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

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.

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 [2–4]). 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 [5–11]). 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 [13–15]).

Let Π be the algebra of polynomials in a single variable x over ℂ 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 [16–19]). 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 [16–19]). 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 ℋ. Therefore, ℋ is thought of as a set of both formal power series and linear functionals. We call ℋ 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 $n\geq0$,

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