*Article* **A Parametric Kind of the Degenerate Fubini Numbers and Polynomials**

**Sunil Kumar Sharma 1,\*, Waseem A. Khan <sup>2</sup> and Cheon Seoung Ryoo <sup>3</sup>**


Received: 26 February 2020; Accepted: 7 March 2020; Published: 12 March 2020

**Abstract:** In this article, we introduce the parametric kinds of degenerate type Fubini polynomials and numbers. We derive recurrence relations, identities and summation formulas of these polynomials with the aid of generating functions and trigonometric functions. Further, we show that the parametric kind of the degenerate type Fubini polynomials are represented in terms of the Stirling numbers.

**Keywords:** Fubini polynomials; degenerate Fubini polynomials; Stirling numbers

**MSC:** 11B83; 11C08; 11Y35

#### **1. Introduction**

In the last decade, many mathematicians, namely, Kargin [1], Duran and Acikgoz [2], Kim et al. [3,4], Kilar and Simsek [5], Su and He [6] have been studied in the area of the Fubini polynomials and numbers, degenerate Fubini polynomials and numbers. The range of Appell polynomials sequences is one of the important classes of polynomial sequences. The Appell polynomials are very frequently used in various problems in pure and applied mathematics related to functional equations in differential equations, approximation theories, interpolation problems, summation methods, quadrature rules, and their multidimensional extensions (see [7,8]). The sequence of Appell polynomials *Aj*(*w*) can be signified by means either following equivalent conditions

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

and satisfying the generating function

$$A(z)e^{wz} = A\_0(w) + A\_1(w)\frac{z}{1!} + A\_2(w)\frac{z^2}{2!} + \cdots + A\_n(w)\frac{z^n}{n!} + \cdots = \sum\_{j=0}^{\infty} A\_j(w)\frac{z^j}{j!},\tag{2}$$

where *A*(*w*) is an entirely real power series with Taylor expansion given by

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

The well known degenerate exponential function [9] is defined by

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

Since

$$\lim\_{\mu \to 0} (1 + \mu z)^{\frac{\eta}{\mu}} = e^{\eta z}.$$

In [10,11], Carlitz introduced the degenerate Bernoulli polynomials which are defined by

$$\frac{z}{(1+\mu z)^{\frac{1}{\mu}}-1}(1+\mu z)^{\frac{z}{\mu}}=\sum\_{j=0}^{\infty}\beta\_{j}(\eta;\mu)\frac{z^{j}}{j!},(\mu\in\mathbb{C}),\tag{4}$$

so that

$$\beta\_{\dot{\boldsymbol{\beta}}}(\boldsymbol{\eta};\boldsymbol{\mu}) = \sum\_{w=0}^{\dot{\boldsymbol{\beta}}} \binom{\boldsymbol{\beta}}{w} \beta\_w(\boldsymbol{\mu}) \left(\frac{\boldsymbol{\eta}}{\mu}\right)\_{\dot{\boldsymbol{\beta}}-w}.\tag{5}$$

When *η* = 0, *βj*(*μ*) = *βj*(0; *μ*) are called the degenerate Bernoulli numbers, (see [12–15]).

From Equation (4), we get

$$\begin{split} \sum\_{j=0}^{\infty} \lim\_{\mu \longrightarrow 0} \beta\_j(\eta; \mu) \frac{z^j}{\overline{\gamma}!} &= \lim\_{(1+\mu z)^{\frac{1}{\overline{\gamma}}} -1} \frac{z}{(1+\mu z)^{\frac{\overline{\gamma}}{\overline{\gamma}}}} (1+\mu z)^{\frac{\overline{\gamma}}{\overline{\mu}}} \\ &= \frac{z}{t^{\overline{\alpha}-1}} e^{\eta z} = \sum\_{j=0}^{\infty} B\_j(\eta) \frac{z^j}{j!}, \end{split} \tag{6}$$

where *Bj*(*η*) are called the Bernoulli polynomials, (see [9,16]).

The Stirling numbers of the first kind [3,14,17]) are defined by

$$\eta^j = \sum\_{i=0}^j \mathcal{S}\_1(j, i)(\eta)\_{i\prime}(j \ge 0),\tag{7}$$

where (*η*)<sup>0</sup> = 1,(*η*)*<sup>j</sup>* = *η*(*η* − 1)···(*η* − *j* + 1),(*j* ≥ 1). Alternatively, the Stirling numbers of the second kind are defined by following generating function (see [4,5])

$$\frac{(e^z - 1)^j}{j!} = \sum\_{i=j}^{\infty} S\_2(j, i) \frac{z^j}{j!}. \tag{8}$$

The degenerate Stirling numbers of the second kind [17] are defined by means of the following generating function

$$\frac{1}{i!} \left( (1 + \mu z)^{\frac{1}{\mu}} - 1 \right)^{i} = \sum\_{j=i}^{\infty} S\_{2,\mu}(j, i) \frac{z^{j}}{j!}. \tag{9}$$

It is clear that

$$\lim\_{\mu \to 0} S\_{2,\mu}(j, i) = S\_2(j, i).$$

The generating function of 2-variable degenerate Fubini polynomials [3] are defined by

$$\frac{1}{1 - \xi((1 + \mu z)^{\frac{1}{\mu}} - 1)}(1 + \mu z)^{\frac{\eta}{\mu}} = \sum\_{j=0}^{\infty} F\_{j,\mu}(\eta; \xi) \frac{z^j}{j!},\tag{10}$$

so that

$$F\_{j,\mu}(\eta;\xi) = \sum\_{i=0}^{j} \binom{j}{i} \, F\_{i,\mu}(\xi)(\eta)\_{j-i,\mu}.$$

When *η* = 0, *Fj*,*μ*(0; *ξ*) = *Fj*,*μ*(*ξ*), *Fj*,*μ*(0; 1) = *Fj*,*<sup>μ</sup>* are called the degenerate Fubini polynomials and degenerate Fubini numbers.

Note that

$$\begin{split} \lim\_{\mu \to 0} \sum\_{j=0}^{\infty} F\_{j,\mu}(\eta; \underline{\xi}) \frac{\underline{z}^j}{\underline{\tau}^j} &= \lim\_{\mu \to 0} \frac{1}{1 - \underline{\xi}((1 + \mu z)^{\frac{1}{\underline{\tau}}} - 1)} (1 + \mu z)^{\frac{\underline{\tau}}{\underline{\tau}}} \\ &= \frac{1}{1 - \underline{\xi}(\epsilon z^2 - 1)} \mathfrak{e}^{\eta z} = \sum\_{j=0}^{\infty} F\_j(\eta; \underline{\xi}) \frac{\underline{z}^j}{\underline{\tau}^j} . \end{split} \tag{11}$$

where *Fj*(*η*; *ξ*) are called the 2-variable Fubini polynomials, (see, [1,18]).

The two trigonometric functions *eη<sup>z</sup>* cos *ξz* and *eη<sup>z</sup>* sin *ξz* are defined as follows (see [19,20]):

$$e^{\eta z} \cos \xi z = \sum\_{k=0}^{\infty} \mathcal{C}\_k(\eta, \xi) \frac{z^k}{k!} \, , \tag{12}$$

and

$$
\epsilon^{\eta z} \sin \xi z = \sum\_{k=0}^{\infty} S\_k(\eta, \xi) \frac{z^k}{k!} \tag{13}
$$

where

$$\mathbb{C}\_{k}(\eta,\xi) = \sum\_{j=0}^{\left[\frac{k}{2}\right]} \binom{k}{2j} (-1)^{j} \eta^{k-2j} \xi^{2j} \,\_{\prime} \tag{14}$$

and

$$S\_k(\eta,\xi) = \sum\_{j=0}^{\lfloor \frac{k-1}{2} \rfloor} \binom{k}{2j+1} (-1)^j \eta^{k-2j-1} \xi^{2j+1}.\tag{15}$$

Recently, Kim et al. [16] introduced the degenerate cosine-polynomials and degenerate sine-polynomials are respectively, as follows

$$\varepsilon^{\eta}\_{\mu}(z)\cos^{\overline{\xi}}\_{\lambda}(z) = \sum\_{j=0}^{\infty} \mathbb{C}\_{j,\mu}(\eta\_{\prime}\overline{\xi})\frac{z^{j}}{j!},\tag{16}$$

and

$$a\_{\mu}^{\eta}(z)\sin\_{\mu}^{\overline{z}}(z) = \sum\_{j=0}^{\infty} S\_{j,\mu}(\eta, \xi) \frac{z^j}{j!},\tag{17}$$

where

$$\mathbb{C}\_{j,\mu}(\eta,\xi) = \sum\_{k=0}^{\lfloor \frac{j}{2} \rfloor} \sum\_{i=2k}^{j} \binom{j}{i} \mu^{i-2k} (-1)^k \xi^{2k} S^1(i, 2k)(\eta)\_{j-i,\mu} \tag{18}$$

and

$$\mathcal{S}\_{j,\mu}(\eta,\xi) = \sum\_{k=0}^{\lfloor \frac{j-1}{2} \rfloor} \sum\_{i=2k+1}^{j} \binom{j}{i} \mu^{i-2k-1} (-1)^k \xi^{2k+1} S^1(i, 2k+1)(\eta)\_{j-i,\mu}. \tag{19}$$

This paper is organized as follows: In Section 2, we introduce degenerate complex Fubini polynomials with degenerate cosine-Fubini and degenerate sine-Fubini polynomials and present some properties and their relations. In Section 3, we derive partial differentiation, recurrence relations and summation formulas, Stirling numbers of the second kind of degenerate type Fubini numbers and polynomials by using a generating function, respectively.

#### **2. A Parametric Kind of the Degenerate Fubini Polynomials**

In this section, we study the parametric kind of degenerate Fubini polynomials by employing the real and imaginary parts separately and introduce the degenerate Fubini polynomials in terms of degenerate complex polynomials.

The well known degenerate Euler's formula is defined as follows (see [16])

$$\varepsilon\_{\mu}^{(\eta+i\overline{\eta})z} = \varepsilon\_{\mu}^{\eta z}\varepsilon\_{\mu}^{\overline{\eta}z} = \varepsilon\_{\mu}^{\eta z}(\cos\_{\mu}\overline{\xi}z + i\sin\_{\mu}\overline{\xi}z),\tag{20}$$

where

$$\cos\_{\mu} z = \frac{e^{i}\_{\mu}(z) + e^{-i}\_{\mu}(z)}{2}, \quad \sin\_{\mu} z = \frac{e^{i}\_{\mu}(z) - e^{-i}\_{\mu}(z)}{2i}. \tag{21}$$

Note that

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

Using (8), we find

$$\frac{1}{1 - \rho(e\_{\mu}(z) - 1)} e\_{\mu}^{\eta + i\frac{\pi}{5}}(z) = \sum\_{j=0}^{\infty} F\_{j,\mu}(\eta + i\xi; \rho) \frac{z^j}{j!} \tag{22}$$

and

$$\frac{1}{1 - \rho(e\_{\mu}(z) - 1)} e\_{\mu}^{\eta - i\_{\mu}^{\overline{x}}}(z) = \sum\_{j=0}^{\infty} F\_{j,\mu}(\eta + i\_{\nu}^{\overline{x}}; \rho) \frac{z^{j}}{j!}. \tag{23}$$

From Equations (22) and (23), we obtain

$$\frac{1}{1 - \rho(\varepsilon\_{\mu}(z) - 1)} \varepsilon\_{\mu}(\eta z) \cos\_{\mu}(\zeta z) = \sum\_{j=0}^{\infty} \frac{F\_{j,\mu}(\eta + i\xi; \rho) + F\_{j,\mu}(\eta - i\xi; \rho)}{2} \frac{z^j}{j!},\tag{24}$$

and

$$\frac{1}{1 - \rho(\varepsilon\_{\mu}(z) - 1)} \varepsilon\_{\mu}(\eta z) \sin\_{\mu}(\xi z) = \sum\_{j=0}^{\infty} \frac{F\_{j,\mu}(\eta + i\_{5}^{x}; \rho) - F\_{j,\mu}(\eta - i\_{5}^{x}; \rho)}{2i} \frac{z^{j}}{j!}. \tag{25}$$

**Definition 1.** *For a non negative integer n, let us define the degenerate cosine-Fubini polynomials <sup>F</sup>*(*c*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *ξ*; *ρ*) *and the degenerate sine-Fubini polynomials F*(*s*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *ξ*; *ρ*) *by the generating functions, respectively, as follows*

$$\frac{1}{1 - \rho(\mathfrak{e}\_{\mu}(z) - 1)} e\_{\mu}(\eta z) \cos\_{\mu}(\mathfrak{z}z) = \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta, \mathfrak{z}; \rho) \frac{z^j}{j!},\tag{26}$$

*and*

$$\frac{1}{1 - \rho(e\_{\varmu}(z) - 1)} \epsilon\_{\varmu}(\eta z) \sin\_{\varmu}(\ulcorner z) = \sum\_{j=0}^{\infty} F\_{j,\varmu}^{(s)}(\eta, \ulcorner \xi; \rho) \frac{z^j}{j!}.\tag{27}$$

*It is noted that*

$$F^{(c)}\_{j,\mu}(0,0;1) = F\_{j,\mu}, \; F^{(s)}\_{j,\mu}(0,0;1) = 0, (j \ge 0).$$

*The first few of them are:*

$$\begin{aligned} F\_{0,\mu}^{(c)}(\eta,\xi;\rho)&=1, \\ F\_{1,\mu}^{(c)}(\eta,\xi;\rho)&=\eta+\rho, \\ F\_{2,\mu}^{(c)}(\eta,\xi;\rho)&=-\mu\eta+\eta^2-\xi^2+\rho-\mu\rho+2\eta\rho+2\rho^2, \\ F\_{3,\mu}^{(c)}(\eta,\xi;\rho)&=2\mu^2\eta-3\mu\eta^2+\eta^3-3\eta\xi^2+3\mu\xi^3+\rho-3\mu\rho+2\mu^2\rho+3\eta\rho-6\mu\eta\rho, \\ &\quad +3\eta^2\rho-3\xi^2\rho+6\rho^2-6\mu\rho^2+6\eta\rho^2+6\rho^3, \end{aligned}$$

*and*

$$\begin{split} F^{(s)}\_{0,\mu}(\eta,\underline{\mathfrak{x}};\rho) &= 0, \\ F^{(s)}\_{1,\mu}(\eta,\underline{\mathfrak{x}};\rho) &= \underline{\mathfrak{x}}, \\ F^{(s)}\_{2,\mu}(\eta,\underline{\mathfrak{x}};\rho) &= 2\eta\underline{\mathfrak{x}} - \mu\underline{\mathfrak{x}}^{2} + 2\underline{\mathfrak{x}}\rho, \\ F^{(s)}\_{3,\mu}(\eta,\underline{\mathfrak{x}};\rho) &= -3\mu\eta\underline{\mathfrak{x}} + 3\eta^{2}\underline{\mathfrak{x}} - 3\mu\eta\underline{\mathfrak{x}}^{2} - \underline{\mathfrak{x}}^{3} + 2\mu^{2}\underline{\mathfrak{x}}^{3} + 3\underline{\mathfrak{x}}\rho - 3\mu\underline{\mathfrak{x}}\rho, \\ &\quad + 6\eta\underline{\mathfrak{x}}\rho - 3\mu\underline{\mathfrak{x}}^{2}\rho + 6\underline{\mathfrak{x}}\rho^{2}. \end{split}$$

*Note that* lim*μ*−→<sup>0</sup> *<sup>F</sup>*(*c*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *<sup>ξ</sup>*; *<sup>ρ</sup>*) = *<sup>F</sup>*(*c*) *<sup>j</sup>* (*η*, *<sup>ξ</sup>*; *<sup>ρ</sup>*)*,* lim*μ*−→<sup>0</sup> *<sup>F</sup>*(*s*) *<sup>j</sup>*,*<sup>μ</sup>* (*η*, *<sup>ξ</sup>*; *<sup>ρ</sup>*) = *<sup>F</sup>*(*s*) *<sup>j</sup>* (*η*, *ξ*; *ρ*)*,* (*j* ≥ 0)*, where F*(*c*) *<sup>j</sup>* (*η*, *<sup>ξ</sup>*; *<sup>ρ</sup>*) *and F*(*s*) *<sup>j</sup>* (*η*, *ξ*; *ρ*) *are the new type of Fubini polynomials.*

*From Equations (24)–(27), we determine*

$$F\_{j,\mu}^{(c)}(\eta,\xi;\rho) = \frac{F\_{j,\mu}(\eta+i\xi;\rho) + F\_{j,\mu}(\eta-i\xi;\rho)}{2},\tag{28}$$

*and*

$$F\_{j,\mu}^{(s)}(\eta,\xi;\rho) = \frac{F\_{\bar{j},\mu}(\eta+i\xi;\rho) - F\_{\bar{j},\mu}(\eta-i\xi;\rho)}{2i}.\tag{29}$$

**Theorem 1.** *The following result holds true*

$$\begin{aligned} F\_{j,\mu}(\eta + i\xi; \rho) &= \sum\_{r=0}^{j} \binom{j}{r} (i\xi)\_{j-r,\mu} F\_{r,\mu}(\eta; \rho) \\ &= \sum\_{r=0}^{j} \binom{j}{r} (\eta + i\xi)\_{j-r,\mu} F\_{r,\mu}(\rho), \end{aligned} \tag{30}$$

*and*

$$\begin{aligned} F\_{\vec{j},\mu}(\eta - i\xi; \rho) &= \sum\_{r=0}^{\dot{j}} \binom{\dot{j}}{r} (-1)^{j-r} (i\xi)\_{\dot{j}-r,\mu} F\_{\vec{r},\mu}(\eta; \rho) \\ &= \sum\_{r=0}^{\dot{j}} \binom{\dot{j}}{r} (-1)^{j-r} (i\xi - \eta)\_{\dot{j}-r,\mu} F\_{\vec{r},\mu}(\rho), \end{aligned} \tag{31}$$
  $where \ (\eta\)\_{0,\mu} = 1$ ,  $\langle \eta\rangle\_{\dot{j},\mu} = \eta(\eta + \mu) \cdot \cdots (\eta + \mu(\dot{j}-1)), (j \ge 1).$ 

.

**Proof.** From Equation (26), we derive

$$\sum\_{j=0}^{\infty} F\_{\vec{j}, \mu}(\eta + i\xi; \rho) \frac{z^j}{j!} = \frac{1}{1 - \rho(c\_{\mu}(z) - 1)} \varepsilon\_{\mu}^{\eta}(z) \varepsilon\_{\mu}^{i\vec{\zeta}}(z)$$

$$= \sum\_{r=0}^{\infty} F\_{r, \mu}(\eta; \rho) \frac{z^r}{r!} \sum\_{j=0}^{\infty} (i\xi)\_{j, \mu} \frac{z^j}{j!}$$

$$= \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^{j} \binom{j}{r} (i\xi)\_{j-r, \mu} F\_{r, \mu}(\eta; \rho) \right) \frac{z^j}{j!}$$

Similarly, we find

$$\sum\_{j=0}^{\infty} F\_{j,\mu}(\eta + i\xi; \rho) \frac{z^j}{j!} = \sum\_{r=0}^{\infty} F\_{r,\mu}(\rho) \frac{z^r}{r!} \sum\_{j=0}^{\infty} (\eta + i\xi)\_{j,\mu} \frac{z^j}{j!}$$

$$= \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^{j} \binom{j}{r} (\eta + i\xi)\_{j-r,\mu} F\_{r,\mu}(\eta; \rho) \right) \frac{z^j}{j!}$$

which implies the asserted result (30). The proof of (31) is similar.

**Theorem 2.** *The following result holds true*

$$\begin{aligned} F\_{j,\mu}^{(c)}(\eta,\xi;\rho) &= \sum\_{r=0}^{j} \binom{j}{r} F\_{r,\mu}^{(c)} \rho) \mathbb{C}\_{j-r,\mu}(\eta,\xi) \\ &= \sum\_{q=0}^{\left[\frac{j}{2}\right]} \sum\_{r=2q}^{j} \binom{j}{r} \mu^{r-2q} (-1)^q \xi^{2q} S^{(1)}(r, 2q) F\_{j-r,\mu}(\eta;\rho), \end{aligned} \tag{32}$$

*and*

$$\begin{aligned} F\_{j,\mu}^{(s)}(\eta,\xi;\rho) &= \sum\_{r=0}^{j} \binom{j}{r} F\_{r,\mu}^{(s)}(\rho) S\_{j-r,\mu}(\eta,\xi) \\ = \sum\_{q=0}^{\lfloor \frac{j-1}{2} \rfloor} \sum\_{r=2q+1}^{j} \binom{j}{r} \mu^{r-2q-1}(-1)^q \xi^{2q+1} S^{(1)}(r, 2q+1) F\_{j-r,\mu}(\eta;\rho). \end{aligned} \tag{33}$$

**Proof.** From Equations (26) and (16), we find

$$\begin{split} \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\xi;\rho) \frac{z^j}{j!} &= \frac{1}{1 - \rho(c\_{\mu}(z) - 1)} e\_{\mu}(\eta z) \cos \mu(\xi z) \\ &= \left( \sum\_{r=0}^{\infty} F\_{r,\mu}^{(c)}(\rho) \frac{z^{r}}{j!} \right) \left( \sum\_{j=0}^{\infty} \mathcal{C}\_{j,\mu}(\eta,\xi) \frac{z^j}{j!} \right) \\ &= \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^{j} \binom{j}{r} F\_{r,\mu}^{(c)}(\rho) \mathcal{C}\_{j-r,\mu}(\eta,\xi) \right) \frac{z^j}{j!} . \end{split} \tag{34}$$

On the other hand, we find

$$\begin{split} \frac{1}{1-\rho(\varepsilon\_{\mathbb{P}}(z)-1)}\varepsilon\_{\mathbb{A}}(\eta z)\cos\_{\mu}(\mathbb{X}z) &= \sum\_{j=0}^{\infty}F\_{j,\mu}^{(c)}(\eta;\rho)\frac{z^{j}}{\overline{\gamma}^{\prime}}\sum\_{r=0}^{\infty}\frac{\left[\frac{j}{\tau}\right]\_{r}}{\mu-\mathbb{Q}}(-1)^{q}\xi^{2q}S^{(1)}(r,q)\frac{z^{r}}{\overline{\gamma}^{\prime}}\\ &= \sum\_{j=0}^{\infty}\left(\sum\_{r=0}^{j}\sum\_{q=0}^{\left[\frac{j}{\tau}\right]}{\mu-\mathbb{Q}}(-1)^{q}\xi^{2q}S^{(1)}(r,2q)F\_{j-r,\mu}^{(c)}(\eta;\rho)\right)\frac{z^{j}}{\overline{\gamma}^{\prime}}\\ &= \sum\_{j=0}^{\infty}\left(\sum\_{q=0}^{\left[\frac{j}{\tau}\right]}\sum\_{r=2q}^{n}\binom{j}{r}\mu^{r-2q}(-1)^{q}\xi^{2q}S^{(1)}(r,2q)F\_{j-r,\mu}(\eta;\rho)\right)\frac{z^{j}}{\overline{\gamma}^{\prime}}. \end{split} \tag{35}$$

Therefore, by Equations (34) and (35), we obtain (32). The proof of (33) is similar.

**Theorem 3.** *The following relation holds true*

$$C\_{\vec{j},\mu}(\eta,\xi) = F\_{\vec{j},\mu}^{(\varepsilon)}(\eta,\xi;\rho) - \rho \sum\_{r=0}^{j} \binom{j}{r} (1)\_{r,\mu} F\_{\vec{j}-r,\mu}^{(\varepsilon)}(\eta,\xi;\rho) + \rho F\_{\vec{j},\mu}^{(\varepsilon)}(\eta,\xi;\rho), \tag{36}$$

*and*

$$S\_{\vec{j},\mu}(\eta,\xi) = F\_{\vec{j},\mu}^{(s)}(\eta,\xi;\rho) - \rho \sum\_{r=0}^{j} \binom{j}{r} \left(1\right)\_{r,\mu} F\_{\vec{j}-r,\mu}^{(s)}(\eta,\xi;\rho) + \rho F\_{\vec{j},\mu}^{(s)}(\eta,\xi;\rho). \tag{37}$$

**Proof.** In view of (16) and (26), we have

$$\mathcal{L}\_{\mu}(\eta z)\cos\_{\mu}(\pounds z) = \left[1 - \rho(\mathfrak{e}\_{\mu}(z) - 1)\right] \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta, \pounds; \rho) \frac{z^{j}}{j!}$$

$$\begin{split} \sum\_{j=0}^{\infty} \mathcal{C}\_{j,\mu}(\eta,\mathfrak{z}) \frac{z^j}{j!} &= \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\mathfrak{z};\rho) \frac{z^j}{j!} - \rho \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\mathfrak{z};\rho) \frac{z^j}{j!} \sum\_{r=0}^{\infty} (1)\_{r,\mu} \frac{z^r}{r!} \\ &+ \rho \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\mathfrak{z};\rho) \frac{z^j}{j!} \\ &= \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\mathfrak{z};\rho) \frac{z^j}{j!} - \rho \sum\_{j=0}^{\infty} \sum\_{r=0}^{j} \binom{j}{r} (1)\_{r,\mu} F\_{j-r,\mu}^{(c)}(\eta,\mathfrak{z};\rho) \frac{z^j}{j!} \\ &+ \rho \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\mathfrak{z};\rho) \frac{z^j}{j!} .\end{split}$$

On comparing the coefficients of both sides, we get (36). The proof of (37) is similar.

#### **3. Main Results**

In this section, we derive partial differentiation, recurrence relations, explicit and implicit summation formulae and Stirling numbers of the second kind by using the summation technique series method. We start by the following theorem.

**Theorem 4.** *For every j* <sup>∈</sup> <sup>N</sup>*, the following equations for partial derivatives hold true:*

$$\frac{\partial}{\partial \eta} F\_{j,\mu}^{(c)}(\eta, \xi; \rho) = j F\_{j-1,\mu}^{(c)}(\eta, \xi; \rho)\_{\prime} \tag{38}$$

$$\frac{\partial}{\partial \xi} F\_{j,\mu}^{(\varepsilon)}(\eta, \xi; \rho) = -j F\_{j-1,\mu}^{(s)}(\eta, \xi; \rho), \tag{39}$$

$$\frac{\partial}{\partial \eta} F\_{j,\mu}^{(s)}(\eta, \xi; \rho) = j F\_{j-1,\mu}^{(s)}(\eta, \xi; \rho), \tag{40}$$

$$\frac{\partial}{\partial \xi^{\varepsilon}} F^{(\varsigma)}\_{j,\mu}(\eta, \xi; \rho) = j F^{(\varsigma)}\_{j-1,\mu}(\eta, \xi; \rho). \tag{41}$$

**Proof.** Using Equation (26), we see

$$\sum\_{j=1}^{\infty} \frac{\partial}{\partial \eta} F\_{j,\mu}^{(c)}(\eta, \not\!\!z; \rho) \frac{z^j}{j!} = \frac{\partial}{\partial \eta} \frac{e\_{\mu}(\eta z) \cos\_{\mu}(\not\!\!z)}{1 - \rho(e\_{\mu}(z) - 1)} = \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta, \not\!\!\!z; \rho) \frac{z^{j+1}}{j!}$$

$$= \sum\_{j=0}^{\infty} F\_{j-1,\mu}^{(c)}(\eta, \not\!\!z; \rho) \frac{z^j}{(j-1)!} = \sum\_{j=1}^{\infty} n F\_{j-1,\mu}^{(c)}(\eta, \not\!\!z; \rho) \frac{z^j}{j!},$$

proving (38). Other (39), (40) and (41) can be similarly derived.

**Theorem 5.** *For j* ≥ 0*, the following formula holds true:*

$$\frac{1}{1-\rho} \sum\_{r=0}^{j} \binom{j}{r} \,^r \mathbb{F}\_{r,\mu} \left( \frac{\rho}{1-\rho} \right) \mathbb{C}\_{j-r,\mu}(\eta, \lhd\_{\ell} \mathfrak{F}) = \sum\_{r=0}^{j} \binom{j}{r} \sum\_{q=0}^{\infty} z^q (q)\_{r,\mu} \mathbb{C}\_{j-r,\mu}(\eta, \lhd\_{\ell} \mathfrak{F}),\tag{42}$$

*and*

$$\frac{1}{1-\rho} \sum\_{r=0}^{j} \binom{j}{r} F\_{r,\mu} \left( \frac{\rho}{1-\rho} \right) S\_{j-r,\mu}(\eta\_{\prime} \underline{\mathfrak{z}}) = \sum\_{r=0}^{j} \binom{j}{r} \sum\_{q=0}^{\infty} z^q (q)\_{r,\mu} S\_{j-r,\mu}(\eta\_{\prime} \underline{\mathfrak{z}}).\tag{43}$$

**Proof.** We begin with the definition (26) and write

$$\sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta,\xi;\rho) \frac{z^j}{j!} = \frac{1}{1 - \rho(\varepsilon\_{\mu}(z) - 1)} \varepsilon\_{\mu}(\eta z) \cos\_{\mu}(\xi z).$$

Let

$$\begin{split} \frac{1}{1-\rho} \left( \frac{1}{1-\frac{\rho}{1-\rho}[\varepsilon\_{\mu}(z)-1]} \right) &= \frac{1}{1-\rho \epsilon\_{\mu}(z)} = \sum\_{q=0}^{\infty} \rho^{q} (1+\mu z)^{\frac{q}{\mu}} \\ &= \sum\_{r=0}^{\infty} \left( \sum\_{k=0}^{\infty} z^{k} (k)\_{r,\lambda} \right) \frac{t^{r}}{r!} \end{split} \tag{44}$$

$$\begin{split} \sum\_{j=0}^{\infty} \mathcal{F}\_{j,\mu}^{(c)}(\eta,\xi;\rho) \frac{z^j}{\mathcal{I}^l} &= \sum\_{r=0}^{\infty} \left( \sum\_{q=0}^{\infty} \rho^q(q)\_{r,\mu} \right) \frac{z^r}{\mathcal{I}^l} \left( \sum\_{j=0}^{\infty} \mathcal{C}\_{j,\mu}(\eta,\xi) \frac{z^j}{\mathcal{I}^l} \right) \\ &= \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^{j} \binom{j}{r} \sum\_{q=0}^{\infty} \rho^q(q)\_{r,\mu} \mathcal{C}\_{j-r,\mu}(\eta,\xi) \right) \frac{z^j}{\mathcal{I}^l}. \end{split} \tag{45}$$

Now, we observe that, by (44), we get

$$\frac{1}{1-\rho} \left( \frac{1}{1 - \frac{\rho}{1-\rho}(1+\mu z)^{\frac{1}{\mu}} - 1} \right) = \frac{1}{1-\rho} \sum\_{j=0}^{\infty} F\_{j,\mu} \left( \frac{\rho}{1-\rho} \right) \frac{z^j}{j!}.$$

Then, we have

$$\begin{split} \sum\_{j=0}^{\infty} \, \mathcal{F}^{(c)}\_{j,\mu} (\eta, \underline{\chi}; \rho) \frac{z^j}{\overline{\gamma}!} &= \frac{1}{1-\rho} \sum\_{r=0}^{\infty} \, \mathcal{F}\_{r,\mu} \left( \frac{\rho}{1-\rho} \right) \frac{z^r}{\overline{\gamma}!} \left( \sum\_{j=0}^{\infty} \mathcal{C}\_{j,\mu} (\eta, \underline{\xi}) \frac{z^j}{\overline{\gamma}!} \right) \\ &= \frac{1}{1-\rho} \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^{j} \binom{j}{r} \, \mathcal{F}\_{r,\mu} \left( \frac{\rho}{1-\rho} \right) \mathcal{C}\_{j-r,\mu} (\eta, \underline{\xi}) \right) \frac{z^j}{\overline{\gamma}!} . \end{split} \tag{46}$$

Therefore, by Equations (45) and (46), we get (42). The proof of (43) is similar.

**Theorem 6.** *For j* ≥ 0*, the following formula holds true:*

$$C\_{j,\mu}(\eta,\xi) = F\_{j,\mu}^{(c)}(\eta,\xi;\rho) - \rho F\_{j,\mu}^{(c)}(\eta+1,\xi;\rho) + \rho F\_{j,\mu}^{(c)}(\eta,\xi;\rho),\tag{47}$$

*and*

$$S\_{\vec{\jmath},\mu}(\eta,\xi) = F\_{\vec{\jmath},\mu}^{(s)}(\eta,\xi;\rho) - \rho F\_{\vec{\jmath},\mu}^{(s)}(\eta+1,\xi;\rho) + \rho F\_{\vec{\jmath},\mu}^{(s)}(\eta,\xi;\rho). \tag{48}$$

**Proof.** We begin with the definition (26) and write

$$\begin{split} \mathcal{e}\_{\mu}(\eta z) \cos\_{\mu}(\xi z) &= \frac{1 - \rho(\epsilon\_{\mu}(z) - 1)}{1 - \rho(\epsilon\_{\mu}(z) - 1)} \mathcal{e}\_{\mu}(\eta z) \cos\_{\mu}(\xi z) \\ &= \frac{\epsilon\_{\mu}(\eta z) \cos\_{\mu}(\xi z)}{1 - \rho(\epsilon\_{\mu}(z) - 1)} - \frac{\rho(\epsilon\_{\mu}(z) - 1)}{1 - \rho(\epsilon\_{\mu}(z) - 1)} \mathcal{e}\_{\mu}(\eta z) \cos\_{\mu}(\xi z) .\end{split}$$
 
$$\sum\_{j=0}^{\infty} \mathcal{C}\_{j,\mu}(\eta, \xi) \frac{z^{j}}{j!} = \sum\_{j=0}^{\infty} \left[ F\_{j,\mu}^{(c)}(\eta, \xi; \rho) - \rho F\_{j,\mu}^{(c)}(\eta + 1, \xi; \rho) + \rho F\_{j,\mu}^{(c)}(\eta, \xi; \rho) \right] \frac{z^{j}}{j!}.$$

Finally, comparing the coefficients of *<sup>z</sup><sup>j</sup> <sup>j</sup>*! , we get (47). The proof of (48) is similar. **Theorem 7.** *For j* ≥ 0 *and ρ*<sup>1</sup> = *ρ*2*, the following formula holds true:*

$$\begin{split} \sum\_{q=0}^{j} \binom{j}{q} F\_{j-q,\mu}^{(c)}(\eta\_1, \underline{\chi}\_1; \rho\_1) F\_{q,\mu}^{(c)}(\eta\_2, \underline{\chi}\_2; \rho\_2) \\ = \frac{\rho\_2 F\_{j,\mu}^{(c)}(\eta\_1 + \eta\_2, \underline{\xi}\_1 + \underline{\xi}\_2 \underline{\rho}\_2) - \rho\_1 F\_{n,\mu}^{(c)}(\eta\_1 + \eta\_2, \underline{\xi}\_1 + \underline{\xi}\_2; \rho\_1)}{\rho\_2 - \rho\_1} \end{split} \tag{49}$$

*and*

$$\begin{split} \sum\_{q=0}^{j} \binom{j}{q} \, & F\_{j-q, \mu}^{(s)} \left( \eta\_1, \zeta\_1; \rho\_1 \right) F\_{q, \mu}^{(s)} \left( \eta\_2, \zeta\_2; \rho\_2 \right) \\ = \, & \frac{\rho\_2 F\_{j, \mu}^{(s)} \left( \eta\_1 + \eta\_2 \zeta\_1 + \zeta\_2 \rho\_2 \right) - \rho\_1 F\_{n, \mu}^{(s)} \left( \eta\_1 + \eta\_2 \zeta\_1 + \zeta\_2 \rho\_1 \right)}{\rho\_2 - \rho\_1} . \end{split} \tag{50}$$

**Proof.** The products of (26) can be written as

$$\sum\_{j=0}^{\infty} \sum\_{q=0}^{\infty} F\_{n,\mu}^{(c)}(\eta\_1, \xi\_1; \rho\_1 \rho\_1) \frac{z^j}{j!} F\_{q,\mu}^{(c)}(\eta\_2, \xi\_2; \rho\_2) \frac{z^q}{q!} = \frac{e\_{\mu}(\eta\_1 z) \cos\_{\mu}(\xi\_1 z) e\_{\mu}(\eta\_2 z) \cos\_{\mu}(\xi\_2 z)}{(1 - \rho\_1(\varepsilon\_{\mu}(z) - 1))(1 - \rho\_2(\varepsilon\_{\mu}(z) - 1))}$$

$$\sum\_{j=0}^{\infty} \left( \sum\_{q=0}^{j} \binom{j}{q} F\_{j-q,\mu}^{(c)}(\eta\_1, \xi\_1; \rho\_1) F\_{q,\mu}^{(c)}(\eta\_2, \xi\_2; \rho\_2) \right) \frac{z^j}{j!}$$

$$= \frac{\rho\_2}{\rho\_2 - \rho\_1} \frac{e\_{\mu}((\eta\_1 + \eta\_2)z) \cos\_{\mu}((\xi\_1 + \xi\_2)z)}{1 - \rho\_1(\varepsilon\_{\mu}(z) - 1)} - \frac{\rho\_1}{\rho\_2 - \rho\_1} \frac{e\_{\mu}((\eta\_1 + \eta\_2)z) \cos\_{\mu}((\xi\_1 + \xi\_2)z)}{1 - z\_2(\varepsilon\_{\lambda}(z) - 1)}$$

$$= \left( \frac{\rho\_2 F\_{j,\mu}^{(c)}(\eta\_1 + \eta\_2, \xi\_1 + \tilde{\xi}\_2; \rho\_2) - \rho\_1 F\_{j,\mu}^{(c)}(\eta\_1 + \eta\_2, \xi\_1 + \tilde{\xi}\_2; \rho\_1)}{\rho\_2 - \rho\_1} \right) \frac{z^j}{j!}.$$

By equating the coefficients of *<sup>z</sup><sup>j</sup> <sup>j</sup>*! on both sides, we get (49). The proof of (50) is similar.

**Theorem 8.** *For j* ≥ 0*, the following formula holds true:*

$$
\rho F\_{j,\mu}^{(c)}(\eta+1,\xi;\rho) = (1+\rho)F\_{j,\mu}^{(c)}(\eta,\xi;\rho) - C\_{j,\mu}(\eta,\xi),\tag{51}
$$

*and*

$$
\rho F\_{j,\mu}^{(s)}(\eta+1,\xi;\rho) = (1+\rho)F\_{j,\mu}^{(s)}(\eta,\xi;\rho) - S\_{j,\mu}(\eta,\xi). \tag{52}
$$

**Proof.** Equation (26), we see

$$\sum\_{j=0}^{\infty} \left[ F\_{j,\mu}^{(c)}(\eta+1, \underline{\mathfrak{z}}; \rho) - F\_{j,\mu}^{(c)}(\eta, \underline{\mathfrak{z}}; \rho) \right] \frac{z^j}{\overline{\mathfrak{z}}^j} = \frac{\epsilon\_{\mu}(\eta z) \cos\_{\mu}(\underline{\mathfrak{z}} z)}{1 - \rho(\epsilon\_{\mu}(z) - 1)} (\epsilon\_{\mu}(z) - 1)$$

$$= \frac{1}{\rho} \left[ \frac{\epsilon\_{\mu}(\eta z) \cos\_{\mu}(\underline{\mathfrak{z}} z)}{1 - \rho(\epsilon\_{\mu}(z) - 1)} - \epsilon\_{\mu}(\eta z) \cos\_{\mu}(\underline{\mathfrak{z}} z) \right]$$

$$= \frac{1}{\rho} \sum\_{j=0}^{\infty} \left[ F\_{j,\mu}^{(c)}(x, y; z) - C\_{j,\mu}(\eta, \underline{\mathfrak{z}}) \right] \frac{z^j}{j!}.$$

Comparing the coefficients of *<sup>z</sup><sup>j</sup> <sup>j</sup>*! on both sides, we obtain (51). The proof of (52) is similar.

**Corollary 1.** *The following summation formula holds true*

$$F\_{j,\mu}^{(c)}(\eta+1,\xi;\rho) = \sum\_{r=0}^{j} \binom{j}{r} F\_{j-r,\mu}^{(c)}(\eta,\xi;\rho)(1)\_{r,\mu,\xi}$$

*Mathematics* **2020**, *8*, 405

*and*

$$F\_{j,\mu}^{(s)}(\eta+1,\xi;\rho) = \sum\_{r=0}^{j} \binom{j}{r} F\_{j-r,\mu}^{(s)}(\eta,\xi;\rho)(1)\_{r,\mu}.$$

**Theorem 9.** *For j* ≥ 0*, then*

$$F\_{j,\mu}^{(c)}(\eta+\mathfrak{a},\mathfrak{z};\rho) = \sum\_{r=0}^{j} \binom{j}{r} F\_{j-r,\mu}^{(c)}(\eta,\mathfrak{z};\rho)(\mathfrak{a})\_{r,\mu} \tag{53}$$

*and*

$$F\_{j,\mu}^{(s)}(\eta+\alpha,\xi;\rho) = \sum\_{r=0}^{j} \binom{j}{r} F\_{j-r,\mu}^{(s)}(\eta,\xi;\rho)(\alpha)\_{r,\mu}.\tag{54}$$

**Proof.** Replacing *η* by *η* + *α* in (26), we have

$$\sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta+\alpha, \underline{\mathfrak{z}}; \rho) \frac{z^j}{j!} = \frac{1}{1 - \rho(c\_{\mu}(z) - 1)} c\_{\mu}((\eta+\alpha)z) \cos\_{\mu}(\underline{\mathfrak{z}}z)$$

$$= \frac{1}{1 - \rho(c\_{\mu}(z) - 1)} c\_{\mu}(\eta z) \cos\_{\mu}(\underline{\mathfrak{z}}z) c\_{\mu}(\alpha z)$$

$$= \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta, \underline{\mathfrak{z}}; \rho) \frac{z^j}{j!} \sum\_{r=0}^{\infty} (a)\_{r,\mu} \frac{z^j}{j!}$$

$$= \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^{j} \binom{j}{r} F\_{j-r,\mu}^{(c)}(\eta, \underline{\mathfrak{z}}; \rho)(a)\_{r,\mu} \right) \frac{z^j}{j!}.$$

On comparing the coefficients of *z* in both sides, we get (53). The proof of (54) is similar.

**Theorem 10.** *For j* ≥ 0*, the following formula holds true:*

$$F\_{j,\mu}^{(c)}(\eta,\xi;\rho) = \sum\_{k=0}^{j} \sum\_{q=0}^{k} \binom{j}{k} (\eta)\_q S\_{\mu}^{(2)}(k,q) F\_{j-k,\mu}^{(c)}(0,\xi;\rho),\tag{55}$$

*and*

$$F\_{j,\mu}^{(s)}(\eta,\xi;\rho) = \sum\_{k=0}^{j} \sum\_{q=0}^{k} \binom{j}{k} (\eta)\_q S\_{\mu}^{(2)}(k,q) F\_{j-k,\mu}^{(s)}(0,\xi;\rho). \tag{56}$$

**Proof.** Consider (26), we find

$$\sum\_{j=0}^{\infty} F\_{j,\mu}^{(\epsilon)}(\eta,\xi;\rho) \frac{z^j}{j!} = \frac{1}{1 - \rho(\epsilon\_{\mu}(z) - 1)} \left[\epsilon\_{\mu}(z) - 1 + 1\right]^{\eta} \cos\_{\mu}(\xi z)$$

$$= \frac{1}{1 - \rho(\epsilon\_{\mu}(z) - 1)} \sum\_{q=0}^{\infty} \binom{\eta}{q} \left(\epsilon\_{\mu}(z) - 1\right)^q \cos\_{\mu}(\xi z)$$

$$= \frac{1}{1 - \rho(\epsilon\_{\mu}(z) - 1)} \cos\_{\mu}(\xi z) \sum\_{q=0}^{\infty} (\eta)\_{q} \sum\_{k=q}^{\infty} S\_{\mu}^{(2)}(k, q) \frac{z^k}{k!}$$

$$= \sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(0, \xi; \rho) \frac{z^j}{j!} \sum\_{k=0}^{\infty} \left(\sum\_{q=0}^{k} (\eta)\_{q} S\_{\mu}^{(2)}(k, q)\right) \frac{z^k}{k!}$$

$$= \sum\_{j=0}^{\infty} \left(\sum\_{k=0}^{j} \sum\_{q=0}^{k} \left(\frac{j}{k}\right) (\eta)\_{q} S\_{\mu}^{(2)}(k, q) F\_{j-k,\mu}^{(c)}(0, \xi; \rho)\right) \frac{z^j}{j!}!$$

On comparing the coefficients of *z* in both sides, we get (55). The proof of (56) is similar.

**Theorem 11.** *Let j* ≥ 0*, then*

$$F^{(\varepsilon)}\_{j,\mu}(\eta\_{\prime}\mathfrak{z};\rho) = \sum\_{r=0}^{j} \binom{j}{r} \, \mathrm{C}\_{j-r,\mu}(\eta\_{\prime}\mathfrak{z}) \sum\_{k=0}^{r} \rho^{k} k! S\_{2,\mu}(r,k),\tag{57}$$

*and*

$$F\_{j,\mu}^{(s)}(\eta,\xi;\rho) = \sum\_{r=0}^{j} \binom{j}{r} \mathcal{S}\_{j-r,\mu}(\eta,\xi) \sum\_{k=0}^{r} \rho^k k! \mathcal{S}\_{2,\mu}(r,k). \tag{58}$$

**Proof.** Using definition (26), we find

$$\sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta, \xi; \rho) \frac{z^j}{j!} = \frac{1}{1 - \rho(c\_{\mu}(z) - 1)} e\_{\mu}(\eta z) \cos\_{\mu}(\xi z)$$

$$= e\_{\mu}(\eta z) \cos\_{\mu}(\xi z) \sum\_{k=0}^{\infty} \rho^k (c\_{\mu}(z) - 1)^k$$

$$= e\_{\mu}(\eta z) \cos\_{\mu}(\xi z) \sum\_{k=0}^{\infty} \rho^k k! \sum\_{r=k}^{\infty} S\_{2,\mu}(r, k) \frac{z^r}{r!}$$

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

$$L.H.S = \sum\_{j=0}^{\infty} \left( \sum\_{r=0}^j \binom{j}{r} \right) C\_{j-r,\mu}(\eta, \xi) \sum\_{k=0}^r \rho^k k! S\_{2,\mu}(r, k) \Big) \frac{z^j}{j!}.$$

Equating the coefficients of *<sup>z</sup><sup>j</sup> <sup>j</sup>*! in both sides, we get (57). The proof of (58) is similar.

**Theorem 12.** *For j* ≥ 0*, the following formula holds true:*

$$F\_{j,\mu}^{(c)}(\eta+a,\xi;\rho) = \sum\_{q=0}^{j} \binom{j}{q} \mathcal{C}\_{j-q,\mu}(\eta,\xi) \sum\_{k=0}^{q} \rho^{k} k! S\_{2,\mu}(q+a,k+a),\tag{59}$$

*and*

$$F\_{j,\mu}^{(s)}(\eta+a,\xi;\rho) = \sum\_{q=0}^{j} \binom{j}{q} S\_{j-q,\mu}(\eta,\xi) \sum\_{k=0}^{q} \rho^{k} k! S\_{2,\mu}(q+a,k+a). \tag{60}$$

**Proof.** Replacing *η* by *η* + *α* in (26), we see

$$\sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta+\alpha, \underline{\mathfrak{z}}; \rho) \frac{z^j}{j!} = \frac{1}{1 - \rho(\varepsilon\_{\mu}(z) - 1)} c\_{\mu}((\eta+\alpha)z) \cos\_{\mu}(\underline{\mathfrak{z}} z)$$

$$= c\_{\mu}((\eta+\alpha)z) \cos\_{\mu}(\underline{\mathfrak{z}} z) c\_{\mu}(rt) \sum\_{k=0}^{\infty} \rho^{k}(e\_{\mu}(z) - 1)^{k}$$

$$= c\_{\mu}((\eta+\alpha)z) \cos\_{\mu}(\underline{\mathfrak{z}} z) c\_{\mu}(rt) \sum\_{k=0}^{\infty} \rho^{k} \sum\_{q=k}^{\infty} k! S\_{2,\mu}(q,k) \frac{z^{q}}{q!}$$

$$= \sum\_{j=0}^{\infty} \mathcal{C}\_{j,\mu}(\eta, \underline{\mathfrak{z}}^{\*}) \frac{z^{j}}{j!} \sum\_{q=0}^{\infty} \rho^{k} \sum\_{k=0}^{q} k! S\_{2,\mu}(q+\alpha, k+\alpha) \frac{z^{q}}{q!}.$$

$$\sum\_{j=0}^{\infty} F\_{j,\mu}^{(c)}(\eta+\alpha,\xi;\rho) \frac{z^j}{j!}$$

$$=\sum\_{n=0}^{\infty} \left( \sum\_{q=0}^{j} \binom{j}{q} \mathcal{C}\_{j-q,\mu}(\eta,\xi) \sum\_{k=0}^{q} \rho^k k! \mathcal{S}\_{2,\mu}(q+\alpha,k+\alpha) \right) \frac{z^j}{j!}$$

.

Comparing the coefficients of *<sup>z</sup><sup>j</sup> <sup>j</sup>*! in both sides, we get (59). The proof of (60) is similar.

#### **4. Conclusions**

In this paper, we study the general properties and identities of the degenerate Fubini polynomials by treating the real and imaginary parts separately, which provide the degenerate cosine Fubini polynomials and degenerate sine Fubini polynomials. These presented results can be applied to any complex Appell type polynomials such as complex Bernoulli and complex Euler polynomials. Furthermore, we show that the degenerate cosine Fubini polynomials and degenerate sine Fubini polynomials can be expressed in terms of the Stirling numbers of the second kind.

**Author Contributions:** S.K.S., W.A.K., C.S.R. contributed equally to the manuscript and typed, read, and approved final manuscript. All authors have read and agreed to the published version of the manuscript.

**Funding:** The author would like to thank Deanship of Scientific Research at Majmaah University for supporting this work under Project Number No. R-1441-93.

**Acknowledgments:** The authors would like to thank the referees for their valuable comments and suggestions.

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

#### **References**


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