On generalized degenerate Euler–Genocchi polynomials

ABSTRACT

We introduce the generalized degenerate Euler–Genocchi polynomials as a degenerate version of the Euler–Genocchi polynomials. In addition, we introduce their higher-order version, namely the generalized degenerate Euler–Genocchi polynomials of order α, as a degenerate version of the generalized Euler–Genocchi polynomials of order α. The aim of this paper is to study certain properties and identities involving those polynomials, the generalized falling factorials, the degenerate Euler polynomials of order α, the degenerate Stirling numbers of the second kind, and the ‘alternating degenerate power sum of integers’.

KEYWORDS:

• Generalized degenerate Euler–Genocchi polynomials
• generalized degenerate Euler–Genocchi polynomials of order α
• the alternating degenerate power sum of integers

AMS 2010:

• 11B68
• 11B73
• 11B83

1. Introduction

Explorations for various degenerate versions of some special numbers and polynomials have drawn the attention of many mathematicians in recent years, which were initiated by Carlitz's work in [1,2]. Many interesting results were obtained by exploiting different tools such as generating functions, combinatorial methods, p-adic analysis, umbral calculus, differential equations, probability theory, operator theory, analytic number theory and quantum physics (see [3–14]).

Belbachir et al. introduced the Euler–Genocchi polynomials in [15,16] and Goubi generalized them to the generalized Euler–Genocchi polynomials of order α in [17]. Here we introduce degenerate versions for both of them. Namely, we introduce the generalized degenerate Euler–Genocchi polynomials as a degenerate version of the Euler–Genocchi polynomials. In addition, we introduce their higher-order version, namely the generalized degenerate Euler–Genocchi polynomials of order α, as a degenerate version of the generalized Euler–Genocchi polynomials of order α. The aim of this paper is to study certain properties and identities involving those polynomials, the generalized falling factorials, the degenerate Euler polynomials of order α, the degenerate Stirling numbers of the second kind, and the ‘alternating degenerate power sum of integers’ (see (30(30) $\begin{array}{r}{T}_{k,\lambda }\left(n\right)=\sum _{i=0}^n(-1{)}^i(i{)}_{k,\lambda },(k\in \mathbb{N}\cup \{0\left\}\right).\end{array}$(30) )). The novelty of this paper is that it is the first paper that introduces the generalized degenerate Euler–Genocchi polynomials and the generalized degenerate Euler–Genocchi polynomials of order α, as degenerate versions of the polynomials introduced earlier in [15–17].

The outline of this paper is as follows. In Section 1, we recall the degenerate exponentials, the degenerate Euler polynomials, the degenerate Euler polynomials of order α, and the degenerate Genocchi polynomials of order α. Also, we remind the reader of the degenerate Stirling numbers of the second and the incomplete Bell polynomials. Section 2 is the main result of this paper. We introduce the generalized degenerate Euler–Genocchi polynomials $A_{n,\lambda }^{\left(r\right)}\left(x\right)$, as a generalization of both the degenerate Euler polynomials and the degenerate Genocchi polynomials. In Theorem 2.1, the generalized falling factorials $(x{)}_{n,\lambda }$ are expressed in terms of $A_{n,\lambda }^{\left(r\right)}\left(x\right)$. A distribution property is derived for $A_{n,\lambda }^{\left(r\right)}\left(x\right)$ in Theorem 2.2. Then the generalized degenerate Euler–Genocchi polynomials of order α, $A_{n,\lambda }^{(r,\alpha )}\left(x\right)$, are introduced as a higher-order version of $A_{n,\lambda }^{\left(r\right)}\left(x\right)$. We deduce a simple expression for $A_{n,\lambda }^{(r,-m)}\left(x\right)$, with m a positive integer in Theorem 2.3. In Theorems 2.4 and 2.5, we observe certain relations between $A_{n,\lambda }^{(r,\alpha )}\left(x\right)$ and the degenerate Euler polynomials of order α, ${\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}\left(x\right)$. In Theorem 2.6, ${\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}\left(x\right)$ are expressed in terms of the degenerate Stirling numbers of the second kind $S_{2,\lambda }(n,k)$. In Theorem 2.7, $(x{)}_{n,\lambda }$ are represented in terms of $A_{n,\lambda }^{(r,\alpha )}\left(x\right)$ and $S_{2,\lambda }(n,k)$. In Theorem 2.8, we obtain an identity involving $A_{n,\lambda }^{(r,\alpha )}\left(x\right)$, $A_{n,\lambda }^{\left(r\right)}\left(x\right)$, and ${\mathcal{E}}_{n,\lambda }^{(\alpha -1)}\left(x\right)$. Let $T_{k,\lambda }\left(n\right)$ denote the ‘alternating degenerate power sum of integers.’ In Theorem 2.9, we get a representation of $T_{k,\lambda }\left(n\right)$ in terms of $S_{2,\lambda }(n,k)$. Finally, we get an expression of $A_{n,\lambda }^{\left(r\right)}\left(x\right)$ in terms of $T_{k,\lambda }\left(n\right)$ and $S_{2,\lambda }(n,k)$ in Theorem 2.10. In the rest of this section, we recall the facts that are needed throughout this paper.

It is well known that the Euler polynomials are defined by (1) $\begin{array}{r}\frac{2}{{\mathrm{e}}^t+1}{\mathrm{e}}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{E}_n\left(x\right)\frac{t^n}{n!},\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[1-22\left]\right).\end{array}$(1) The Genocchi polynomials are given by (2) $\begin{array}{r}\frac{2t}{{\mathrm{e}}^t+1}{\mathrm{e}}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{G}_n\left(x\right)\frac{t^n}{n!},\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[1-22\left]\right),\end{array}$(2) When x = 0, $G_n=G_n\left(0\right)$ are called the Genocchi numbers.

For any nonzero $\lambda \in \mathbb{R}$, the degenerate exponentials are defined by $\begin{array}{r}{e}_{\lambda }^x\left(t\right)=(1+\lambda t{)}^{\frac{x}{\lambda }}=\sum _{n=0}^{\mathrm{\infty }}(x{)}_{n,\lambda }\frac{t^n}{n!},e_{\lambda }\left(t\right)=e_{\lambda }^1\left(t\right),\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[6\left]\right),\end{array}$ where the generalized falling factorials are given by $\begin{array}{r}(x{)}_{0,\lambda }=1,(x{)}_{n,\lambda }=x(x-\lambda )\cdots (x-(n-1\left)\lambda \right),(n\ge 1).\end{array}$ In [1,2], Carlitz introduced the degenerate Euler polynomials which are given by $\begin{array}{r}\frac{2}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_{n,\lambda }\left(x\right)\frac{t^n}{n!}.\end{array}$ When x = 0, ${\mathcal{E}}_{n,\lambda }={\mathcal{E}}_{n,\lambda }\left(0\right)$ are called the degenerate Euler numbers.

For any nonzero $\alpha \in \mathbb{C}$, the degenerate Euler polynomials of order α are defined by $\begin{array}{r}{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}\left(x\right)\frac{t^n}{n!},\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[1,2\left]\right).\end{array}$ When x = 0, ${\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}={\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}\left(0\right)$ are called the degenerate Euler numbers of order α.

Recently, the degenerate Genocchi polynomials of order α are defined by $\begin{array}{r}{\left(\frac{2t}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{G}_{n,\lambda }^{\left(\alpha \right)}\left(x\right)\frac{t^n}{n!},\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[11-14,22\left]\right).\end{array}$ When x = 0, $G_{n,\lambda }^{\left(\alpha \right)}=G_{n,\lambda }^{\left(\alpha \right)}\left(0\right)$ are called the degenerate Genocchi numbers of order α.

In particular, $\alpha =1$, $G_{n,\lambda }\left(x\right)=G_{n,\lambda }^{\left(1\right)}\left(x\right)$ are called the degenerate Genocchi polynomials.

For $n\ge 0$, the degenerate Stirling numbers of the second kind are introduced by Kim–Kim as $\begin{array}{r}(x{)}_{n,\lambda }=\sum _{k=0}^n{S}_{2,\lambda }(n,k\left)\right(x{)}_k,\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[4,9\left]\right),\end{array}$ where $\begin{array}{r}(x{)}_0=1,(x{)}_n=x(x-1)\cdots (x-n+1),(n\ge 1).\end{array}$ For $k\ge 0$, the incomplete Bell polynomials are defined by $\begin{array}{r}\frac{1}{k!}{\left(\sum _{i=1}^{\mathrm{\infty }}{x}_i\frac{t^i}{i}\right)}^k=\sum _{n=k}^{\mathrm{\infty }}{B}_{n,k}(x_1,x_2,\dots ,x_{n-k+1})\frac{t^n}{n!},\left(\mathrm{s}\mathrm{e}\mathrm{e}\right[8,10,19,21\left]\right),\end{array}$ where $\begin{array}{rl}& B_{n,k}(x_1,x_2,\dots ,x_{n-k+1})\\ & =\sum _{\begin{array}{c}{l}_1+\cdots +l_{n-k+1}=k\\ l_1+2l_2+\cdots +(n-k+1)l_{n-k+1}=n\end{array}}\frac{n!}{l_1!l_2!\cdots l_{n-k+1}!}{\left(\frac{x_1}{1!}\right)}^{l_1}\cdots {\left(\frac{x_{n-k+1}}{(n-k+1)!}\right)}^{l_{n-k+1}}.\end{array}$

2. Generalized degenerate Euler–Genocchi numbers and polynomials

For $r\in \mathbb{Z}$ with $r\ge 0$, we consider the generalized degenerate Euler–Genocchi polynomials given by (3) $\begin{array}{r}\frac{2t^r}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)\frac{t^n}{n!}.\end{array}$(3) Note that $A_{0,\lambda }^{\left(r\right)}=A_{1,\lambda }^{\left(r\right)}\left(x\right)=\cdots =A_{r-1,\lambda }^{\left(r\right)}\left(x\right)=0$.

When x = 0, $A_{n,\lambda }^{\left(r\right)}=A_{n,\lambda }^{\left(r\right)}\left(0\right)$ are called the generalized degenerate Euler–Genocchi numbers. Observe that (4) $\begin{array}{r}{A}_{n,\lambda }^{\left(0\right)}\left(x\right)={\mathcal{E}}_{n,\lambda }\left(x\right),A_{n,\lambda }^{\left(1\right)}\left(x\right)=G_{n,\lambda }\left(x\right),(n\ge 0).\end{array}$(4) From (3(3) $\begin{array}{r}\frac{2t^r}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)\frac{t^n}{n!}.\end{array}$(3) ), we have (5) $\begin{array}{r}{A}_{n,\lambda }^{\left(r\right)}(x+1)=\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})\left(1{)}_{n-l,\lambda }{A}_{l,\lambda }^{\left(r\right)}\right(x),(n\ge 0).\end{array}$(5) By (3(3) $\begin{array}{r}\frac{2t^r}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)\frac{t^n}{n!}.\end{array}$(3) ), we get (6) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}(x{)}_{n,\lambda }\frac{t^n}{n!}\\ & =e_{\lambda }^x\left(t\right)=\frac{1}{2t^r}\sum _{l=0}^{\mathrm{\infty }}{A}_{l,\lambda }^{\left(r\right)}\left(x\right)\frac{t^l}{l!}\left(e_{\lambda }\right(t)+1)\\ & =\frac{1}{2}\sum _{l=0}^{\mathrm{\infty }}{A}_{l+r,\lambda }^{\left(r\right)}\left(x\right)\frac{t^l}{(l+r)!}(\sum _{m=0}^{\mathrm{\infty }}\frac{(1{)}_{m,\lambda }}{m!}{t}^m+1)\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n+r}{l+r})\frac{n!}{(n+r)!}\left(1{)}_{n-l,\lambda }{A}_{l+r,\lambda }^{\left(r\right)}\right(x)\frac{t^n}{n!}+\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{A}_{n+r,\lambda }^{\left(r\right)}(x)\frac{n!}{(n+r)!}\frac{t^n}{n!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2}\left(\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n+r}{l+r})\right(1{)}_{n-l,\lambda }\frac{A_{l+r}^{\left(r\right)}\left(x\right)}{(n+r{)}_r}+\frac{A_{n+r,\lambda }^{\left(r\right)}\left(x\right)}{(n+r{)}_r})\frac{t^n}{n!}.\end{array}$(6) Therefore, by comparing the coefficients on both sides of (6(6) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}(x{)}_{n,\lambda }\frac{t^n}{n!}\\ & =e_{\lambda }^x\left(t\right)=\frac{1}{2t^r}\sum _{l=0}^{\mathrm{\infty }}{A}_{l,\lambda }^{\left(r\right)}\left(x\right)\frac{t^l}{l!}\left(e_{\lambda }\right(t)+1)\\ & =\frac{1}{2}\sum _{l=0}^{\mathrm{\infty }}{A}_{l+r,\lambda }^{\left(r\right)}\left(x\right)\frac{t^l}{(l+r)!}(\sum _{m=0}^{\mathrm{\infty }}\frac{(1{)}_{m,\lambda }}{m!}{t}^m+1)\\ & =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n+r}{l+r})\frac{n!}{(n+r)!}\left(1{)}_{n-l,\lambda }{A}_{l+r,\lambda }^{\left(r\right)}\right(x)\frac{t^n}{n!}+\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{A}_{n+r,\lambda }^{\left(r\right)}(x)\frac{n!}{(n+r)!}\frac{t^n}{n!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2}\left(\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n+r}{l+r})\right(1{)}_{n-l,\lambda }\frac{A_{l+r}^{\left(r\right)}\left(x\right)}{(n+r{)}_r}+\frac{A_{n+r,\lambda }^{\left(r\right)}\left(x\right)}{(n+r{)}_r})\frac{t^n}{n!}.\end{array}$(6) ), we obtain the following theorem.

Theorem 2.1

For $n\ge 0$, we have $\begin{array}{r}(x{)}_{n,\lambda }=\frac{1}{2(n+r{)}_r}\left(\sum _{l=0}^{n-1}(\genfrac{}{}{0ex}{}{n+r}{l+r})\right(1{)}_{n-l,\lambda }{A}_{l+r,\lambda }^{\left(r\right)}\left(x\right)+2A_{n+r,\lambda }^{\left(r\right)}\left(x\right)).\end{array}$

For $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$, we have (7) $\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)\frac{t^n}{n!}& =\frac{2t^r}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\frac{2t^r}{e_{\lambda }^m\left(t\right)+1}\sum _{l=0}^{m-1}(-1{)}^l{e}_{\lambda }^{l+x}(t)\\ & =\frac{2(mt{)}^r}{\left(e_{\lambda /m}\right(mt)+1)}\frac{1}{m^r}\sum _{l=0}^{m-1}(-1{)}^l{e}_{\lambda /m}^{\frac{l+x}{m}}(mt)\\ & =\frac{1}{m^r}\sum _{l=0}^{m-1}(-1{)}^l\frac{2(mt{)}^r}{e_{\lambda /m}\left(mt\right)+1}{e}_{\lambda /m}^{\frac{l+x}{m}}(mt)\\ & =\frac{1}{m^r}\sum _{l=0}^{m-1}(-1{)}^l\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right)m^n\frac{t^n}{n!}\\ & =\sum _{n=0}^{\mathrm{\infty }}{m}^{n-r}\sum _{l=0}^{m-1}(-1{)}^l{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right)\frac{t^n}{n!}.\end{array}$(7) Therefore, by comparing the coefficients on both sides of (7(7) $\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)\frac{t^n}{n!}& =\frac{2t^r}{e_{\lambda }\left(t\right)+1}{e}_{\lambda }^x\left(t\right)=\frac{2t^r}{e_{\lambda }^m\left(t\right)+1}\sum _{l=0}^{m-1}(-1{)}^l{e}_{\lambda }^{l+x}(t)\\ & =\frac{2(mt{)}^r}{\left(e_{\lambda /m}\right(mt)+1)}\frac{1}{m^r}\sum _{l=0}^{m-1}(-1{)}^l{e}_{\lambda /m}^{\frac{l+x}{m}}(mt)\\ & =\frac{1}{m^r}\sum _{l=0}^{m-1}(-1{)}^l\frac{2(mt{)}^r}{e_{\lambda /m}\left(mt\right)+1}{e}_{\lambda /m}^{\frac{l+x}{m}}(mt)\\ & =\frac{1}{m^r}\sum _{l=0}^{m-1}(-1{)}^l\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right)m^n\frac{t^n}{n!}\\ & =\sum _{n=0}^{\mathrm{\infty }}{m}^{n-r}\sum _{l=0}^{m-1}(-1{)}^l{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right)\frac{t^n}{n!}.\end{array}$(7) ), we obtain the following theorem.

Theorem 2.2

For $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$, we have $\begin{array}{r}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)=m^{n-r}\sum _{l=0}^{m-1}(-1{)}^l{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right).\end{array}$

For nonzero $\alpha \in \mathbb{C}$, and $r\in \mathbb{Z}$ with $r\ge 0$, we consider the generalized degenerate Euler–Genocchi polynomials of order α which are given by (8) $\begin{array}{r}{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)\frac{t^n}{n!}.\end{array}$(8) When x = 0, $A_{n,\lambda }^{(r,\alpha )}=A_{n,\lambda }^{(r,\alpha )}\left(0\right)$ are called the generalized degenerate Euler–Genocchi numbers of order α.

We mention here that these polynomials can be viewed as a special case of polynomials $L_n^{(f,g,h)}$ defined by the generating function $\begin{array}{r}f\left(t\right)g\circ h\left(t\right)=\sum _{n\ge 0}{L}_n^{(f,g,h)}\frac{t^n}{n!},\end{array}$ which are recently studied by Goubi [23]. For this, one can take $\begin{array}{r}f\left(t\right)=t^r{e}_{\lambda }^x\left(t\right),g\left(t\right)=t^{\alpha },h\left(t\right)=\frac{2}{e_{\lambda }\left(t\right)+1}.\end{array}$ From (8(8) $\begin{array}{r}{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)\frac{t^n}{n!}.\end{array}$(8) ), we note that (9) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{k,\lambda }^{(r,\alpha )}(x{)}_{n-k,\lambda }\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda },(n\ge 0).\end{array}$(9) Let $\alpha =-m(m\in \mathbb{N})$. Then, by (8(8) $\begin{array}{r}{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)\frac{t^n}{n!}.\end{array}$(8) ), we get (10) $\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,-m)}\left(x\right)\frac{t^n}{n!}& =\frac{t^r}{2^m}\left(e_{\lambda }\right(t)+1{)}^m{e}_{\lambda }^x(t)=\frac{t^r}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})e_{\lambda }^{k+x}(t)\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k+x{)}_{n,\lambda }\frac{t^{n+r}}{n!}\\ & =\sum _{n=r}^{\mathrm{\infty }}\frac{1}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k+x{)}_{n-r,\lambda }(n{)}_r\frac{t^n}{n!}.\end{array}$(10) Therefore, by comparing the coefficients on both sides of (10(10) $\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,-m)}\left(x\right)\frac{t^n}{n!}& =\frac{t^r}{2^m}\left(e_{\lambda }\right(t)+1{)}^m{e}_{\lambda }^x(t)=\frac{t^r}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})e_{\lambda }^{k+x}(t)\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k+x{)}_{n,\lambda }\frac{t^{n+r}}{n!}\\ & =\sum _{n=r}^{\mathrm{\infty }}\frac{1}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k+x{)}_{n-r,\lambda }(n{)}_r\frac{t^n}{n!}.\end{array}$(10) ), we obtain the following theorem.

Theorem 2.3

For $m\in \mathbb{N}$, we have $\begin{array}{r}{A}_{n,\lambda }^{(r,-m)}\left(x\right)=\frac{(n{)}_r}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k+x{)}_{n-r,\lambda }.\end{array}$ When x = 0, we have (11) $\begin{array}{r}{A}_{n,\lambda }^{(r,-m)}=\frac{(n{)}_r}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k{)}_{n-r,\lambda }.\end{array}$(11)

From (9(9) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{k,\lambda }^{(r,\alpha )}(x{)}_{n-k,\lambda }\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda },(n\ge 0).\end{array}$(9) ) and (11(11) $\begin{array}{r}{A}_{n,\lambda }^{(r,-m)}=\frac{(n{)}_r}{2^m}\sum _{k=0}^m(\genfrac{}{}{0ex}{}{m}{k})(k{)}_{n-r,\lambda }.\end{array}$(11) ), we have (12) $\begin{array}{rl}{A}_{n,\lambda }^{(r,-m)}\left(x\right)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,-m)}(x{)}_{k,\lambda }\\ & =\frac{1}{2^m}\sum _{k=0}^n\sum _{j=0}^m(\genfrac{}{}{0ex}{}{n}{k})(\genfrac{}{}{0ex}{}{m}{j})\left(j{)}_{n-k-r,\lambda }\right(n-k{)}_r(x{)}_{k,\lambda }.\end{array}$(12) By (8(8) $\begin{array}{r}{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)\frac{t^n}{n!}.\end{array}$(8) ), we get (13) $\begin{array}{rl}\sum _{n=r}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)\frac{t^n}{n!}& =t^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^x\left(t\right)\\ & =t^r\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}\left(x\right)\frac{t^n}{n!}=\sum _{n=r}^{\mathrm{\infty }}{\mathcal{E}}_{n-r,\lambda }^{\left(\alpha \right)}\left(x\right)\frac{n!}{(n-r)!}\frac{t^n}{n!}\\ & =\sum _{n=r}^{\mathrm{\infty }}\left(n{)}_r{\mathcal{E}}_{n-r,\lambda }^{\left(\alpha \right)}\right(x)\frac{t^n}{n!}.\end{array}$(13)

Theorem 2.4

For $n,r\ge 0$ with $n\ge r$, we have $\begin{array}{r}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)=\left(n{)}_r{\mathcal{E}}_{n-r,\lambda }^{\left(\alpha \right)}\right(x).\end{array}$ In particular, for x = 0, we get (14) $\begin{array}{r}{A}_{n,\lambda }^{(r,\alpha )}=(n{)}_r{\mathcal{E}}_{n-r,\lambda }^{\left(\alpha \right)}.\end{array}$(14)

By (9(9) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{k,\lambda }^{(r,\alpha )}(x{)}_{n-k,\lambda }\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda },(n\ge 0).\end{array}$(9) ) and (14(14) $\begin{array}{r}{A}_{n,\lambda }^{(r,\alpha )}=(n{)}_r{\mathcal{E}}_{n-r,\lambda }^{\left(\alpha \right)}.\end{array}$(14) ), we get (15) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda }=\sum _{k=0}^{n-r}(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda }\end{array}$(15) (16) $\begin{array}{rl}& =\sum _{k=0}^{n-r}(\genfrac{}{}{0ex}{}{n}{k})(n-k{)}_r{\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}(x{)}_{k,\lambda }=\left(n{)}_r\sum _{k=0}^{n-r}(\genfrac{}{}{0ex}{}{n-r}{k}){\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}\right(x{)}_{k,\lambda }\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k}){\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}\right(x{)}_{k,\lambda }.\end{array}$(16) Therefore, by (15(15) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda }=\sum _{k=0}^{n-r}(\genfrac{}{}{0ex}{}{n}{k})A_{n-k,\lambda }^{(r,\alpha )}(x{)}_{k,\lambda }\end{array}$(15) ), we obtain the following theorem.

Theorem 2.5

For any nonzero $\alpha \in \mathbb{C}$ and $n,r\ge 0$ with $n\ge r$, we have $\begin{array}{r}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)=\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k}){\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}\right(x{)}_{k,\lambda }.\end{array}$

Let $f\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n\frac{t^n}{n!}\in \mathbb{C}\left[\right[t\left]\right]$, where $a_0\ne 0$.

Let $0\ne \alpha \in \mathbb{C}$. We quote from [24] the identity (17) $\begin{array}{r}{\left(\sum _{k=0}^{\mathrm{\infty }}{a}_k\frac{t^k}{k!}\right)}^{\alpha }=a_0^{\alpha }+\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^n\left(\alpha {)}_k{a}_0^{\alpha -k}{B}_{n,k}\right(a_1,\dots ,a_{n-k+1})\frac{t^n}{n!},\end{array}$(17) and for the proof, we refer to [25].

From the definition of degenerate Stirling numbers of the second kind, we have (18) $\begin{array}{rl}\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k)\frac{t^n}{n!}& =\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k\\ & =\frac{1}{k!}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{k-j}\sum _{n=0}^{\mathrm{\infty }}(j{)}_{n,\lambda }\frac{t^n}{n!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{k!}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{k-j}(j{)}_{n,\lambda }\frac{t^n}{n!}.\end{array}$(18) Comparing the coefficients on both sides of (18(18) $\begin{array}{rl}\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k)\frac{t^n}{n!}& =\frac{1}{k!}\left(e_{\lambda }\right(t)-1{)}^k\\ & =\frac{1}{k!}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{k-j}\sum _{n=0}^{\mathrm{\infty }}(j{)}_{n,\lambda }\frac{t^n}{n!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{k!}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{k-j}(j{)}_{n,\lambda }\frac{t^n}{n!}.\end{array}$(18) ), we have $\begin{array}{r}{S}_{2,\lambda }(n,k)=\frac{1}{k!}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{k-j}(j{)}_{n,\lambda },(n\ge k\ge 0).\end{array}$ Also, we have (19) $\begin{array}{rl}\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k)\frac{t^n}{n!}& =\frac{1}{k!}{\left(e_{\lambda }\right(t)-1)}^k=\frac{1}{k!}{\left(\sum _{i=1}^{\mathrm{\infty }}\frac{(1{)}_{i,\lambda }}{i!}{t}^i\right)}^k\\ & =\sum _{n=k}^{\mathrm{\infty }}{B}_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda })\frac{t^n}{n!}.\end{array}$(19) Comparing the coefficients on both sides of (19(19) $\begin{array}{rl}\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k)\frac{t^n}{n!}& =\frac{1}{k!}{\left(e_{\lambda }\right(t)-1)}^k=\frac{1}{k!}{\left(\sum _{i=1}^{\mathrm{\infty }}\frac{(1{)}_{i,\lambda }}{i!}{t}^i\right)}^k\\ & =\sum _{n=k}^{\mathrm{\infty }}{B}_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda })\frac{t^n}{n!}.\end{array}$(19) ), we have (20) $\begin{array}{r}{S}_{2,\lambda }(n,k)=B_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda }).\end{array}$(20) We note that (21) $\begin{array}{r}\frac{1}{2}\left(e_{\lambda }\right(t)+1)=\frac{1}{2}(\sum _{l=0}^{\mathrm{\infty }}\frac{(1{)}_{l,\lambda }}{l!}{t}^l+1)=1+\frac{1}{2}\sum _{l=1}^{\mathrm{\infty }}(1{)}_{l,\lambda }\frac{t^l}{l!},\end{array}$(21) and (22) $\begin{array}{rl}{B}_{n,k}(\frac{(1{)}_{1,\lambda }}{2},\frac{(1{)}_{2,\lambda }}{2},\dots ,\frac{(1{)}_{n-k+1,\lambda }}{2})& =\frac{1}{2^k}{B}_{n,k}\left(\right(1{)}_{1,\lambda },\dots ,\left(1{)}_{n-k+1,\lambda }\right)\\ & ={\left(\frac{1}{2}\right)}^k{S}_{2,\lambda }(n,k).\end{array}$(22) Since we have $\begin{array}{r}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}\frac{t^n}{n!}={(1+\sum _{i=1}^{\mathrm{\infty }}\frac{1}{2}(1{)}_{i,\lambda }\frac{t^i}{i!})}^{-\alpha },\end{array}$ by taking $a_0=1$ and $a_i=\frac{1}{2}(1{)}_{i,\lambda },(i\ge 1)$ we obtain the following from (17(17) $\begin{array}{r}{\left(\sum _{k=0}^{\mathrm{\infty }}{a}_k\frac{t^k}{k!}\right)}^{\alpha }=a_0^{\alpha }+\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^n\left(\alpha {)}_k{a}_0^{\alpha -k}{B}_{n,k}\right(a_1,\dots ,a_{n-k+1})\frac{t^n}{n!},\end{array}$(17) ) (23) $\begin{array}{r}{\mathcal{E}}_{0,\lambda }^{\left(\alpha \right)}=1,{\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}=\sum _{k=1}^n(-\alpha {)}_k{\left(\frac{1}{2}\right)}^k{B}_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda }),(n\ge 1).\end{array}$(23) Now, from (20(20) $\begin{array}{r}{S}_{2,\lambda }(n,k)=B_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda }).\end{array}$(20) ) and (23(23) $\begin{array}{r}{\mathcal{E}}_{0,\lambda }^{\left(\alpha \right)}=1,{\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}=\sum _{k=1}^n(-\alpha {)}_k{\left(\frac{1}{2}\right)}^k{B}_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda }),(n\ge 1).\end{array}$(23) ) we obtain the following theorem.

Theorem 2.6

For $n\ge 1$, we have $\begin{array}{rl}{\mathcal{E}}_{n,\lambda }^{\left(\alpha \right)}& =\sum _{k=1}^n(-\alpha {)}_k{\left(\frac{1}{2}\right)}^k{B}_{n,k}\left(\right(1{)}_{1,\lambda },(1{)}_{2,\lambda },\dots ,(1{)}_{n-k+1,\lambda })\\ & =\sum _{k=1}^n(-\alpha {)}_k{\left(\frac{1}{2}\right)}^k{S}_{2,\lambda }(n,k).\end{array}$

From Theorem 2.5, we have (24) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k}){\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}\right(x{)}_{k,\lambda }\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k})\right(x{)}_{k,\lambda }\sum _{j=1}^{n-k-r}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-k}{k})\right(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)(x{)}_{k,\lambda },\end{array}$(24) and (25) $\begin{array}{r}{A}_{n,\lambda }^{(r,\alpha )}=\left(n{)}_r\sum _{j=1}^{n-r}\right(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-r,j),\end{array}$(25) where $n,r\in \mathbb{Z}$ with $n>r\ge 0$.

Replacing n by n + r in (24(24) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k}){\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}\right(x{)}_{k,\lambda }\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k})\right(x{)}_{k,\lambda }\sum _{j=1}^{n-k-r}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-k}{k})\right(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)(x{)}_{k,\lambda },\end{array}$(24) ), we get (26) $\begin{array}{rl}{A}_{n+r,\lambda }^{(r,\alpha )}\left(x\right)& =(n+r{)}_r(x{)}_{n,\lambda }+(n+r{)}_r\sum _{k=0}^{n-1}\sum _{j=1}^{n-k}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j(\genfrac{}{}{0ex}{}{n+r-k}{k})\\ & \times S_{2,\lambda }(n-k,j)(x{)}_{k,\lambda },(n\ge 1).\end{array}$(26) Thus, we have (27) $\begin{array}{rl}(x{)}_{n,\lambda }& =\frac{1}{(n+r{)}_r}{A}_{n+r,\lambda }^{(r,\alpha )}\left(x\right)-\sum _{k=0}^{n-1}\sum _{j=1}^{n-k}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j(\genfrac{}{}{0ex}{}{n+r-k}{k})\\ & \times S_{2,\lambda }(n-k,j)(x{)}_{k,\lambda },(n\ge 1).\end{array}$(27) Therefore, by (27(27) $\begin{array}{rl}(x{)}_{n,\lambda }& =\frac{1}{(n+r{)}_r}{A}_{n+r,\lambda }^{(r,\alpha )}\left(x\right)-\sum _{k=0}^{n-1}\sum _{j=1}^{n-k}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j(\genfrac{}{}{0ex}{}{n+r-k}{k})\\ & \times S_{2,\lambda }(n-k,j)(x{)}_{k,\lambda },(n\ge 1).\end{array}$(27) ), we obtain the following theorem.

Theorem 2.7

For $n\ge 1$, we have $\begin{array}{r}(x{)}_{n,\lambda }=\frac{1}{(n+r{)}_r}{A}_{n+r,\lambda }^{(r,\alpha )}-\sum _{k=0}^{n-1}\sum _{j=1}^{n-k}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j(\genfrac{}{}{0ex}{}{n+r-k}{k})S_{2,\lambda }(n-k,j)(x{)}_{k,\lambda }.\end{array}$

For $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$, we have (28) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{m-1}(-1{)}^k{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}=\sum _{k=0}^{m-1}(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}\\ & =\sum _{k=0}^{m-1}(-1{)}^k{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^{\frac{x+k}{m}}(t)={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{t}^r{e}_{\lambda }^{\frac{x}{m}}(t\left)\sum _{k=0}^{m-1}\right(-1{)}^k{e}_{\lambda }^{\frac{k}{m}}\left(t\right)\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{1+e_{\lambda }\left(t\right)}{1+e_{\lambda }^{1/m}\left(t\right)}\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha -1}{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{2}{1+e_{m\lambda }\left(\frac{t}{m}\right)}\\ & =\sum _{j=0}^{\mathrm{\infty }}{\mathcal{E}}_{j,\lambda }^{(\alpha -1)}\frac{t^j}{j!}\sum _{l=0}^{\mathrm{\infty }}{A}_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}\frac{t^l}{l!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})A_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}{\mathcal{E}}_{n-l,\lambda }^{(\alpha -1)}\frac{t^n}{n!}.\end{array}$(28) Here we note that the condition $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$ is used in the rightmost sum of the second line of (28(28) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{m-1}(-1{)}^k{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}=\sum _{k=0}^{m-1}(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}\\ & =\sum _{k=0}^{m-1}(-1{)}^k{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^{\frac{x+k}{m}}(t)={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{t}^r{e}_{\lambda }^{\frac{x}{m}}(t\left)\sum _{k=0}^{m-1}\right(-1{)}^k{e}_{\lambda }^{\frac{k}{m}}\left(t\right)\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{1+e_{\lambda }\left(t\right)}{1+e_{\lambda }^{1/m}\left(t\right)}\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha -1}{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{2}{1+e_{m\lambda }\left(\frac{t}{m}\right)}\\ & =\sum _{j=0}^{\mathrm{\infty }}{\mathcal{E}}_{j,\lambda }^{(\alpha -1)}\frac{t^j}{j!}\sum _{l=0}^{\mathrm{\infty }}{A}_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}\frac{t^l}{l!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})A_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}{\mathcal{E}}_{n-l,\lambda }^{(\alpha -1)}\frac{t^n}{n!}.\end{array}$(28) ). Therefore, by comparing the coefficients on both sides of (28(28) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{m-1}(-1{)}^k{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}=\sum _{k=0}^{m-1}(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}\\ & =\sum _{k=0}^{m-1}(-1{)}^k{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^{\frac{x+k}{m}}(t)={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{t}^r{e}_{\lambda }^{\frac{x}{m}}(t\left)\sum _{k=0}^{m-1}\right(-1{)}^k{e}_{\lambda }^{\frac{k}{m}}\left(t\right)\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{1+e_{\lambda }\left(t\right)}{1+e_{\lambda }^{1/m}\left(t\right)}\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha -1}{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{2}{1+e_{m\lambda }\left(\frac{t}{m}\right)}\\ & =\sum _{j=0}^{\mathrm{\infty }}{\mathcal{E}}_{j,\lambda }^{(\alpha -1)}\frac{t^j}{j!}\sum _{l=0}^{\mathrm{\infty }}{A}_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}\frac{t^l}{l!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})A_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}{\mathcal{E}}_{n-l,\lambda }^{(\alpha -1)}\frac{t^n}{n!}.\end{array}$(28) ), we obtain the following theorem.

Theorem 2.8

For $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$, we have $\begin{array}{r}\sum _{k=0}^{m-1}(-1{)}^k{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)=\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})A_{l,m\lambda }^{\left(r\right)}(x)\frac{1}{m^{l-r}}{\mathcal{E}}_{n-l,\lambda }^{(\alpha -1)}.\end{array}$

Note that ${\mathcal{E}}_{0,\lambda }^{\left(0\right)}=1$, and ${\mathcal{E}}_{n,\lambda }^{\left(0\right)}=0,(n\in \mathbb{N})$.

Let us take $\alpha =1$ in (28(28) $\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{m-1}(-1{)}^k{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}=\sum _{k=0}^{m-1}(-1{)}^k\sum _{n=0}^{\mathrm{\infty }}{A}_{n,\lambda }^{(r,\alpha )}\left(\frac{x+k}{m}\right)\frac{t^n}{n!}\\ & =\sum _{k=0}^{m-1}(-1{)}^k{t}^r{\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{e}_{\lambda }^{\frac{x+k}{m}}(t)={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{t}^r{e}_{\lambda }^{\frac{x}{m}}(t\left)\sum _{k=0}^{m-1}\right(-1{)}^k{e}_{\lambda }^{\frac{k}{m}}\left(t\right)\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha }{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{1+e_{\lambda }\left(t\right)}{1+e_{\lambda }^{1/m}\left(t\right)}\\ & ={\left(\frac{2}{e_{\lambda }\left(t\right)+1}\right)}^{\alpha -1}{m}^r{\left(\frac{t}{m}\right)}^r{e}_{m\lambda }^x\left(\frac{t}{m}\right)\frac{2}{1+e_{m\lambda }\left(\frac{t}{m}\right)}\\ & =\sum _{j=0}^{\mathrm{\infty }}{\mathcal{E}}_{j,\lambda }^{(\alpha -1)}\frac{t^j}{j!}\sum _{l=0}^{\mathrm{\infty }}{A}_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}\frac{t^l}{l!}\\ & =\sum _{n=0}^{\mathrm{\infty }}\sum _{l=0}^n(\genfrac{}{}{0ex}{}{n}{l})A_{l,m\lambda }^{\left(r\right)}\left(x\right)\frac{1}{m^{l-r}}{\mathcal{E}}_{n-l,\lambda }^{(\alpha -1)}\frac{t^n}{n!}.\end{array}$(28) ). Then we have (29) $\begin{array}{r}\sum _{k=0}^{m-1}(-1{)}^k{A}_{n,\lambda }^{\left(r\right)}\left(\frac{x+k}{m}\right)=A_{n,m\lambda }^{\left(r\right)}(x){\left(\frac{1}{m}\right)}^{n-r},\end{array}$(29) where $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$.

The alternating degenerate power sum of integers $T_{k,\lambda }\left(n\right)$ is defined by (30) $\begin{array}{r}{T}_{k,\lambda }\left(n\right)=\sum _{i=0}^n(-1{)}^i(i{)}_{k,\lambda },(k\in \mathbb{N}\cup \{0\left\}\right).\end{array}$(30)

Remark

$T_k\left(n\right)=\sum _{i=0}^n(-1{)}^i{i}^k$ is defined in [3].

Note that (31) $\begin{array}{r}\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\sum _{k=0}^{\mathrm{\infty }}\sum _{i=0}^n(-1{)}^i(i{)}_{k,\lambda }\frac{t^k}{k!}=\sum _{k=0}^{\mathrm{\infty }}{T}_{k,\lambda }(n)\frac{t^k}{k!},\end{array}$(31) and (32) $\begin{array}{r}\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\frac{1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)}{e_{\lambda }\left(t\right)+1}=\frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}({\mathcal{E}}_{k,\lambda }-(-1{)}^{n+1}{\mathcal{E}}_{k,\lambda }(n+1))\frac{t^k}{k!}.\end{array}$(32) By (31(31) $\begin{array}{r}\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\sum _{k=0}^{\mathrm{\infty }}\sum _{i=0}^n(-1{)}^i(i{)}_{k,\lambda }\frac{t^k}{k!}=\sum _{k=0}^{\mathrm{\infty }}{T}_{k,\lambda }(n)\frac{t^k}{k!},\end{array}$(31) ) and (32(32) $\begin{array}{r}\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\frac{1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)}{e_{\lambda }\left(t\right)+1}=\frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}({\mathcal{E}}_{k,\lambda }-(-1{)}^{n+1}{\mathcal{E}}_{k,\lambda }(n+1))\frac{t^k}{k!}.\end{array}$(32) ), we get (33) $\begin{array}{r}{T}_{k,\lambda }\left(n\right)=\frac{1}{2}({\mathcal{E}}_{k,\lambda }+(-1{)}^n{\mathcal{E}}_{k,\lambda }(n+1)),(n,k\ge 0).\end{array}$(33) It is easy to show that (34) $\begin{array}{rl}1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)& =1-(-1{)}^{n+1}\sum _{l=0}^{\mathrm{\infty }}\frac{(n+1{)}_{l,\lambda }}{l!}{t}^l\\ & =(1+(-1{)}^n)+(-1{)}^n\sum _{i=1}^{\mathrm{\infty }}\frac{(n+1{)}_{i,\lambda }}{i!}{t}^i.\end{array}$(34) For a fixed n, let ${\alpha }_{k,\lambda }$ be the sequence with (35) $\begin{array}{r}{\alpha }_{0,\lambda }=1+(-1{)}^n,{\alpha }_{k,\lambda }=(-1{)}^n(n+1{)}_{k,\lambda },(k\ge 1).\end{array}$(35) Then we have (36) $\begin{array}{r}1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)=\sum _{k=0}^{\mathrm{\infty }}{\alpha }_{k,\lambda }\frac{t^k}{k!}.\end{array}$(36) Now, we observe that (37) $\begin{array}{rl}\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_{n,\lambda }\frac{t^n}{n!}& =\frac{1}{2}{(1+\frac{1}{2}\sum _{i=1}^{\mathrm{\infty }}\frac{(1{)}_{i,\lambda }}{i!}{t}^i)}^{-1}\\ & =\frac{1}{2}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^n(-1{)}^{k}k!{\left(\frac{1}{2}\right)}^k{B}_{n,k}\left(\right(1{)}_{1,\lambda },\dots ,\left(1{)}_{n-k+1,\lambda }\right)\frac{t^n}{n!}\\ & =\frac{1}{2}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^n(-1{)}^{k}k!{\left(\frac{1}{2}\right)}^k{S}_{2,\lambda }(n,k)\frac{t^n}{n!}.\end{array}$(37) From (37(37) $\begin{array}{rl}\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}{\mathcal{E}}_{n,\lambda }\frac{t^n}{n!}& =\frac{1}{2}{(1+\frac{1}{2}\sum _{i=1}^{\mathrm{\infty }}\frac{(1{)}_{i,\lambda }}{i!}{t}^i)}^{-1}\\ & =\frac{1}{2}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^n(-1{)}^{k}k!{\left(\frac{1}{2}\right)}^k{B}_{n,k}\left(\right(1{)}_{1,\lambda },\dots ,\left(1{)}_{n-k+1,\lambda }\right)\frac{t^n}{n!}\\ & =\frac{1}{2}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\sum _{k=1}^n(-1{)}^{k}k!{\left(\frac{1}{2}\right)}^k{S}_{2,\lambda }(n,k)\frac{t^n}{n!}.\end{array}$(37) ), we see that (38) $\begin{array}{r}{\mathcal{E}}_{0,\lambda }=1,{\mathcal{E}}_{n,\lambda }=\sum _{k=1}^n(-1{)}^{k}k!{\left(\frac{1}{2}\right)}^k{S}_{2,\lambda }(n,k),(n\ge 1).\end{array}$(38) From (31(31) $\begin{array}{r}\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\sum _{k=0}^{\mathrm{\infty }}\sum _{i=0}^n(-1{)}^i(i{)}_{k,\lambda }\frac{t^k}{k!}=\sum _{k=0}^{\mathrm{\infty }}{T}_{k,\lambda }(n)\frac{t^k}{k!},\end{array}$(31) ) and (36(36) $\begin{array}{r}1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)=\sum _{k=0}^{\mathrm{\infty }}{\alpha }_{k,\lambda }\frac{t^k}{k!}.\end{array}$(36) ), we note that (39) $\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}{T}_{k,\lambda }\left(n\right)\frac{t^k}{k!}& =\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\frac{1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)}{1+e_{\lambda }\left(t\right)}\\ & =\sum _{l=0}^{\mathrm{\infty }}{\alpha }_{l,\lambda }\frac{t^l}{l!}\sum _{j=0}^{\mathrm{\infty }}\frac{1}{2}{\mathcal{E}}_{j,\lambda }\frac{t^j}{j!}=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{2}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j}){\mathcal{E}}_{j,\lambda }{\alpha }_{k-j,\lambda }\frac{t^k}{k!}.\end{array}$(39) Using (39(39) $\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}{T}_{k,\lambda }\left(n\right)\frac{t^k}{k!}& =\sum _{i=0}^n(-1{)}^i{e}_{\lambda }^i(t)=\frac{1-(-1{)}^{n+1}{e}_{\lambda }^{n+1}(t)}{1+e_{\lambda }\left(t\right)}\\ & =\sum _{l=0}^{\mathrm{\infty }}{\alpha }_{l,\lambda }\frac{t^l}{l!}\sum _{j=0}^{\mathrm{\infty }}\frac{1}{2}{\mathcal{E}}_{j,\lambda }\frac{t^j}{j!}=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{2}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j}){\mathcal{E}}_{j,\lambda }{\alpha }_{k-j,\lambda }\frac{t^k}{k!}.\end{array}$(39) ), for $k\ge 1$, we have (40) $\begin{array}{rl}{T}_{k,\lambda }\left(n\right)& =\frac{1}{2}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j}){\mathcal{E}}_{j,\lambda }{\alpha }_{k-j,\lambda }=\frac{1}{2}{\mathcal{E}}_{0,\lambda }{\alpha }_{k,\lambda }+\frac{1}{2}{\mathcal{E}}_{k,\lambda }{\alpha }_{0,\lambda }+\frac{1}{2}\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j}){\mathcal{E}}_{j,\lambda }{\alpha }_{k-j,\lambda }\\ & =\frac{(-1{)}^n}{2}(n+1{)}_{k,\lambda }+(1+(-1{)}^n)\sum _{j=1}^k(-1{)}^{j}j!{\left(\frac{1}{2}\right)}^{j+1}{S}_{2,\lambda }(k,j)\\ & +\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j})\sum _{i=1}^j(-1{)}^{i}i!{\left(\frac{1}{2}\right)}^{i+1}{S}_{2,\lambda }(j,i\left)\right(-1{)}^n(n+1{)}_{k-j,\lambda }\\ & =\frac{(-1{)}^n}{2}(n+1{)}_{k,\lambda }+(1+(-1{)}^n)\sum _{j=1}^k(-1{)}^{j}j!{\left(\frac{1}{2}\right)}^{j+1}{S}_{2,\lambda }(k,j)\\ & +(-1{)}^n\sum _{j=1}^{k-1}\sum _{i=1}^j(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{i}i!{\left(\frac{1}{2}\right)}^{i+1}{S}_{2,\lambda }(j,i)(n+1{)}_{k-j,\lambda }.\end{array}$(40) Therefore, by (40(40) $\begin{array}{rl}{T}_{k,\lambda }\left(n\right)& =\frac{1}{2}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j}){\mathcal{E}}_{j,\lambda }{\alpha }_{k-j,\lambda }=\frac{1}{2}{\mathcal{E}}_{0,\lambda }{\alpha }_{k,\lambda }+\frac{1}{2}{\mathcal{E}}_{k,\lambda }{\alpha }_{0,\lambda }+\frac{1}{2}\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j}){\mathcal{E}}_{j,\lambda }{\alpha }_{k-j,\lambda }\\ & =\frac{(-1{)}^n}{2}(n+1{)}_{k,\lambda }+(1+(-1{)}^n)\sum _{j=1}^k(-1{)}^{j}j!{\left(\frac{1}{2}\right)}^{j+1}{S}_{2,\lambda }(k,j)\\ & +\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j})\sum _{i=1}^j(-1{)}^{i}i!{\left(\frac{1}{2}\right)}^{i+1}{S}_{2,\lambda }(j,i\left)\right(-1{)}^n(n+1{)}_{k-j,\lambda }\\ & =\frac{(-1{)}^n}{2}(n+1{)}_{k,\lambda }+(1+(-1{)}^n)\sum _{j=1}^k(-1{)}^{j}j!{\left(\frac{1}{2}\right)}^{j+1}{S}_{2,\lambda }(k,j)\\ & +(-1{)}^n\sum _{j=1}^{k-1}\sum _{i=1}^j(\genfrac{}{}{0ex}{}{k}{j})(-1{)}^{i}i!{\left(\frac{1}{2}\right)}^{i+1}{S}_{2,\lambda }(j,i)(n+1{)}_{k-j,\lambda }.\end{array}$(40) ), we obtain the following theorem.

Theorem 2.9

For $n\ge 0$ and $k\ge 1$, we have $\begin{array}{rl}{T}_{k,\lambda }\left(n\right)& =\frac{(-1{)}^n}{2}(n+1{)}_{k,\lambda }+(1+(-1{)}^n)\sum _{j=1}^k(-1{)}^{j}j!{\left(\frac{1}{2}\right)}^{j+1}{S}_{2,\lambda }(k,j)\\ & +(-1{)}^n\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j})(n+1{)}_{k-j,\lambda }\sum _{i=1}^j(-1{)}^{i}i!{\left(\frac{1}{2}\right)}^{i+1}{S}_{2,\lambda }(j,i).\end{array}$

From Theorem 2.9, we note that (41) $\begin{array}{rl}{T}_{k,\lambda }\left(2n\right)& =\frac{1}{2}(2n+1{)}_{k,\lambda }+2\sum _{j=1}^k(-1{)}^{j}j!{\left(\frac{1}{2}\right)}^{j+1}{S}_{2,\lambda }(k,j)\\ & +\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j})(2n+1{)}_{k-j,\lambda }\sum _{i=1}^j(-1{)}^{i}i!{\left(\frac{1}{2}\right)}^{i+1}{S}_{2,\lambda }(j,i),\end{array}$(41) and (42) $\begin{array}{r}{T}_{k,\lambda }(2n+1)=-2^{k-1}(n+1{)}_{k,\lambda /2}-\sum _{j=1}^{k-1}(\genfrac{}{}{0ex}{}{k}{j})(n+1{)}_{k-j,\lambda /2}\sum _{i=1}^j(-1{)}^{i}i!2^{k-j-i-1}{S}_{2,\lambda }(j,i).\end{array}$(42) From (24(24) $\begin{array}{rl}{A}_{n,\lambda }^{(r,\alpha )}\left(x\right)& =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k}){\mathcal{E}}_{n-k-r,\lambda }^{\left(\alpha \right)}\right(x{)}_{k,\lambda }\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}(\genfrac{}{}{0ex}{}{n-r}{k})\right(x{)}_{k,\lambda }\sum _{j=1}^{n-k-r}(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-k}{k})\right(-\alpha {)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)(x{)}_{k,\lambda },\end{array}$(24) ), we have (43) $\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)& =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-r}{k})\right(-1{)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)(x{)}_{k,\lambda }\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-r}{k})j!{(-\frac{1}{2})}^j{S}_{2,\lambda }(n-k-r,j\left)\right(x{)}_{k,\lambda }.\end{array}$(43) Thus, by (43(43) $\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)& =\left(n{)}_r\right(x{)}_{n-r,\lambda }+\left(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-r}{k})\right(-1{)}_j{\left(\frac{1}{2}\right)}^j{S}_{2,\lambda }(n-k-r,j)(x{)}_{k,\lambda }\\ & =\left(n{)}_r\right(x{)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-r}{k})j!{(-\frac{1}{2})}^j{S}_{2,\lambda }(n-k-r,j\left)\right(x{)}_{k,\lambda }.\end{array}$(43) ), we get (44) $\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)& =(n{)}_r{\left(\frac{x+i}{m}\right)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-r}{k})j!{(-\frac{1}{2})}^j\\ & \times S_{2,\lambda }(n-k-r,j){\left(\frac{x+i}{m}\right)}_{k,\lambda }.\end{array}$(44) From (44(44) $\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)& =(n{)}_r{\left(\frac{x+i}{m}\right)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}(\genfrac{}{}{0ex}{}{n-r}{k})j!{(-\frac{1}{2})}^j\\ & \times S_{2,\lambda }(n-k-r,j){\left(\frac{x+i}{m}\right)}_{k,\lambda }.\end{array}$(44) ), we can derive the following Equation (45(45) $\begin{array}{rl}\sum _{i=0}^{m-1}(-1{)}^i{A}_{m,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)& =\left(n{)}_r\sum _{i=0}^{m-1}\right(-1{)}^i{\left(\frac{x+i}{m}\right)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}j!{(-\frac{1}{2})}^j\\ & \times (\genfrac{}{}{0ex}{}{n-r}{k})S_{2,\lambda }(n-k-r,j)\sum _{i=0}^{m-1}(-1{)}^i{\left(\frac{x+i}{m}\right)}_{k,\lambda }.\end{array}$(45) ). (45) $\begin{array}{rl}\sum _{i=0}^{m-1}(-1{)}^i{A}_{m,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)& =\left(n{)}_r\sum _{i=0}^{m-1}\right(-1{)}^i{\left(\frac{x+i}{m}\right)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}j!{(-\frac{1}{2})}^j\\ & \times (\genfrac{}{}{0ex}{}{n-r}{k})S_{2,\lambda }(n-k-r,j)\sum _{i=0}^{m-1}(-1{)}^i{\left(\frac{x+i}{m}\right)}_{k,\lambda }.\end{array}$(45) Now, we observe that (46) $\begin{array}{rl}\sum _{i=0}^{m-1}(-1{)}^i{\left(\frac{x+i}{m}\right)}_{k,\lambda }& =\frac{1}{m^k}\sum _{i=0}^{m-1}(-1{)}^i(x+i{)}_{k,m\lambda }\\ & =\frac{1}{m^k}\sum _{i=0}^{m-1}(-1{)}^i\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(i{)}_{l,m\lambda }(x{)}_{k-l,m\lambda }\\ & =\frac{1}{m^k}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\left(x{)}_{k-l,m\lambda }{T}_{l,m\lambda }\right(m-1)\end{array}$(46) From (45(45) $\begin{array}{rl}\sum _{i=0}^{m-1}(-1{)}^i{A}_{m,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)& =\left(n{)}_r\sum _{i=0}^{m-1}\right(-1{)}^i{\left(\frac{x+i}{m}\right)}_{n-r,\lambda }+(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}j!{(-\frac{1}{2})}^j\\ & \times (\genfrac{}{}{0ex}{}{n-r}{k})S_{2,\lambda }(n-k-r,j)\sum _{i=0}^{m-1}(-1{)}^i{\left(\frac{x+i}{m}\right)}_{k,\lambda }.\end{array}$(45) ) and (46(46) $\begin{array}{rl}\sum _{i=0}^{m-1}(-1{)}^i{\left(\frac{x+i}{m}\right)}_{k,\lambda }& =\frac{1}{m^k}\sum _{i=0}^{m-1}(-1{)}^i(x+i{)}_{k,m\lambda }\\ & =\frac{1}{m^k}\sum _{i=0}^{m-1}(-1{)}^i\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})(i{)}_{l,m\lambda }(x{)}_{k-l,m\lambda }\\ & =\frac{1}{m^k}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\left(x{)}_{k-l,m\lambda }{T}_{l,m\lambda }\right(m-1)\end{array}$(46) ), we have (47) $\begin{array}{rl}& \sum _{i=0}^{m-1}(-1{)}^i{A}_{n,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)=(n{)}_r\frac{1}{m^{n-r}}\sum _{l=0}^{n-r}(\genfrac{}{}{0ex}{}{n-r}{l})\left(x{)}_{n-l-r,m\lambda }{T}_{l,m\lambda }\right(m-1)\\ & +(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}j!{(-\frac{1}{2})}^j(\genfrac{}{}{0ex}{}{n-r}{k})S_{2,\lambda }(n-k-r,j)\frac{1}{m^k}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\\ & \times T_{l,m\lambda }(m-1)(x{)}_{k-l,m\lambda }.\end{array}$(47) For $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$, by Theorem 2.2 and (47(47) $\begin{array}{rl}& \sum _{i=0}^{m-1}(-1{)}^i{A}_{n,\lambda }^{\left(r\right)}\left(\frac{x+i}{m}\right)=(n{)}_r\frac{1}{m^{n-r}}\sum _{l=0}^{n-r}(\genfrac{}{}{0ex}{}{n-r}{l})\left(x{)}_{n-l-r,m\lambda }{T}_{l,m\lambda }\right(m-1)\\ & +(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}j!{(-\frac{1}{2})}^j(\genfrac{}{}{0ex}{}{n-r}{k})S_{2,\lambda }(n-k-r,j)\frac{1}{m^k}\sum _{l=0}^k(\genfrac{}{}{0ex}{}{k}{l})\\ & \times T_{l,m\lambda }(m-1)(x{)}_{k-l,m\lambda }.\end{array}$(47) ), we get (48) $\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)& =m^{n-r}\sum _{l=0}^{m-1}(-1{)}^l{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right)=(n{)}_r\sum _{l=0}^{n-r}(\genfrac{}{}{0ex}{}{n-r}{l})\left(x{)}_{n-l-r,\lambda }{T}_{l,\lambda }\right(m-1)\\ & +(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}\sum _{l=0}^{k}j!{(-\frac{1}{2})}^j(\genfrac{}{}{0ex}{}{n-r}{k})(\genfrac{}{}{0ex}{}{k}{l})m^{n-r-k}\\ & \times S_{2,\lambda }(n-k-r,j)T_{l,\lambda }(m-1)(x{)}_{k-l,\lambda }.\end{array}$(48) Therefore, by (48(48) $\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)& =m^{n-r}\sum _{l=0}^{m-1}(-1{)}^l{A}_{n,\lambda /m}^{\left(r\right)}\left(\frac{l+x}{m}\right)=(n{)}_r\sum _{l=0}^{n-r}(\genfrac{}{}{0ex}{}{n-r}{l})\left(x{)}_{n-l-r,\lambda }{T}_{l,\lambda }\right(m-1)\\ & +(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}\sum _{l=0}^{k}j!{(-\frac{1}{2})}^j(\genfrac{}{}{0ex}{}{n-r}{k})(\genfrac{}{}{0ex}{}{k}{l})m^{n-r-k}\\ & \times S_{2,\lambda }(n-k-r,j)T_{l,\lambda }(m-1)(x{)}_{k-l,\lambda }.\end{array}$(48) ), we obtain the following theorem.

Theorem 2.10

For $m\in \mathbb{N}$ with $m\equiv 1\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)$, we have

$\begin{array}{rl}{A}_{n,\lambda }^{\left(r\right)}\left(x\right)& =\left(n{)}_r\sum _{l=0}^{n-r}(\genfrac{}{}{0ex}{}{n-r}{l})\right(x{)}_{n-l-r,\lambda }{T}_{l,\lambda }(m-1)\\ & +(n{)}_r\sum _{k=0}^{n-r-1}\sum _{j=1}^{n-k-r}\sum _{l=0}^{k}j!{(-\frac{1}{2})}^j(\genfrac{}{}{0ex}{}{n-r}{k})(\genfrac{}{}{0ex}{}{k}{l})\\ & \times m^{n-r-k}{S}_{2,\lambda }(n-k-r,j)T_{l,\lambda }(m-1)(x{)}_{k-l,\lambda }.\end{array}$

3. Conclusion

In recent years, various degenerate versions of many special numbers and polynomials have been explored by using different methods as aforementioned in the introduction.

In this paper, we introduced the generalized degenerate Euler–Genocchi polynomials as a degenerate version of the Euler–Genocchi polynomials. In addition, we introduced their higher-order version, namely the generalized degenerate Euler–Genocchi polynomials of order α, as a degenerate version of the generalized Euler–Genocchi polynomials of order α. Then we studied certain properties and identities involving those polynomials, the generalized falling factorials, the degenerate Euler polynomials of order α, the degenerate Stirling numbers of the second kind, and the alternating degenerate power sum of integers.

It is one of our future projects to continue to study various degenerate versions of some special numbers and polynomials and to find their applications to physics, science and engineering as well as to mathematics.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

This work was supported by the National Research Foundation of Korea [grant number NRF-2021R1F1A1050151].