{e} (-n\boldsymbol{\alpha}) .$$

For *j* = 1, 2, set

$$I\_j = \int\_{\mathfrak{m}\_j} \left( \prod\_{k=1}^4 f\_k(\alpha) \right) \mathcal{K}\_j(\alpha) \mathrm{d}\alpha.$$

By (23)–(25), we derive that

$$I\_j \gg \sum\_{n \in \mathbb{Z}\_j(N)} \frac{n^{\frac{13}{12}}}{\log^{\frac{5}{6}} n} \gg \frac{Z\_j N^{\frac{13}{12}}}{\log^5 N}, \qquad j = 1, 2. \tag{26}$$

By Lemma 2.1 of Wooley [24] with *k* = 2, we know that, for *j* = 1, 2, there holds

$$\int\_0^1 \left| f\_2(a) \mathcal{K}\_{\dot{\jmath}}(a) \right|^2 \mathrm{d}a \ll N^{\varepsilon} \left( \mathcal{Z}\_{\dot{\jmath}} N^{\frac{1}{2}} + \mathcal{Z}\_{\dot{\jmath}}^2 \right). \tag{27}$$

It follows from Cauchy's inequality, Lemma 4 and (27) that

$$\begin{split} \left| I\_{1} \ll \left( \sup\_{a \in \mathfrak{m}\_{1}} |f\_{3}(a)| \right) \left( \sup\_{a \in \mathfrak{m}\_{1}} |f\_{4}(a)| \right) \left( \int\_{0}^{1} |f\_{2}(a) \mathbb{K}\_{1}(a)|^{2} \mathrm{d}a \right)^{\frac{1}{2}} \left( \int\_{0}^{1} |f\_{1}(a)|^{2} \mathrm{d}a \right)^{\frac{1}{2}} \\ \ll N^{\frac{11}{30} + \varepsilon} \cdot N^{\frac{23}{30} + \varepsilon} \cdot \left( N^{\mathbb{E}} \left( \mathcal{Z}\_{1} N^{\frac{1}{2}} + \mathcal{Z}\_{1}^{2} \right) \right)^{\frac{1}{2}} \cdot N^{\frac{1}{2}} \\ \ll N^{\frac{301}{30} + \varepsilon} \left( \mathcal{Z}\_{1}^{\frac{1}{2}} N^{\frac{1}{2}} + \mathcal{Z}\_{1} \right) \ll \mathcal{Z}\_{1}^{\frac{1}{2}} N^{\frac{32}{30} + \varepsilon} + \mathcal{Z}\_{1} N^{\frac{301}{30} + \varepsilon}. \end{split} \tag{28}$$

Combining (26) and (28), we get

$$\mathcal{Z}\_1 N^{\frac{13}{12}} \log^{-5} N \ll I\_1 \ll Z\_1^{\frac{1}{2}} N^{\frac{373}{288} + \varepsilon} + \mathcal{Z}\_1 N^{\frac{301}{288} + \varepsilon}.$$

which implies

$$
\mathbb{Z}\_1 \ll N^{\frac{61}{144} + \varepsilon}.\tag{29}
$$

Next, we give the upper bound for Z2. By (9), we obtain

$$\begin{split} I\_2 &\ll \int\_{\mathfrak{m}\_2} \left| f\_1(\alpha) f\_2(\alpha) V\_3(\alpha) f\_4(\alpha) \mathcal{K}\_2(\alpha) \right| \mathrm{d}\alpha \\ &+ N^{\frac{4}{15} + \varepsilon} \cdot \int\_{\mathfrak{m}\_2} \left| f\_1(\alpha) f\_2(\alpha) f\_4(\alpha) \mathcal{K}\_2(\alpha) \right| \mathrm{d}\alpha \\ &= I\_{21} + I\_{22} \end{split} \tag{30}$$

say. For *<sup>α</sup>* ∈ <sup>m</sup>2, we have either *<sup>Q</sup>*<sup>100</sup> <sup>0</sup> < *<sup>q</sup>* < *<sup>Q</sup>*<sup>1</sup> or *<sup>Q</sup>*<sup>100</sup> <sup>0</sup> <sup>&</sup>lt; *<sup>N</sup>*|*q<sup>α</sup>* <sup>−</sup> *<sup>a</sup>*<sup>|</sup> <sup>&</sup>lt; *NQ*−<sup>1</sup> <sup>2</sup> = *Q*1. Therefore, by Lemma 1, we get

$$\sup\_{\alpha \in \mathfrak{m}\_2} |f\_4(\alpha)| \ll \frac{N^{\frac{1}{4}}}{\log^{40A} N}.\tag{31}$$

In view of the fact that m<sup>2</sup> ⊆ I, where I is defined by (8), Hölder's inequality, the trivial estimate K2(*α*)  Z<sup>2</sup> and Theorem 4 of Hua (See [7], p. 19), we obtain

$$I\_{21} \ll Z\_2 \sup\_{a \in \mathfrak{m}\_2} |f\_4(a)| \times \left( \int\_0^1 |f\_1(a)|^2 \mathrm{d}a \right)^{\frac{1}{2}} \left( \int\_0^1 |f\_2(a)|^4 \mathrm{d}a \right)^{\frac{1}{4}} \left( \int\_{\mathcal{X}} |V\_3(a)|^4 \mathrm{d}a \right)^{\frac{1}{4}}$$

$$\ll Z\_2 \cdot \frac{N^{\frac{1}{4}}}{\log^{40A}N} \cdot N^{\frac{1}{4}} \cdot (N \log^{\varepsilon} N)^{\frac{1}{4}} \cdot (N^{\frac{1}{3}} \log^{\varepsilon} N)^{\frac{1}{4}} \ll \frac{Z\_2 N^{\frac{1}{12}}}{\log^{30A}N}.\tag{32}$$

Moreover, it follows from (27), (31) and Cauchy's inequality that

$$I\_{22} \ll N^{\frac{4}{15} + \varepsilon} \cdot \sup\_{a \in \mathfrak{m}\_2} |f\_4(a)| \times \left(\int\_0^1 |f\_1(a)|^2 \mathrm{d}a\right)^{\frac{1}{2}} \left(\int\_0^1 |f\_2(a)\mathcal{K}\_2(a)|^2 \mathrm{d}a\right)^{\frac{1}{2}}$$

$$\ll N^{\frac{4}{15} + \varepsilon} \cdot \frac{N^{\frac{1}{2}}}{\log^{40A}N} \cdot N^{\frac{1}{2}} \cdot \left(N^{\varepsilon}(\mathcal{Z}\_2 N^{\frac{1}{2}} + \mathcal{Z}\_2^2)\right)^{\frac{1}{2}}$$

$$\ll N^{\frac{61}{10} + \varepsilon} \left(\mathcal{Z}\_2^{\frac{1}{2}} N^{\frac{1}{2}} + \mathcal{Z}\_2\right) \ll \mathcal{Z}\_2^{\frac{1}{2}} N^{\frac{20}{15} + \varepsilon} + \mathcal{Z}\_2 N^{\frac{61}{10} + \varepsilon}. \tag{33}$$

Combining (26), (30), (32) and (33), we deduce that

$$\frac{Z\_2 N^{\frac{13}{12}}}{\log^5 N} \ll I\_2 = I\_{21} + I\_{22} \ll \frac{Z\_2 N^{\frac{13}{12}}}{\log^{30A} N} + Z\_2^{\frac{1}{2}} N^{\frac{19}{15} + \varepsilon} + Z\_2 N^{\frac{61}{60} + \varepsilon},$$

which implies

$$
\mathcal{Z}\_2 \ll N^{\frac{11}{30} + \varepsilon}.\tag{34}
$$

From (29) and (34), we have

$$Z(N) \ll Z\_1 + Z\_2 \ll N^{\frac{6}{3}n^{+\varepsilon}},$$

which completes the proof of Proposition 2.

**Author Contributions:** All authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research work is supported by National Natural Science Foundation of China (Grant No. 11901566, 11971476), the Fundamental Research Funds for the Central Universities (Grant No. 2019QS02), and National Training Program of Innovation and Entrepreneurship for Undergraduates (Grant No. 201911413071, C201907622).

**Acknowledgments:** The authors would like to express the most sincere gratitude to the referee for his/her patience in refereeing this paper.

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

### **References**


© 2020 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 and Inequalities Involving Symmetry Sums of Legendre Polynomials**

### **Tingting Wang \* and Liang Qiao**

College of Science, Northwest A&F University, Yangling 712100, China; 18309225762@163.com **\*** Correspondence: ttwang@nwsuaf.edu.cn

Received: 13 November 2019; Accepted: 12 December 2019; Published: 16 December 2019 -

**Abstract:** By using the analysis methods and the properties of Chebyshev polynomials of the first kind, this paper studies certain symmetry sums of the Legendre polynomials, and gives some new and interesting identities and inequalities for them, thus improving certain existing results.

**Keywords:** Legendre polynomials; Chebyshev polynomials of the first kind; power series; symmetry sums; polynomial identities; polynomial inequalities

### **1. Introduction**

For any integer *n* ≥ 0, the Legendre polynomials {*Pn*(*x*)} are defined as follows:

$$P\_n(\mathbf{x}) = \frac{2n-1}{n} \mathbf{x} P\_{n-1}(\mathbf{x}) - \frac{n-1}{n} P\_{n-2}(\mathbf{x})$$

for all *n* ≥ 2, with *P*0(*x*) = 1 and *P*1(*x*) = *x*, see [1,2] for more information.

The first few terms of *Pn*(*x*) are *P*2(*x*) = <sup>1</sup> 2 <sup>3</sup>*x*<sup>2</sup> − <sup>1</sup> , *P*3(*x*) = <sup>1</sup> 2 <sup>5</sup>*x*<sup>3</sup> − <sup>3</sup>*<sup>x</sup>* , *P*4(*x*) = <sup>1</sup> 8 <sup>35</sup>*x*<sup>4</sup> − <sup>30</sup>*x*<sup>2</sup> + <sup>3</sup> , *P*5(*x*) = <sup>1</sup> 8 <sup>63</sup>*x*<sup>5</sup> − <sup>70</sup>*x*<sup>3</sup> + <sup>15</sup>*<sup>x</sup>* , ··· .

In fact, the general term of *Pn*(*x*) is given by the formula

$$P\_{\rm ll}(\mathbf{x}) = \frac{1}{2^n} \cdot \sum\_{k=0}^{\left[\frac{n}{2}\right]} \frac{(-1)^k \cdot (2n - 2k)!}{k! \cdot (n - k)! \cdot (n - 2k)!} \cdot \mathbf{x}^{n - 2k},$$

where [*y*] denotes the greatest integer less than or equal to *y*.

It is clear that *Pn*(*x*) is an orthogonal polynomial (see [1,2]). That is,

$$\int\_{-1}^{1} P\_m(\mathbf{x}) P\_n(\mathbf{x}) d\mathbf{x} = \begin{cases} 0, & \text{if } m \neq n; \\\ \frac{2}{2n+1}, & \text{if } m = n. \end{cases}$$

The generating function of *Pn*(*x*) is

$$\frac{1}{\sqrt{1 - 2\mathbf{x}t + t^2}} = \sum\_{n=0}^{\infty} P\_n(\mathbf{x}) \cdot t^n, \ |\mathbf{x}| \le 1, \ |t| < 1. \tag{1}$$

These polynomials play a vital role in the study of function orthogonality and approximation theory, as a result, some scholars have dedicated themselves to studying their various natures and obtained a series of meaningful research results. The studies that are concerned with this content can be found in [1–20]. Recently, Shen Shimeng and Chen Li [3] give certain symmetry sums of *Pn*(*x*), and proved the following result:

For any positive integer *k* and integer *n* ≥ 0, one has the identity

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

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

where (2*k* −1)!! = (2*k* −1)·(2*k* −3)··· 3 · 1, 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.

The calculation formula for the sum of Legendre polynomials given above is virtually a linear combination of some *Pn*(*x*), and the coefficients *C*(*k*, *i*) are very regular. However, the result is in the form of a recursive formula, in other words, especially when *k* is relatively large, the formula is not actually easy to use for calculating specific values.

In an early paper, Zhou Yalan and Wang Xia [4] obtained some special cases with *k* = 3 and *k* = 5. It is even harder to calculate their exact values for the general positive integer *k*, especially if *k* is large enough.

Naturally, we want to ask a question: Is there a more concise and specific formula for the calculation of the above problems? This is the starting point of this paper. We used the different methods to come up with additional simpler identities. It is equal to saying that we have used the analysis method and the properties of the first kind of Chebyshev polynomials, thereby establishing the symmetry of the Legendre polynomial and symmetry relationship with the first kind of Chebyshev polynomial, and proved the following three results:

**Theorem 1.** *For any integers k* ≥ 1 *and n* ≥ 0*, we have the identity*

$$\sum\_{a\_1+a\_2+\cdots+a\_k=n} P\_{\mathfrak{a}\_1}(\mathbf{x}) \cdot P\_{\mathfrak{a}\_2}(\mathbf{x}) \cdot \cdots \cdot P\_{\mathfrak{a}\_k}(\mathbf{x}) = \sum\_{i=0}^n \frac{\le \frac{k}{2} > i}{i!} \cdot \frac{< \frac{k}{2} >\_{n-i}}{(n-i)!} \cdot T\_{n-2i}(\mathbf{x}),$$

*where* < *x* >0= 1*,* < *x* >*k*= *x*(*x* + 1)(*x* + 2)···(*x* + *k* − 1) *for all integers k* ≥ 1*, and Tn*(*x*) = *T*−*n*(*x*) = 1 2 *<sup>x</sup>* <sup>+</sup> <sup>√</sup>*x*<sup>2</sup> <sup>−</sup> <sup>1</sup> *n* + *<sup>x</sup>* <sup>+</sup> <sup>√</sup>*x*<sup>2</sup> <sup>−</sup> <sup>1</sup> *n denotes Chebyshev polynomials of the first kind.*

**Theorem 2.** *Let q* > 1 *is an integer, χ is any primitive character* mod *q. Then for any integers k* ≥ 1 *and n* ≥ 0*, we have the inequality*

$$\begin{aligned} & \left| \sum\_{\substack{a\_1 + a\_2 + \cdots + a\_k = n \ a = 1}} \sum\_{a=1}^q \chi(a) P\_{a\_1} \left( \cos \frac{2 \pi a}{q} \right) \cdot P\_{a\_2} \left( \cos \frac{2 \pi a}{q} \right) \cdot \cdots \cdot P\_{a\_k} \left( \cos \frac{2 \pi a}{q} \right) \right| \\ & \le \sqrt{q} \cdot \binom{n+k-1}{k-1} . \end{aligned}$$

**Theorem 3.** *For any integer n* ≥ 0 *with* 2 *n, we have the identity*

$$\begin{aligned} &\int\_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum\_{a\_1+a\_2+\cdots+a\_k=n} P\_{a\_1} \left( \sin \theta \right) \cdot P\_{a\_2} \left( \sin \theta \right) \cdot \cdots \cdot P\_{a\_k} \left( \sin \theta \right) \right)^2 d\theta \\ &= \quad 2\pi \cdot \sum\_{i=0}^{\left[ \frac{\pi}{2} \right]} \left( \frac{< \frac{k}{2} >\_i}{i!} \cdot \frac{< \frac{k}{2} >\_{n-i}}{(n-i)!} \right)^2; \end{aligned}$$

*If n* = 2*m, then we have*

$$\int\_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum\_{a\_1 + a\_2 + \cdots + a\_k = n} P\_{a\_1} \left( \sin \theta \right) \cdot P\_{a\_2} \left( \sin \theta \right) \cdot \cdots \cdot P\_{a\_k} \left( \sin \theta \right) \right)^2 d\theta$$

$$= -2\pi \cdot \sum\_{i=0}^m \left( \frac{< \frac{k}{2} > i\_i}{i!} \cdot \frac{< \frac{k}{2} >\_{2m-i}}{(2m-i)!} \right)^2 - \pi \cdot \left( \frac{< \frac{k}{2} >\_m}{m!} \right)^4.$$

Essentially, the main result of this paper is Theorem 1, which not only reveals the profound properties of Legendre polynomials and Chebyshev polynomials, but also greatly simplifies the calculation of the symmetry sum of Legendre polynomials in practice. We can replace the calculation of the symmetric sum of the Legendre polynomial with the first single Chebyshev polynomial calculation, which can greatly simplify the calculation of the symmetric sum.

Theorem 2 gives an upper bound estimate of the character sum of Legendre polynomials. Theorem 3 reveals the orthogonality of the symmetry sum of Legendre polynomials, which is a generalization of the orthogonality of functions. Of course, Theorems 2 and 3 can also be seen as the direct application of Theorem 1 in analytical number theory and the orthogonality of functions. This is of great significance in analytic number theory, and it has also made new contributions to the study of Gaussian sums.

In fact if we taking *<sup>k</sup>* <sup>=</sup> 1, and note that the identity <sup>&</sup>lt; <sup>1</sup> <sup>2</sup> >*<sup>h</sup> <sup>h</sup>*! <sup>=</sup> <sup>1</sup> <sup>4</sup>*<sup>h</sup>* · ( 2*h <sup>h</sup>* ), then from our theorems we may immediately deduce the following three corollaries.

**Corollary 1.** *For any integer n* ≥ 0*, we have the identity*

$$P\_{\mathbb{II}}(\mathbf{x}) = \frac{1}{4^n} \sum\_{i=0}^n \binom{2i}{i} \binom{2n-2i}{n-i} \cdot T\_{n-2i}(\mathbf{x})\_i$$

*where Tn*(*x*) *denotes Chebyshev polynomials of the first kind.*

**Corollary 2.** *Let q* > 1 *is an integer, χ is any primitive character* mod*q. Then for any integer n* ≥ 0*, we have the inequality*

$$\left| \sum\_{a=1}^{q} \chi(a) P\_n \left( \cos \frac{2\pi a}{q} \right) \right| \le \sqrt{q}.$$

**Corollary 3.** *For any integer n* ≥ 0 *with* 2 *n, we have the identity*

$$\int\_{-\frac{\pi}{2}}^{\frac{\pi}{2}} P\_n^2 \left(\sin \theta\right) d\theta = \frac{2\pi}{4^{2n}} \sum\_{i=0}^{\left[\frac{\pi}{2}\right]} \binom{2i}{i}^2 \binom{2n-2i}{n-i}^2;$$

*If n* = 2*m, then we have the identity*

$$\int\_{-\frac{\pi}{2}}^{\frac{\pi}{2}} P\_n^2 \left(\sin \theta\right) d\theta = \frac{2\pi}{4^{2n}} \sum\_{i=0}^{\left[\frac{\pi}{2}\right]} \binom{2i}{i}^2 \binom{2n-2i}{n-i}^2 - \frac{\pi}{4^{2n}} \cdot \binom{2m}{m}^4.$$

### **2. Proofs of the Theorems**

In this section, we will directly prove the main results in this paper by by means of the properties of characteristic roots.

**Proof of Theorem 1.** First we prove Theorem 1. Let *<sup>α</sup>* <sup>=</sup> *<sup>x</sup>* <sup>+</sup> <sup>√</sup>*x*<sup>2</sup> <sup>−</sup> <sup>1</sup> and *<sup>β</sup>* <sup>=</sup> *<sup>x</sup>* <sup>−</sup> <sup>√</sup>*x*<sup>2</sup> <sup>−</sup> <sup>1</sup> be two characteristic roots of the characteristic equation *<sup>λ</sup>*<sup>2</sup> − <sup>2</sup>*x<sup>λ</sup>* + <sup>1</sup> = 0. Then from the definition and properties of Chebyshev polynomials *Tn*(*x*) of the first kind, we have

$$T\_n(\mathfrak{x}) = T\_{-n}(\mathfrak{x}) = \frac{1}{2} \left( \mathfrak{a}^n + \mathfrak{z}^n \right), \ n \ge 0.$$

For any positive integer *k*, combining properties of power series and Formula (1) we have the identity

$$\left(\frac{1}{\sqrt{1-2xt+t^2}}\right)^k = \frac{1}{\left(1-2xt+t^2\right)^{\frac{k}{2}}} = \frac{1}{\left(1-at\right)^{\frac{k}{2}}\left(1-\beta t\right)^{\frac{k}{2}}}$$

$$=\sum\_{n=0}^{\infty}\left(\sum\_{a\_1+a\_2+\cdots+a\_k=n}P\_{a\_1}(\mathbf{x})\cdot P\_{a\_2}(\mathbf{x})\cdot P\_{a\_3}(\mathbf{x})\cdot\cdots P\_{a\_k}(\mathbf{x})\right)\cdot t^n.\tag{2}$$

At the same time, we focus on the power series

$$\frac{1}{(1-\mathbf{x})^{\frac{1}{2}}} = \sum\_{n=0}^{\infty} \frac{<\frac{k}{2} >\_n}{n!} \cdot \mathbf{x}^n, \ |\mathbf{x}| < 1,\tag{3}$$

where < *x* >0= 1, < *x* >*h*= *x*(*x* + 1)(*x* + 2)···(*x* + *h* − 1) for all integers *h* ≥ 1.

So for any positive integer *k*, note that *α* · *β* = 1, from (3) and the symmetry properties of *α* and *β* we have


Combining (2) and (4), and then by comparing the coefficients on both sides of the power series, we can find

$$\sum\_{a\_1+a\_2+\cdots+a\_k=n} P\_{a\_1}(\mathbf{x}) \cdot P\_{a\_2}(\mathbf{x}) \cdot \cdots \cdot P\_{a\_k}(\mathbf{x}) = \sum\_{i=0}^n \frac{\frac{1}{2} \times \frac{1}{i}}{i!} \cdot \frac{< \frac{1}{2} \times\_{n-i}}{(n-i)!} \cdot T\_{n-2i}(\mathbf{x}) .$$

This proves Theorem 1.

**Proof of Theorem 2.** The proof of Theorem 2 is next. Let *q* > 1 be any integer, *χ* denotes any primitive character mod *<sup>q</sup>*. Then from Theorem 1 with *<sup>x</sup>* <sup>=</sup> cos 2*π<sup>a</sup> q* and the identity *Tn* (cos *θ*) = cos(*nθ*), we have

$$\sum\_{a\_1 + a\_2 + \dots + a\_k = n} \sum\_{a=1}^q \chi(a) P\_{a\_1} \left( \cos \frac{2\pi a}{q} \right) \cdot P\_{a\_2} \left( \cos \frac{2\pi a}{q} \right) \cdots P\_{a\_k} \left( \cos \frac{2\pi a}{q} \right)$$

$$\begin{aligned} &= \sum\_{i=0}^n \frac{\left\langle \frac{k}{2} \right\rangle\_i}{i!} \cdot \frac{\left\langle \frac{k}{2} \right\rangle\_i}{(n-i)!} \cdot \sum\_{a=1}^q \chi(a) \cdot \cos \left( \frac{2\pi a (n-2i)}{q} \right) \\ &= \frac{1}{2} \sum\_{i=0}^n \frac{\left\langle \frac{k}{2} \right\rangle\_i}{i!} \cdot \frac{\left\langle \frac{k}{2} \right\rangle\_i}{(n-i)!} \cdot \sum\_{a=1}^q \chi(a) \left( c \left( \frac{a(n-2i)}{q} \right) + e \left( \frac{-a(n-2i)}{q} \right) \right) \end{aligned}$$

$$\left( \begin{aligned} &= \frac{\pi(\chi)}{2} \sum\_{i=0}^n \frac{\left\langle \frac{k}{2} \right\rangle\_i}{i!} \cdot \frac{\left\langle \frac{k}{2} \right\rangle\_i}{(n-i)!} \cdot \left( \overline{\chi}(n-2i) + \overline{\chi} \left( - (n-2i) \right) \right) \end{aligned} \tag{5}$$

where *e*(*y*) = *e*2*πiy*, and *q* ∑ *a*=1 *χ*(*a*)*e na q* = *χ*(*n*)*τ*(*χ*).

Note that for any primitive character *χ* mod *q*, from the properties of Gauss sums, we have <sup>|</sup>*τ*(*χ*)<sup>|</sup> <sup>=</sup> <sup>√</sup>*q*, and for any positive integer *<sup>k</sup>* <sup>≥</sup> 1, we have

$$\begin{aligned} \frac{1}{(1-x)^k} &= \sum\_{n=0}^{\infty} \binom{n+k-1}{k-1} \cdot x^n = \frac{1}{(1-x)^{\frac{k}{2}}} \cdot \frac{1}{(1-x)^{\frac{k}{2}}} \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{i=0}^n \frac{< \frac{k}{2} > i}{i!} \cdot \frac{< \frac{k}{2} >\_{n-i}}{(n-i)!} \right) \cdot x^n \end{aligned}$$

or

$$
\binom{n+k-1}{k-1} = \sum\_{i=0}^{n} \frac{<\frac{k}{2} >\_i}{i!} \cdot \frac{<\frac{k}{2} >\_{n-i}}{(n-i)!}.\tag{6}
$$

Combining (5) and (6), there will be an estimation formula immediately deduced

$$\begin{split} & \left| \sum\_{a\_1 + a\_2 + \dots + a\_k = n} \sum\_{a=1}^q \chi(a) P\_{a\_1} \left( \cos \frac{2 \pi a}{q} \right) \cdot P\_{a\_2} \left( \cos \frac{2 \pi a}{q} \right) \cdot \cdots \cdot P\_{a\_k} \left( \cos \frac{2 \pi a}{q} \right) \right| \\ & = \frac{\sqrt{q}}{2} \cdot \left| \sum\_{i=0}^n \frac{\le \frac{k}{2} >\_i}{i!} \cdot \frac{\le \frac{k}{2} >\_{n-i}}{(n-i)!} \cdot \left[ \overline{\chi}(n-2i) + \overline{\chi} \left( -(n-2i) \right) \right] \right| \\ & \le \sqrt{q} \cdot \sum\_{i=0}^n \frac{\le \frac{k}{2} >\_i}{i!} \cdot \frac{\le \frac{k}{2} >\_{n-i}}{(n-i)!} = \sqrt{q} \cdot \binom{n+k-1}{k-1} . \end{split}$$

Theorem 2 is proven completely.

**Proof of Theorem 3.** We prove Theorem 3 below. From the orthogonality of Chebyshev polynomials of the first kind we know that

$$\int\_{-1}^{1} \frac{T\_m(\mathbf{x})T\_n(\mathbf{x})}{\sqrt{1 - \mathbf{x}^2}} d\mathbf{x} = \begin{cases} 0, & \text{if } m \neq n; \\ \frac{\pi}{2}, & \text{if } m = n > 0, \\ \pi, & \text{if } m = n = 0. \end{cases} \tag{7}$$

If integer *n* ≥ 1 with 2 *n*, then for any integer 0 ≤ *i* ≤ *n*, we have *n* − 2*i* = 0, note that *Tn*(*x*) = *T*−*n*(*x*), so from (7) and Theorem 1 we have

$$\int\_{-1}^{1} \frac{1}{\sqrt{1-\mathbf{x}^2}} \left( \sum\_{a\_1+a\_2+\cdots+a\_k=n} P\_{a\_1}(\mathbf{x}) \cdot P\_{a\_2}(\mathbf{x}) \cdot \cdots \cdot P\_{a\_k}(\mathbf{x}) \right)^2 d\mathbf{x}$$

$$=\int\_{-1}^{1} \frac{1}{\sqrt{1-\mathbf{x}^2}} \left( \sum\_{i=0}^n \frac{\le \frac{k}{2} \ge i}{i!} \cdot \frac{\le \frac{k}{2} \ge\_{n-i}}{(n-i)!} \cdot T\_{n-2i}(\mathbf{x}) \right)^2 d\mathbf{x}$$

$$=4 \int\_{-1}^{1} \frac{1}{\sqrt{1-\mathbf{x}^2}} \left( \sum\_{i=0}^{\frac{k}{2}} \frac{\le \frac{k}{2} \ge i}{i!} \cdot \frac{\le \frac{k}{2} \ge\_{n-i}}{(n-i)!} \cdot T\_{n-2i}(\mathbf{x}) \right)^2 d\mathbf{x}$$

$$=2\pi \cdot \sum\_{i=0}^{\left[\frac{k}{2}\right]} \left( \frac{\le \frac{k}{2} \ge i}{i!} \cdot \frac{\le \frac{k}{2} \ge\_{n-i}}{(n-i)!} \right)^2. \tag{8}$$

For *n* = 2*m*, if *n* − 2*i* = 0, then *i* = *m*. So from (7), Theorem 1 and the methods of proving (8) we have

$$\begin{split} &\int\_{-1}^{1} \frac{1}{\sqrt{1-\mathbf{x}^{2}}} \left( \sum\_{a\_{1}+a\_{2}+\cdots+a\_{n}=n} P\_{a\_{1}}(\mathbf{x}) \cdot P\_{a\_{2}}(\mathbf{x}) \cdot \cdots \cdot P\_{a\_{n}}(\mathbf{x}) \right)^{2} d\mathbf{x} \\ &= \int\_{-1}^{1} \frac{1}{\sqrt{1-\mathbf{x}^{2}}} \left( \left(\frac{<\frac{k}{2}>\_{m}}{m!}\right)^{2} + 2\sum\_{i=0}^{m-1} \frac{<\frac{k}{2}>\_{i}}{i!} \cdot \frac{<\frac{k}{2}>\_{2m-i}}{(2m-i)!} \cdot T\_{2m-2i}(\mathbf{x}) \right)^{2} d\mathbf{x} \\ &= \pi \cdot \left(\frac{<\frac{k}{2}>\_{m}}{m!}\right)^{4} + 2\pi \cdot \sum\_{i=0}^{m-1} \left(\frac{<\frac{k}{2}>\_{i}}{i!} \cdot \frac{<\frac{k}{2}>\_{i}}{(n-i)!} \right)^{2} \\ &= 2\pi \cdot \sum\_{i=0}^{m} \left(\frac{<\frac{k}{2}>\_{i}}{i!} \cdot \frac{<\frac{k}{2}>\_{2m-i}}{(2m-i)!} \right)^{2} - \pi \cdot \left(\frac{<\frac{k}{2}>\_{m}}{m!} \right)^{4} . \end{split}$$

Let *x* = sin *θ*, then we have

$$\int\_{-1}^{1} \frac{1}{\sqrt{1-\mathbf{x}^{2}}} \left( \sum\_{a\_{1}+a\_{2}+\cdots+a\_{k}=n} P\_{a\_{1}}(\mathbf{x}) \cdot P\_{a\_{2}}(\mathbf{x}) \cdot \cdots \cdot P\_{a\_{k}}(\mathbf{x}) \right)^{2} d\mathbf{x}$$

$$\int\_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum\_{a\_{1}+a\_{2}+\cdots+a\_{k}=n} P\_{a\_{1}}(\sin \theta) \cdot P\_{a\_{2}}(\sin \theta) \cdot \cdots \cdot P\_{a\_{k}}(\sin \theta) \right)^{2} d\theta. \tag{10}$$

Now Theorem 3 follows from (8), (9), and (10).

### **3. Conclusions**

Three theorems and three inferences are the main results in the paper. Theorem 1 gives proof of the symmetry of Legendre polynomials and the symmetry relationship with Chebyshev polynomials of the first kind. This conclusion also improves the early results in [4], and also gives us a different representation for the result in [3]. Theorem 2 obtained an inequality involving Dirichlet characters and Legendre polynomials; this is actually a new contribution to the study of Legendre polynomials and character sums mod *q*. Theorem 3 established an integral identity involving the symmetry sums of the Legendre polynomials. The three corollaries are some special cases of our three theorems for *k* = 1, and can not only enrich the research content of the Legendre polynomials, but also promote its research development.

**Author Contributions:** All authors have equally contributed to this work. All authors read and approved the final manuscript.

**Funding:** This work is supported by the Y. S. T. N. S. P (2019KJXX-076) and N. S. B. R. P in Shaanxi Province (2019JM-207).

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

**Conflicts of Interest:** The authors declare that there are no conflicts of interest regarding 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* **A New Identity Involving Balancing Polynomials and Balancing Numbers**

### **Yuanyuan Meng**

School of Mathematics, Northwest University, Xi'an 710127, Shaanxi, China; yymeng@stumail.nwu.edu.cn

Received: 13 August 2019 ; Accepted: 5 September 2019 ; Published: 7 September 2019

**Abstract:** In this paper, a second-order nonlinear recursive sequence *M*(*h*, *i*) is studied. By using this sequence, the properties of the power series, and the combinatorial methods, some interesting symmetry identities of the structural properties of balancing numbers and balancing polynomials are deduced.

**Keywords:** balancing numbers; balancing polynomials; combinatorial methods; symmetry sums

### **1. Introduction**

For any positive integer *n* ≥ 2, we denote the balancing number by *Bn* and the balancer corresponding to it by *r*(*n*) if

$$1 + 2 + \cdots + (B\_n - 1) = (B\_n + 1) + (B\_n + 2) + \cdots + (B\_n + r(n))$$

holds for some positive integer *<sup>r</sup>*(*n*) and *Bn*. It is clear that *<sup>r</sup>*(*n*) = *Bn*−*Bn*−1−<sup>1</sup> <sup>2</sup> , for example, *r*(2) = 2, *r*(3) = 14, *r*(4) = 84, *r*(5) = 492. . .

It is found that the balancing numbers satisfy the second order linear recursive sequence *Bn*+<sup>1</sup> = 6*Bn* − *Bn*−<sup>1</sup> (*n* ≥ 1), providing *B*<sup>0</sup> = 0 and *B*<sup>1</sup> = 1 [1].

The balancing polynomials *Bn*(*x*) are defined by *<sup>B</sup>*0(*x*) = 1, *<sup>B</sup>*1(*x*) = <sup>6</sup>*x*, *<sup>B</sup>*2(*x*) = <sup>36</sup>*x*<sup>2</sup> − 1, *<sup>B</sup>*3(*x*) = <sup>216</sup>*x*<sup>3</sup> − <sup>12</sup>*x*, *<sup>B</sup>*4(*x*) = <sup>1296</sup>*x*<sup>4</sup> − <sup>108</sup>*x*<sup>2</sup> + 1, and the second-order linear difference equation:

$$B\_{n+1}(\mathbf{x}) = \mathfrak{G}\mathfrak{X}B\_n(\mathbf{x}) - B\_{n-1}(\mathbf{x}), n \ge 1, \mathbf{y}$$

where *x* is any real number. While *n* ≥ 1, we get *Bn*+<sup>1</sup> = 6*Bn* − *Bn*−<sup>1</sup> with *Bn*(1) = *Bn*+1. Such balancing numbers have been widely studied in recent years. G. K. Panda and T. Komatsu [2] studied the reciprocal sums of the balancing numbers and proved the following inequation holds for any positive integer *n*:

$$\frac{1}{B\_{\text{fl}} - B\_{\text{n-1}}} < \sum\_{k=n}^{\infty} \frac{1}{B\_k} < \frac{1}{B\_{\text{fl}} - B\_{\text{n-1}} - 1}.$$

G. K. Panda [3] studied some fascinating properties of balancing numbers and gave the following result for any natural numbers *m* > *n*:

$$(B\_m + B\_n)(B\_m - B\_n) = B\_{m+n} \cdot B\_{m-n}$$

Other achievements related to balancing numbers can be found in [4–7].

It is found that the balancing polynomials *Bn*(*x*) can be generally expressed as

$$B\_n(\mathbf{x}) = \frac{1}{2\sqrt{9\mathbf{x}^2 - 1}} \left[ \left( 3\mathbf{x} + \sqrt{9\mathbf{x}^2 - 1} \right)^{n+1} - \left( 3\mathbf{x} - \sqrt{9\mathbf{x}^2 - 1} \right)^{n+1} \right],$$

and the generating function of the balancing polynomials *Bn*(*x*) is given by

$$\frac{1}{1 - 6\mathbf{x}t + t^2} = \sum\_{n=0}^{\infty} B\_n(\mathbf{x}) \cdot t^n. \tag{1}$$

Recently, our attention was drawn to the sums of polynomials calculating problem [8–11], which is important in mathematical application. We are going to study the computational problem of the symmetry summation:

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

where *h* is any positive integer. We shall prove the following theorem holds.

**Theorem 1.** *For any specific positive integer h and any integer n* ≥ 0*, the following identity stands:*

$$\sum\_{a\_1 + a\_2 + \dots + a\_{h+1} = n} B\_{a\_1}(\mathbf{x}) B\_{a\_2}(\mathbf{x}) \cdot \dots \cdot B\_{a\_{h+1}}(\mathbf{x})$$

$$= \frac{1}{2^{h} \cdot h!} \cdot \sum\_{j=1}^{h} \frac{M(h, j)}{(3\mathbf{x})^{2h-j}} \sum\_{i=0}^{n} \frac{(n-i+j)!}{(n-i)!} \cdot \frac{B\_{n-i+j}(\mathbf{x})}{(3\mathbf{x})^{i}} \cdot \binom{2h+i-j-1}{i} \cdot \binom{2h+i-j-1}{i}$$

*where M*(*h*, *i*) *is defined by M*(*h*, 0) = 0*, M*(*h*, *i*) = (2*h*−*i*−1)! <sup>2</sup>*h*−*i*·(*h*−*i*)!·(*i*−1)! *for all positive integers* <sup>1</sup> <sup>≤</sup> *<sup>i</sup>* <sup>≤</sup> *h.*

In particular, for *n* = 0, the following corollary can be deduced.

**Corollary 1.** *For any positive integer h* ≥ 1*, the following formula holds:*

$$\sum\_{j=1}^{h} M(h,j) \cdot j! \cdot (3\mathfrak{x})^j \cdot B\_j(\mathfrak{x}) = 2^h \cdot h! \cdot (3\mathfrak{x})^{2h}.$$

The formula in Corollary 1 shows the close relationship among the balancing polynomials. For *h* = 2, the following corollary can be inferred by Theorem 1.

**Corollary 2.** *For any integer n* ≥ 0*, we obtain*

$$\begin{aligned} \sum\_{a+b+c=n} B\_a(\mathbf{x}) \cdot B\_b(\mathbf{x}) \cdot B\_c(\mathbf{x}) &= \frac{1}{216\pi^3} \sum\_{i=0}^n (n-i+1)(i+1)(i+2) \cdot \frac{B\_{n-i+2}}{(3\mathbf{x})^i} \\ &+ \quad \frac{1}{72\pi^2} \sum\_{i=0}^n (n-i+1)(n-i+2)(i+1) \cdot \frac{B\_{n-i+3}}{(3\mathbf{x})^i} .\end{aligned}$$

For *x* = 1, *h* = 2 *and* 3, according to Theorem 1 we can also infer the following corollaries:

**Corollary 3.** *For any integer n* ≥ 0*, we obtain*

$$\begin{aligned} \sum\_{a+b+c=n} B\_{a+1} \cdot B\_{b+1} \cdot B\_{c+1} &= \frac{1}{216} \sum\_{i=0}^{n} (n-i+1)(i+1)(i+2) \cdot \frac{B\_{n-i+2}}{3^i} \\ &+ \quad \frac{1}{72} \sum\_{i=0}^{n} (n-i+1)(n-i+2)(i+1) \cdot \frac{B\_{n-i+3}}{3^i} .\end{aligned}$$

**Corollary 4.** *For any integer n* ≥ 0*, we obtain:*

$$\begin{split} &\sum\_{a+b+c+d=n} B\_{a+1} \cdot B\_{b+1} \cdot B\_{c+1} \cdot B\_{d+1} \\ &= \quad \frac{1}{3888} \sum\_{i=0}^{n} (n-i+1)(i+1)(i+2)(i+3)(i+4) \cdot \frac{B\_{n-i+2}}{3^i} \\ &+ \quad \frac{1}{1296} \sum\_{i=0}^{n} (n-i+1)(n-i+2)(i+1)(i+2)(i+3) \cdot \frac{B\_{n-i+3}}{3^i} \\ &+ \quad \frac{1}{1296} \sum\_{i=0}^{n} (n-i+1)(n-i+2)(n-i+3)(i+1)(i+2) \cdot \frac{B\_{n-i+4}}{3^i} .\end{split}$$

**Corollary 5.** *For any odd prime p, we have the congruence M*(*p*, *i*) ≡ 0(*modp*)*,* 0 ≤ *i* ≤ *p* − 1*.*

**Corollary 6.** *The balancing polynomials are essentially Chebyshev polynomials of the second kind, specifically Bn*(*x*) = *Un*(3*x*)*. Taking x* = <sup>1</sup> <sup>3</sup> *x in Theorem 1, we can get the following:*

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

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

*Compared with [8], we give a more precise result for* <sup>∑</sup>*a*1+*a*2+···+*ah*+1=*<sup>n</sup> Ua*<sup>1</sup> (*x*)*Ua*<sup>2</sup> (*x*)··· *Uah*<sup>+</sup><sup>1</sup> (*x*) *with the specific expressions of M*(*h*, *i*)*. This shows our novelty.*

Here, we list the first several terms of *M*(*h*, *i*) in Table 1 in order to demonstrate the properties of the sequence *M*(*h*, *i*) clearly.


**Table 1.** Values of *M*(*h*, *i*).

### **2. Several Lemmas**

For the sake of clarity, several lemmas that are necessary for proving our theorem will be given in this section.

**Lemma 1.** *For the sequence M*(*n*, *i*)*, the following identity holds for all* 1 ≤ *i* ≤ *n:*

$$M(n,i) = \frac{(2n-i-1)!}{2^{n-i} \cdot (n-i)! \cdot (i-1)!}.$$

**Proof.** We present a straightforward proof of this lemma by using mathematical introduction. It is obvious that

$$M(1,1) = \frac{0!}{1 \cdot 0! \cdot 0!} = 1... $$

This means Lemma 1 is valid for *n* = 1. Without loss of generality, we assume that Lemma 1 holds for 1 ≤ *n* = *h* and all 1 ≤ *i* ≤ *h*. Then, we have

$$\begin{aligned} M(h,i) &= \frac{(2h-i-1)!}{2^{h-i} \cdot (h-i)! \cdot (i-1)!}, \\\\ M(h,i+1) &= \frac{(2h-i-2)!}{2^{h-i-1} \cdot (h-i-1)! \cdot i!} \end{aligned}$$

.

According to the definitions of *M*(*n*, *i*), it is easy to find that

$$\begin{aligned} M(h+1,i+1) &= \quad (2h-1-i) \cdot M(h,i+1) + M(h,i) \\ &= \quad (2h-1-i) \cdot \frac{2(h-i)}{(2h-i-1)i} \cdot M(h,i) + M(h,i) \\ &= \quad \frac{2h-i}{i} M(h,i) = \frac{(2h-i)!}{2^{h-i} \cdot (h-i)! \cdot i!} \\ &= \quad \frac{(2(h+1)-(i+1)-1)!}{2^{h-i} \cdot (h-i)! \cdot i!} .\end{aligned}$$

Thus, Lemma 1 is also valid for *n* = *h* + 1. From now on, Lemma 1 has been proved.

**Lemma 2.** *If we have a function f*(*t*) = <sup>1</sup> <sup>1</sup>−6*xt*+*t*<sup>2</sup> *, then for any positive integer n, real numbers <sup>x</sup> and <sup>t</sup> with* |*t*| < |3*x*|*, the following identity holds:*

$$2^n \cdot n! \cdot f^{n+1}(t) = \sum\_{i=1}^n M(n, i) \cdot \frac{f^{(i)}(t)}{(3\pi - t)^{2n-i}},$$

*where f* (*i*)(*t*) *denotes the i-th order derivative of f*(*t*)*, with respect to variable t and M*(*n*, *i*)*, which is defined in the theorem.*

**Proof.** Similarly, Lemma 2 will be proved by mathematical induction. We start by showing that Lemma 2 is valid for *n* = 1. Using the properties of the derivative, we have:

$$f'(t) = (6x - 2t) \cdot f^2(t),$$

or

$$2f^2(t) = \frac{f'(t)}{3\mathbf{x} - t} = M(\mathbf{1}, \mathbf{1}) \cdot \frac{f'(t)}{3\mathbf{x} - t}.$$

This is in fact true and provides the main idea to show the following steps. Without loss of generality, we assume that Lemma 2 holds for 1 ≤ *n* = *h*. Then, we have

$$2^h \cdot h! \cdot f^{h+1}(t) = \sum\_{i=1}^h M(h, i) \cdot \frac{f^{(i)}(t)}{(3x - t)^{2h - i}}.\tag{2}$$

As an immediate consequence, we can tell by (2), the properties of *M*(*n*, *i*), and the derivative, we get

$$\begin{aligned} 2^h \cdot (h+1)! \cdot f^h(t) \cdot f'(t) &= 2^{h+1} \cdot (h+1)! \cdot (3\mathbf{x} - t) \cdot f^{h+2}(t) \\ &= \sum\_{i=1}^h \frac{M(h,i)}{(3\mathbf{x}-t)^{2h-i}} \cdot f^{(i+1)}(t) + \sum\_{i=1}^h \frac{(2h-i)M(h,i)}{(3\mathbf{x}-t)^{2h-i+1}} \cdot f^{(i)}(t) \end{aligned}$$

$$\begin{aligned} \label{eq:10} &= \quad \frac{M(h,h)}{(3\mathbf{x}-t)^h} \cdot f^{(h+1)}(t) + \sum\_{i=1}^{h-1} \frac{M(h,i)}{(3\mathbf{x}-t)^{2h-i}} \cdot f^{(i+1)}(t) + \frac{(2h-1)M(h,1)}{(3\mathbf{x}-t)^{2h}} \cdot f'(t) \\ &+ \sum\_{i=1}^{h-1} \frac{(2h-i-1)M(h,i+1)}{(3\mathbf{x}-t)^{2h-i}} \cdot f^{(i+1)}(t) \\ &= \quad \frac{M(h+1,h+1)}{(3\mathbf{x}-t)^h} \cdot f^{(h+1)}(t) + \frac{M(h+1,1)}{(3\mathbf{x}-t)^{2h}} \cdot f'(t) + \sum\_{i=1}^{h-1} \frac{M(h+1,i+1)}{(3\mathbf{x}-t)^{2h-i}} \cdot f^{(i+1)}(t) \\ &= \quad \frac{M(h+1,h+1)}{(3\mathbf{x}-t)^h} \cdot f^{(h+1)}(t) + \frac{M(h+1,1)}{(3\mathbf{x}-t)^{2h}} \cdot f'(t) + \sum\_{i=2}^{h} \frac{M(h+1,i)}{(3\mathbf{x}-t)^{2h+1-i}} \cdot f^{(i)}(t) \\ &= \quad \sum\_{i=1}^{h+1} M(h+1,i) \cdot \frac{f^{(i)}(t)}{(3\mathbf{x}-t)^{2h+1-i}} .\end{aligned}$$

Then, it is deduced that

$$2^{h+1} \cdot (h+1)! \cdot (3x-t) \cdot f^{h+2}(t) = \sum\_{i=1}^{h+1} M(h+1, i) \cdot \frac{f^{(i)}(t)}{(3x-t)^{2h+1-i}}$$

,

.

or

$$2^{h+1} \cdot (h+1)! \cdot f^{h+2}(t) = \sum\_{i=1}^{h+1} M(h+1, i) \cdot \frac{f^{(i)}(t)}{(3x-t)^{2h+2-i}}$$

Thus, Lemma 2 is also valid for *n* = *h* + 1. From now on, Lemma 2 has been proved.

**Lemma 3.** *The following power series expansion holds for arbitrary positive integers h and k:*

$$\frac{f^{(h)}(t)}{(3\mathbf{x}-t)^{k}} = \frac{1}{(3\mathbf{x})^{k}} \sum\_{n=0}^{\infty} \left( \sum\_{i=0}^{n} \frac{(n-i+h)!}{(n-i)!} \cdot \frac{B\_{n-i+h}(\mathbf{x})}{(3\mathbf{x})^{i}} \cdot \binom{i+k-1}{i} \right) t^{n} \zeta\_{n}$$

*where t and x are any real numbers with* |*t*| < |3*x*|*.*

**Proof.** According to the definition of the balancing polynomials *Bn*(*x*), we have:

$$f(t) = \frac{1}{1 - 6\mathfrak{x}t + t^2} = \sum\_{n=0}^{\infty} B\_n(\mathfrak{x}) \cdot t^n.$$

For any positive integer *h*, from the properties of the power series, we can obtain

$$f^{(h)}(t) = \sum\_{n=0}^{\infty} (n+h)(n+h-1)\cdots(n+1) \cdot B\_{n+h}(\mathbf{x}) \cdot t^n$$

$$= \sum\_{n=0}^{\infty} \frac{(n+h)!}{n!} \cdot B\_{n+h}(\mathbf{x}) \cdot t^n. \tag{4}$$

For all real *t* and *x* with |*t*| < |3*x*|, we have the following power series expansion:

$$\frac{1}{3x-t} = \frac{1}{3x} \cdot \sum\_{n=0}^{\infty} \frac{t^n}{(3x)^n},$$

and

$$\frac{1}{(3x-t)^k} = \frac{1}{(3x)^k} \cdot \sum\_{n=0}^{\infty} \binom{n+k-1}{n} \cdot \frac{t^n}{(3x)^n} \tag{5}$$

with any positive integer *k*. Then, it is found that

$$\begin{split} &\quad \frac{f^{(h)}(t)}{(3x-t)^{k}}\\ &= \quad \frac{1}{(3x)^{k}}\cdot\left(\sum\_{n=0}^{\infty}\frac{(n+h)!}{n!}\cdot B\_{n+h}(\mathbf{x})\cdot t^{n}\right)\left(\sum\_{n=0}^{\infty}\binom{n+k-1}{n}\cdot\frac{t^{n}}{(3x)^{n}}\right)\\ &= \quad \frac{1}{(3x)^{k}}\sum\_{n=0}^{\infty}\left(\sum\_{i+j=n}\frac{(j+h)!}{j!}\cdot B\_{j+h}(\mathbf{x})\cdot\left(\sum\_{i}^{i}+k-1\right)\cdot\frac{1}{(3x)^{i}}\right)t^{n}\\ &= \quad \frac{1}{(3x)^{k}}\sum\_{n=0}^{\infty}\left(\sum\_{i=0}^{n}\frac{(n-i+h)!}{(n-i)!}\cdot B\_{n-i+h}(\mathbf{x})\cdot\left(\sum\_{i}^{i}+k-1\right)\cdot\frac{1}{(3x)^{i}}\right)t^{n}.\end{split}$$

where we have used the multiplicative of the power series. Lemma 3 has been proved.

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

Based on the lemmas in the above section, it is easy to deduce the proof of Theorem 1. For any positive integer *h*, we can derive

$$\mathcal{Z}^{\mathrm{h}} \cdot h! \cdot f^{h+1}(t) = \mathcal{Z}^{\mathrm{h}} \cdot h! \cdot \left(\sum\_{n=0}^{\infty} B\_{n}(\mathbf{x}) \cdot t^{n}\right)^{h+1}$$

$$= \quad \mathcal{Z}^{\mathrm{h}} \cdot h! \cdot \sum\_{n=0}^{\infty} \left(\sum\_{a\_{1}+a\_{2}+\cdots+a\_{h+1}=n} B\_{a\_{1}}(\mathbf{x}) B\_{a\_{2}}(\mathbf{x}) \cdot \cdots B\_{a\_{h+1}}(\mathbf{x})\right) \cdot t^{n}.\tag{6}$$

On the other hand, by the observation made in Lemma 3, it is deduced that

$$2^h \cdot h! \cdot f^{h+1}(t) = \sum\_{j=1}^h M(h,j) \cdot \frac{f^{(j)}(t)}{(3x-t)^{2h-j}}$$

$$= \sum\_{j=1}^h \frac{M(h,j)}{(3x)^{2h-j}} \cdot \left(\sum\_{n=0}^\infty \left(\sum\_{i=0}^n \frac{(n-i+j)!}{(n-i)!} \cdot B\_{n-i+j}(x) \cdot \left(\begin{array}{c} 2h+i-j-1\\i \end{array}\right) \cdot \frac{1}{(3x)^i}\right) t^n\right)$$

$$= \sum\_{n=0}^\infty \left(\sum\_{j=1}^h \frac{M(h,j)}{(3x)^{2h-j}} \sum\_{i=0}^n \frac{(n-i+j)!}{(n-i)!} \cdot \frac{B\_{n-i+j}(x)}{(3x)^i} \cdot \left(\begin{array}{c} 2h+i-j-1\\i \end{array}\right)\right) \cdot t^n. \tag{7}$$

Altogether, we obtain the identity:

$$2^h \cdot h! \sum\_{a\_1 + a\_2 + \dots + a\_{h+1} = n} B\_{a\_1}(\mathbf{x}) B\_{a\_2}(\mathbf{x}) \cdot \dots \cdot B\_{a\_{h+1}}(\mathbf{x})$$

$$= \sum\_{j=1}^h \frac{M(h, j)}{(3\mathbf{x})^{2h-j}} \sum\_{i=0}^n \frac{(n-i+j)!}{(n-i)!} \cdot \frac{B\_{n-i+j}(\mathbf{x})}{(3\mathbf{x})^i} \cdot \binom{2h+i-j-1}{i}$$

.

This proves Theorem 1.

### **4. Conclusions**

In this paper, a representation of a linear combination of balancing polynomials *Bi*(*x*) (see Theorem 1) is obtained. Moreover, the specific expressions of *M*(*h*, *i*) is given by using mathematical induction (see Lemma 1).

Theorem 1 can be reduced to various studies for the specific values of *x*, *n*, and *h* in the literature. For example, if *n* = 0, our results reduce to Corollary 1. Taking *h* = 2, our results reduce to Corollary 2. Taking *x* = 1, *h* = 2, 3, our results reduce to Corollary 3 and Corollary 4, respectively.

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

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

**Conflicts of Interest:** The author declares that there are no conflicts of interest regarding 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/).
