**Differential Equations Arising from the Generating Function of the (***r***,** *β***)-Bell Polynomials and Distribution of Zeros of Equations**

#### **Kyung-Won Hwang 1, Cheon Seoung Ryoo 2,\* and Nam Soon Jung <sup>3</sup>**


Received: 2 July 2019; Accepted: 9 August 2019; Published: 12 August 2019

**Abstract:** In this paper, we study differential equations arising from the generating function of the (*r*, *β*)-Bell polynomials. We give explicit identities for the (*r*, *β*)-Bell polynomials. Finally, we find the zeros of the (*r*, *β*)-Bell equations with numerical experiments.

**Keywords:** differential equations; Bell polynomials; *r*-Bell polynomials; (*r*, *β*)-Bell polynomials; zeros

**MSC:** 05A19; 11B83; 34A30; 65L99

#### **1. Introduction**

The moments of the Poisson distribution are a well-known connecting tool between Bell numbers and Stirling numbers. As we know, the Bell numbers *Bn* are those using generating function

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

The Bell polynomials *Bn*(*λ*) are this formula using the generating function

*e λ*(*e<sup>t</sup>* <sup>−</sup>1) = ∞ ∑ *n*=0 *Bn*(*λ*) *t n n*! , (1)

(see [1,2]).

Observe that

$$B\_n(\lambda) = \sum\_{i=0}^n \lambda^i S\_2(n, i),$$

where *<sup>S</sup>*2(*n*, *<sup>i</sup>*) = <sup>1</sup> *i*! ∑*i l*=0(−1)*i*−*<sup>l</sup>* ( *i l* )*l <sup>n</sup>* denotes the second kind Stirling number. The generalized Bell polynomials *Bn*(*x*, *λ*) are these formula using the generating function:

$$\sum\_{n=0}^{\infty} B\_n(x,\lambda) \frac{t^n}{n!} = e^{\mathbf{x}t - \lambda(e^t - t - 1)}, \text{ (see [2])}.$$

In particular, the generalized Bell polynomials *Bn*(*x*, <sup>−</sup>*λ*) = *<sup>E</sup>λ*[(*<sup>Z</sup>* <sup>+</sup> *<sup>x</sup>* <sup>−</sup> *<sup>λ</sup>*)*n*], *<sup>λ</sup>*, *<sup>x</sup>* <sup>∈</sup> <sup>R</sup>, *<sup>n</sup>* <sup>∈</sup> <sup>N</sup>, where *Z* is a Poission random variable with parameter *λ* > 0 (see [1–3]). The (*r*, *β*)-Bell polynomials *Gn*(*x*,*r*, *β*) are this formula using the generating function:

$$F(t, \mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{n=0}^{\infty} G\_{\text{fl}}(\mathbf{x}, r, \boldsymbol{\beta}) \frac{t^n}{n!} = e^{rt + (e^{\beta t} - 1)\frac{\mathbf{x}}{\beta}},\tag{2}$$

(see [3]), where, *β* and *r* are real or complex numbers and (*r*, *β*) = (0, 0). Note that *Bn*(*x* + *r*, −*x*) = *Gn*(*x*,*r*, 1) and *Bn*(*x*) = *Gn*(*x*, 0, 1). The first few examples of (*r*, *β*)-Bell polynomials *Gn*(*x*,*r*, *β*) are

$$\begin{aligned} G\_0(\mathbf{r}, r, \boldsymbol{\beta}) &= 1, \\ G\_1(\mathbf{r}, r, \boldsymbol{\beta}) &= r + \mathbf{x}, \\ G\_2(\mathbf{r}, r, \boldsymbol{\beta}) &= r^2 + \beta \mathbf{x} + 2r\mathbf{x} + \mathbf{x}^2, \\ G\_3(\mathbf{r}, r, \boldsymbol{\beta}) &= r^3 + \beta^2 \mathbf{x} + 3\beta r \mathbf{x} + 3r^2 \mathbf{x} + 3\beta \mathbf{x}^2 + 3r \mathbf{x}^2 + \mathbf{x}^3, \\ G\_4(\mathbf{r}, r, \boldsymbol{\beta}) &= r^4 + \beta^3 \mathbf{x} + 4\beta^2 r \mathbf{x} + 6\beta r^2 \mathbf{x} + 4r^3 \mathbf{x} + 7\beta^2 \mathbf{x}^2 + 12\beta r \mathbf{x}^2 \\ &\quad + 6r^2 \mathbf{x}^2 + 6\beta \mathbf{x}^3 + 4r \mathbf{x}^3 + \mathbf{x}^4, \\ G\_5(\mathbf{r}, r, \boldsymbol{\beta}) &= r^5 + \beta^4 \mathbf{x} + 5\beta^3 r \mathbf{x} + 10\beta^2 r^2 \mathbf{x} + 10\beta r^3 \mathbf{x} + 5r^4 \mathbf{x} + 15\beta^3 \mathbf{x}^2 + 35\beta^2 r \mathbf{x}^2 \\ &\quad + 30\theta r^2 \mathbf{x}^2 + 10r^3 \mathbf{x}^2 + 25\theta^2 \mathbf{x}^3 + 30\theta r \mathbf{x}^3 + 10r^2 \mathbf{x}^4 + 10\theta \mathbf{x}^4 + 5r \mathbf{x}^4 + \mathbf{x}^5. \end{aligned}$$

From (1) and (2), we see that

$$\begin{split} \sum\_{n=0}^{\infty} G\_n(\mathbf{x}, r, \boldsymbol{\beta}) \frac{t^n}{n!} &= e^{(e^{\beta t} - 1)\frac{t}{\beta}} e^{rt} \\ &= \left( \sum\_{k=0}^{\infty} B\_k(\mathbf{x}/\beta) \boldsymbol{\beta}^k \frac{t^k}{k!} \right) \left( \sum\_{m=0}^{\infty} r^m \frac{t^m}{m!} \right) \\ &= \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^n \binom{n}{k} B\_k(\mathbf{x}/\beta) \boldsymbol{\beta}^k r^{n-k} \right) \frac{t^n}{n!} . \end{split} \tag{3}$$

Compare the coefficients in Formula (3). We can get

$$G\_n(\mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{k=0}^n \binom{n}{k} \beta^k B\_k(\mathbf{x}/\boldsymbol{\beta}) r^{n-k}, \quad (n \ge 0).$$

Similarly we also have

$$G\_n(x+y,r,\beta) = \sum\_{k=0}^n \binom{n}{k} G\_k(x,r,\beta) B\_{n-k}(y/\beta) \beta^{n-k}.$$

Recently, many mathematicians have studied the differential equations arising from the generating functions of special polynomials (see [4–8]). Inspired by their work, we give a differential equations by generation of (*r*, *β*)-Bell polynomials *Gn*(*x*,*r*, *β*) as follows. Let *D* denote differentiation with respect to *t*, *D*<sup>2</sup> denote differentiation twice with respect to *t*, and so on; that is, for positive integer *N*,

$$D^N F = \left(\frac{\partial}{\partial t}\right)^N F(t, \mathbf{x}, r, \boldsymbol{\beta}).$$

We find differential equations with coefficients *ai*(*N*, *x*,*r*, *β*), which are satisfied by

$$\left(\frac{\partial}{\partial t}\right)^N F(t, \mathbf{x}, r, \boldsymbol{\beta}) - a\_0(N, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r, \boldsymbol{\beta}) - \dots - a\_N(N, \mathbf{x}, r, \boldsymbol{\beta}) e^{\beta t N} F(t, \mathbf{x}, r, \boldsymbol{\beta}) = 0.$$

Using the coefficients of this differential equation, we give explicit identities for the (*r*, *β*)-Bell polynomials. In addition, we investigate the zeros of the (*r*, *β*)-Bell equations with numerical methods. Finally, we observe an interesting phenomena of 'scattering' of the zeros of (*r*, *β*)-Bell equations. Conjectures are also presented through numerical experiments.

#### **2. Differential Equations Related to (***R***,** *β***)-Bell Polynomials**

Differential equations arising from the generating functions of special polynomials are studied by many authors to give explicit identities for special polynomials (see [4–8]). In this section, we study differential equations arising from the generating functions of (*r*, *β*)-Bell polynomials.

Let

$$F = F(t, \mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{n=0}^{\infty} \mathcal{G}\_{\mathbb{H}}(\mathbf{x}, r, \boldsymbol{\beta}) \frac{t^n}{n!} = e^{\mathbf{f}t + (e^{\boldsymbol{\beta}t} - 1)\frac{\mathbf{r}}{\mathbb{H}}}, \quad \mathbf{x}, r, \boldsymbol{\beta} \in \mathbb{C}. \tag{4}$$

Then, by (4), we have

$$\begin{split} DF &= \frac{\partial}{\partial t} F(t, \mathbf{x}, r, \boldsymbol{\beta}) = \frac{\partial}{\partial t} \left( \boldsymbol{\varepsilon}^{rt + (\boldsymbol{\varepsilon}^{\partial t} - 1)\frac{\boldsymbol{x}}{\boldsymbol{\beta}}} \right) \\ &= \boldsymbol{\varepsilon}^{rt + (\boldsymbol{\varepsilon}^{\partial t} - 1)\frac{\boldsymbol{x}}{\boldsymbol{\beta}}} (r + \mathbf{x} \boldsymbol{\varepsilon}^{\partial t}) \\ &= r \boldsymbol{\varepsilon}^{rt + (\boldsymbol{\varepsilon}^{\partial t} - 1)\frac{\boldsymbol{x}}{\boldsymbol{\beta}}} + \mathbf{x} \boldsymbol{\varepsilon}^{(r + \boldsymbol{\beta})t + (\boldsymbol{\varepsilon}^{\partial t} - 1)\frac{\boldsymbol{x}}{\boldsymbol{\beta}}} \\ &= r F(t, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x} F(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}), \end{split} \tag{5}$$

$$\begin{split} D^2F &= rDF(t, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x}DF(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}) \\ &= r^2 F(t, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x}(2r + \boldsymbol{\beta})F(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}) + \mathbf{x}^2 F(t, \mathbf{x}, r + 2\boldsymbol{\beta}, \boldsymbol{\beta}), \end{split} \tag{6}$$

and

$$\begin{split} D^3F &= r^2DF(t, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x}(2r + \boldsymbol{\beta})DF(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}) + \mathbf{x}^2 DF(t, \mathbf{x}, r + 2\boldsymbol{\beta}, \boldsymbol{\beta}) \\ &= r^3F(t, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x}\left(r^2 + (2r + \boldsymbol{\beta})(r + \boldsymbol{\beta})\right)F(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}) \\ &\quad + \mathbf{x}^2(3r + 3\boldsymbol{\beta})F(t, \mathbf{x}, r + 2\boldsymbol{\beta}, \boldsymbol{\beta}) + \mathbf{x}^3 F(t, \mathbf{x}, r + 3\boldsymbol{\beta}, \boldsymbol{\beta}). \end{split}$$

We prove this process by induction. Suppose that

$$D^N F = \sum\_{i=0}^N a\_i(N, x, r, \beta) F(t, x, r + i\beta, \beta)\_\prime (N = 0, 1, 2, \dots). \tag{7}$$

is true for N. From (7), we get

$$\begin{split} D^{N+1}F &= \sum\_{i=0}^{N} a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) DF(t, \mathbf{x}, r + i\boldsymbol{\beta}, \boldsymbol{\beta}) \\ &= \sum\_{i=0}^{N} a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) \left\{ (r + i\boldsymbol{\beta}) F(t, \mathbf{x}, r + i\boldsymbol{\beta}, \boldsymbol{\beta}) + \mathbf{x} F(t, \mathbf{x}, r + (i + 1)\boldsymbol{\beta}, \boldsymbol{\beta}) \right\} \\ &= \sum\_{i=0}^{N} a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) (r + i\boldsymbol{\beta}) F(t, \mathbf{x}, r + i\boldsymbol{\beta}, \boldsymbol{\beta}) \\ &\qquad + \mathbf{x} \sum\_{i=0}^{N} a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r + (i + 1)\boldsymbol{\beta}, \boldsymbol{\beta}) \\ &= \sum\_{i=0}^{N} (r + i\boldsymbol{\beta}) a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r + i\boldsymbol{\beta}, \boldsymbol{\beta}) \\ &\qquad + \mathbf{x} \sum\_{i=1}^{N+1} a\_{i-1}(N, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r + i\boldsymbol{\beta}, \boldsymbol{\beta}). \end{split} (8)$$

From (8), we get

$$D^{N+1}F = \sum\_{i=0}^{N+1} a\_i (N+1, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r+i\boldsymbol{\beta}, \boldsymbol{\beta}) . \tag{9}$$

We prove that

$$D^{k+1}F = \sum\_{i=0}^{k+1} a\_i(k+1, \mathfrak{x}, r, \mathfrak{z}) F(t, \mathfrak{x}, r+i\mathfrak{z}, \mathfrak{z}).$$

If we compare the coefficients on both sides of (8) and (9), then we get

$$a\_0(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = ra\_0(N, \mathbf{x}, r, \boldsymbol{\beta}), \quad a\_{N+1}(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x}a\_N(N, \mathbf{x}, r, \boldsymbol{\beta}), \tag{10}$$

and

$$a\_i(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = (r+i\beta)a\_{i-1}(N, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x}a\_{i-1}(N, \mathbf{x}, r, \boldsymbol{\beta}), (1 \le i \le N). \tag{11}$$

In addition, we get

$$F(t, \mathbf{x}, r, \boldsymbol{\beta}) = a\_0(0, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r, \boldsymbol{\beta}). \tag{12}$$

Now, by (10), (11) and (12), we can obtain the coefficients *ai*(*j*, *x*,*r*, *β*)0≤*i*,*j*≤*N*+<sup>1</sup> as follows. By (12), we get

$$a\_0(0, \ge, r, \mathcal{G}) = 1.\tag{13}$$

It is not difficult to show that

$$\begin{aligned} &rF(t, \mathbf{x}, r, \boldsymbol{\beta}) + \mathbf{x}F(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}) \\ &= DF(t, \mathbf{x}, r, \boldsymbol{\beta}) \\ &= \sum\_{i=0}^{1} a\_{i}(1, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}) \\ &= a\_{0}(1, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r, \boldsymbol{\beta}) + a\_{1}(1, \mathbf{x}, r, \boldsymbol{\beta}) F(t, \mathbf{x}, r + \boldsymbol{\beta}, \boldsymbol{\beta}). \end{aligned} \tag{14}$$

Thus, by (14), we also get

$$a\_0(\mathbf{1}, \mathbf{x}, r, \boldsymbol{\beta}) = r, \quad a\_1(\mathbf{1}, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x}. \tag{15}$$

From (10), we have that

$$a\_0(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = ra\_0(N, \mathbf{x}, r, \boldsymbol{\beta}) = \dots = r^N a\_0(1, \mathbf{x}, r, \boldsymbol{\beta}) = r^{N+1},\tag{16}$$

and

$$a\_{N+1}(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x} \\ a\_N(N, \mathbf{x}, r, \boldsymbol{\beta}) = \dots = \mathbf{x}^N \\ a\_1(1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x}^{N+1}.\tag{17}$$

For *i* = 1, 2, 3 in (11), we have

$$a\_1(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x} \sum\_{k=0}^{N} (r+\beta)^k a\_0(N-k, \mathbf{x}, r, \boldsymbol{\beta}), \tag{18}$$

$$a\_2(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x} \sum\_{k=0}^{N-1} (r+2\beta)^k a\_1(N-k, \mathbf{x}, r, \boldsymbol{\beta}), \tag{19}$$

and

$$a\_3(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x} \sum\_{k=0}^{N-2} (r+3\beta)^k a\_2(N-k, \mathbf{x}, r, \boldsymbol{\beta}). \tag{20}$$

By induction on *i*, we can easily prove that, for 1 ≤ *i* ≤ *N*,

$$a\_i(N+1, \mathbf{x}, r, \boldsymbol{\beta}) = \mathbf{x} \sum\_{k=0}^{N-i+1} (r+i\boldsymbol{\beta})^k a\_{i-1}(N-k, \mathbf{x}, r, \boldsymbol{\beta}). \tag{21}$$

Here, we note that the matrix *ai*(*j*, *x*,*r*, *β*)0≤*i*,*j*≤*N*+<sup>1</sup> is given by


Now, we give explicit expressions for *ai*(*N* + 1, *x*,*r*, *β*). By (18), (19), and (20), we get

$$\begin{aligned} a\_1(N+1, \mathbf{x}, r, \boldsymbol{\beta}) &= \mathbf{x} \sum\_{k\_1=0}^N (r+\boldsymbol{\beta})^{k\_1} a\_0(N-k\_1, \mathbf{x}, r, \boldsymbol{\beta}) \\ &= \sum\_{k\_1=0}^N (r+\boldsymbol{\beta})^{k\_1} r^{N-k\_1} \\ a\_2(N+1, \mathbf{x}, r, \boldsymbol{\beta}) &= \mathbf{x} \sum\_{k\_2=0}^{N-1} (r+2\boldsymbol{\beta})^{k\_2} a\_1(N-k\_2, \mathbf{x}, r, \boldsymbol{\beta}) \\ &= \mathbf{x}^2 \sum\_{k\_2=0}^{N-1} \sum\_{k\_1=0}^{N-1-k\_2} (r+\boldsymbol{\beta})^{k\_1} (r+2\boldsymbol{\beta})^{k\_2} r^{N-k\_2-k\_1-1} \end{aligned}$$

and

$$\begin{aligned} &a\_3(N+1, \mathbf{x}, r, \boldsymbol{\beta}) \\ &= \mathbf{x} \sum\_{k\_3=0}^{N-2} (r+3\boldsymbol{\beta})^{k\_3} a\_2(N-k\_3, \mathbf{x}, r, \boldsymbol{\beta}) \\ &= \mathbf{x}^3 \sum\_{k\_3=0}^{N-2} \sum\_{k\_2=0}^{N-2-k\_3} \sum\_{k\_1=0}^{N-2-k\_3-k\_2} (r+3\boldsymbol{\beta})^{k\_3} (r+2\boldsymbol{\beta})^{k\_2} (r+\boldsymbol{\beta})^{k\_1} r^{N-k\_3-k\_2-k\_1-2} .\end{aligned}$$

By induction on *i*, we have

$$\begin{split} &a\_i(N+1, \mathbf{x}, r, \boldsymbol{\beta}) \\ &= \mathbf{x}^i \sum\_{k\_i=0}^{N-i+1} \sum\_{k\_{i-1}=0}^{N-i+1-k\_i} \cdots \sum\_{k\_1=0}^{N-i+1-k\_i-\cdots-k\_2} \left( \prod\_{l=1}^i (r+l\boldsymbol{\beta})^{k\_l} \right) r^{N-i+1-\sum\_{l=1}^i k\_l} . \end{split} \tag{22}$$

Finally, by (22), we can derive a differential equations with coefficients *ai*(*N*, *x*,*r*, *β*), which is satisfied by

$$\left(\frac{\partial}{\partial t}\right)^{N}F(t,\mathbf{x},r,\beta) - a\_{0}(N,\mathbf{x},r,\beta)F(t,\mathbf{x},r,\beta) - \cdots - a\_{N}(N,\mathbf{x},r,\beta)e^{\beta tN}F(t,\mathbf{x},r,\beta) = 0.$$

**Theorem 1.** *For same as below N* = 0, 1, 2, . . . , *the differential equation*

$$D^N F = \sum\_{i=0}^N a\_i(N, \ge, r, \beta) e^{i\beta t} F(t, \ge, r, \beta)$$

*has a solution*

$$F = F(t, \mathbf{x}, r, \boldsymbol{\beta}) = e^{rt + (e^{\beta t} - 1)\frac{\mathbf{x}}{\beta}}r$$

*where*

$$\begin{aligned} a\_0(N, \mathbf{x}, r, \boldsymbol{\beta}) &= r^N, \\ a\_N(N, \mathbf{x}, r, \boldsymbol{\beta}) &= \mathbf{x}^N, \\ a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) &= \mathbf{x}^i \sum\_{k\_i=0}^{N-i} \sum\_{k\_{i-1}=0}^{i-k\_i} \cdots \sum\_{k\_1=0}^{N-i-k\_i-\cdots-k\_2} \left( \prod\_{l=1}^i (r+l\boldsymbol{\beta})^{k\_l} \right) r^{N-i-\sum\_{l=1}^i k\_l}, \\ & (1 \le i \le N). \end{aligned}$$

From (4), we have this

$$D^N F = \left(\frac{\partial}{\partial t}\right)^N F(t, \mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{k=0}^{\infty} \mathcal{G}\_{k+N}(\mathbf{x}, r, \boldsymbol{\beta}) \frac{t^k}{k!}. \tag{23}$$

By using Theorem 1 and (23), we can get this equation:

$$\begin{split} \sum\_{k=0}^{\infty} G\_{k+N}(x, r, \boldsymbol{\beta}) \frac{k^k}{k!} &= D^N F \\ &= \left( \sum\_{i=0}^N a\_i(N, x, r, \boldsymbol{\beta}) e^{i\boldsymbol{\beta}t} \right) F(t, x, r, \boldsymbol{\beta}) \\ &= \sum\_{i=0}^N a\_i(N, x, r, \boldsymbol{\beta}) \left( \sum\_{l=0}^{\infty} (i\boldsymbol{\beta})^l \frac{l^l}{l!} \right) \left( \sum\_{m=0}^{\infty} G\_m(x, r, \boldsymbol{\beta}) \frac{t^m}{m!} \right) \\ &= \sum\_{i=0}^N a\_i(N, x, r, \boldsymbol{\beta}) \left( \sum\_{k=0}^{\infty} \sum\_{m=0}^k \binom{k}{m} (i\boldsymbol{\beta})^{k-m} G\_m(x, r, \boldsymbol{\beta}) \frac{t^k}{k!} \right) \\ &= \sum\_{k=0}^{\infty} \left( \sum\_{i=0}^N \sum\_{m=0}^k \binom{k}{m} (i\boldsymbol{\beta})^{k-m} a\_i(N, x, r, \boldsymbol{\beta}) G\_m(x, r, \boldsymbol{\beta}) \right) \frac{t^k}{k!} .\end{split} \tag{24}$$

Compare coefficients in (24). We get the below theorem.

**Theorem 2.** *For k*, *N* = 0, 1, 2, . . . , *we have*

$$\mathbf{G}\_{k+N}(\mathbf{x}, r, \boldsymbol{\uprho}) = \sum\_{i=0}^{N} \sum\_{m=0}^{k} \binom{k}{m} \mathbf{i}^{k-m} \boldsymbol{\uprho}^{k-m} a\_i(N, \mathbf{x}, r, \boldsymbol{\uprho}) \mathbf{G}\_m(\mathbf{x}, r, \boldsymbol{\uprho}), \tag{25}$$

*where*

$$\begin{aligned} a\_0(N, \mathbf{x}, r, \boldsymbol{\beta}) &= r^N, \\ a\_N(N, \mathbf{x}, r, \boldsymbol{\beta}) &= \mathbf{x}^N, \\ a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}) &= \mathbf{x}^i \sum\_{k\_i=0}^{N-i} \sum\_{k\_{i-1}=0}^{N-i-k\_i} \cdots \sum\_{k\_1=0}^{N-i-k\_i-\cdots-k\_2} \left( \prod\_{l=1}^i (r+l\beta)^{k\_l} \right) r^{N-i-\sum\_{l=1}^i k\_l}, \\ & (1 \le i \le N). \end{aligned}$$

By using the coefficients of this differential equation, we give explicit identities for the (*r*, *β*)-Bell polynomials. That is, in (25) if *k* = 0, we have corollary.

**Corollary 1.** *For N* = 0, 1, 2, . . . , *we have*

$$G\_N(\mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{i=0}^{N} a\_i(N, \mathbf{x}, r, \boldsymbol{\beta}).$$

For *N* = 0, 1, 2, . . . , it follows that equation

$$(D^N F - \sum\_{i=0}^N a\_i(N, x, r, \beta)e^{i\beta t}F(t, x, r, \beta) = 0$$

has a solution

$$F = F(t, \mathbf{x}, r, \boldsymbol{\beta}) = e^{rt + (e^{\beta t} - 1)\frac{\mathbf{x}}{\beta}}.$$

In Figure 1, we have a sketch of the surface about the solution *F* of this differential equation. On the left of Figure 1, we give −3 ≤ *x* ≤ 3, −1 ≤ *t* ≤ 1, and *r* = 2, *β* = 5. On the right of Figure 1, we give −3 ≤ *x* ≤ 3, −1 ≤ *t* ≤ 1, and *r* = −3, *β* = 2.

**Figure 1.** The surface for the solution *F*(*t*, *x*,*r*, *β*).

Making *N*-times derivative for (4) with respect to *t*, we obtain

$$\left(\frac{\partial}{\partial t}\right)^{N} F(t, \mathbf{x}, r, \boldsymbol{\beta}) = \left(\frac{\partial}{\partial t}\right)^{N} \boldsymbol{\varepsilon}^{\mathrm{tr} + (\boldsymbol{\varepsilon}^{\mathrm{fit}} - 1)\frac{\boldsymbol{x}}{\boldsymbol{\beta}}} = \sum\_{m=0}^{\infty} \mathcal{G}\_{m+N}(\mathbf{x}, r, \boldsymbol{\beta}) \frac{\boldsymbol{t}^{m}}{m!}. \tag{26}$$

By multiplying the exponential series *ext* = ∑<sup>∞</sup> *<sup>m</sup>*=<sup>0</sup> *<sup>x</sup><sup>m</sup> <sup>t</sup> m m*! in both sides of (26) and Cauchy product, we derive

$$\begin{split} e^{-mt} \left( \frac{\partial}{\partial t} \right)^{N} F(t, \mathbf{x}, r, \boldsymbol{\beta}) &= \left( \sum\_{m=0}^{\infty} (-n)^{m} \frac{t^{m}}{m!} \right) \left( \sum\_{m=0}^{\infty} G\_{m+N}(\mathbf{x}, r, \boldsymbol{\beta}) \frac{t^{m}}{m!} \right) \\ &= \sum\_{m=0}^{\infty} \left( \sum\_{k=0}^{m} \binom{m}{k} (-n)^{m-k} G\_{N+k}(\mathbf{x}, r, \boldsymbol{\beta}) \right) \frac{t^{m}}{m!} . \end{split} \tag{27}$$

By using the Leibniz rule and inverse relation, we obtain

$$\begin{split} e^{-nt} \left( \frac{\partial}{\partial t} \right)^{N} F(t, x, y) &= \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} \left( \frac{\partial}{\partial t} \right)^{k} \left( e^{-nt} F(t, x, r, \beta) \right) \\ &= \sum\_{m=0}^{\infty} \left( \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} G\_{m+k}(x - n, r, \beta) \right) \frac{t^{m}}{m!} . \end{split} \tag{28}$$

So using (27) and (28), and using the coefficients of *<sup>t</sup> m m*! gives the below theorem.

**Theorem 3.** *Let m*, *n*, *N be nonnegative integers. Then*

$$\sum\_{k=0}^{m} \binom{m}{k} (-n)^{m-k} \mathbf{G}\_{N+k}(\mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} \mathbf{G}\_{m+k}(\mathbf{x} - n, r, \boldsymbol{\beta}). \tag{29}$$

When we give *m* = 0 in (29), then we get corollary.

**Corollary 2.** *For N* = 0, 1, 2, . . . , *we have*

$$G\_N(\mathbf{x}, r, \boldsymbol{\beta}) = \sum\_{k=0}^{N} \binom{N}{k} n^{N-k} G\_k(\mathbf{x} - n, r, \boldsymbol{\beta}) \dots$$

#### **3. Distribution of Zeros of the (***R***,** *β***)-Bell Equations**

This section aims to demonstrate the benefit of using numerical investigation to support theoretical prediction and to discover new interesting patterns of the zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0. We investigate the zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 with numerical experiments. We plot the zeros of the *Bn*(*x*, *<sup>λ</sup>*) = 0 for *<sup>n</sup>* <sup>=</sup> 16,*<sup>r</sup>* <sup>=</sup> <sup>−</sup>5, <sup>−</sup>3, 3, 5, *<sup>β</sup>* <sup>=</sup> 2, 3 and *<sup>x</sup>* <sup>∈</sup> <sup>C</sup> (Figure 2).

In top-left of Figure 2, we choose *n* = 16 and *r* = −5, *β* = 2. In top-right of Figure 2, we choose *n* = 16 and *r* = −3, *β* = 3. In bottom-left of Figure 2, we choose *n* = 16 and *r* = 3, *β* = 2 . In bottom-right of Figure 2, we choose *n* = 16 and *r* = 5, *β* = 3.

Prove that *Gn*(*x*,*r*, *<sup>β</sup>*), *<sup>x</sup>* <sup>∈</sup> <sup>C</sup>, has *Im*(*x*) = 0 reflection symmetry analytic complex functions (see Figure 3). Stacks of zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 for 1 ≤ *n* ≤ 20 from a 3-D structure are presented (Figure 3).

On the left of Figure 3, we choose *r* = −5 and *β* = 2. On the right of Figure 3, we choose *r* = 5 and *β* = 2. In Figure 3, the same color has the same degree *n* of (*r*, *β*)-Bell polynomials *Gn*(*x*,*r*, *β*). For example, if *n* = 20, zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 is red.

**Figure 2.** Zeros of *Gn*(*x*,*r*, *β*) = 0.

**Figure 3.** Stacks of zeros of *Gn*(*x*,*r*, *β*) = 0, 1 ≤ *n* ≤ 20.

Our numerical results for approximate solutions of real zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 are displayed (Tables 1 and 2).


**Table 1.** Numbers of real and complex zeros of *Gn*(*x*,*r*, *β*) = 0

**Table 2.** Approximate solutions of *Gn*(*x*,*r*, *<sup>β</sup>*) = 0, *<sup>x</sup>* <sup>∈</sup> <sup>R</sup>.


Plot of real zeros of *Gn*(*x*,*r*, *β*) = 0 for 1 ≤ *n* ≤ 20 structure are presented (Figure 4).

In Figure 4 (left), we choose *r* = 5 and *β* = −2. In Figure 4 (right), we choose *r* = 5 and *β* = 2. In Figure 4, the same color has the same degree *n* of (*r*, *β*)-Bell polynomials *Gn*(*x*,*r*, *β*). For example, if *n* = 20, real zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 is blue.

Next, we calculated an approximate solution satisfying *Gn*(*x*,*r*, *<sup>β</sup>*) = 0,*<sup>r</sup>* <sup>=</sup> 5, *<sup>β</sup>* <sup>=</sup> 2, *<sup>x</sup>* <sup>∈</sup> <sup>R</sup>. The results are given in Table 2.

**Figure 4.** Stacks of zeros of *Gn*(*x*,*r*, *β*) = 0, 1 ≤ *n* ≤ 20.

#### **4. Conclusions**

We constructed differential equations arising from the generating function of the (*r*, *β*)-Bell polynomials. This study obtained the some explicit identities for (*r*, *β*)-Bell polynomials *Gn*(*x*,*r*, *β*) using the coefficients of this differential equation. The distribution and symmetry of the roots of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 were investigated. We investigated the symmetry of the zeros of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 for various variables *r* and *β*, but, unfortunately, we could not find a regular pattern. We make the following series of conjectures with numerical experiments:

Let us use the following notations. *RGn*(*x*,*r*,*β*) denotes the number of real zeros of *Gn*(*x*,*r*, *β*) = 0 lying on the real plane *Im*(*x*) = 0 and *CGn*(*x*,*r*,*β*) denotes the number of complex zeros of *Gn*(*x*,*r*, *β*) = 0. Since *n* is the degree of the polynomial *Gn*(*x*,*r*, *β*), we have *RGn*(*x*,*r*,*β*) = *n* − *CGn*(*x*,*r*,*β*)(see Table 1).

We can see a good regular pattern of the complex roots of the (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 for *r* > 0 and *β* > 0. Therefore, the following conjecture is possible.

**Conjecture 1.** *For r* > 0 *and β* > 0*, prove or disprove that*

$$
\mathbb{C}\_{H\_n(x,y)} = 0.
$$

*As a result of investigating more r* > 0 *and β* > 0 *variables, it is still unknown whether the conjecture 1 is true or false for all variables r* > 0 *and β* > 0 *(see Figure 1 and Table 1).*

We observe that solutions of (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *β*) = 0 has *Im*(*x*) = 0, reflecting symmetry analytic complex functions. It is expected that solutions of (*r*, *β*)-Bell equations *Gn*(*x*,*r*, *<sup>β</sup>*) = 0, has not *Re*(*x*) = *<sup>a</sup>* reflection symmetry for *<sup>a</sup>* <sup>∈</sup> <sup>R</sup> (see Figures 2–4).

**Conjecture 2.** *Prove or disprove that solutions of* (*r*, *β*)*-Bell equations Gn*(*x*,*r*, *β*) = 0*, has not Re*(*x*) = *a reflection symmetry for a* <sup>∈</sup> <sup>R</sup>*.*

Finally, how many zeros do *Gn*(*x*,*r*, *β*) = 0 have? We are not able to decide if *Gn*(*x*,*r*, *β*) = 0 has *n* distinct solutions (see Tables 1 and 2). We would like to know the number of complex zeros *CGn*(*x*,*r*,*β*) of *Gn*(*x*,*r*, *β*) = 0, *Im*(*x*) = 0.

**Conjecture 3.** *Prove or disprove that Gn*(*x*,*r*, *β*) = 0 *has n distinct solutions.*

As a result of investigating more *n* variables, it is still unknown whether the conjecture is true or false for all variables *n* (see Tables 1 and 2). We expect that research in these directions will make a new approach using the numerical method related to the research of the (*r*, *β*)-Bell numbers and polynomials which appear in mathematics, applied mathematics, statistics, and mathematical physics. The reader may refer to [5–10] for the details.

**Author Contributions:** These authors contributed equally to this work.

**Funding:** This work was supported by the Dong-A university research fund.

**Acknowledgments:** The authors would like to thank the referees for their valuable comments, which improved the original manuscript in its present form.

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

#### **References**


c 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* **Truncated Fubini Polynomials**

**Ugur Duran 1,\* and Mehmet Acikgoz <sup>2</sup>**


Received: 22 April 2019; Accepted: 9 May 2019; Published: 15 May 2019

**Abstract:** In this paper, we introduce the two-variable truncated Fubini polynomials and numbers and then investigate many relations and formulas for these polynomials and numbers, including summation formulas, recurrence relations, and the derivative property. We also give some formulas related to the truncated Stirling numbers of the second kind and Apostol-type Stirling numbers of the second kind. Moreover, we derive multifarious correlations associated with the truncated Euler polynomials and truncated Bernoulli polynomials.

**Keywords:** Fubini polynomials; Euler polynomials; Bernoulli polynomials; truncated exponential polynomials; Stirling numbers of the second kind

**MSC:** Primary 11B68; Secondary 11B83, 11B37, 05A19

#### **1. Introduction**

The classical Bernoulli and Euler polynomials are defined by means of the following generating functions: <sup>∞</sup>

$$\sum\_{n=0}^{\infty} B\_n \left( x \right) \frac{t^n}{n!} = \frac{t}{c^t - 1} e^{xt} \quad \left( |t| < 2\pi \right) \tag{1}$$

and: <sup>∞</sup>

$$\sum\_{n=0}^{\infty} E\_{\text{fl}}\left(\mathbf{x}\right) \frac{t^n}{n!} = \frac{2}{\varepsilon^t + 1} e^{\mathbf{x}t} \quad \left(|t| < \pi\right), \tag{2}$$

see [1–10] for details about the aforesaid polynomials. The Bernoulli numbers *Bn* and Euler numbers *En* are obtained by the special cases of the corresponding polynomials at *x* = 0, namely:

$$B\_n\left(0\right) := B\_n \text{ and } E\_n\left(0\right) := E\_n. \tag{3}$$

The truncated exponential polynomials have played a role of crucial importance to evaluate integrals including products of special functions; cf. [11], and also see the references cited therein. Recently, several mathematicians have studied truncated-type special polynomials such as truncated Bernoulli polynomials and truncated Euler polynomials; cf. [1,4,7,9,11,12].

*Mathematics* **2019**, *7*, 431; doi:10.3390/math7050431 www.mdpi.com/journal/mathematics

For non-negative integer *m*, the truncated Bernoulli and truncated Euler polynomials are introduced as follows: <sup>∞</sup>

$$\sum\_{n=0}^{\infty} B\_{m,n} \left( x \right) \frac{t^n}{n!} = \frac{\frac{t^n}{m!}}{e^t - \sum\_{j=0}^{m-1} \frac{t^j}{j!}} e^{xt} \qquad \text{(cf. [11])}\tag{4}$$

and: <sup>∞</sup>

$$\sum\_{n=0}^{\infty} E\_{m,n}\left(x\right) \frac{t^n}{n!} = \frac{2\frac{t^m}{m!}}{e^t + 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!}} e^{xt} \qquad \text{(cf. [7]) }. \tag{5}$$

Upon setting *x* = 0 in (4) and (5), the mentioned polynomials (*Bm*,*<sup>n</sup>* (*x*) and *Em*,*<sup>n</sup>* (*x*)), reduce to the corresponding numbers:

$$B\_{m,n}\left(0\right) := B\_{m,n} \text{ and } E\_{m,n}\left(0\right) := E\_{m,n} \tag{6}$$

termed as the truncated Bernoulli numbers and truncated Euler numbers, respectively.

**Remark 1.** *Setting m* = 0 *in (4) and m* = 1 *(5), then the truncated Bernoulli and truncated Euler polynomials reduce to the classical Bernoulli and Euler polynomials in (1) and (2).*

The Stirling numbers of the second kind are given by the following exponential generating function:

$$\sum\_{n=0}^{\infty} S\_2\left(n,k\right) \frac{t^n}{n!} = \frac{\left(t^t - 1\right)^k}{k!} \quad \text{(cf. [2-5.7,8,10,13])}\tag{7}$$

or by the recurrence relation for a fixed non-negative integer *ζ*,

$$\mathbf{x}^{\mathbb{Z}} = \sum\_{\mu=0}^{\mathbb{Z}} \text{S2}\left(\zeta, \mu\right) \left(\mathbf{x}\right)\_{\mu\text{\textquotedblleft}\mu} \tag{8}$$

where the notation (*x*)*<sup>μ</sup>* called the falling factorial equals *x* (*x* − 1)···(*x* − *μ* + 1); cf. [2–5,7–10,13], and see also the references cited therein.

The Apostol-type Stirling numbers of the second kind is defined by (cf. [8]):

$$\sum\_{n=0}^{\infty} S\_2\left(n, k: \lambda\right) \frac{t^n}{n!} = \frac{\left(\lambda x^l - 1\right)^k}{k!} \quad \left(\lambda \in \mathbb{C} / \{1\}\right). \tag{9}$$

The following sections are planned as follows: the second section includes the definition of the two-variable truncated Fubini polynomials and provides several formulas and relations including Stirling numbers of the second kind with several extensions. The third part covers the correlations for the two-variable truncated Fubini polynomials associated with the truncated Euler polynomials and the truncated Bernoulli polynomials. The last part of this paper analyzes the results acquired in this paper.

#### **2. Two-Variable Truncated Fubini Polynomials**

In this part, we define the two-variable truncated Fubini polynomials and numbers. We investigate several relations and identities for these polynomials and numbers.

We firstly remember the classical two-variable Fubini polynomials by the following generating function (cf. [2,3,5,6,10,13]): <sup>∞</sup>

$$\sum\_{n=0}^{\infty} F\_{\mathbb{II}}\left(x, y\right) \frac{t^n}{n!} = \frac{e^{xt}}{1 - y\left(e^t - 1\right)}.\tag{10}$$

When *x* = 0 in (10), the two-variable Fubini polynomials *Fn* (*x*, *y*) reduce to the usual Fubini polynomials given by (cf. [2,3,5,6,10,13]):

$$\sum\_{n=0}^{\infty} F\_{\mathbb{II}}\left(y\right) \frac{t^n}{n!} = \frac{1}{1 - y\left(\varepsilon^t - 1\right)}.\tag{11}$$

It is easy to see that for a non-negative integer *n* (cf. [2]):

$$F\_{\rm li}\left(\mathbf{x}, -\frac{1}{2}\right) = E\_{\rm il}\left(\mathbf{x}\right), \; F\_{\rm li}\left(-\frac{1}{2}\right) = E\_{\rm il} \tag{12}$$

and (cf. [3,5,6,10,13]):

$$F\_n\left(y\right) = \sum\_{\mu=0}^n \mathcal{S}\_2\left(n,\mu\right)\mu!y^{\mu}.\tag{13}$$

Substituting *y* by 1 in (11), we have the familiar Fubini numbers *Fn* (1) := *Fn* as follows (cf. [2,3,5,6,10,13]): <sup>∞</sup>

$$\sum\_{n=0}^{\infty} F\_n \frac{t^n}{n!} = \frac{1}{2 - e^t}.\tag{14}$$

For more information about the applications of the usual Fubini polynomials and numbers, cf. [2,3,5,6,10,13], and see also the references cited therein.

We now define the two-variable truncated Fubini polynomials as follows.

**Definition 1.** *For non-negative integer m, the two-variable truncated Fubini polynomials are defined via the following exponential generating function:*

$$\sum\_{n=0}^{\infty} F\_{\text{m,n}}(\mathbf{x}, \boldsymbol{y}) \frac{t^n}{n!} = \frac{\frac{t^{\text{m}}}{\text{n}!} e^{\mathbf{x} \boldsymbol{t}}}{1 - \mathbf{y} \left(e^{\mathbf{t}} - 1 - \sum\_{j=0}^{\text{m}-1} \frac{t^j}{j!} \right)} \,. \tag{15}$$

In the case *x* = 0 in (15), we then get a new type of Fubini polynomial, which we call the truncated Fubini polynomials given by:

$$\sum\_{n=0}^{\infty} F\_{m,n} \left( y \right) \frac{t^n}{n!} = \frac{\frac{t^m}{m!}}{1 - y \left( e^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)}. \tag{16}$$

Upon setting *x* = 0 and *y* = 1 in (15), we then attain the truncated Fubini numbers *Fm*,*<sup>n</sup>* defined by the following Taylor series expansion about *t* = 0:

$$\sum\_{n=0}^{\infty} F\_{m,n} \frac{t^n}{n!} = \frac{\frac{t^m}{m!}}{2 + \sum\_{j=m}^{\infty} \frac{t^j}{j!}}. \tag{17}$$

The two-variable truncated Fubini polynomials *Fm*,*<sup>n</sup>* (*x*, *y*) cover generalizations of some known polynomials and numbers that we discuss below.

**Remark 2.** *Setting m* = 0 *in (15), the polynomials Fm*,*<sup>n</sup>* (*x*, *y*) *reduce to the two-variable Fubini polynomials Fn* (*x*, *y*) *in (10).*

**Remark 3.** *When m* = 0 *and x* = 0 *in (15), the polynomials Fm*,*<sup>n</sup>* (*x*, *y*) *become the usual Fubini polynomials Fn* (*y*) *in (11).*

**Remark 4.** *In the special cases m* = 0*, y* = 1*, and x* = 0 *in (15), the polynomials Fm*,*<sup>n</sup>* (*x*, *y*) *reduce to the familiar Fubini numbers Fn in (14).*

We now are ready to examine the relations and properties for the two-variable Fubini polynomials *Fn* (*x*, *y*), and so, we firstly give the following theorem.

**Theorem 1.** *The following summation formula:*

$$F\_{m,n}\left(\mathbf{x},\mathbf{y}\right) = \sum\_{k=0}^{n} \binom{n}{k} F\_{m,k}\left(\mathbf{y}\right) \mathbf{x}^{n-k} \tag{18}$$

*holds true for non-negative integers m and n.*

**Proof.** By (15), using the Cauchy product in series, we observe that:

$$\begin{array}{rcl} \sum\_{n=0}^{\infty} F\_{\text{m,n}} \left( \mathbf{x}, y \right) \frac{t^n}{n!} &=& \frac{\frac{t^m}{m!}}{1 - y \left( \varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \varepsilon^{\mathbf{x} \mathbf{t}} \\ &=& \sum\_{n=0}^{\infty} F\_{\text{m,n}} \left( y \right) \frac{t^n}{n!} \sum\_{n=0}^{\infty} \mathbf{x}^n \frac{t^n}{n!} \\ &=& \sum\_{n=0}^{\infty} \sum\_{k=0}^n \binom{n}{k} F\_{\text{m,k}} \left( y \right) \mathbf{x}^{n-k} \frac{t^n}{n!} . \end{array}$$

which provides the asserted result (18).

We now provide another summation formula for the polynomials *Fm*,*<sup>n</sup>* (*x*, *y*) as follows.

**Theorem 2.** *The following summation formulas:*

$$F\_{\rm{W,n}}\left(\mathbf{x}+\mathbf{z},\mathbf{y}\right) = \sum\_{k=0}^{n} \binom{n}{k} F\_{\rm{W,k}}\left(\mathbf{x},\mathbf{y}\right) z^{n-k} \tag{19}$$

*and:*

$$F\_{m,n}\left(\mathbf{x}+z,\mathbf{y}\right) = \sum\_{k=0}^{n} \binom{n}{k} F\_{m,k}\left(\mathbf{y}\right) \left(\mathbf{x}+z\right)^{n-k} \tag{20}$$

*are valid for non-negative integers m and n.*

**Proof.** From (15), we obtain:

$$\begin{array}{rcl} \sum\_{n=0}^{\infty} F\_{m,n} \left( x, y \right) \frac{t^n}{n!} &=& \frac{\frac{t^n}{m!}}{1 - y \left( \varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \varepsilon^{(x+z)t} \\ &=& \frac{\frac{t^m}{m!} \varepsilon^{xt}}{1 - y \left( \varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \varepsilon^{xt} \\ &=& \sum\_{n=0}^{\infty} F\_{m,n} \left( x, y \right) \frac{t^n}{n!} \sum\_{n=0}^{\infty} z^n \frac{t^n}{n!} \\ &=& \sum\_{n=0}^{\infty} \sum\_{k=0}^n \binom{n}{k} F\_{m,k} \left( x, y \right) z^{n-k} \frac{t^n}{n!} \end{array}$$

and similarly:

$$\begin{array}{rcl} \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, \boldsymbol{y} \right) \frac{t^n}{n!} &=& \frac{\frac{t^m}{m!}}{1 - \mathbf{y} \left( \mathbf{c}^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \mathbf{c}^{(x+z)t} \\ &=& \sum\_{n=0}^{\infty} F\_{m,n} \left( \boldsymbol{y} \right) \frac{t^n}{n!} \sum\_{n=0}^{\infty} \left( \mathbf{x} + z \right)^n \frac{t^n}{n!} \\ &=& \sum\_{n=0}^{\infty} \sum\_{k=0}^n \binom{n}{k} F\_{m,k} \left( \boldsymbol{y} \right) \left( \mathbf{x} + z \right)^{n-k} \frac{t^n}{n!} \end{array}$$

which yield the desired results (19) and (20).

We here define the truncated Stirling numbers of the second kind as follows:

$$\sum\_{n=0}^{\infty} S\_{2,m}\left(n,k\right) \frac{t^n}{n!} = \frac{\left(e^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!}\right)^k}{k!}.\tag{21}$$

**Remark 5.** *Upon setting m* = 0 *in (21), the truncated Stirling numbers of the second kind S*2,*<sup>m</sup>* (*n*, *k*) *reduce to the classical Stirling numbers of the second kind in (8).*

The truncated Stirling numbers of the second kind satisfy the following relationship.

**Proposition 1.** *The following correlation:*

$$\mathcal{S}\_{2,m}\left(n,k+l\right) = \frac{l!k!}{(k+l)!} \sum\_{s=0}^{n} \binom{n}{s} \mathcal{S}\_{2,m}\left(s,k\right) \mathcal{S}\_{2,m}\left(n-s,l\right) \tag{22}$$

*holds true for non-negative integers m and n.*

**Proof.** In view of (8) and (21), we have:

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

which gives the claimed result (22).

We present the following correlation between two types of Stirling numbers of the second kind.

**Proposition 2.** *The following correlation:*

$$S\_{2,1}\left(n,k\right) = 2^k S\_2\left(n,k:\frac{1}{2}\right) \tag{23}$$

,

*holds true for non-negative integers m and n.*

**Proof.** In view of (8) and (21), we have:

$$\begin{aligned} \sum\_{n=0}^{\infty} S\_{2,1} \left( n,k \right) \frac{t^n}{n!} &=& \frac{\left(e^t - 1 - 1\right)^k}{k!} \\ &=& \frac{2^k \left(\frac{1}{2}e^t - 1\right)^k}{k!} \\ &=& 2^k \sum\_{n=0}^{\infty} S\_2 \left( n,k : \frac{1}{2} \right) \frac{t^n}{n!} \end{aligned}$$

which presents the desired result (23).

A relation that includes *Fm*,*<sup>n</sup>* (*x*) and *S*2,*<sup>m</sup>* (*n*, *k*) is given by the following theorem.

**Theorem 3.** *The following relation:*

$$F\_{\rm H,n+m}\left(\mathbf{x}\right) = \sum\_{k=0}^{n} \binom{n+m}{m} \mathbf{x}^{k} k! S\_{2,m}\left(n,k\right) \tag{24}$$

*is valid for a complex number x with* |*x*| < 1 *and non-negative integers m and n.*

**Proof.** By (16) and (21), we see that:

$$\begin{array}{rcl} \sum\_{n=0}^{\infty} \mathsf{F}\_{m,n} \left( \mathbf{x} \right) \frac{t^n}{n!} &=& \frac{\frac{t^m}{m!}}{1 - \mathbf{x} \left( \mathbf{c}^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \\ &=& \frac{t^m}{m!} \sum\_{k=0}^{\infty} \mathbf{x}^k \left( \mathbf{c}^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)^k \\ &=& \frac{t^m}{m!} \sum\_{k=0}^{\infty} \mathbf{x}^k k! \sum\_{n=0}^{\infty} S\_{2,m} \left( n, k \right) \frac{t^n}{n!} \\ &=& \sum\_{n=0}^{\infty} \sum\_{k=0}^{\infty} \mathbf{x}^k k! S\_{2,m} \left( n, k \right) \frac{t^{n+m}}{m! n!} \\ \end{array}$$

which implies the desired result (24).

We now state the following theorem.

**Theorem 4.** *The following identity:*

$$F\_{1,n+1}\left(\mathbf{x}\right) = n \sum\_{k=1}^{\infty} \mathbf{x}^k k! S\_2\left(n, k: \frac{1}{2}\right) \tag{25}$$

*holds true for a complex number x with* |*x*| < 1 *and a positive integer n.*

**Proof.** By (9) an (16), using the Cauchy product in series, we observe that:

$$\begin{aligned} \sum\_{n=0}^{\infty} \,^k \,^l \mathbf{r}\_{l,n} \left( \mathbf{x} \right) \frac{t^n}{n!} &= \quad \frac{t}{1 - \mathbf{x} \left( \mathbf{c}^t - \mathbf{2} \right)} \\ &= \quad t \sum\_{k=0}^{\infty} \mathbf{x}^k \left( \mathbf{c}^t - \mathbf{2} \right)^k \\ &= \quad t \sum\_{k=0}^{\infty} \mathbf{x}^k k! \frac{\left( \frac{1}{2} \mathbf{c}^t - \mathbf{1} \right)^k}{k!} \mathbf{2}^k \\ &= \quad \sum\_{n=0}^{\infty} \sum\_{k=0}^{\infty} \mathbf{x}^k k! S\_2 \left( n, k : \frac{1}{2} \right) \frac{t^{n+1}}{n!} . \end{aligned}$$

which provides the asserted result (25).

We now provide the derivative property for the polynomials *Fm*,*<sup>n</sup>* (*x*, *y*) as follows.

**Theorem 5.** *The derivative formula:*

$$\frac{\partial}{\partial \mathbf{x}} F\_{\mathbf{m},n}(\mathbf{x}, \mathbf{y}) = n F\_{\mathbf{m},n-1}(\mathbf{x}, \mathbf{y}) \tag{26}$$

*holds true for non-negative integers m and a positive integer n.*

**Proof.** Applying the derivative operator with respect to *x* to both sides of the equation (15), we acquire:

$$\frac{\partial}{\partial \mathbf{x}} \left( \sum\_{n=0}^{\infty} F\_{\text{H}, \text{ul}} \left( \mathbf{x}\_{\prime} y \right) \frac{t^{n}}{n!} \right) = \frac{\partial}{\partial \mathbf{x}} \left( \frac{\frac{t^{m}}{m!} e^{\mathbf{x} t}}{1 - y \left( e^{t} - 1 - \sum\_{j=0}^{m-1} \frac{t^{j}}{j!} \right)} \right)$$

and then:

$$\begin{split} \sum\_{n=0}^{\infty} \frac{\partial}{\partial \mathbf{x}} F\_{m,n} \left( \mathbf{x}, \mathbf{y} \right) \frac{t^n}{n!} &=& \frac{\frac{t^m}{m!}}{1 - y \left( \mathbf{c}^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \frac{\partial}{\partial \mathbf{x}} \mathbf{c}^{\mathbf{x}t} \\ &=& \frac{\frac{t^m}{m!} \mathbf{c}^{\mathbf{x}t}}{1 - y \left( \mathbf{c}^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} t^j \\ &=& \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, \mathbf{y} \right) \frac{t^{n+1}}{n!} . \end{split}$$

which means the claimed result (26).

A recurrence relation for the two-variable truncated Fubini polynomials is given by the following theorem.

**Theorem 6.** *The following equalities:*

$$F\_{m,n}(\mathbf{x}, \mathbf{y}) = 0 \quad (n = 0, 1, 2, \cdots, m - 1)$$

*and:*

$$F\_{m,n+m}(\mathbf{x},\mathbf{y}) = \frac{y}{1+y} \sum\_{s=0}^{n} \binom{n+m}{s} F\_{m,s}(\mathbf{x},\mathbf{y}) - \frac{\mathbf{x}^{n}}{1+y} \frac{(n+m)!}{n!m!} \tag{27}$$

*hold true for non-negative integers m and n.*

**Proof.** Using Definition 1, we can write:

$$\begin{split} \frac{t^{n}}{n!} ^{\mathrm{wt}}{}{\!} &= \quad \left( 1 - y \left( \sum\_{j=m}^{\infty} \frac{t^{j}}{j!} - 1 \right) \right) \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} \\ &= \quad \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} - y \left[ \sum\_{j=m}^{\infty} \frac{t^{j}}{j!} \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} - \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} \right] \\ &= \quad \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} - y \left[ \sum\_{j=0}^{\infty} \frac{t^{j+m}}{(j+m)!} \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} - \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, y \right) \frac{t^{n}}{n!} \right]. \end{split}$$

Because of: <sup>∞</sup>

$$\sum\_{j=0}^{\infty} \frac{t^{j+m}}{(j+m)!} \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}, \mathbf{y} \right) \frac{t^n}{n!} = \sum\_{n=0}^{\infty} \sum\_{j=0}^n \binom{n+m}{j} F\_{m,j} \left( \mathbf{x}, \mathbf{y} \right) \frac{t^{n+m}}{(n+m)!},$$

we obtain:

$$\begin{split} \sum\_{n=0}^{\infty} \mathbf{x}^{n} \frac{t^{n+m}}{n!m!} &= \sum\_{n=0}^{\infty} F\_{\mathbf{m},n} \left( \mathbf{x}, \boldsymbol{\mathcal{Y}} \right) \frac{t^{n}}{n!} - \boldsymbol{\mathcal{Y}} \sum\_{n=0}^{\infty} \sum\_{j=0}^{n} \binom{n+m}{j} F\_{\mathbf{m},j} \left( \mathbf{x}, \boldsymbol{\mathcal{Y}} \right) \frac{t^{n+m}}{(n+m)!} \\ &+ \boldsymbol{\mathcal{Y}} \sum\_{n=0}^{\infty} F\_{\mathbf{m},n} \left( \mathbf{x}, \boldsymbol{\mathcal{Y}} \right) \frac{t^{n}}{n!} . \end{split}$$

Thus, we arrive at the following equality:

$$\sum\_{n=0}^{\infty} F\_{\mathfrak{M}, \mathfrak{U}} \left( x, y \right) \frac{t^n}{n!} = \frac{1}{1+y} \sum\_{n=0}^{\infty} \left( y \sum\_{j=0}^n \binom{n+m}{j} \frac{F\_{\mathfrak{M}, j} \left( x, y \right)}{(n+m)!} - \frac{x^n}{n! m!} \right) t^{n+m}.$$

Comparing the coefficients of both sides of the last equality, the proof is completed.

Theorem 6 can be used to determine the two-variable truncated Fubini polynomials. Thus, we provide some examples as follows.

**Example 1.** *Choosing m* = 1*, then we have F*1,0 (*x*, *y*) = 0. *Utilizing the recurrence formula (27), we derive:*

$$F\_{1,n+1}\left(\mathbf{x},\mathbf{y}\right) = \frac{y}{1-y} \sum\_{s=0}^{n} \binom{n+1}{s} F\_{1,s}\left(\mathbf{x},\mathbf{y}\right) - \frac{\mathbf{x}^n}{1-y} \left(n+1\right) \dots$$

*Thus, we subsequently acquire:*

$$\begin{aligned} F\_{1,1} \left( \mathbf{x}, y \right) &= -\frac{1}{1+y}, \\ F\_{1,2} \left( \mathbf{x}, y \right) &= -\frac{2y}{\left( 1+y \right)^2} - \frac{2x}{1+y}, \\ F\_{1,3} \left( \mathbf{x}, y \right) &= \frac{3}{1+y} \left( \frac{2y^2}{\left( 1+y \right)^2} - \frac{2xy}{1+y} - \mathbf{x}^2 \right). \end{aligned}$$

*Furthermore, choosing m* = 2*, we then obtain the following recurrence relation:*

$$F\_{2,n+2}\left(\mathbf{x},\mathbf{y}\right) = \frac{y}{1+y} \sum\_{s=0}^{n} \binom{n+2}{s} F\_{2,s}\left(\mathbf{x},\mathbf{y}\right) - \frac{\mathbf{x}^{n}}{1+y} \frac{\left(n+2\right)\left(n+1\right)}{2}$$

*which yields the following polynomials:*

$$\begin{aligned} F\_{2,0} \left( \mathbf{x}, y \right) &= \quad F\_{2,1} \left( \mathbf{x}, y \right) = 0, \\ F\_{2,2} \left( \mathbf{x}, y \right) &= \quad -\frac{1}{1 + y}, \\ F\_{2,3} \left( \mathbf{x}, y \right) &= \quad -\frac{3\mathbf{x}}{1 + y}, \\ F\_{2,4} \left( \mathbf{x}, y \right) &= \quad \frac{6\mathbf{x}}{y + 1} \left( \frac{3\mathbf{y}}{1 + y} + \mathbf{x} \right) \end{aligned}$$

By applying a similar method used above, one can derive the other two-variable truncated Fubini polynomials.

.

Here is a correlation that includes the truncated Fubini polynomials and Stirling numbers of the second kind.

**Theorem 7.** *For non-negative integers n and m, we have:*

$$F\_{\mathfrak{M},\mathfrak{n}}\left(\mathbf{x},\mathbf{y}\right) = \sum\_{\mathfrak{u}=0}^{n} \sum\_{k=0}^{\mathfrak{u}} \binom{\mathfrak{n}}{\mathfrak{u}} F\_{\mathfrak{M},\mathfrak{n}-\mathfrak{u}}\left(\mathbf{y}\right) \operatorname{S}\_{2}\left(\mathfrak{u},k\right)\left(\mathbf{x}\right)\_{k} \,. \tag{28}$$

**Proof.** By means of Theorem 1 and Formula (8), we get:

$$\begin{aligned} F\_{\mathfrak{m},\mathfrak{u}}\left(\mathbf{x},\mathbf{y}\right) &= \sum\_{\mathfrak{u}=0}^{\mathfrak{n}} \binom{\mathfrak{n}}{\mathfrak{u}} F\_{\mathfrak{m},\mathfrak{n}-\mathfrak{u}}\left(\mathbf{y}\right) \mathbf{x}^{\mathfrak{u}} \\ &= \sum\_{\mathfrak{u}=0}^{\mathfrak{n}} \binom{\mathfrak{n}}{\mathfrak{u}} F\_{\mathfrak{m},\mathfrak{n}-\mathfrak{u}}\left(\mathbf{y}\right) \sum\_{k=0}^{\mathfrak{u}} S\_2\left(\mathfrak{u},k\right) \left(\mathbf{x}\right)\_{k'} \end{aligned}$$

which completes the proof of this theorem.

The rising factorial number *x* is defined by (*x*) (*n*) <sup>=</sup> *<sup>x</sup>* (*<sup>x</sup>* <sup>+</sup> <sup>1</sup>) (*<sup>x</sup>* <sup>+</sup> <sup>2</sup>)···(*<sup>x</sup>* <sup>+</sup> *<sup>n</sup>* <sup>−</sup> <sup>1</sup>) for a positive integer *n*. We also note that the negative binomial expansion is given as follows:

$$\mathbf{x}^{\prime}(\mathbf{x}+\mathbf{a})^{-n} = \sum\_{k=0}^{\infty} (-1)^{k} \binom{n+k-1}{k} \mathbf{x}^{k} a^{-n-k} \tag{29}$$

for negative integer −*n* and |*x*| < *a*; cf. [7].

Here, we give the following theorem.

**Theorem 8.** *The following relationship:*

$$F\_{m,n}\left(\mathbf{x},\mathbf{y}\right) = \sum\_{k=0}^{\infty} \sum\_{l=-k}^{n} \binom{n}{l} S\_2\left(l,k\right) F\_{n,n-l}\left(-k,\mathbf{y}\right) \left(\mathbf{x}\right)^{(k)}\tag{30}$$

*holds true for non-negative integers n and m.*

**Proof.** By means of Definition 1 and using Equations (7) and (29), we attain:

$$\begin{split} \sum\_{n=0}^{\infty} \mathsf{F}\_{\mathsf{H},n}(\mathbf{x},\mathbf{y}) \frac{t^{n}}{n!} &=& \frac{\frac{t^{n}}{\mathsf{m}!}}{1 - y\left(\epsilon^{t} - 1 - \sum\_{j=0}^{m-1} \frac{t^{j}}{j!}\right)} \left(\epsilon^{-t}\right)^{-x} \\ &=& \frac{\frac{t^{m}}{\mathsf{m}!}}{1 - y\left(\epsilon^{t} - 1 - \sum\_{j=0}^{m-1} \frac{t^{j}}{j!}\right)} \sum\_{k=0}^{\infty} \binom{\mathsf{x} + k - 1}{k} \left(1 - \epsilon^{-t}\right)^{-k} \\ &=& \frac{\frac{t^{m}}{\mathsf{m}!}}{1 - y\left(\epsilon^{t} - 1 - \sum\_{j=0}^{m-1} \frac{t^{j}}{j!}\right)} \sum\_{k=0}^{\infty} \left(\mathbf{x}\right)^{(k)} \frac{\left(\epsilon^{t} - 1\right)^{k}}{k!} \epsilon^{-kt} \\ &=& \sum\_{k=0}^{\infty} \left(\mathbf{x}\right)^{(k)} \sum\_{n=0}^{\infty} \mathsf{F}\_{\mathsf{m},n}\left(-k, y\right) \frac{t^{n}}{n!} \sum\_{n=0}^{\infty} \mathcal{S}\_{2}\left(n, k\right) \frac{t^{n}}{n!} \\ &=& \sum\_{k=0}^{\infty} \left(\mathbf{x}\right)^{(k)} \sum\_{n=0}^{\infty} \left(\sum\_{l=0}^{n} \binom{n}{l} F\_{\mathsf{m},n-l}\left(-k, y\right) \mathcal{S}\_{2}\left(l, k\right)\right) \frac{t^{n}}{n!}. \end{split}$$

which gives the asserted result (30).

Therefore, we give the following theorem.

**Theorem 9.** *The following relationship:*

$$\begin{split} \left(y \sum\_{k=0}^{n} \binom{n}{k} F\_{m,n-k} \left(z, y \right) F\_{m+1,k} \left(x, y \right) = \sum\_{k=0}^{n} \binom{n}{k} F\_{m+1,n-k} \left(x, y \right) z^k \\ - \frac{n}{m+1} \sum\_{k=0}^{n-1} \binom{n-1}{k} F\_{m,n-1-k} \left(z, y \right) x^k \end{split} \tag{31}$$

*holds true for non-negative integers n and m.*

**Proof.** By means of Definition 1, we see that:

$$\begin{split} \epsilon^{\operatorname{xt}} \frac{t^{\operatorname{m}+1}}{(m+1)!} &= \quad \left(1 - y \left(\epsilon^{l} - 1 - \sum\_{j=0}^{m} \frac{t^{j}}{j!} \right) \right) \sum\_{n=0}^{\infty} F\_{m+1, \operatorname{\boldsymbol{n}}} \left(\mathbf{x}, y\right) \frac{t^{n}}{n!} \\ &= \quad \left(1 - y \left(\epsilon^{l} - 1 - \sum\_{j=0}^{m-1} \frac{t^{j}}{j!} \right) \right) \sum\_{n=0}^{\infty} F\_{m+1, \boldsymbol{n}} \left(\mathbf{x}, y\right) \frac{t^{n}}{n!} \\ &- y \frac{t^{m}}{m!} \sum\_{n=0}^{\infty} F\_{m+1, \boldsymbol{n}} \left(\mathbf{x}, y\right) \frac{t^{n}}{n!} .\end{split}$$

Thus, we get:

$$\begin{aligned} \varepsilon^{xt} \frac{t^{m+1}}{(m+1)!} \sum\_{n=0}^{\infty} F\_{\text{H},\text{N}} \left( z, \underline{y} \right) \frac{t^{\text{n}}}{n!} &= \frac{t^{m}}{m!} \varepsilon^{zt} \sum\_{n=0}^{\infty} F\_{m+1,\text{n}} \left( x, \underline{y} \right) \frac{t^{\text{n}}}{n!} \\ -\underline{y} \frac{t^{\text{m}}}{m!} \sum\_{n=0}^{\infty} F\_{\text{H},\text{n}} \left( z, \underline{y} \right) \frac{t^{\text{n}}}{n!} \sum\_{n=0}^{\infty} F\_{m+1,\text{n}} \left( x, \underline{y} \right) \frac{t^{\text{n}}}{n!} \end{aligned}$$

and then:

$$\begin{aligned} \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} \binom{n}{k} F\_{m,n-k} \left( z, \underline{y} \right) \ge \frac{t^{n+1}}{n! \left( m+1 \right)} &= \sum\_{n=0}^{\infty} \sum\_{k=0}^{n} \binom{n}{k} F\_{m+1,n-k} \left( x, \underline{y} \right) z^k \frac{t^n}{n!} \\ -\underline{y} \sum\_{n=0}^{\infty} \underline{y} \sum\_{k=0}^{n} \binom{n}{k} F\_{m,n-k} \left( z, \underline{y} \right) F\_{m+1,k} \left( x, \underline{y} \right) \frac{t^n}{n!} \end{aligned}$$

which provides the claimed result in (31).

Here, we investigate a linear combination for the two-variable truncated Fubini polynomials for different *y* values in the following theorem.

**Theorem 10.** *Let the numbers m and n be non-negative integers and y*<sup>1</sup> = *y*2*. We then have:*

$$\frac{m!n!}{(n+m)!} \sum\_{k=0}^{n+m} \binom{n+m}{k} F\_{m,n+m-k} \left(\mathbf{x}\_1, y\_1\right) F\_{m,k} \left(\mathbf{x}\_2, y\_2\right) = \frac{y\_2 F\_{m,n-k} \left(\mathbf{x}\_1 + \mathbf{x}\_2, y\_2\right) - y\_1 F\_{m,n-k} \left(\mathbf{x}\_1 + \mathbf{x}\_2, y\_1\right)}{y\_2 - y\_1}.\tag{32}$$

**Proof.** By Definition 1, we consider the following product:

$$\frac{\frac{t^m}{m!}e^{x\_1 t}}{1 - y\_1 \left(\varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \frac{\frac{t^m}{m!}e^{x\_2 t}}{1 - y\_2 \left(\varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)}$$

$$= \frac{y\_2}{y\_2 - y\_1} \frac{\frac{t^{2m}}{(m!)^2}e^{(x\_1 + x\_2)t}}{1 - y\_2 \left(\varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} - \frac{y\_1}{y\_2 - y\_1} \frac{\frac{t^{2m}}{(m!)^2}e^{(x\_1 + x\_2)t}}{1 - y\_1 \left(\varepsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)}$$

which yields

$$\begin{aligned} &\sum\_{n=0}^{\infty} \sum\_{k=0}^{n} \binom{n}{k} F\_{m,n-k} \left( \mathbf{x}\_1, \mathbf{y}\_1 \right) F\_{m,k} \left( \mathbf{x}\_2, \mathbf{y}\_2 \right) \frac{t^n}{n!} \\ &= \quad \frac{\mathcal{Y}\_2}{\mathcal{Y}\_2 - \mathcal{Y}\_1} \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}\_1 + \mathbf{x}\_2, \mathbf{y}\_2 \right) \frac{t^{n+m}}{n! m!} - \frac{\mathcal{Y}\_1}{\mathcal{Y}\_2 - \mathcal{Y}\_1} \sum\_{n=0}^{\infty} F\_{m,n} \left( \mathbf{x}\_1 + \mathbf{x}\_2, \mathbf{y}\_1 \right) \frac{t^{n+m}}{n! m!} \end{aligned}$$

Thus, we get:

$$\begin{aligned} &\sum\_{n=0}^{\infty} \left( \sum\_{k=0}^{n} \binom{n}{k} F\_{m,n-k} \left( \mathbf{x}\_1, y\_1 \right) F\_{m,k} \left( \mathbf{x}\_2, y\_2 \right) \right) \frac{t^n}{n!} \\ &= \quad \sum\_{n=0}^{\infty} \left( \frac{y\_2}{y\_2 - y\_1} F\_{m,n} \left( \mathbf{x}\_1 + \mathbf{x}\_2, y\_2 \right) - \frac{y\_1}{y\_2 - y\_1} F\_{m,n} \left( \mathbf{x}\_1 + \mathbf{x}\_2, y\_1 \right) \right) \frac{t^{n+m}}{n! m!}, \end{aligned}$$

which gives the desired result (32).

#### **3. Correlations with Truncated Euler and Bernoulli Polynomials**

In this section, we investigate several correlations for the two-variable truncated Fubini polynomials *Fm*,*<sup>n</sup>* (*x*, *y*) related to the truncated Euler polynomials *Em*,*<sup>n</sup>* (*x*) and numbers *Em*,*<sup>n</sup>* and the truncated Bernoulli polynomials *Bm*,*<sup>n</sup>* (*x*) and numbers *Bm*,*n*.

Here is a relation between the truncated Euler polynomials and two-variable truncated Fubini polynomials at the special value *<sup>y</sup>* <sup>=</sup> <sup>−</sup><sup>1</sup> 2 .

**Theorem 11.** *We have:*

$$F\_{\mathfrak{m},\mathfrak{n}}\left(\mathbf{x},-\frac{1}{2}\right) = E\_{\mathfrak{m},\mathfrak{n}}\left(\mathbf{x}\right). \tag{33}$$

.

**Proof.** In terms of (5) and (15), we get:

$$\begin{aligned} \sum\_{n=0}^{\infty} \mathbb{F}\_{m,n} \left( \mathbf{x} , -\frac{1}{2} \right) \frac{t^n}{n!} &=& \frac{\frac{t^m}{m!} e^{\mathbf{x} t}}{1 + \frac{1}{2} \left( \mathbf{c}^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)^2} \\ &=& \frac{2 \frac{t^m}{m!} \mathbf{c}^{\mathbf{x} t}}{\mathbf{c}^t + 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!}} \\ &=& \sum\_{n=0}^{\infty} E\_{m,n} \left( \mathbf{x} \right) \frac{t^n}{n!} \end{aligned}$$

which implies the asserted result (33).

**Corollary 1.** *Taking x* = 0*, we then get a relation between the truncated Euler numbers and truncated Fubini polynomials at the special value y* <sup>=</sup> <sup>−</sup><sup>1</sup> <sup>2</sup> *, namely:*

$$F\_{\mathfrak{N},\mathfrak{n}}\left(-\frac{1}{2}\right) = E\_{\mathfrak{N},\mathfrak{n}}.\tag{34}$$

**Remark 6.** *The relations (33) and (34) are extensions of the relations in (12).*

We now state the following theorem, which includes a correlation for *Fm*,*<sup>n</sup>* (*x*, *y*), *Fm*,*<sup>n</sup>* (*y*) and *Em*,*<sup>n</sup>* (*x*).

**Theorem 12.** *The following formula:*

$$F\_{m,n}\left(x,y\right) \quad = \frac{n!m!}{(n+m)!} \sum\_{l=0}^{n+m} \frac{1}{2} \binom{n+m}{l} F\_{m,l}\left(y\right) E\_{m,n+m-l}\left(x\right) \tag{35}$$

$$+ \frac{n!m!}{(n+m)!} \sum\_{j=0}^{n} \frac{1}{2} \binom{n+m}{j} \sum\_{l=0}^{j} \binom{j}{l} F\_{m,l}\left(y\right) E\_{m,j-l}\left(x\right)$$

*is valid for non-negative integers m and n.*

**Proof.** By (5) and (15), we acquire that:

∞ ∑ *n*=0 *Fm*,*<sup>n</sup>* (*x*, *<sup>y</sup>*) *<sup>t</sup> n <sup>n</sup>*! <sup>=</sup> *t m m*!*ext* 1 − *y <sup>e</sup><sup>t</sup>* <sup>−</sup> <sup>1</sup> <sup>−</sup> <sup>∑</sup>*m*−<sup>1</sup> *<sup>j</sup>*=<sup>0</sup> *<sup>t</sup><sup>j</sup> j*! 2 *t m m*! *<sup>e</sup><sup>t</sup>* <sup>+</sup> <sup>1</sup> <sup>−</sup> <sup>∑</sup>*m*−<sup>1</sup> *<sup>j</sup>*=<sup>0</sup> *<sup>t</sup><sup>j</sup> j*! *<sup>e</sup><sup>t</sup>* <sup>+</sup> <sup>1</sup> <sup>−</sup> <sup>∑</sup>*m*−<sup>1</sup> *<sup>j</sup>*=<sup>0</sup> *<sup>t</sup><sup>j</sup> j*! 2 *<sup>t</sup><sup>m</sup> m*! <sup>=</sup> <sup>1</sup> 2 *m*! *tm* ∞ ∑ *n*=0 *Fm*,*<sup>n</sup>* (*y*) *<sup>t</sup> n n*! ∞ ∑ *n*=0 *Em*,*<sup>n</sup>* (*x*) *<sup>t</sup> n n*! - ∞ ∑ *j*=*m tj j*! + 1 <sup>=</sup> *<sup>m</sup>*! 2 ∞ ∑ *n*=0 - *n* ∑ *l*=0 *n l Fm*,*<sup>l</sup>* (*y*) *Em*,*n*−*<sup>l</sup>* (*x*) *t n*−*m n*! - ∞ ∑ *j*=0 *tj*+*<sup>m</sup>* (*j* + *m*)! + 1 <sup>=</sup> *<sup>m</sup>*! 2 ∞ ∑ *n*=0 - *n* ∑ *l*=0 *n l Fm*,*<sup>l</sup>* (*y*) *Em*,*n*−*<sup>l</sup>* (*x*) *t n*−*m n*! <sup>+</sup>*m*! 2 ∞ ∑ *n*=0 *n* ∑ *j*=0 *n* + *m j j* ∑ *l*=0 *j l Fm*,*<sup>l</sup>* (*y*) *Em*,*j*−*<sup>l</sup>* (*x*) *t n* (*n* + *m*)! ,

which completes the proof of the theorem.

We finally state the relations for the truncated Bernoulli and Fubini polynomials as follows.

**Theorem 13.** *The following relation:*

$$F\_{m,n}\left(\mathbf{x},\mathbf{y}\right) = \frac{n!m!}{(n+m)!} \sum\_{l=0}^{n} \binom{n+m}{l} \sum\_{k=0}^{l} \binom{l}{k} F\_{m,l}\left(\mathbf{y}\right) B\_{m,l-k}\left(\mathbf{x}\right) \tag{36}$$

*is valid for non-negative integers m and n.*

**Proof.** By (5) and (15), we acquire that:

$$\begin{split} \sum\_{n=0}^{\infty} \, ^t\_{m,n} \left( \mathbf{x}, y \right) \frac{t^n}{n!} &= \quad \frac{t^n\_m e^{xt}}{1 - y \left( \epsilon^t - 1 - \sum\_{j=0}^{m-1} \frac{t^j}{j!} \right)} \frac{t^n}{\epsilon^t - \sum\_{j=0}^{m-1} \frac{t^j}{j!}} \frac{\epsilon^t - \sum\_{j=0}^{m-1} \frac{t^j}{j!}}{\frac{t^m}{m!}} \\ &= \quad \frac{m!}{t^m} \sum\_{n=0}^{\infty} F\_{m,n} \left( y \right) \frac{t^n}{n!} \sum\_{n=0}^{\infty} B\_{m,n} \left( \mathbf{x} \right) \frac{t^n}{n!} \sum\_{j=m}^{\infty} \frac{t^j}{j!} \\ &= \quad m! \sum\_{n=0}^{\infty} \left( \sum\_{k=0}^n \binom{n}{k} F\_{m,k} \left( \mathbf{y} \right) B\_{m,n-k} \left( \mathbf{x} \right) \right) \frac{t^n}{n!} \sum\_{j=0}^{\infty} \frac{t^j}{(j+m)!} \\ &= \quad \frac{n! m!}{(n+m)!} \sum\_{n=0}^{\infty} \sum\_{l=0}^n \binom{n+m}{l} \left( \sum\_{k=0}^l \binom{l}{k} F\_{m,l} \left( \mathbf{y} \right) B\_{m,l-k} \left( \mathbf{x} \right) \right) \frac{t^n}{(n+m)!} . \end{split}$$

which means the asserted result (36).

#### **4. Conclusions**

In this paper, we firstly considered two-variable truncated Fubini polynomials and numbers, and we then obtained some identities and properties for these polynomials and numbers, involving summation formulas, recurrence relations, and the derivative property. We also proved some formulas related to the truncated Stirling numbers of the second kind and Apostol-type Stirling numbers of the second kind. Furthermore, we gave some correlations including the two-variable truncated Fubini polynomials, the truncated Euler polynomials, and truncated Bernoulli polynomials.

**Author Contributions:** Both authors have equally contributed to this work. Both authors read and approved the final manuscript.

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

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

#### **References**



c 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/).

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