Open Access

On Poly-Bernoulli polynomials of the second kind with umbral calculus viewpoint

  • Dae San Kim1,
  • Taekyun Kim2Email author,
  • Toufik Mansour3 and
  • Dmitry V Dolgy4
Advances in Difference Equations20152015:27

https://doi.org/10.1186/s13662-015-0362-5

Received: 19 November 2014

Accepted: 7 January 2015

Published: 31 January 2015

Abstract

Poly-Bernoulli polynomials of the second kind were introduced in Kim et al. (Adv. Differ. Equ. 2014:219, 2014) as a generalization of the Bernoulli polynomial of the second kind. Here we investigate those polynomials and derive further results about them by using umbral calculus.

Keywords

Bernoulli polynomialspoly-Bernoulli polynomialsStirling numbersumbral calculus

1 Introduction

Following Kaneko [1], the poly-Bernoulli polynomials have been studied by many researchers in recent decades. Poly-Bernoulli polynomials \(B_{n}^{(k)}(x)\) were defined as \(\frac {Li_{k}(1-e^{-t})}{1-e^{-t}}e^{xt}=\sum_{n\geq0}B_{n}^{(k)}(x)\frac {t^{n}}{n!}\), where \(Li_{k}(x)=\sum_{r\geq1} \frac{x^{r}}{r^{k}}\) is the classical polylogarithm function, which satisfies \(\frac{d}{dx}Li_{k}(x)=\frac{1}{x}Li_{k-1}(x)\). The poly-Bernoulli polynomials have wide-ranging applications in mathematics and applied mathematics (see [24]). For \(k\in \mathbb{Z}\), the poly-Bernoulli polynomials \(b_{n}^{(k)}(x)\) of the second kind are given by the generating function
$$\begin{aligned} \frac{Li_{k}(1-e^{-t})}{\log(1+t)}(1+t)^{x}=\sum _{n\geq0}b_{n}^{(k)}(x)\frac{t^{n}}{n!}. \end{aligned}$$
(1.1)
When \(x=0\), \(b_{n}^{(k)}=b_{n}^{(k)}(0)\) are called the poly-Bernoulli numbers of the second kind. When \(k=1\), \(b_{n}(x)=b_{n}^{(1)}(x)\) are called the Bernoulli polynomial of the second kind (see [511]). Poly-Bernoulli polynomials of the second kind were introduced as a generalization of the Bernoulli polynomial of the second kind (see [12]). The aim of this paper is to use umbral calculus to obtain several new and interesting explicit formulas, recurrence relations and identities of poly-Bernoulli polynomials of the second kind. Umbral calculus has been used in numerous problems of mathematics. Umbral techniques have been of use in different areas of physics; for example it is used in group theory and quantum mechanics by Biedenharn et al. (see [1315]).
Let Π be the algebra of polynomials in a single variable x over and let \(\Pi^{*}\) be the vector space of all linear functionals on Π. We denote the action of a linear functional L on a polynomial \(p(x)\) by \(\langle L|p(x)\rangle\). Define the vector space structure on \(\Pi^{*}\) by \(\langle cL+c'L'|p(x)\rangle=c\langle L|p(x)\rangle+c'\langle L'|p(x)\rangle\), where \(c,c'\in\mathbb{C}\) (see [1619]). We define the algebra of a formal power series in a single variable t to be
$$\begin{aligned} \mathcal{H}= \biggl\{ f(t)=\sum_{k\geq0} a_{k} \frac{t^{k}}{k!}\Big| a_{k}\in \mathbb{C} \biggr\} . \end{aligned}$$
(1.2)
The formal power series in the variable t defines a linear functional on Π by setting \(\langle f(t)|x^{n}\rangle=a_{n}\), for all \(n\geq0\) (see [1619]). Thus
$$\begin{aligned} \bigl\langle t^{k}|x^{n}\bigr\rangle =n!\delta_{n,k}, \quad\mbox{for all }n,k\geq0\ (\mbox{see [16--19]}), \end{aligned}$$
(1.3)
where \(\delta_{n,k}\) is the Kronecker symbol. Let \(f_{L}(t)=\sum_{n\geq 0}\langle L|x^{n}\rangle\frac{t^{n}}{n!}\). By (1.3), we have \(\langle f_{L}(t)|x^{n}\rangle=\langle L|x^{n}\rangle\). Thus, the map \(L\mapsto f_{L}(t)\) is a vector space isomorphism from \(\Pi^{*}\) onto . Therefore, is thought of as a set of both formal power series and linear functionals. We call the umbral algebra. The umbral calculus is the study of the umbral algebra.
Let \(f(t)\) be a non-zero power series, the order \(O(f(t))\) is the smallest integer k for which the coefficient of \(t^{k}\) does not vanish. If \(O(f(t))=1\) (respectively, \(O(f(t))=0\)), then \(f(t)\) is called a delta (respectively, an invertable) series. Suppose that \(f(t)\) is a delta series and \(g(t)\) is an invertable series, then there exists a unique sequence \(s_{n}(x)\) of polynomials such that \(\langle g(t)(f(t))^{k}|s_{n}(x)\rangle=n!\delta_{n,k}\), 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 [18, 19]). For \(f(t)\in\mathcal{H}\) and \(p(x)\in\Pi\), we have \(\langle e^{yt}|p(x)\rangle=p(y)\), \(\langle f(t)g(t)|p(x)\rangle =\langle g(t)|f(t)p(x)\rangle\), and \(f(t)=\sum_{n\geq0}\langle f(t)|x^{n}\rangle\frac{t^{n}}{n!}\) and \(p(x)=\sum_{n\geq0}\langle t^{n}|p(x)\rangle\frac{x^{n}}{n!}\) (see [18, 19]). Thus, we obtain \(\langle t^{k}|p(x)\rangle=p^{(k)}(0)\) and \(\langle1|p^{(k)}(x)\rangle =p^{(k)}(0)\), where \(p^{(k)}(0)\) denotes the kth derivative of \(p(x)\) with respect to x at \(x=0\). Therefore, we get \(t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}}{dx^{k}}p(x)\), for all \(k\geq0\) (see [18, 19]). Thus, for \(s_{n}(x)\sim(g(t),f(t))\), we have
$$\begin{aligned} \frac{1}{g(\bar{f}(t))}e^{y\bar{f}(t)}=\sum_{n\geq0}s_{n}(y) \frac{t^{n}}{n!}, \end{aligned}$$
(1.4)
for all \(y\in\mathbb{C}\), where \(\bar{f}(t)\) is the compositional inverse of \(f(t)\) (see [18, 19]). 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
$$\begin{aligned} c_{n,k}=\frac{1}{k!} \biggl\langle \frac{h(\bar{f}(t))}{g(\bar{f}(t))}\bigl(\ell \bigl(\bar{f}(t)\bigr)\bigr)^{k}\Big| x^{n} \biggr\rangle \end{aligned}$$
(1.5)
(see [18, 19]).
It is immediate from (1.1) and (1.4) to see that \(b_{n}^{(k)}(x)\) is the Sheffer polynomial for the pair \(g(x)=\frac {t}{Li_{k}(1-e^{1-e^{t}})}\) and \(f(t)=e^{t}-1\), that is,
$$\begin{aligned} b_{n}^{(k)}(x)\sim \biggl(\frac{t}{Li_{k}(1-e^{1-e^{t}})},e^{t}-1 \biggr). \end{aligned}$$
(1.6)
The aim of the present paper is to present several new identities for the poly-Bernoulli polynomials by the use of umbral calculus.

2 Explicit expressions

Before proceeding, we observe that
$$\begin{aligned} Li_{k}\bigl(1-e^{-t}\bigr)&=\sum _{r\geq1}\frac{1}{r^{k}}\bigl(1-e^{-t} \bigr)^{r}=\sum_{r\geq 1}\frac{(-1)^{r}}{r^{k}} \bigl(e^{-t}-1\bigr)^{r} \\ &=\sum_{r\geq1}\frac{(-1)^{r}r!}{r^{k}}\sum _{\ell\geq r}S_{2}(\ell,r)\frac {(-t)^{\ell}}{\ell!}=\sum _{r\geq1}\sum_{\ell\geq r} \frac{(-1)^{r+\ell }r!}{r^{k}}S_{2}(\ell,r)\frac{t^{\ell}}{\ell!} \\ &=\sum_{\ell\geq1}\sum_{r=1}^{\ell}\frac{(-1)^{r+\ell}r!}{r^{k}}S_{2}(\ell ,r)\frac{t^{\ell}}{\ell!}, \end{aligned}$$
(2.1)
where \(S_{2}(n,k)\) is the Stirling number of the second kind, which is defined by the identity \(x^{n}=\sum_{k=0}^{n}S_{2}(n,k)(x)_{k}\) with \((x)_{0}=1\) and \((x)_{k}=x(x-1)\cdots(x-k+1)\). This shows
$$\begin{aligned} \frac{1}{t}Li_{k}\bigl(1-e^{-t}\bigr) &= \sum_{\ell\geq0}\sum_{r=1}^{\ell+1} \frac{(-1)^{r+\ell+1}r!}{r^{k}}\frac {S_{2}(\ell+1,r)}{\ell+1}\frac{t^{\ell}}{\ell!}. \end{aligned}$$
(2.2)
Thus,
$$\begin{aligned} Li_{k}\bigl(1-e^{1-e^{t}}\bigr)&=\sum _{r\geq1}\sum_{\ell\geq r} \frac{(-1)^{r+\ell }r!}{r^{k}}S_{2}(\ell,r)\frac{(e^{t}-1)^{\ell}}{\ell!}\\ &=\sum _{r\geq1}\sum_{\ell \geq r}\sum _{m\geq\ell}\frac{(-1)^{r+\ell}r!}{r^{k}}S_{2}(\ell,r)S_{2}(m, \ell )\frac{t^{m}}{m!} \\ &=\sum_{m\geq1}\sum_{r=1}^{m} \sum_{\ell=r}^{m}\frac{(-1)^{r+\ell }r!}{r^{k}}S_{2}( \ell,r)S_{2}(m,\ell)\frac{t^{m}}{m!}, \end{aligned}$$
which implies that
$$\begin{aligned} \frac{1}{t}Li_{k}\bigl(1-e^{1-e^{t}}\bigr)&=\sum _{m\geq0}\sum_{r=1}^{m+1}\sum _{\ell =r}^{m+1}\frac{(-1)^{r+\ell}r!}{r^{k}}S_{2}( \ell,r)S_{2}(m+1,\ell)\frac {t^{m}}{(m+1)!}. \end{aligned}$$
(2.3)

Now, we are ready to present several formulas for the nth poly-Bernoulli polynomials of the second kind.

Theorem 2.1

For all \(n\geq1\),
$$\begin{aligned} b_{n}^{(k)}(x) &=\sum_{m=0}^{n} \sum_{j=m}^{n}\sum _{r=1}^{j-m+1}\sum_{\ell=r}^{j-m+1} \frac {(-1)^{r+\ell}}{j-m+1}\frac{r!}{r^{k}}\binom{n-1}{j-1}\binom{j}{m}S_{2}( \ell ,r)S_{2}(j-m+1,\ell)B_{n-j}^{(n)}x^{m}. \end{aligned}$$

Proof

Since \(x^{n}\sim(1,t)\) and \(\frac{t}{Li_{k}(1-e^{1-e^{t}})}b_{n}^{(k)}(x)\sim (1,e^{t}-1)\) (see (1.6)), we obtain
$$\begin{aligned} \frac{t}{Li_{k}(1-e^{1-e^{t}})}b_{n}^{(k)}(x)&=x \biggl( \frac{t}{e^{t}-1} \biggr)^{n}x^{n-1}=x \biggl(\sum _{j\geq0}B_{j}^{(n)}\frac{t^{j}}{j!} \biggr)x^{n-1} \\ &=x\sum_{j=0}^{n-1}\binom{n-1}{j} B_{j}^{(n)}x^{n-1-j}=\sum _{j=1}^{n}\binom{n-1}{j-1}B_{n-j}^{(n)}x^{j}. \end{aligned}$$
Thus, by (2.3) we have
$$\begin{aligned} b_{n}^{(k)}(x)&=\sum_{j=1}^{n} \binom{n-1}{j-1}B_{n-j}^{(n)}\frac {Li_{k}(1-e^{1-e^{t}})}{t}x^{j} \\ &=\sum_{j=1}^{n}\binom{n-1}{j-1} B_{n-j}^{(n)} \Biggl(\sum_{m=0}^{j} \sum_{r=1}^{m+1}\sum _{\ell=r}^{m+1}\frac{(-1)^{r+\ell}r!}{r^{k}}S_{2}(\ell ,r)S_{2}(m+1,\ell)\frac{t^{m}}{(m+1)!} \Biggr)x^{j} \\ &=\sum_{j=1}^{n}\sum _{m=0}^{j}\sum_{r=1}^{m+1} \sum_{\ell=r}^{m+1}\frac {(-1)^{r+\ell}}{m+1} \frac{r!}{r^{k}}\binom{n-1}{j-1}\binom{j}{m}S_{2}(\ell ,r)S_{2}(m+1,\ell)B_{n-j}^{(n)}x^{j-m} \\ &=\sum_{j=1}^{n}\sum _{m=0}^{j}\sum_{r=1}^{j-m+1} \sum_{\ell=r}^{j-m+1}\frac {(-1)^{r+\ell}}{j-m+1} \frac{r!}{r^{k}}\binom{n-1}{j-1}\binom{j}{m}S_{2}(\ell ,r)S_{2}(j-m+1,\ell)B_{n-j}^{(n)}x^{m}, \end{aligned}$$
which completes the proof. □

Let \(S_{1}(n,k)\) be the Stirling number of the first kind, which is defined by the identity \((x)_{n}=\sum_{j=0}^{n}S_{1}(n,k)x^{k}\). Now, we are ready to present our second explicit formula.

Theorem 2.2

For all \(n\geq0\),
$$\begin{aligned} b_{n}^{(k)}(x)&=\sum_{m=0}^{n} \sum_{j=m}^{n}\sum _{r=1}^{j-m+1}\sum_{\ell =r}^{j-m+1} \frac{(-1)^{r+\ell}}{j-m+1}\frac{r!}{r^{k}}\binom {j}{m}S_{1}(n,j)S_{2}( \ell,r)S_{2}(j-m+1,\ell)x^{m}. \end{aligned}$$

Proof

Note that \((x)_{n}=\sum_{j=0}^{n}S_{1}(n,j)x^{j}\sim(1,e^{t}-1)\). So, by (1.6) we have \(\frac{t}{Li_{k}(1-e^{1-e^{t}})}b_{n}^{(k)}(x)\sim(1,e^{t}-1)\), which implies that
$$\begin{aligned} b_{n}^{(k)}(x)=\sum_{j=0}^{n}S_{1}(n,j) \frac{Li_{k}(1-e^{1-e^{t}})}{t}x^{j}. \end{aligned}$$
(2.4)
Thus, by (2.3) and using the arguments in the proof of Theorem 2.1, we obtain the required formula. □

For the next explicit formula, we use the conjugation representation, namely (1.5).

Theorem 2.3

For all \(n\geq0\),
$$\begin{aligned} b_{n}^{(k)}(x)&=b_{n}^{(k)}+\sum _{j=1}^{n}\frac{1}{j} \Biggl(\sum _{m=j-1}^{n-1}\sum_{r=1}^{n-m} \frac{(-1)^{r+n-m}r!}{r^{k}}\binom {n}{m}S_{1}(m,j-1)S_{2}(n-m,r) \Biggr)x^{j}. \end{aligned}$$

Proof

By (1.5) and (1.6), we have \(b_{n}^{(k)}(x)=\sum_{j=0}^{n}c_{n,j}x^{j}\), where
$$\begin{aligned} j!c_{n,j}= \bigl\langle \bigl(g\bigl(\bar{f}(t)\bigr) \bigr)^{-1}\bar{f}^{j}(t)| x^{n} \bigr\rangle = \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\bigl(\log (1+t)\bigr)^{j}\Big| x^{n} \biggr\rangle . \end{aligned}$$
If \(j=0\), then \(c_{n,0}=b_{n}^{(k)}\). Thus, assume now that \(1\leq j\leq n\). So
$$\begin{aligned}[b] j!c_{n,j}&= \bigl\langle Li_{k}\bigl(1-e^{-t}\bigr) \bigl(\log(1+t)\bigr)^{j-1}| x^{n} \bigr\rangle \\ &= \biggl\langle Li_{k}\bigl(1-e^{-t}\bigr)\Big| (j-1)!\sum _{m\geq j-1}S_{1}(m,j-1)\frac {t^{m}}{m!}x^{n} \biggr\rangle \\ &=(j-1)!\sum_{m=j-1}^{n}\binom{n}{m}S_{1}(m,j-1) \bigl\langle Li_{k}\bigl(1-e^{-t}\bigr)| x^{n-m} \bigr\rangle , \end{aligned} $$
which, by (2.1), implies that
$$\begin{aligned} j!c_{n,j}&=(j-1)!\sum_{m=j-1}^{n} \binom{n}{m}S_{1}(m,j-1) \Biggl\langle \sum _{\ell\geq1}\sum_{r=1}^{\ell}\frac{(-1)^{r+\ell}r!}{r^{k}}S_{2}(\ell,r)\frac {t^{\ell}}{\ell!}\bigg| x^{n-m} \Biggr\rangle \\ &=(j-1)!\sum_{m=j-1}^{n}\binom{n}{m}S_{1}(m,j-1) \Biggl(\sum_{r=1}^{n-m}\frac {(-1)^{r+n-m}r!}{r^{k}}S_{2}(n-m,r) \Biggr), \end{aligned}$$
which completes the proof. □

In order to state our next formula, we recall that \(b_{n}(x)=b_{n}^{(1)}(x)\) is the Bernoulli polynomial of the second kind, which is given by the generating function \(\frac{t}{\log(1+t)}(1+t)^{x}=\sum_{n\geq 0}b_{n}(x)\frac{t^{n}}{n!}\).

Theorem 2.4

For all \(n\geq0\),
$$\begin{aligned} b_{n}^{(k)}(x)&=\frac{1}{n+1}\sum _{j=0}^{n}\binom{n+1}{j}\bigl(B_{n+1-j}^{(k)}-B_{n+1-j}^{(k)}(-1)\bigr) b_{\ell}(x), \end{aligned}$$
where \(B_{n}^{(k)}(x)\) is the nth poly-Bernoulli polynomial.

Proof

From the definitions, we have
$$\begin{aligned} b_{n}^{(k)}(y)&= \biggl\langle \sum _{\ell\geq0}b_{\ell}^{(k)}(y) \frac {t^{\ell}}{\ell!}\Big| x^{n} \biggr\rangle = \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log (1+t)}(1+t)^{y}\Big| x^{n} \biggr\rangle \\ &= \biggl\langle \frac{e^{-t}-1}{-t}\frac{Li_{k}(1-e^{-t})}{1-e^{-t}}\frac {t}{\log(1+t)}(1+t)^{y}\Big| x^{n} \biggr\rangle \\ &= \biggl\langle \frac{e^{-t}-1}{-t}\frac{Li_{k}(1-e^{-t})}{1-e^{-t}}\Big| \frac {t}{\log(1+t)}(1+t)^{y}x^{n} \biggr\rangle \\ &= \biggl\langle \frac {e^{-t}-1}{-t}\frac{Li_{k}(1-e^{-t})}{1-e^{-t}}\Big| \sum _{\ell\geq0}b_{\ell}(y)\frac{t^{\ell}}{\ell!}x^{n} \biggr\rangle . \end{aligned}$$
Since \(B_{n}^{(k)}(x)\) is the poly-Bernoulli polynomial given by the generating function \(\frac{Li_{k}(1-e^{-t})}{1-e^{-t}} e^{xt}=\sum_{n\geq0}B_{n}^{(k)}(x)\frac {t^{n}}{n!}\), we have \(\frac{Li_{k}(1-e^{-t})}{1-e^{-t}}x^{n}=B_{n}^{(k)}(x)\) and \(\frac{d}{dx}B_{n}^{(k)}(x)=nB_{n-1}^{(k)}(x)\). Thus \(b_{n}^{(k)}(y)=\sum_{\ell=0}^{n}\binom{n}{\ell} b_{\ell}(y) \langle \frac{e^{-t}-1}{-t}|B_{n-\ell}^{(k)}(x) \rangle\). By the fact that \(\langle f(at)|p(x)\rangle=\langle f(t)|p(ax)\rangle\) for constant a (see Proposition 2.1.11 in [19]), we obtain
$$\begin{aligned} b_{n}^{(k)}(y)=\sum_{\ell=0}^{n} \binom{n}{\ell} b_{\ell}(y) \biggl\langle \frac{e^{t}-1}{t}\Big| B_{n-\ell}^{(k)}(-x) \biggr\rangle . \end{aligned}$$
Note that \(\langle\frac{e^{t}-1}{t}|B_{n-\ell}^{(k)}(-x) \rangle=\int_{0}^{1}B_{n-\ell}^{(k)}(-u)\,du=\frac{1}{n+1-\ell}(B_{n+1-\ell}^{(k)}-B_{n+1-\ell}^{(k)}(-1))\), which leads to
$$\begin{aligned} b_{n}^{(k)}(y)&=\sum_{\ell=0}^{n} \binom{n}{\ell} b_{\ell}(y)\frac {1}{n+1-\ell}\bigl( B_{n+1-\ell}^{(k)}-B_{n+1-\ell}^{(k)}(-1)\bigr) \\ &=\frac{1}{n+1}\sum_{j=0}^{n} \binom{n+1}{j}\bigl(B_{n+1-j}^{(k)}-B_{n+1-j}^{(k)}(-1) \bigr)b_{\ell}(y), \end{aligned}$$
which completes the proof. □

Theorem 2.5

For all \(n\geq0\),
$$b_{n}^{(k)}(x)=\sum_{m=0}^{n} \binom{n}{m} \Biggl[\sum_{r=1}^{m+1}(-1)^{r+m+1} \frac{r!S_{2}(m+1,r)}{r^{k}(m+1)} \Biggr]b_{n-m}(x). $$

Proof

By using a similar argument as in the proof of Theorem 2.4, we obtain
$$\begin{aligned} b_{n}^{(k)}(y)&= \biggl\langle \frac{Li_{k}(1-e^{-t})}{t} \frac{t}{\log (1+t)}(1+t)^{y}\Big| x^{n} \biggr\rangle = \biggl\langle \frac {Li_{k}(1-e^{-t})}{t}\Big|\sum_{m\geq0} b_{m}(y)\frac{t^{m}}{m!}x^{n} \biggr\rangle \\ &=\sum_{m=0}^{n}\binom{n}{m} b_{m}(y) \biggl\langle \frac {Li_{k}(1-e^{-t})}{t}\Big|x^{n-m} \biggr\rangle , \end{aligned}$$
which, by (2.2), gives
$$\begin{aligned} b_{n}^{(k)}(y)&=\sum_{m=0}^{n} \binom{n}{m}b_{m}(y) \Biggl\langle \sum _{\ell\geq0}\sum_{r=1}^{\ell+1} \frac{(-1)^{r+\ell+1}r!}{r^{k}}\frac {S_{2}(\ell+1,r)}{\ell+1}\frac{t^{\ell}}{\ell!}\bigg|x^{n-m} \Biggr\rangle \\ &=\sum_{m=0}^{n}\binom{n}{m} b_{m}(y) \Biggl(\sum_{r=1}^{n-m+1} \frac {(-1)^{r+n-m+1}r!}{r^{k}}\frac{S_{2}(n-m+1,r)}{n-m+1} \Biggr) \\ &=\sum_{m=0}^{n}\binom{n}{m} \Biggl[ \sum_{r=1}^{m+1}(-1)^{r+m+1} \frac {r!S_{2}(m+1,r)}{r^{k}(m+1)} \Biggr]b_{n-m}(y), \end{aligned}$$
as required. □

Note that the statement of Theorem 2.5 has been obtained in Theorem 2.2 of [12].

3 Recurrence relations

By (1.6) we have \(b_{n}^{(k)}(x)\sim (\frac {t}{Li_{k}(1-e^{1-e^{t}})},e^{t}-1 )\) with \(P_{n}(x)=\frac {t}{Li_{k}(1-e^{1-e^{t}})}b_{n}^{(k)}(x)=(x)_{n}=x(x-1)\cdots(x-n+1)\sim (1,e^{t}-1)\). Thus,
$$b_{n}^{(k)}(x+y)=\sum_{j=0}^{n} \binom{n}{j}b_{j}^{(k)}(x) (y)_{n-j}. $$
The aim of this section is to derive recurrence relations for the poly-Bernoulli polynomials of the second kind. As first trivial recurrence, by using the fact that if \(S_{n}(x)\sim(g(t),f(t))\) then \(f(t)S_{n}(x)=nS_{n-1}(x)\), we derive that \((e^{t}-1)b_{n}^{(k)}(x)=nb_{n-1}^{(k)}(x)\), and hence \(b_{n}^{(k)}(x+1)=b_{n}^{(k)}(x)+nb_{n-1}^{(k)}(x)\). Our next results establish other types of recurrence relations.

Theorem 3.1

For all \(n\geq0\),
$$\begin{aligned} b_{n+1}^{(k)}(x) ={}&xb_{n}^{(k)}(x-1) \\ &{}+\sum_{j=0}^{n}\sum _{\ell=0}^{j+1}\sum_{m=0}^{j+1-\ell} \frac{1}{m}\binom {j}{m-1}S_{1}(n,j)S_{2}(j+1-m, \ell) \bigl(B_{\ell}^{(k-1)}(-1)x^{m}- b_{\ell}^{(k)}(x-1)^{m}\bigr). \end{aligned}$$

Proof

It is well known that if \(S_{n}(x)\sim(g(t),f(t))\) then \(S_{n+1}(x)=(x-\frac{g'(t)}{g(t)})\frac{1}{f'(t)}S_{n}(x)\). Hence, by (1.6), we have
$$b_{n+1}^{(k)}(x)=xb_{n}^{(k)}(x-1)-e^{-t} \frac{g'(t)}{g(t)}b_{n}^{(k)}(x) $$
with
$$\begin{aligned} \frac{g'(t)}{g(t)}&=\bigl(\log\bigl(g(t)\bigr)\bigr)'=\bigl(\log t-\log Li_{k}\bigl(1-e^{1-e^{t}}\bigr)\bigr)'= \frac{1}{t} \biggl(1-\frac {te^{t}e^{1-e^{t}}Li_{k-1}(1-e^{1-e^{t}})}{(1-e^{1-e^{t}})Li_{k}(1-e^{1-e^{t}})} \biggr), \end{aligned}$$
where \(1-\frac {te^{t}e^{1-e^{t}}Li_{k-1}(1-e^{1-e^{t}})}{(1-e^{1-e^{t}})Li_{k}(1-e^{1-e^{t}})}\) has order at least one. Thus, by (2.4), we get
$$\begin{aligned} -e^{-t}\frac{g'(t)}{g(t)}b_{n}^{(k)}(x) &= \frac{-e^{-t}}{t} \biggl(1-\frac {te^{t}e^{1-e^{t}}Li_{k-1}(1-e^{1-e^{t}})}{(1-e^{1-e^{t}})Li_{k}(1-e^{1-e^{t}})} \biggr)\sum _{j=0}^{n}S_{1}(n,j)\frac{Li_{k}(1-e^{1-e^{t}})}{t}x^{j} \\ &=-\sum_{j=0}^{n}\frac{S_{1}(n,j)}{j+1} \biggl( \frac {e^{-t}Li_{k}(1-e^{1-e^{t}})}{\log(1+e^{t}-1)}-\frac {e^{1-e^{t}}Li_{k-1}(1-e^{1-e^{t}})}{(1-e^{1-e^{t}})} \biggr)x^{j+1} \\ &=-\sum_{j=0}^{n}\frac{S_{1}(n,j)}{j+1} \biggl(e^{-t}\sum_{\ell\geq0}b_{\ell}^{(k)}\frac{(e^{t}-1)^{\ell}}{\ell!}-\sum_{\ell\geq0} B_{\ell}^{(k-1)}(-1)\frac{(e^{t}-1)^{\ell}}{\ell!} \biggr)x^{j+1}, \end{aligned}$$
where
$$\begin{aligned} &e^{-t}\sum_{\ell\geq0}b_{\ell}^{(k)} \frac{(e^{t}-1)^{\ell}}{\ell !}x^{j+1}\\ &\quad=e^{-t}\sum_{\ell=0}^{j+1} b_{\ell}^{(k)}\sum_{m=\ell}^{j+1}S_{2}(m, \ell )\frac{t^{m}}{m!}x^{j+1} =e^{-t}\sum _{\ell=0}^{j+1}\sum_{m=\ell }^{j+1} \binom{j+1}{m}b_{\ell}^{(k)}S_{2}(m, \ell)x^{j+1-m} \\ &\quad=e^{-t}\sum_{\ell=0}^{j+1}\sum _{m=0}^{j+1-\ell}\binom{j+1}{m} b_{\ell}^{(k)}S_{2}(j+1-m,\ell)x^{m} \\ &\quad= \sum_{\ell=0}^{j+1}\sum _{m=0}^{j+1-\ell}\binom{j+1}{m}b_{\ell}^{(k)}S_{2}(j+1-m,\ell) (x-1)^{m} \end{aligned}$$
and
$$\begin{aligned} \sum_{\ell\geq0}B_{\ell}^{(k-1)}(-1) \frac{(e^{t}-1)^{\ell}}{\ell !}x^{j+1}&=\sum_{\ell=0}^{j+1} B_{\ell}^{(k-1)}(-1)\sum_{m=\ell }^{j+1}S_{2}(m, \ell)\frac{t^{m}}{m!}x^{j+1} \\ &=\sum_{\ell=0}^{j+1}\sum _{m=\ell}^{j+1}\binom{j+1}{m}S_{2}(m,\ell) B_{\ell}^{(k-1)}(-1)x^{j+1-m}\\ &=\sum _{\ell=0}^{j+1}\sum_{m=0}^{j+1-\ell } \binom{j+1}{m}S_{2}(j+1-m,\ell)B_{\ell}^{(k-1)}(-1)x^{m}. \end{aligned}$$
Thus,
$$\begin{aligned} b_{n+1}^{(k)}(x)={}&xb_{n}^{(k)}(x-1) \\ &{}+\sum_{j=0}^{n}\frac{S_{1}(n,j)}{j+1}\sum _{\ell=0}^{j+1}\sum_{m=0}^{j+1-\ell} \binom{j+1}{m}S_{2}(j+1-m,\ell) \bigl(B_{\ell}^{(k-1)}(-1)x^{m}-b_{\ell}^{(k)}(x-1)^{m} \bigr) \\ ={}&xb_{n}^{(k)}(x-1) \\ &{}+\sum_{j=0}^{n}\sum _{\ell=0}^{j+1}\sum_{m=0}^{j+1-\ell} \frac{1}{m}\binom {j}{m-1}S_{1}(n,j)S_{2}(j+1-m, \ell) \bigl(B_{\ell}^{(k-1)}(-1)x^{m}- b_{\ell}^{(k)}(x-1)^{m}\bigr), \end{aligned}$$
which completes the proof. □

Theorem 3.2

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

Proof

We proceed in the proof by using the fact that if \(S_{n}(x)\sim (g(t),f(t))\) then
$$\frac{d}{dx}S_{n}(x)=\sum_{\ell=0}^{n-1} \binom{n}{\ell}\bigl\langle \bar {f}(t)|x^{n-\ell}\bigr\rangle S_{\ell}(x). $$
By (1.6), we have \(\langle\bar{f}(t)|x^{n-\ell}\rangle=\langle \log(1+t)|x^{n-\ell}\rangle\), which leads to
$$\bigl\langle \bar{f}(t)|x^{n-\ell}\bigr\rangle = \biggl\langle \sum _{m\geq 1}(-1)^{m-1}(m-1)!\frac{t^{m}}{m!}\Big|x^{n-\ell} \biggr\rangle =(-1)^{n-1-\ell }(n-1-\ell)!. $$
Thus \(\frac{d}{dx}b_{n}^{(k)}(x)=n!\sum_{\ell=0}^{n-1}\frac{(-1)^{n-1-\ell }}{\ell!(n-\ell)}b_{\ell}^{(k)}(x)\), as required. □

Theorem 3.3

For all \(n\geq1\),
$$\begin{aligned} b_{n}^{(k)}(x)=xb_{n-1}^{(k)}(x-1)+ \frac{1}{n}\sum_{\ell=0}^{n} \binom {n}{\ell}\bigl(B_{\ell}^{(k-1)}(-1)b_{n-\ell}(x)- b_{\ell}^{(k)}b_{n-\ell}(x-1)\bigr). \end{aligned}$$

Proof

Let \(n\geq1\). Then (1.6), we have
$$ \begin{aligned}[b] b_{n}^{(k)}(y)&= \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log (1+t)}(1+t)^{y}\Big|x^{n} \biggr\rangle = \biggl\langle \frac{d}{dt} \biggl[\frac {Li_{k}(1-e^{-t})}{\log(1+t)}(1+t)^{y} \biggr]\Big|x^{n-1} \biggr\rangle \\ &= \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\frac{d}{dt} \bigl[(1+t)^{y} \bigr]\Big|x^{n-1} \biggr\rangle + \biggl\langle \frac{d}{dt} \biggl[ \frac{Li_{k}(1-e^{-t})}{\log(1+t)} \biggr](1+t)^{y}\Big|x^{n-1} \biggr\rangle . \end{aligned} $$
(3.1)
The first term in (3.1) is given by
$$\begin{aligned} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\frac{d}{dt} \bigl[(1+t)^{y} \bigr]\Big|x^{n-1} \biggr\rangle =y \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}(1+t)^{y-1}\Big|x^{n-1} \biggr\rangle =yb_{n-1}^{(k)}(y-1). \end{aligned}$$
(3.2)
For the second term in (3.1), we note that
$$\begin{aligned} \frac{d}{dt} \biggl[\frac{Li_{k}(1-e^{-t})}{\log(1+t)} \biggr](1+t)^{y} &= \frac{1}{t}\frac{t}{\log(1+t)} \biggl(\frac {Li_{k-1}(1-e^{-t})}{1-e^{-t}}e^{-t}- \frac{Li_{k}(1-e^{-t})}{\log (1+t)}\frac{1}{1+t} \biggr) (1+t)^{y} \\ &=\frac{1}{t} \biggl(\frac{t(1+t)^{y}}{\log(1+t)}\frac {Li_{k-1}(1-e^{-t})}{1-e^{-t}}e^{-t}- \frac{t(1+t)^{y-1}}{\log(1+t)}\frac {Li_{k}(1-e^{-t})}{\log(1+t)} \biggr), \end{aligned}$$
which has order at least zero. So, the second term in (3.1) is given by
$$\begin{aligned} & \biggl\langle \frac{d}{dt} \biggl[\frac{Li_{k}(1-e^{-t})}{\log(1+t)} \biggr](1+t)^{y}\Big|x^{n-1} \biggr\rangle \\ &\quad=\frac{1}{n} \biggl( \biggl\langle \frac{t}{\log(1+t)}(1+t)^{y}\Big| \frac {Li_{k-1}(1-e^{-t})}{1-e^{-t}}e^{-t}x^{n} \biggr\rangle \\ &\qquad{}- \biggl\langle \frac{t}{\log(1+t)}(1+t)^{y-1}\Big|\frac{Li_{k}(1-e^{-t})}{\log (1+t)}x^{n} \biggr\rangle \biggr) \\ &\quad=\frac{1}{n} \biggl( \biggl\langle \frac{t}{\log(1+t)}(1+t)^{y}\Big| \sum_{\ell\geq 0}B_{\ell}^{(k-1)}(-1) \frac{t^{\ell}}{\ell!}x^{n} \biggr\rangle \\ &\qquad{}- \biggl\langle \frac{t}{\log(1+t)}(1+t)^{y-1}\Big|\sum_{\ell\geq0} b_{\ell}^{(k)}\frac{t^{\ell}}{\ell!}x^{n} \biggr\rangle \biggr) \\ &\quad=\frac{1}{n} \Biggl(\sum_{\ell=0}^{n} \binom{n}{\ell} B_{\ell}^{(k-1)}(-1) \biggl\langle \frac{t}{\log(1+t)}(1+t)^{y}\Big|x^{n-\ell} \biggr\rangle \\\ &\qquad{}-\sum _{\ell=0}^{n}\binom{n}{\ell} b_{\ell}^{(k)} \biggl\langle \frac{t}{\log (1+t)}(1+t)^{y-1}\Big|x^{n-\ell} \biggr\rangle \Biggr) \\ &\quad=\frac{1}{n} \Biggl(\sum_{\ell=0}^{n} \binom{n}{\ell} B_{\ell}^{(k-1)}(-1)b_{n-\ell}(y) - \sum_{\ell=0}^{n}\binom{n}{\ell} b_{\ell}^{(k)}b_{n-\ell}(y-1) \Biggr) \\ &\quad=\frac{1}{n}\sum_{\ell=0}^{n} \binom{n}{\ell}\bigl(B_{\ell}^{(k-1)}(-1)b_{n-\ell}(y)-b_{\ell}^{(k)}b_{n-\ell}(y-1) \bigr). \end{aligned}$$
(3.3)
By substituting (3.2) and (3.3) into (3.1), we complete the proof. □

4 Identities

In this section we present some identities related to poly-Bernoulli numbers of the second kind.

Theorem 4.1

For all \(n\geq0\),
$$\sum_{\ell=0}^{n}(-1)^{n-\ell}(n-\ell)! \binom{n+1}{\ell} b_{\ell}^{(k)}=\sum _{m=0}^{n} (-1)^{n-m}\binom{n}{m} B_{m}^{(k-1)}. $$

Proof

We compute \(A=\langle Li_{k}(1-e^{-t})|x^{n+1}\rangle\) in two different ways. On the one hand, by (1.6), it is
$$\begin{aligned} A&= \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\Big|\log(1+t)x^{n+1} \biggr\rangle = \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\Big|\sum_{\ell\geq1} \frac {(-1)^{\ell-1}t^{\ell}}{\ell}x^{n+1} \biggr\rangle \\ &=\sum_{\ell=0}^{n}(-1)^{n-\ell}(n- \ell)!\binom{n+1}{\ell} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\Big|x^{\ell}\biggr\rangle \\ &=\sum_{\ell=0}^{n}(-1)^{n-\ell}(n- \ell)!\binom{n+1}{\ell} b_{\ell}^{(k)}. \end{aligned}$$
(4.1)
On the other hand, by (1.6), it is
$$\begin{aligned} A&= \bigl\langle Li_{k}\bigl(1-e^{-t}\bigr)|x^{n+1} \bigr\rangle = \biggl\langle \int_{0}^{t} \frac{d}{ds}Li_{k}\bigl(1-e^{-s}\bigr)\,ds\Big|x^{n+1} \biggr\rangle = \biggl\langle \int_{0}^{t} e^{-s}\frac{Li_{k-1}(1-e^{-s})}{1-e^{-s}}\,ds\Big|x^{n+1} \biggr\rangle \\ &= \biggl\langle \int_{0}^{t} \sum _{a\geq0}\frac{(-s)^{a}}{a!}\sum_{m\geq0} B_{m}^{(k-1)}\frac{s^{m}}{m!}\,ds\Big|x^{n+1} \biggr\rangle = \Biggl\langle \sum_{\ell \geq0}\sum _{m=0}^{\ell}(-1)^{\ell-m}\binom{\ell}{m} B_{m}^{(k-1)}\frac {t^{\ell+1}}{(\ell+1)!}\Big|x^{n+1} \Biggr\rangle \\ &=\sum_{m=0}^{n} (-1)^{n-m} \binom{n}{m}B_{m}^{(k-1)}. \end{aligned}$$
(4.2)
By comparing (4.1) and (4.2), we obtain the required identity. □
By using similar techniques as in the proof of Theorem 4.1 with computing
$$\biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\bigl(\log(1+t)\bigr)^{m}\Big|x^{n} \biggr\rangle $$
in two different ways, we obtain the following result (we leave the proof as an exercise to the interested reader).

Theorem 4.2

For all \(n-1\geq m\geq1\),
$$\begin{aligned} &\sum_{\ell=0}^{n-m}\binom{n}{\ell}S_{1}(n- \ell,m)b_{\ell}^{(k)} \\ &\quad =\sum_{\ell=0}^{n-m}\binom{n-1}{\ell}S_{1}(n-1- \ell,m-1)b_{\ell}^{(k)}(-1) \\ &\qquad{} +\frac{1}{n}\sum_{\ell=0}^{n-1-m}\sum _{j=0}^{\ell+1}\binom {n}{\ell+1} \binom{\ell+1}{j}S_{1}(n-1-\ell,m) \bigl(b_{\ell+1-j}B_{j}^{(k-1)}(-1)-b_{\ell+1-j}(-1)b_{j}^{(k)} \bigr). \end{aligned}$$
Let \(b_{n}^{(k)}(x)=\sum_{m=0}^{n} c_{n,m}(x)_{m}\). By (1.5), (1.6) and the fact that \((x)_{m}\sim(1,e^{t}-1)\), we obtain
$$\begin{aligned} c_{n,m}&=\frac{1}{m!} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)} \Big|t^{m}x^{n} \biggr\rangle =\binom{n}{m} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)} \Big|x^{n-m} \biggr\rangle =\binom{n}{m} b_{n-m}^{(k)}, \end{aligned}$$
which leads to the following identity.

Theorem 4.3

For all \(n\geq0\),
$$b_{n}^{(k)}(x)=\sum_{m=0}^{n} \binom{n}{m}b_{n-m}^{(k)}(x)_{m}. $$
Let \(\mathbb{B}_{n}^{(s)}(x)\) be the nth Bernoulli polynomial of order s. Then \(\mathbb{B}_{n}^{(s)}(x)\sim(((e^{t}-1)/t)^{s},t)\). Also, the Bernoulli numbers of the second kind of order s are given by \(\frac{t^{s}}{\log^{s}(1+t)}=\sum_{j\geq0}\mathbf{b}_{j}^{(s)}\frac{t^{j}}{j!}\) and let \(b_{n}^{(k)}(x)=\sum_{m=0}^{n} c_{n,m}\mathbb{B}_{m}^{(s)}(x)\). By (1.5) and (1.6), we obtain
$$\begin{aligned} c_{n,m}&=\frac{1}{m!} \biggl\langle \frac{\frac{t^{s}}{\log^{s}(1+t)}}{\frac {\log(1+t)}{Li_{k}(1-e^{-t})}} \log^{m}(1+t) \Big|x^{n} \biggr\rangle =\frac{1}{m!} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\frac{t^{s}}{\log ^{s}(1+t)}\Big|\log^{m}(1+t)x^{n} \biggr\rangle \\ &=\frac{1}{m!} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\frac {t^{s}}{\log^{s}(1+t)}\Big|m!\sum _{\ell\geq m}S_{1}(\ell,m)\frac{t^{\ell}}{\ell !}x^{n} \biggr\rangle \\ &=\sum_{\ell=m}^{n}\binom{n}{\ell}S_{1}( \ell,m) \biggl\langle \frac {Li_{k}(1-e^{-t})}{\log(1+t)}\frac{t^{s}}{\log^{s}(1+t)}\Big|x^{n-\ell} \biggr\rangle \\ &=\sum_{\ell=0}^{n-m}\binom{n}{\ell}S_{1}(n- \ell,m) \biggl\langle \frac {Li_{k}(1-e^{-t})}{\log(1+t)}\Big|\frac{t^{s}}{\log^{s}(1+t)}x^{\ell} \biggr\rangle \\ &=\sum_{\ell=0}^{n-m}\binom{n}{\ell}S_{1}(n- \ell,m) \biggl\langle \frac {Li_{k}(1-e^{-t})}{\log(1+t)}\Big|\sum_{j\geq0} \mathbf{b}_{j}^{(s)}\frac {t^{j}}{j!}x^{\ell} \biggr\rangle \\ &=\sum_{\ell=0}^{n-m}\sum _{j=0}^{\ell}\binom{n}{\ell}\binom{\ell }{j}S_{1}(n- \ell,m)\mathbf{b}_{j}^{(s)} \biggl\langle \frac {Li_{k}(1-e^{-t})}{\log(1+t)}\Big|x^{\ell-j} \biggr\rangle \\ &=\sum_{\ell=0}^{n-m}\sum _{j=0}^{\ell}\binom{n}{\ell}\binom{\ell }{j}S_{1}(n- \ell,m)\mathbf{b}_{j}^{(s)}b_{\ell-j}^{(k)}, \end{aligned}$$
which gives the following identity.

Theorem 4.4

For all \(n\geq0\),
$$b_{n}^{(k)}(x)=\sum_{m=0}^{n} \Biggl(\sum_{\ell=0}^{n-m}\sum _{j=0}^{\ell}\binom{n}{\ell}\binom{\ell}{j}S_{1}(n- \ell,m)\mathbf{b}_{j}^{(s)}b_{\ell -j}^{(k)} \Biggr)\mathbb{B}_{m}^{(s)}(x). $$
Define \(H_{n}^{(s)}(\lambda,x)\) to be the nth Frobenius-Euler polynomials of order s. Note that these polynomial satisfy \(H_{n}^{(s)}(\lambda,x)\sim(((e^{t}-\lambda)/(1-\lambda))^{s},t)\). Let \(b_{n}^{(k)}(x)=\sum_{m=0}^{n} c_{n,m}H_{m}^{(s)}(\lambda,x)\). By (1.5) and (1.6), we obtain
$$\begin{aligned} c_{n,m}&=\frac{1}{m!} \biggl\langle \frac{\frac{(1+t-\lambda)^{s}}{(1-\lambda )^{s}}}{\frac{\log(1+t)}{Li_{k}(1-e^{-t})}} \log^{m}(1+t) \Big|x^{n} \biggr\rangle =\frac{1}{m!(1-\lambda)^{s}} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log (1+t)}\log^{m}(1+t)\Big|(1-\lambda+t)^{s}x^{n} \biggr\rangle \\ &=\frac{1}{m!(1-\lambda)^{s}}\sum_{j=0}^{n-m} \binom{s}{j}(1-\lambda )^{s-j}(n)_{j} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\Big|\log ^{m}(1+t)x^{n-j} \biggr\rangle \\ &=\frac{1}{m!(1-\lambda)^{s}}\sum_{j=0}^{n-m} \binom{s}{j}(1-\lambda )^{s-j}(n)_{j} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log(1+t)}\Big|m!\sum_{\ell \geq m}S_{1}( \ell,m)\frac{t^{\ell}}{\ell!}x^{n-j} \biggr\rangle \\ &=\sum_{j=0}^{n-m}\sum _{\ell=m}^{n-j}\binom{s}{j}\binom{n-j}{\ell }S_{1}( \ell,m) (1-\lambda)^{-j}(n)_{j} \biggl\langle \frac{Li_{k}(1-e^{-t})}{\log (1+t)}\Big|x^{n-j-\ell} \biggr\rangle \\ &=\sum_{j=0}^{n-m}\sum _{\ell=m}^{n-j}\binom{s}{j}\binom{n-j}{\ell }S_{1}( \ell,m) (1-\lambda)^{-j}(n)_{j}b_{n-j-\ell}^{(k)} \\ &=\sum_{j=0}^{n-m}\sum _{\ell=0}^{n-m-j}\binom{s}{j}\binom{n-j}{\ell }S_{1}(n-j- \ell,m) (1-\lambda)^{-j}(n)_{j}b_{\ell}^{(k)}, \end{aligned}$$
which gives the following identity.

Theorem 4.5

For all \(n\geq0\),
$$b_{n}^{(k)}(x)=\sum_{m=0}^{n} \Biggl(\sum_{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\binom{s}{j}\binom{n-j}{\ell}S_{1}(n-j- \ell,m) (1-\lambda )^{-j}(n)_{j}b_{\ell}^{(k)} \Biggr)H_{m}^{(s)}(\lambda,x). $$

Declarations

Acknowledgements

This work was partially supported by Kwangwoon University in 2014.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Sogang University
(2)
Department of Mathematics, Kwangwoon University
(3)
Department of Mathematics, University of Haifa
(4)
Institute of Natural Sciences, Far Eastern Federal University

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