• Research
• Open Access

# Degenerate poly-Bernoulli polynomials with umbral calculus viewpoint

Journal of Inequalities and Applications20152015:228

https://doi.org/10.1186/s13660-015-0748-7

• Received: 7 April 2015
• Accepted: 2 July 2015
• Published:

## Abstract

In this paper, we consider the degenerate poly-Bernoulli polynomials. We present several explicit formulas and recurrence relations for these polynomials. Also, we establish a connection between our polynomials and several known families of polynomials.

## Keywords

• degenerate poly-Bernoulli polynomials
• umbral calculus

• 05A19
• 05A40
• 11B83

## 1 Introduction

The degenerate Bernoulli polynomials $$\beta_{n}(\lambda,x)$$ ($$\lambda\neq0$$) were introduced by Carlitz  and rediscovered by Ustinov  under the name Korobov polynomials of the second kind. They are given by the generating function
$$\frac{t}{(1+\lambda t)^{1/\lambda}-1}(1+\lambda t)^{x/\lambda}=\sum _{n\geq0}\beta_{n}(\lambda,x)\frac{t^{n}}{n!}.$$
When $$x=0$$, $$\beta_{n}(\lambda)=\beta_{n}(\lambda,0)$$ are called the degenerate Bernoulli numbers (see ). We observe that $$\lim_{\lambda\rightarrow0}\beta_{n}(\lambda,x)=B_{n}(x)$$, where $$B_{n}(x)$$ is the nth ordinary Bernoulli polynomial (see the references).
The poly-Bernoulli polynomials $$PB_{n}^{(k)}(x)$$ are defined by
$$\frac{Li_{k}(1-e^{-t})}{e^{t}-1}e^{xt}=\sum_{n\geq0}PB_{n}^{(k)}(x) \frac{t^{n}}{n!},$$
where $$Li_{k}(x)$$ ($$k\in\mathbb{Z}$$) is the classical polylogarithm function given by $$Li_{k}(x)=\sum_{n\geq1}\frac{x^{n}}{n^{k}}$$ (see ).
For $$0\neq\lambda\in\mathbb{C}$$ and $$k\in\mathbb{Z}$$, the degenerate poly-Bernoulli polynomials $$P\beta_{n}^{(k)}(\lambda,x)$$ are defined by Kim and Kim to be
\begin{aligned} \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}(1+\lambda t)^{x/\lambda} =\sum _{n\geq0}P\beta_{n}^{(k)}(\lambda,x) \frac{t^{n}}{n!} \quad\mbox{(see )}. \end{aligned}
(1.1)
When $$x=0$$, $$P\beta_{n}^{(k)}(\lambda)=P\beta_{n}^{(k)}(\lambda,0)$$ are called degenerate poly-Bernoulli numbers. We observe that $$\lim_{\lambda\rightarrow0}P\beta_{n}^{(k)}(\lambda,x)=PB_{n}^{(k)}(x)$$.
The goal of this paper is to use umbral calculus to obtain several new and interesting identities of degenerate poly-Bernoulli polynomials. To do that we recall the umbral calculus as given in [7, 8]. We denote the algebra of polynomials in a single variable x over $$\mathbb {C}$$ by Π and the vector space of all linear functionals on Π by $$\Pi^{*}$$. The action of a linear functional L on a polynomial $$p(x)$$ is denoted by $$\langle L\mid p(x)\rangle$$. We define the vector space structure on $$\Pi^{*}$$ by $$\langle cL+c'L'\mid p(x)\rangle=c\langle L\mid p(x)\rangle+c'\langle L'\mid p(x)\rangle$$, where $$c,c'\in\mathbb{C}$$. We define the algebra of formal power series in a single variable t to be
\begin{aligned} \mathcal{H}= \biggl\{ f(t)=\sum_{k\geq0} a_{k} \frac{t^{k}}{k!}\Bigm| a_{k}\in \mathbb{C} \biggr\} . \end{aligned}
(1.2)
A power series $$f(t)\in\mathcal{H}$$ defines a linear functional on Π by setting
\begin{aligned} \bigl\langle f(t)\mid x^{n}\bigr\rangle =a_{n}, \quad\mbox{for all }n \geq0\ \mbox{(see [6, 8--10])}. \end{aligned}
(1.3)
By (1.2) and (1.3), we have
\begin{aligned} \bigl\langle t^{k}\mid x^{n}\bigr\rangle =n!\delta_{n,k},\quad \mbox{for all }n,k\geq0, \end{aligned}
(1.4)
where $$\delta_{n,k}$$ is the Kronecker symbol. Let $$f_{L}(t)=\sum_{n\geq0}\langle L\mid x^{n}\rangle\frac{t^{n}}{n!}$$. From (1.4), we have $$\langle f_{L}(t)\mid x^{n}\rangle=\langle L\mid x^{n}\rangle$$. So, the map $$L\mapsto f_{L}(t)$$ is a vector space isomorphism from $$\Pi ^{*}$$ onto $$\mathcal{H}$$. Thus, $$\mathcal{H}$$ is thought of as set of both formal power series and linear functionals. We call $$\mathcal{H}$$ the umbral algebra. The umbral calculus is the study of umbral algebra.
The order $$O(f(t))$$ of the non-zero power series $$f(t)\in\mathcal {H}$$ is the smallest integer k for which the coefficient of $$t^{k}$$ does not vanish. Suppose that $$f(t),g(t)\in\mathcal{H}$$ such that $$O(f(t))=1$$ and $$O(g(t))=0$$, then there exists a unique sequence $$s_{n}(x)$$ of polynomials such that
\begin{aligned} \bigl\langle g(t) \bigl(f(t)\bigr)^{k}\mid s_{n}(x)\bigr\rangle =n!\delta_{n,k}, \end{aligned}
(1.5)
where $$n,k\geq0$$. The sequence $$s_{n}(x)$$ is called the Sheffer sequence for $$(g(t),f(t))$$, which is denoted by $$s_{n}(x)\sim(g(t),f(t))$$ (see [7, 8]). For $$f(t)\in\mathcal{H}$$ and $$p(x)\in\Pi$$, we have $$\langle e^{yt}\mid p(x)\rangle=p(y)$$, $$\langle f(t)g(t)\mid p(x)\rangle =\langle g(t)\mid f(t)p(x)\rangle$$, and
\begin{aligned} f(t)=\sum_{n\geq0}\bigl\langle f(t)\mid x^{n} \bigr\rangle \frac{t^{n}}{n!},\qquad p(x)=\sum_{n\geq0}\bigl\langle t^{n}\mid p(x)\bigr\rangle \frac{x^{n}}{n!} \end{aligned}
(1.6)
(see [7, 8]). From (1.6), we obtain $$\langle t^{k}\mid p(x)\rangle=p^{(k)}(0)$$ and $$\langle1\mid p^{(k)}(x)\rangle =p^{(k)}(0)$$, where $$p^{(k)}(0)$$ denotes the kth derivative of $$p(x)$$ with respect to x at $$x=0$$. So, we get $$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}}{dx^{k}}p(x)$$, for all $$k\geq0$$. Let $$s_{n}(x)\sim(g(t),f(t))$$, then we have
\begin{aligned} \frac{1}{g(\bar{f}(t))}e^{y\bar{f}(t)}=\sum_{n\geq0}s_{n}(y) \frac {t^{n}}{n!}, \end{aligned}
(1.7)
for all $$y\in\mathbb{C}$$, where $$\bar{f}(t)$$ is the compositional inverse of $$f(t)$$ (see [7, 8]). For $$s_{n}(x)\sim(g(t),f(t))$$ and $$r_{n}(x)\sim(h(t),\ell(t))$$, let $$s_{n}(x)=\sum_{k=0}^{n} c_{n,k}r_{k}(x)$$, then we have
\begin{aligned} c_{n,k}=\frac{1}{k!} \biggl\langle \frac{h(\bar{f}(t))}{g(\bar{f}(t))}\bigl(\ell \bigl(\bar{f}(t)\bigr)\bigr)^{k}\Bigm|x^{n} \biggr\rangle \end{aligned}
(1.8)
(see [7, 8]).
From (1.1), we see that $$P\beta _{n}^{(k)}(\lambda,x)$$ is the Sheffer sequence for the pair
\begin{aligned} \bigl(g(t),f(t)\bigr)= \biggl(\frac{e^{t}-1}{Li_{k}(1-e^{-\frac{1}{\lambda}(e^{\lambda t}-1)})},\frac{1}{\lambda} \bigl(e^{\lambda t}-1\bigr) \biggr). \end{aligned}
(1.9)

In this paper, we will use umbral calculus in order to derive some properties, explicit formulas, recurrence relations, and identities as regards the degenerate poly-Bernoulli polynomials. Also, we establish a connection between our polynomials and several known families of polynomials.

## 2 Explicit formulas

In this section we present several explicit formulas for the degenerate poly-Bernoulli polynomials, namely $$P\beta _{n}^{(k)}(\lambda,x)$$. To do so, we recall that Stirling numbers $$S_{1}(n,k)$$ of the first kind can be defined by means of exponential generating functions as $$\sum_{\ell\geq j}S_{1}(\ell,j)\frac{t^{\ell}}{\ell!}=\frac{1}{j!}\log^{j}(1+t)$$ and can be defined by means of ordinary generating functions as
\begin{aligned} (x)_{n}=\sum_{m=0}^{n}S_{1}(n,m)x^{m} \sim\bigl(1,e^{t}-1\bigr), \end{aligned}
(2.1)
where $$(x)_{n}=x(x-1)(x-2)\cdots(x-n+1)$$ with $$(x)_{0}=1$$. For $$\lambda\neq 0$$, we define $$(x\mid \lambda)_{n}=\lambda^{n}(x/\lambda)_{n}$$. Sometimes, for simplicity, we denote the function $$\frac{e^{t}-1}{Li_{k}(1-e^{-\frac {1}{\lambda}(e^{\lambda t}-1)})}$$ by $$G_{k}(t)$$.

First, we express the degenerate poly-Bernoulli polynomials in terms of degenerate poly-Bernoulli numbers.

### Theorem 2.1

For all $$n\geq0$$,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x)&=\sum_{j=0}^{n} \sum_{\ell=j}^{n}\binom{n}{\ell }S_{1}( \ell,j)\lambda^{\ell-j}P\beta_{n-\ell}^{(k)}( \lambda)x^{j}. \end{aligned}

### Proof

By (1.5), for $$s_{n}(x)\sim(g(t),f(t))$$ we have $$s_{n}(x)=\sum_{j=0}^{n}\frac{1}{j!}\langle g(\bar{f}(t))^{-1}\bar{f}(t)^{j}\mid x^{n}\rangle x^{j}$$. Thus, in the case of degenerate poly-Bernoulli polynomials (see (1.9)), we have
\begin{aligned} &\frac{1}{j!}\bigl\langle g\bigl(\bar{f}(t)\bigr)^{-1} \bar{f}(t)^{j}\mid x^{n}\bigr\rangle \\ &\quad=\frac{1}{j!} \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1} \biggl(\frac{1}{\lambda}\log(1+\lambda t) \biggr)^{j}\Bigm| x^{n} \biggr\rangle \\ &\quad=\lambda^{-j} \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}\Bigm| \frac{\log^{j}(1+\lambda t)}{j!}x^{n} \biggr\rangle \\ &\quad=\lambda^{-j} \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}\Bigm|\sum _{\ell\geq j}S_{1}(\ell,j)\frac{\lambda^{\ell}t^{\ell}}{\ell!}x^{n} \biggr\rangle \\ &\quad=\sum_{\ell=j}^{n}\binom{n}{\ell}S_{1}( \ell,j)\lambda^{\ell-j} \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}\Bigm| x^{n-\ell} \biggr\rangle \\ &\quad=\sum_{\ell=j}^{n}\binom{n}{\ell}S_{1}( \ell,j)\lambda^{\ell-j} \biggl\langle \sum_{m\geq0}P \beta_{m}^{(k)}(\lambda)\frac{t^{m}}{m!}\Bigm| x^{n-\ell} \biggr\rangle \\ &\quad=\sum_{\ell=j}^{n}\binom{n}{\ell}S_{1}( \ell,j)\lambda^{\ell-j}P\beta _{n-\ell}^{(k)}(\lambda), \end{aligned}
which completes the proof. □
Note that Stirling numbers $$S_{2}(n,k)$$ of the second kind can be defined by the exponential generating functions as
\begin{aligned} \sum_{n\geq k}S_{2}(n,k)\frac{x^{n}}{n!}= \frac{(e^{t}-1)^{k}}{k!}. \end{aligned}
(2.2)

### Theorem 2.2

For all $$n\geq0$$,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x) &=\sum_{j=0}^{n} \Biggl(\sum_{m=j}^{n}\sum _{\ell=0}^{m-j}\binom {m}{j}S_{1}(n,m)S_{2}(m-j, \ell)\lambda^{n-\ell-j}P\beta_{\ell}^{(k)}(\lambda ) \Biggr)x^{j}. \end{aligned}

### Proof

By (2.1), we have $$(x\mid\lambda)_{n}=\sum_{m=0}^{n}S_{1}(n,m)\lambda^{n-m}x^{m}\sim(1,\frac {1}{\lambda}(e^{\lambda t}-1))$$, and by (1.9), we have
\begin{aligned} G_{k}(t)P\beta _{n}^{(k)}(\lambda,x)\sim\biggl(1,\frac{1}{\lambda} \bigl(e^{\lambda t}-1\bigr)\biggr), \end{aligned}
(2.3)
which implies $$G_{k}(t)P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n}S_{1}(n,m)\lambda^{n-m}x^{m}$$. Thus,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x)&=\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m}\frac {Li_{k}(1-e^{-\frac{1}{\lambda}(e^{\lambda t}-1)})}{e^{t}-1}x^{m} \\ &=\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m} \frac {Li_{k}(1-e^{-v})}{(1+\lambda v)^{1/\lambda}-1}\Big| _{v=\frac{1}{\lambda }(e^{\lambda t}-1)}x^{m} \\ &=\sum_{m=0}^{n}\sum _{\ell\geq0}S_{1}(n,m)\lambda^{n-m}P \beta_{\ell }^{(k)}(\lambda)\frac{ (\frac{1}{\lambda}(e^{\lambda t}-1) )^{\ell}}{\ell!}x^{m} \\ &=\sum_{m=0}^{n}\sum _{\ell=0}^{m}S_{1}(n,m)\lambda^{n-m-\ell}P \beta_{\ell }^{(k)}(\lambda)\sum_{j\geq\ell}S_{2}(j, \ell)\frac{\lambda ^{j}t^{j}}{j!}x^{m} \\ &=\sum_{m=0}^{n}\sum _{\ell=0}^{m}\sum_{j=\ell}^{m} \binom {m}{j}S_{1}(n,m)S_{2}(j,\ell)\lambda^{n-m-\ell+j}P \beta_{\ell}^{(k)}(\lambda )x^{m-j} \\ &=\sum_{m=0}^{n}\sum _{\ell=0}^{m}\sum_{j=0}^{m-\ell} \binom {m}{j}S_{1}(n,m)S_{2}(m-j,\ell) \lambda^{n-\ell-j}P\beta_{\ell}^{(k)}(\lambda )x^{j} \\ &=\sum_{j=0}^{n} \Biggl(\sum _{m=j}^{n}\sum_{\ell=0}^{m-j} \binom {m}{j}S_{1}(n,m)S_{2}(m-j,\ell) \lambda^{n-\ell-j}P\beta_{\ell}^{(k)}(\lambda ) \Biggr)x^{j}, \end{aligned}
(2.4)
which completes the proof. □

### Theorem 2.3

For all $$n\geq1$$,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x)&=\sum_{j=0}^{n} \Biggl(\sum_{\ell=0}^{n-j}\sum _{m=0}^{n-j-\ell}\binom{n-1}{\ell}\binom{n-\ell}{j} \lambda^{n-m-j} S_{2}(n-j-\ell,m)B_{\ell}^{(n)}P \beta_{m}^{(k)}(\lambda) \Biggr)x^{j}. \end{aligned}

### Proof

Note that $$x^{n}\sim(1,t)$$. Thus, by (2.3) and transfer formula, we have
\begin{aligned} G_{k}(t)P\beta _{n}^{(k)}(\lambda,x)&=x \biggl(\frac{\lambda t}{e^{\lambda t}-1} \biggr)^{n}x^{-1}x^{n}=x \biggl(\frac{\lambda t}{e^{\lambda t}-1} \biggr)^{n}x^{n-1} \\ &=x\sum_{\ell\geq0}B_{\ell}^{(n)} \frac{\lambda^{\ell}t^{\ell}}{\ell!}x^{n-1}= x\sum_{\ell=0}^{n-1} \binom{n-1}{\ell}\lambda^{\ell}B_{\ell}^{(n)}x^{n-1-\ell} \\ &=\sum_{\ell=0}^{n-1}\binom{n-1}{\ell} \lambda^{\ell}B_{\ell}^{(n)}x^{n-\ell}. \end{aligned}
Therefore, $$P\beta _{n}^{(k)}(\lambda,x)=\sum_{\ell=0}^{n-1}\binom{n-1}{\ell}\lambda ^{\ell}B_{\ell}^{(n)}G_{k}(t)^{-1}x^{n-\ell}$$, which, by (2.4), completes the proof. □

### Theorem 2.4

For all $$n\geq0$$,
\begin{aligned} &P\beta _{n}^{(k)}(\lambda,x)=\sum_{\ell=0}^{n} \Biggl(\sum_{m=0}^{\ell }(-1)^{m+\ell} \binom{n}{\ell}\frac{(m+1)!}{(m+1)^{k}(\ell+1)}S_{2}(\ell +1,m+1) \Biggr) \beta_{n-\ell}(\lambda,x). \end{aligned}

### Proof

By (2.3), we have
\begin{aligned} P\beta _{n}^{(k)}(\lambda,y) &= \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda }-1}(1+\lambda t)^{y/\lambda}\Bigm| x^{n} \biggr\rangle \\ &= \biggl\langle \frac{Li_{k}(1-e^{-t})}{t}\Bigm|\frac{t}{(1+\lambda t)^{1/\lambda}-1}(1+\lambda t)^{y/\lambda}x^{n} \biggr\rangle \\ &= \biggl\langle \frac{Li_{k}(1-e^{-t})}{t}\Bigm|\sum_{\ell\geq0} \beta_{\ell}(\lambda,y)\frac{t^{\ell}}{\ell!}x^{n} \biggr\rangle \\ &=\sum_{\ell=0}^{n}\binom{n}{\ell} \beta_{\ell}(\lambda,y) \biggl\langle \frac {1}{t}\sum _{m\geq1}\frac{(1-e^{-t})^{m}}{m^{k}}\Bigm| x^{n-\ell} \biggr\rangle \\ &=\sum_{\ell=0}^{n}\sum _{m=1}^{n-\ell+1}\binom{n}{\ell}\beta_{\ell}( \lambda ,y) \biggl\langle \frac{(-1)^{m}(e^{-t}-1)^{m}}{m^{k}t}\Bigm| x^{n-\ell} \biggr\rangle . \end{aligned}
(2.5)
Thus, by (2.2), we obtain
\begin{aligned} P\beta _{n}^{(k)}(\lambda,y) &=\sum_{\ell=0}^{n}\sum _{m=0}^{n-\ell}\binom{n}{\ell}\beta_{\ell}( \lambda ,y) \Biggl\langle \frac{(-1)^{m+1}(m+1)!}{(m+1)^{k}}\sum_{j=m+1}^{n-\ell +1}S_{2}(j,m+1) \frac{(-1)^{j}}{j!}t^{j-1}\Bigm| x^{n-\ell} \Biggr\rangle \\ &=\sum_{\ell=0}^{n}\sum _{m=0}^{n-\ell}\binom{n}{\ell}\beta_{\ell}( \lambda ,y)\frac{(-1)^{m+1}(m+1)!}{(m+1)^{k}}S_{2}(n-\ell+1,m+1)\frac{(-1)^{n-\ell +1}(n-\ell)!}{(n-\ell+1)!} \\ &=\sum_{\ell=0}^{n}\sum _{m=0}^{n-\ell}(-1)^{n+m-\ell}\binom{n}{\ell} \frac {(m+1)!}{(m+1)^{k}(n-\ell+1)}S_{2}(n-\ell+1,m+1)\beta_{\ell}(\lambda,y), \end{aligned}
which completes the proof. □

Note that the above theorem has been obtained in Theorem 2.2 in .

### Theorem 2.5

For all $$n\geq0$$,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x)=\frac{1}{n+1}\sum_{\ell=0}^{n} \sum_{m=0}^{\ell }\binom{n+1}{n-\ell,m,\ell-m+1}P \beta_{m}^{(k)}\beta_{n-\ell}(\lambda,x), \end{aligned}
where $$\binom{a}{b_{1},b_{2},b_{3}}=\frac{a!}{b_{1}!b_{2}!b_{3}!}$$ is the multinomial coefficient.

### Proof

By (2.5), we have
\begin{aligned} P\beta _{n}^{(k)}(\lambda,y) &=\sum_{\ell=0}^{n} \binom{n}{\ell}\beta_{\ell}(\lambda,y) \biggl\langle \frac {e^{t}-1}{t} \Bigm|\frac{Li_{k}(1-e^{-t})}{e^{t}-1}x^{n-\ell} \biggr\rangle \\ &=\sum_{\ell=0}^{n}\binom{n}{\ell} \beta_{\ell}(\lambda,y) \biggl\langle \frac {e^{t}-1}{t}\Bigm|\sum _{m\geq0}P\beta_{m}^{(k)}\frac{t^{m}}{m!}x^{n-\ell} \biggr\rangle \\ &=\sum_{\ell=0}^{n}\sum _{m=0}^{n-\ell}\binom{n}{\ell}\binom{n-\ell }{m} \beta_{\ell}(\lambda,y)P\beta_{m}^{(k)} \biggl\langle \frac{e^{t}-1}{t}\Bigm| x^{n-\ell-m} \biggr\rangle . \end{aligned}
Note that $$\langle\frac{e^{t}-1}{t}\mid x^{n-\ell-m} \rangle =\int_{0}^{1} u^{n-\ell-m}\,du=\frac{1}{n-\ell-m+1}$$. Thus,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,y) &=\sum_{\ell=0}^{n} \sum_{m=0}^{n-\ell}\frac{1}{n-\ell-m+1} \binom{n}{\ell }\binom{n-\ell}{m}P\beta_{m}^{(k)} \beta_{\ell}(\lambda,y) \\ &=\sum_{\ell=0}^{n}\sum _{m=0}^{\ell}\frac{1}{\ell-m+1}\binom{n}{\ell } \binom{\ell}{m}P\beta_{m}^{(k)}\beta_{n-\ell}( \lambda,y) \\ &=\frac{1}{n+1}\sum_{\ell=0}^{n}\sum _{m=0}^{\ell}\binom{n+1}{n-\ell,m,\ell -m+1}P \beta_{m}^{(k)}\beta_{n-\ell}(\lambda,y), \end{aligned}
which completes the proof. □
Note that $$Li_{2}(1-e^{-t})=\int_{0}^{t}\frac{y}{e^{y}-1}\,dy=\sum_{j\geq0}B_{j}\frac {1}{j!}\int_{0}^{t}y^{j}\,dy=\sum_{j\geq0}\frac{B_{j} t^{j+1}}{j!(j+1)}$$. For general $$k\geq2$$, the function $$Li_{k}(1-e^{-t})$$ has the integral representation
$$Li_{k}\bigl(1-e^{-t}\bigr)=\int_{0}^{t} \underbrace{\frac{1}{e^{y}-1}\int_{0}^{y} \frac {1}{e^{y}-1}\int_{0}^{y}\cdots \frac{1}{e^{y}-1}\int_{0}^{y}}_{(k-2)\ \mathrm{times}} \frac{y}{e^{y}-1}\,dy\cdots \,dy\,dy\,dy,$$
which, by induction on k, implies
\begin{aligned} Li_{k}\bigl(1-e^{-t}\bigr)=\sum _{j_{1}\geq0}\cdots\sum_{j_{k-1}\geq0}t^{j_{1}+\cdots +j_{k-1}+1} \prod_{i=1}^{k-1}\frac{B_{j_{i}}}{j_{i}!(j_{1}+\cdots +j_{i}+1)}. \end{aligned}
(2.6)

### Theorem 2.6

For all $$n\geq0$$ and $$k\geq2$$,
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x)= \sum_{\ell=0}^{n}(n)_{\ell}\beta_{n-\ell}(\lambda,x) \Biggl(\sum_{j_{1}+\cdots +j_{k-1}=\ell} \prod_{i=1}^{k-1}\frac{B_{j_{i}}}{j_{i}!(j_{1}+\cdots +j_{i}+1)} \Biggr). \end{aligned}

### Proof

By (2.5), we have
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x) &=\sum_{\ell=0}^{n} \binom{n}{\ell}\beta_{\ell}(\lambda,x) \biggl\langle \frac {Li_{k}(1-e^{-t})}{t} \Bigm| x^{n-\ell} \biggr\rangle . \end{aligned}
Thus, by (2.6), we obtain
\begin{aligned} P\beta _{n}^{(k)}(\lambda,x)= \sum_{\ell=0}^{n} \frac{n!}{\ell!}\beta_{\ell}(\lambda,x) \Biggl(\sum _{j_{1}+\cdots+j_{k-1}=n-\ell}\prod_{i=1}^{k-1} \frac {B_{j_{i}}}{j_{i}!(j_{1}+\cdots+j_{i}+1)} \Biggr), \end{aligned}
which completes the proof. □
Note that here we compute $$A= \langle Li_{k}(1-e^{-t})\mid x^{n+1} \rangle$$ in several different ways. As for the first way, we have
\begin{aligned} A&= \biggl\langle \int_{0}^{t} \frac{d}{ds} Li_{k}\bigl(1-e^{-s}\bigr)\,ds\Bigm| x^{n+1} \biggr\rangle = \biggl\langle \int_{0}^{t} \frac{e^{-s} Li_{k-1}(1-e^{-s})}{1-e^{-s}}\,ds\Bigm| x^{n+1} \biggr\rangle \\ &= \biggl\langle \int_{0}^{t} \frac{Li_{k-1}(1-e^{-s})}{e^{s}-1}\,ds\Bigm| x^{n+1} \biggr\rangle =\sum_{m\geq0} \frac{PB_{m}^{(k-1)}}{m!} \biggl\langle \int_{0}^{t} s^{m} \,ds\Bigm| x^{n+1} \biggr\rangle \\ &=\sum_{m\geq0}\frac{PB_{m}^{(k-1)}}{(m+1)!} \bigl\langle t^{m+1} \mid x^{n+1} \bigr\rangle =PB_{n}^{(k-1)}. \end{aligned}
As for the second way, we have
\begin{aligned} A&= \biggl\langle \frac{(e^{t}-1)Li_{k}(1-e^{-t})}{e^{t}-1}\Bigm| x^{n+1} \biggr\rangle = \biggl\langle \frac{Li_{k}(1-e^{-t})}{e^{t}-1}\Bigm| \bigl(e^{t}-1\bigr)x^{n+1} \biggr\rangle \\ &= \biggl\langle \frac{Li_{k}(1-e^{-t})}{e^{t}-1}\Bigm| (x+1)^{n+1}-x^{n+1} \biggr\rangle =\sum_{m=0}^{n} \binom{n+1}{m} \biggl\langle \frac{Li_{k}(1-e^{-t})}{e^{t}-1}\Bigm| x^{m} \biggr\rangle \\ &=\sum_{m=0}^{n}\binom{n+1}{m}PB_{m}^{(k)}. \end{aligned}
As for the third way, by (2.6), we have
\begin{aligned} A&=(n+1)!\sum_{j_{1}+\cdots+j_{k-1}=n}\prod _{i=1}^{k-1}\frac {B_{j_{i}}}{j_{i}!(j_{1}+\cdots+j_{i}+1)}. \end{aligned}
Hence, we can state the following result.

### Theorem 2.7

For all $$n\geq0$$,
\begin{aligned} PB_{n}^{(k-1)}&=\sum_{m=0}^{n} \binom{n+1}{m}PB_{m}^{(k)}=(n+1)!\sum _{j_{1}+\cdots+j_{k-1}=n}\prod_{i=1}^{k-1} \frac{B_{j_{i}}}{j_{i}!(j_{1}+\cdots+j_{i}+1)}. \end{aligned}

## 3 Recurrences

In this section, we present several recurrences for the degenerate poly-Bernoulli polynomials, namely $$P\beta _{n}^{(k)}(\lambda,x)$$. Note that, by (1.9) and the fact that $$(x\mid\lambda)_{n}\sim (1,\frac{e^{\lambda t}-1}{\lambda})$$, we obtain the following identity.

### Proposition 3.1

For all $$n\geq0$$, $$P\beta _{n}^{(k)}(\lambda,x+y)=\sum_{j=0}^{n}\binom{n}{j}P\beta _{j}^{(k)}(\lambda ,x)(y\mid\lambda)_{n-j}$$.

It is well known that if $$s_{n}(x)\sim(g(t),f(t))$$, then we have $$f(t)s_{n}(x)=ns_{n-1}(x)$$. Thus, by (1.9), we obtain $$\frac{e^{\lambda t}-1}{\lambda} P\beta _{n}^{(k)}(\lambda,x)=nP\beta _{n-1}^{(k)}(\lambda,x)$$, which implies the following result.

### Proposition 3.2

For all $$n\geq0$$, $$P\beta _{n}^{(k)}(\lambda,x+\lambda)=P\beta _{n}^{(k)}(\lambda,x)+n\lambda P\beta _{n-1}^{(k)}(\lambda,x)$$.

### Theorem 3.3

For all $$n\geq0$$,
\begin{aligned} &P\beta _{n+1}^{(k)}(\lambda,x)-xP\beta _{n}^{(k)}(\lambda,x-\lambda) \\ &\quad=\sum_{i=0}^{m+1}\sum _{\ell=0}^{m+1-i}\binom{m+1}{i}\lambda^{m+1-i-\ell }S_{2}(m+1-i, \ell) \bigl(PB_{\ell}^{(k)}B_{i}(x)-P \beta_{\ell}^{(k)}(\lambda )B_{i}(x+1-\lambda) \bigr). \end{aligned}

### Proof

By applying the fact that $$s_{n+1}(x)=(x-\frac{g'(t)}{g(t)})\frac {1}{f'(t)} s_{n}(x)$$ for all $$s_{n}(x)\sim(g(t),f(t))$$ and (1.9), we obtain
$$P\beta _{n+1}^{(k)}(\lambda,x)=\biggl(x-\frac{g'(t)}{g(t)}\biggr)e^{-\lambda t} P\beta _{n}^{(k)}(\lambda,x) =xP\beta _{n}^{(k)}(\lambda,x-\lambda)-e^{-\lambda t} \frac{g'(t)}{g(t)}P\beta _{n}^{(k)}(\lambda,x),$$
where
\begin{aligned} \frac{g'(t)}{g(t)} &=\bigl(\log\bigl(e^{t}-1\bigr)-\log Li_{k}\bigl(1-e^{-\frac{1}{\lambda}(e^{\lambda t}-1)}\bigr)\bigr)' \\ &=\frac{e^{t}}{e^{t}-1}-\frac{1}{Li_{k}(1-e^{-\frac{1}{\lambda}(e^{\lambda t}-1)})}\frac{Li_{k-1}(1-e^{-\frac{1}{\lambda}(e^{\lambda t}-1)})}{1-e^{-\frac{1}{\lambda}(e^{\lambda t}-1)}}e^{\lambda t}e^{-\frac{1}{\lambda}(e^{\lambda t}-1)}. \end{aligned}
Thus, the expression $$A=e^{-\lambda t}\frac{g'(t)}{g(t)}P\beta _{n}^{(k)}(\lambda,x)$$ is given by
\begin{aligned} \frac{1}{t} \biggl(\frac{te^{(1-\lambda)t}}{e^{t}-1}G_{k}(t)^{-1} - \frac {t}{e^{\frac{1}{\lambda}(e^{\lambda t}-1)}-1}G_{k-1}(t)^{-1} \biggr)G_{k}(t) P\beta _{n}^{(k)}(\lambda,x). \end{aligned}
Note that, by (1.9), we have $$G_{k}(x)P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n}S_{1}(n,m)\lambda^{n-m}x^{m}$$. Therefore,
\begin{aligned} A&=\sum_{m=0}^{n}S_{1}(n,m) \lambda^{n-m}\frac{1}{t} \biggl(\frac {te^{(1-\lambda)t}}{e^{t}-1}G_{k}(t)^{-1}- \frac{t}{e^{\frac{1}{\lambda }(e^{\lambda t}-1)}-1}G_{k-1}(t)^{-1} \biggr)x^{m} \\ &=\sum_{m=0}^{n}\frac{S_{1}(n,m)}{m+1} \lambda^{n-m} \biggl(\frac {te^{(1-\lambda)t}}{e^{t}-1}G_{k}(t)^{-1} -\frac{t}{e^{\frac{1}{\lambda }(e^{\lambda t}-1)}-1}G_{k-1}(t)^{-1} \biggr)x^{m+1}. \end{aligned}
(3.1)
We remark that the expression in the parentheses in (3.1) has order at least one. Now, let us simplify (3.1):
\begin{aligned} &\frac{te^{(1-\lambda)t}}{e^{t}-1}G_{k}(t)^{-1}x^{m+1} \\ &\quad=\frac{te^{(1-\lambda)t}}{e^{t}-1}\frac{Li_{k}(1-e^{-s})}{(1+\lambda s)^{1/\lambda}-1}\Big|_{s=\frac{e^{\lambda t}-1}{\lambda }}x^{m+1} \\ &\quad=\frac{te^{(1-\lambda)t}}{e^{t}-1}\sum_{\ell=0}^{m+1}P \beta_{\ell }^{(k)}(\lambda)\frac{ (\frac{e^{\lambda t}-1}{\lambda} )^{\ell}}{\ell!}x^{m+1} \\ &\quad=\frac{te^{(1-\lambda)t}}{e^{t}-1}\sum_{\ell=0}^{m+1}\sum _{i=\ell }^{m+1}\binom{m+1}{i} \lambda^{i-\ell}S_{2}(i,\ell)P\beta_{\ell }^{(k)}( \lambda)x^{m+1-i} \\ &\quad=\sum_{i=0}^{m+1}\sum _{\ell=0}^{m+1-i}\binom{m+1}{i}\lambda^{m+1-i-\ell }S_{2}(m+1-i, \ell)P\beta_{\ell}^{(k)}(\lambda)\frac{te^{(1-\lambda )t}}{e^{t}-1}x^{i} \\ &\quad=\sum_{i=0}^{m+1}\sum _{\ell=0}^{m+1-i}\binom{m+1}{i}\lambda^{m+1-i-\ell }S_{2}(m+1-i, \ell)P\beta_{\ell}^{(k)}(\lambda)B_{i}(x+1- \lambda) \end{aligned}
(3.2)
and
\begin{aligned} &\frac{t}{e^{\frac{1}{\lambda}(e^{\lambda t}-1)}-1}G_{k-1}(t)^{-1}x^{m+1} \\ &\quad=\frac{t}{e^{t}-1}\frac{Li_{k-1}(1-e^{-s})}{e^{s}-1}\Big|_{s=\frac {e^{\lambda t}-1}{\lambda}}x^{m+1} \\ &\quad= \frac{t}{e^{t}-1}\sum_{\ell =0}^{m+1}PB_{\ell}^{(k)} \frac{ (\frac{e^{\lambda t}-1}{\lambda} )^{\ell}}{\ell!}x^{m+1} \\ &\quad=\frac{t}{e^{t}-1}\sum_{\ell=0}^{m+1}\sum _{i=\ell}^{m+1}\binom {m+1}{i} \lambda^{i-\ell}S_{2}(i,\ell)PB_{\ell}^{(k)}x^{m+1-i} \\ &\quad=\sum_{i=0}^{m+1}\sum _{\ell=0}^{m+1-i}\binom{m+1}{i}\lambda^{m+1-i-\ell }S_{2}(m+1-i, \ell)PB_{\ell}^{(k)}\frac{t}{e^{t}-1}x^{i} \\ &\quad=\sum_{i=0}^{m+1}\sum _{\ell=0}^{m+1-i}\binom{m+1}{i}\lambda^{m+1-i-\ell }S_{2}(m+1-i, \ell)PB_{\ell}^{(k)}B_{i}(x). \end{aligned}
(3.3)
Hence, by (3.1)-(3.3), we complete the proof. □

In the next result we express $$\frac{d}{dx}P\beta _{n}^{(k)}(\lambda,x)$$ in terms of $$P\beta _{n}^{(k)}(\lambda,x)$$.

### Proposition 3.4

For all $$n\geq0$$, $$\frac{d}{dx}P\beta _{n}^{(k)}(\lambda,x)= n!\sum_{\ell=0}^{n-1}\frac{(-\lambda)^{n-\ell-1}}{\ell!(n-\ell)} P\beta _{\ell}^{(k)}(\lambda,x)$$.

### Proof

Note that $$\frac{d}{dx}s_{n}(x)=\sum_{\ell=0}^{n-1}\binom{n}{\ell}\langle \bar{f}(t)\mid x^{n-\ell}\rangle s_{\ell}(x)$$ for all $$s_{n}(x)\sim (g(t),f(t))$$. Thus, by (1.9), we have
\begin{aligned} \bigl\langle \bar{f}(t)\mid x^{n-\ell}\bigr\rangle &=\biggl\langle \frac{1}{\lambda}\log (1+\lambda t)\Bigm| x^{n-\ell}\biggr\rangle = \frac{1}{\lambda}\sum_{m\geq 1}(-1)^{m-1} \lambda^{m}(m-1)!\biggl\langle \frac{x^{m}}{m!}\Bigm|x^{n-\ell}\biggr\rangle \\ &=(-\lambda)^{n-\ell-1}(n-\ell-1)!, \end{aligned}
which completes the proof. □

### Theorem 3.5

For all $$n\geq1$$,
\begin{aligned} &P\beta _{n}^{(k)}(\lambda,x)-xP\beta _{n-1}^{(k)}(\lambda,x-\lambda) \\ & \quad=\frac{1}{n}\sum_{m=0}^{n} \binom{n}{m} \bigl(P\beta _{m}^{(k-1)}(\lambda ,x)B_{n-m}-P\beta _{m}^{(k)}(\lambda,x+1-\lambda)\beta_{n-m}(\lambda) \bigr). \end{aligned}

### Proof

By (1.9), we have
\begin{aligned} P\beta _{n}^{(k)}(\lambda,y) ={}& \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda }-1}(1+\lambda t)^{y/\lambda}\Bigm| x^{n} \biggr\rangle \\ ={}& \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}\frac {d}{dt}(1+\lambda t)^{y/\lambda}\Bigm| x^{n-1} \biggr\rangle \end{aligned}
(3.4)
\begin{aligned} &{}+ \biggl\langle \frac{d}{dt}\frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}(1+\lambda t)^{y/\lambda}\Bigm| x^{n-1} \biggr\rangle . \end{aligned}
(3.5)
The term in (3.4) is given by
\begin{aligned} y \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda }-1}(1+\lambda t)^{(y-\lambda)/\lambda}\Bigm| x^{n-1} \biggr\rangle =yP\beta _{n-1}^{(k)}(\lambda,y-\lambda). \end{aligned}
(3.6)
For the term in (3.5), we observe that $$\frac{d}{dt}\frac {Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}=\frac{1}{t}(A-B)$$, where
\begin{aligned} A=\frac{t}{e^{t}-1}\frac{Li_{k-1}(1-e^{-t})}{(1+\lambda t)^{1/\lambda }-1},\qquad B=\frac{t}{(1+\lambda t)^{1/\lambda}-1} \frac {Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}(1+\lambda t)^{1/\lambda-1}. \end{aligned}
Note that the expression $$A-B$$ has order of at least 1. Now, we are ready to compute the term in (3.5). By (1.9), we have
\begin{aligned} & \biggl\langle \frac{d}{dt}\frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda }-1}(1+\lambda t)^{y/\lambda} \Bigm| x^{n-1} \biggr\rangle \\ &\quad= \biggl\langle \frac{1}{t}(A-B) (1+ \lambda t)^{y/\lambda}\Bigm| x^{n-1} \biggr\rangle \\ &\quad=\frac{1}{n} \bigl\langle A(1+\lambda t)^{y/\lambda}\mid x^{n} \bigr\rangle -\frac{1}{n} \bigl\langle B(1+\lambda t)^{y/\lambda}\mid x^{n} \bigr\rangle \\ &\quad=\frac{1}{n} \biggl\langle \frac{t}{e^{t}-1}\Bigm|\sum _{m\geq0}P\beta _{m}^{(k-1)}(\lambda,y)\frac{t^{m}}{m!}x^{n} \biggr\rangle \\ &\qquad{} -\frac{1}{n} \biggl\langle \frac{t}{(1+\lambda t)^{1/\lambda}-1}\Bigm|\sum _{m\geq0}P\beta _{m}^{(k)}(\lambda,y+1-\lambda)\frac {t^{m}}{m!}x^{n} \biggr\rangle \\ &\quad=\frac{1}{n}\sum_{m=0}^{n} \binom{n}{m}P\beta _{m}^{(k-1)}(\lambda,y) \biggl\langle \frac{t}{e^{t}-1}\Bigm| x^{n-m} \biggr\rangle \\ &\qquad{} -\frac{1}{n}\sum_{m=0}^{n} \binom{n}{m}P\beta _{m}^{(k)}(\lambda,y+1-\lambda) \biggl\langle \frac{t}{(1+\lambda t)^{1/\lambda}-1}\Bigm| x^{n-m} \biggr\rangle \\ &\quad=\frac{1}{n}\sum_{m=0}^{n} \binom{n}{m} \bigl(P\beta _{m}^{(k-1)}(\lambda ,y)B_{n-m}-P\beta _{m}^{(k)}(\lambda,y+1-\lambda)\beta_{n-m}(\lambda) \bigr). \end{aligned}
(3.7)
Thus, if we replace (3.4) by (3.6) and (3.5) by (3.7), we obtain
\begin{aligned} &P\beta _{n}^{(k)}(\lambda,x)-xP\beta _{n-1}^{(k)}(\lambda,x-\lambda) \\ & \quad=\frac{1}{n}\sum_{m=0}^{n} \binom{n}{m} \bigl(P\beta _{m}^{(k-1)}(\lambda ,x)B_{n-m}-P\beta _{m}^{(k)}(\lambda,x+1-\lambda)\beta_{n-m}(\lambda) \bigr), \end{aligned}
as claimed. □

## 4 Connections with families of polynomials

In this section, we present a few examples on the connections with families of polynomials. We start with the connection to Bernoulli polynomials $$B_{n}^{(s)}(x)$$ of order s. Recall that the Bernoulli polynomials $$B_{n}^{(s)}(x)$$ of order s are defined by the generating function $$(\frac{t}{e^{t}-1} )^{s} e^{xt}=\sum_{n\geq0}B_{n}^{(s)}(x)\frac{t^{n}}{n!}$$, equivalently,
\begin{aligned} B_{n}^{(s)}(x)\sim \biggl( \biggl(\frac{e^{t}-1}{t} \biggr)^{s},t \biggr) \end{aligned}
(4.1)
(see ). In the next result, we express our polynomials $$P\beta _{n}^{(k)}(\lambda,x)$$ in terms of Bernoulli polynomials of order s. To do that, we recall that the Bernoulli numbers $$b_{n}^{(s)}$$ of the second kind of order s are defined as
\begin{aligned} \frac{t^{s}}{\log^{s}(1+t)}=\sum_{n\geq0}b_{n}^{(s)} \frac{t^{n}}{n!}. \end{aligned}
(4.2)

### Theorem 4.1

For all $$n\geq0$$,
$$P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n} \Biggl(\sum_{\ell=m}^{n}\sum _{r=0}^{n-\ell}\sum_{j=0}^{n-\ell-r} \sum_{i=0}^{j}\frac{\binom{n}{\ell ,r,j,n-\ell-r-j}}{\binom{j+s}{j}} \lambda^{\ell+r+i-m}c_{n,m}(\ell,r,j,i) \Biggr)B_{m}^{(s)}(x),$$
where $$c_{n,m}(\ell,r,j,i)=S_{1}(\ell ,m)S_{1}(j+s,j-i+s)S_{2}(j-i+s,s)b_{r}^{(s)}P\beta_{n-\ell-r-j}^{(k)}(\lambda)$$ and $$\binom{a}{b_{1},\ldots,b_{m}}=\frac{a!}{b_{1}!\cdots b_{m}!}$$ is the multinomial coefficient.

### Proof

Let $$h_{s}(t)= (\frac{(1+\lambda t)^{1/\lambda}-1}{t} )^{s}$$ and $$P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n}c_{n,m}B_{m}^{(s)}(x)$$. By (1.8), (1.9), and (4.1), we have
\begin{aligned} &m!\lambda^{m}c_{n,m}\\ &\quad= \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1} \biggl(\frac{(1+\lambda t)^{1/\lambda}-1}{t} \biggr)^{s} \biggl(\frac{\lambda t}{\log(1+\lambda t)} \biggr)^{s}\Bigm|\bigl( \log(1+\lambda t)\bigr)^{m}x^{n} \biggr\rangle , \end{aligned}
which, by (4.2), implies
\begin{aligned} \lambda^{m}c_{n,m}&=\sum_{\ell=m}^{n} \binom{n}{\ell}\lambda^{\ell}S_{1}(\ell ,m) \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda }-1}h_{s}(t) \biggl(\frac{\lambda t}{\log(1+\lambda t)} \biggr)^{s} \Bigm| x^{n-\ell} \biggr\rangle \\ &=\sum_{\ell=m}^{n}\binom{n}{\ell} \lambda^{\ell}S_{1}(\ell,m) \biggl\langle \frac{Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}h_{s}(t) \Bigm| \biggl(\frac{\lambda t}{\log(1+\lambda t)} \biggr)^{s} x^{n-\ell} \biggr\rangle \\ &=\sum_{\ell=m}^{n}\sum _{r=0}^{n-\ell}\binom{n}{\ell}\binom{n-\ell }{r} \lambda^{\ell+r} S_{1}(\ell,m)b_{r}^{(s)} \biggl\langle \frac {Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}h_{s}(t) \Bigm| x^{n-\ell-r} \biggr\rangle \\ &=\sum_{\ell=m}^{n}\sum _{r=0}^{n-\ell}\binom{n}{\ell}\binom{n-\ell }{r} \lambda^{\ell+r} S_{1}(\ell,m)b_{r}^{(s)} \biggl\langle \frac {Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}\Bigm| h_{s}(t) x^{n-\ell-r} \biggr\rangle . \end{aligned}
One can show that
\begin{aligned} h_{s}(t)&= \biggl(\frac{e^{\frac{1}{\lambda}\log(1+\lambda t)}-1}{t} \biggr)^{s} \\ &=s!\sum_{j\geq0}\sum_{i=0}^{j}S_{1}(j+s,j-i+s)S_{2}(j-i+s,s) \frac{\lambda ^{i}}{(j+s)!}t^{j}. \end{aligned}
Thus, by (1.9), we have
\begin{aligned} c_{n,m}={}&\sum_{\ell=m}^{n}\sum _{r=0}^{n-\ell}\sum _{j=0}^{n-\ell-r}\sum_{i=0}^{j} \biggl(s!\binom{n}{\ell}\binom{n-\ell}{r}\lambda^{\ell +r-m}S_{1}( \ell,m)b_{r}^{(s)}S_{1}(j+s,j-i+s) \\ &{}\times S_{2}(j-i+s,s)\frac{\lambda^{i}}{(j+s)!}(n-\ell-r)_{j} \biggl\langle \frac {Li_{k}(1-e^{-t})}{(1+\lambda t)^{1/\lambda}-1}\Bigm| x^{n-\ell-r-j} \biggr\rangle \biggr) \\ ={}&\sum_{\ell=m}^{n}\sum _{r=0}^{n-\ell}\sum_{j=0}^{n-\ell-r} \sum_{i=0}^{j} \biggl(\frac{\binom{n}{\ell,r,j,n-\ell-r-j}}{\binom{j+s}{j}} \lambda^{\ell+r+i-m}S_{1}(\ell,m)S_{1}(j+s,j-i+s) \\ &{}\times S_{2}(j-i+s,s)b_{r}^{(s)}P\beta_{n-\ell -r-j}^{(k)}( \lambda) \biggr), \end{aligned}
as required. □

Similar techniques as in the proof of the previous theorem, we can express our polynomials $$P\beta _{n}^{(k)}(\lambda,x)$$ in terms of other families. Below we present three examples, where we leave the proofs to the interested reader.

The first example is to express our polynomials $$P\beta _{n}^{(k)}(\lambda ,x)$$ in terms of Frobenius-Euler polynomials. Note that the Frobenius-Euler polynomials $$H_{n}^{(s)}(x\mid\mu)$$ of order s are defined by the generating function $$(\frac{1-\mu}{e^{t}-\mu} )^{s} e^{xt}=\sum_{n\geq 0}H_{n}^{(s)}(x\mid\mu)\frac{t^{n}}{n!}$$ ($$\mu\neq1$$), or equivalently, $$H_{n}^{(s)}(x\mid\mu)\sim ( (\frac{e^{t}-\mu}{1-\mu} )^{s},t )$$ (see [10, 14]).

### Theorem 4.2

For all $$n\geq0$$,
$$P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n} \Biggl(\sum_{\ell=m}^{n}\sum _{r=0}^{n-\ell}\sum_{i=0}^{s} \binom{n}{\ell}\binom{n-\ell}{r}\binom{s}{i} \frac{\lambda^{\ell-m}(-\mu)^{s-i}}{(1-\mu)^{s}}c_{n,m}( \ell,r,i) \Biggr)H_{m}^{(s)}(x\mid\mu),$$
where $$c_{n,m}(\ell,r,i)=S_{1}(\ell,m)(i\mid\lambda)_{n-\ell-r}P\beta _{r}^{(k)}(\lambda)$$.

If we express our polynomials $$P\beta _{n}^{(k)}(\lambda,x)$$ in terms of falling polynomials $$(x\mid\lambda)_{n}$$, then we get the following result.

### Theorem 4.3

For all $$n\geq0$$, $$P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n}\binom{n}{m}P\beta _{n-m}^{(k)}(\lambda)(x\mid\lambda)_{m}$$.

Our last example is to express our polynomials $$P\beta _{n}^{(k)}(\lambda,x)$$ in terms of degenerate Bernoulli polynomials $$\beta _{n}^{(s)}(\lambda,x)$$ of order s. Note that the degenerate Bernoulli polynomials $$\beta_{n}^{(s)}(\lambda,x)$$ of order s are given by
$$\biggl(\frac{t}{(1+\lambda t)^{1/\lambda}-1} \biggr)^{s}(1+\lambda t)^{x/\lambda}=\sum _{n\geq0}\beta_{n}^{(s)}(\lambda,x) \frac{t^{n}}{n!}.$$

### Theorem 4.4

For all $$n\geq0$$,
$$P\beta _{n}^{(k)}(\lambda,x)=\sum_{m=0}^{n} \binom{n}{m} \Biggl(\sum_{j=0}^{n-m} \sum_{i=0}^{j}\frac{\binom{n-m}{j}}{\binom{j+s}{s}} \lambda^{i} c_{n,m}(j,i) \Biggr)\beta_{m}^{(s)}( \lambda,x),$$
where $$c_{n,m}(j,i)=S_{1}(j+s,j-i+s)S_{2}(j-i+s,s)P\beta _{n-m-j}^{(k)}(\lambda)$$.

## Declarations

### Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MOE) (No. 2012R1A1A2003786).

## Authors’ Affiliations

(1)
Department of Mathematics, Sogang University, Seoul, 121-742, S. Korea
(2)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, S. Korea
(3)
Department of Mathematics, University of Haifa, Haifa, 3498838, Israel

## References 