A note on poly-Bernoulli numbers and polynomials of the second kind

Abstract

In this paper, we consider the poly-Bernoulli numbers and polynomials of the second kind and presents new and explicit formulas for calculating the poly-Bernoulli numbers of the second kind and the Stirling numbers of the second kind.

1 Introduction

As is well known, the Bernoulli polynomials of the second kind are defined by the generating function to be

$$\frac{t}{log(1+t)}{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{b}_n(x)\frac{t^n}{n!}(\text{see [1–3]}).$$

(1)

When $x=0$, $b_n=b_n(0)$ are called the Bernoulli numbers of the second kind. The first few Bernoulli numbers $b_n$ of the second kind are $b_0=1$, $b_1=1/2$, $b_2=-1/12$, $b_3=1/24$, $b_4=-19/720$, $b_5=3/160,\dots$ .

From (1), we have

$$b_n(x)=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)b_l{(x)}_{n-l},$$

(2)

where ${(x)}_n=x(x-1)\cdots (x-n+1)$ ($n\geqq 0$). The Stirling number of the second kind is defined by

$$x^n=\sum _{l=0}^n{S}_2(n,l){(x)}_l(n\geqq 0).$$

(3)

The ordinary Bernoulli polynomials are given by

$$\frac{t}{e^t-1}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x)\frac{t^n}{n!}(\text{see [1–18]}).$$

(4)

When $x=0$, $B_n=B_n(0)$ are called Bernoulli numbers.

It is well known that the classical poly-logarithmic function ${Li}_k(x)$ is given by

$${Li}_k(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n^k}(k\in \mathbb{Z})(\text{see [8–10]}).$$

(5)

For $k=1$, ${Li}_1(x)={\sum }_{n=1}^{\mathrm{\infty }}\frac{x^n}{n}=-log(1-x)$. The Stirling number of the first kind is defined by

$${(x)}_n=\sum _{l=0}^n{S}_1(n,l)x^l(n\ge 0)(\text{see [16]}).$$

(6)

In this paper, we consider the poly-Bernoulli numbers and polynomials of the second kind and presents new and explicit formulas for calculating the poly-Bernoulli number and polynomial and the Stirling number of the second kind.

2 Poly-Bernoulli numbers and polynomials of the second kind

For $k\in \mathbb{Z}$, we consider the poly-Bernoulli polynomials $b_n^{(k)}(x)$ of the second kind:

$$\frac{{Li}_k(1-e^{-t})}{log(1+t)}{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{b}_n^{(k)}(x)\frac{t^n}{n!}.$$

(7)

When $x=0$, $b_n^{(k)}=b_n^{(k)}(0)$ are called the poly-Bernoulli numbers of the second kind.

Indeed, for $k=1$, we have

$$\frac{{Li}_k(1-e^{-t})}{log(1+t)}{(1+t)}^x=\frac{t}{log(1+t)}{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{b}_n(x)\frac{t^n}{n!}.$$

(8)

By (7) and (8), we get

$$b_n^{(1)}(x)=b_n(x)(n\ge 0).$$

(9)

It is well known that

$$\frac{t{(1+t)}^{x-1}}{log(1+t)}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(n)}(x)\frac{t^n}{n!},$$

(10)

where $B_n^{(\alpha )}(x)$ are the Bernoulli polynomials of order α which are given by the generating function to be

$${\left(\frac{t}{e^t-1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(\alpha )}(x)\frac{t^n}{n!}(\text{see [1–18]}).$$

By (1) and (10), we get

$$b_n(x)=B_n^{(n)}(x+1)(n\ge 0).$$

Now, we observe that

(11)

Thus, by (11), we get

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{b}_n^{(2)}(x)\frac{t^n}{n!}& =\frac{{(1+t)}^x}{log(1+t)}{\int }_0^t\frac{x}{e^x-1}dx\\ =\frac{{(1+t)}^x}{log(1+t)}\sum _{l=0}^{\mathrm{\infty }}\frac{B_l}{l!}{\int }_0^t{x}^{l}dx\\ =\left(\frac{t}{log(1+t)}\right){(1+t)}^x\sum _{l=0}^{\mathrm{\infty }}\frac{B_l}{(l+1)}\frac{t^l}{l!}\\ =\sum _{n=0}^{\mathrm{\infty }}\left\{\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{B_l{b}_{n-l}(x)}{l+1}\right\}\frac{t^n}{n!}.\end{array}$$

(12)

Therefore, by (12), we obtain the following theorem.

Theorem 2.1 For $n\ge 0$ we have

$$b_n^{(2)}(x)=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{B_l{b}_{n-l}(x)}{l+1}.$$

From (11), we have

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{b}_n^{(k)}(x)\frac{t^n}{n!}& =\frac{{Li}_k(1-e^{-t})}{log(1+t)}{(1+t)}^x\\ =\frac{t}{log(1+t)}\frac{{Li}_k(1-e^{-t})}{t}{(1+t)}^x.\end{array}$$

(13)

We observe that

$$\begin{array}{rl}\frac{1}{t}{Li}_k(1-e^{-t})& =\frac{1}{t}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^k}{(1-e^{-t})}^n\\ =\frac{1}{t}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^k}n!\sum _{l=n}^{\mathrm{\infty }}{S}_2(l,n)\frac{{(-t)}^l}{l!}\\ =\frac{1}{t}\sum _{l=1}^{\mathrm{\infty }}\sum _{n=1}^l\frac{{(-1)}^{n+l}}{n^k}n!S_2(l,n)\frac{t^l}{l!}\\ =\sum _{l=0}^{\mathrm{\infty }}\sum _{n=1}^{l+1}\frac{{(-1)}^{n+l+1}}{n^k}n!\frac{S_2(l+1,n)}{l+1}\frac{t^l}{l!}.\end{array}$$

(14)

Thus, by (10) and (14), we get

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{b}_n^{(k)}(x)\frac{t^n}{n!}& =(\sum _{m=0}^{\mathrm{\infty }}{b}_m(x)\frac{t^m}{m!})\left\{\sum _{l=0}^{\mathrm{\infty }}(\sum _{p=1}^{l+1}\frac{{(-1)}^{p+l+1}}{p^k}p!\frac{S_2(l+1,p)}{l+1})\frac{t^l}{l!}\right\}\\ =\sum _{n=0}^{\mathrm{\infty }}\{\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\sum _{p=1}^{l+1}\frac{{(-1)}^{p+l+1}p!}{p^k}\frac{S_2(l+1,p)}{l+1}\right)b_{n-l}(x)\}\frac{t^n}{n!}.\end{array}$$

(15)

Therefore, by (15), we obtain the following theorem.

Theorem 2.2 For $n\ge 0$, we have

$$b_n^{(k)}(x)=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)(\sum _{p=1}^{l+1}\frac{{(-1)}^{p+l+1}}{p^k}p!\frac{S_2(l+1,p)}{l+1})b_{n-l}(x).$$

By (7), we get

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}(b_n^{(k)}(x+1)-b_n^{(k)}(x))\frac{t^n}{n!}& =\frac{{Li}_k(1-e^{-t})}{log(1+t)}{(1+t)}^{x+1}-\frac{{Li}_k(1-e^{-t})}{log(1+t)}{(1+t)}^x\\ =\frac{t{Li}_k(1-e^{-t})}{log(1+t)}{(1+t)}^x\\ =\left(\frac{t}{log(1+t)}{(1+t)}^x\right){Li}_k(1-e^{-t})\\ =\left(\sum _{l=0}^{\mathrm{\infty }}\frac{b_l(x)}{l!}{t}^l\right)\left\{\sum _{p=1}^{\mathrm{\infty }}(\sum _{m=1}^p\frac{{(-1)}^{m+p}m!}{m^k}{S}_2(p,m))\right\}\frac{t^p}{p!}\end{array}$$

(16)

$$\begin{array}{r}=\sum _{n=1}^{\mathrm{\infty }}(\sum _{p=1}^n\sum _{m=1}^p\frac{{(-1)}^{m+p}}{m^k}m!S_2(p,m)\frac{b_{n-p}(x)n!}{(n-p)!p!})\frac{t^n}{n!}\\ =\sum _{n=1}^{\mathrm{\infty }}\{\sum _{p=1}^n\sum _{m=1}^p\left(\genfrac{}{}{0ex}{}{n}{p}\right)\frac{{(-1)}^{m+p}m!}{m^k}{S}_2(p,m)b_{n-p}(x)\}\frac{t^n}{n!}.\end{array}$$

(17)

Therefore, by (16), we obtain the following theorem.

Theorem 2.3 For $n\ge 1$, we have

$$b_n^{(k)}(x+1)-b_n^{(k)}(x)=\sum _{p=1}^n\sum _{m=1}^p\left(\genfrac{}{}{0ex}{}{n}{p}\right)\frac{{(-1)}^{m+p}m!}{m^k}{S}_2(p,m)b_{n-p}(x).$$

(18)

From (13), we have

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{b}_n^{(k)}(x+y)\frac{t^n}{n!}& ={\left(\frac{{Li}_k(1-e^{-t})}{log(1+t)}\right)}^k{(1+t)}^{x+y}\\ ={\left(\frac{{Li}_k(1-e^{-t})}{log(1+t)}\right)}^k{(1+t)}^x{(1+t)}^y\\ =(\sum _{l=0}^{\mathrm{\infty }}{b}_l^{(k)}(x)\frac{t^l}{l!})\left(\sum _{m=0}^{\mathrm{\infty }}{(y)}_m\frac{t^m}{m!}\right)\\ =\sum _{n=0}^{\mathrm{\infty }}(\sum _{l=0}^n{(y)}_l{b}_{n-l}^{(k)}(x)\frac{n!}{(n-l)!l!})\frac{t^n}{n!}\\ =\sum _{n=0}^{\mathrm{\infty }}(\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)b_{n-l}^{(k)}(x){(y)}_l)\frac{t^n}{n!}.\end{array}$$

(19)

Therefore, by (17), we obtain the following theorem.

Theorem 2.4 For $n\ge 0$, we have

$$b_n^{(k)}(x+y)=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)b_{n-l}^{(k)}(x){(y)}_l.$$