Theory and Modern Applications

# 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

$\frac{t}{log\left(1+t\right)}{\left(1+t\right)}^{x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1â€“3]}\right).$
(1)

When $x=0$, ${b}_{n}={b}_{n}\left(0\right)$ are called the Bernoulli numbers of the second kind. The first few Bernoulli numbers ${b}_{n}$ of the second kind are ${b}_{0}=1$, ${b}_{1}=1/2$, ${b}_{2}=âˆ’1/12$, ${b}_{3}=1/24$, ${b}_{4}=âˆ’19/720$, ${b}_{5}=3/160,â€¦$â€‰.

From (1), we have

${b}_{n}\left(x\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){b}_{l}{\left(x\right)}_{nâˆ’l},$
(2)

where ${\left(x\right)}_{n}=x\left(xâˆ’1\right)â‹¯\left(xâˆ’n+1\right)$ ($nâ‰§0$). The Stirling number of the second kind is defined by

${x}^{n}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{2}\left(n,l\right){\left(x\right)}_{l}\phantom{\rule{1em}{0ex}}\left(nâ‰§0\right).$
(3)

The ordinary Bernoulli polynomials are given by

$\frac{t}{{e}^{t}âˆ’1}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1â€“18]}\right).$
(4)

When $x=0$, ${B}_{n}={B}_{n}\left(0\right)$ are called Bernoulli numbers.

It is well known that the classical poly-logarithmic function ${Li}_{k}\left(x\right)$ is given by

${Li}_{k}\left(x\right)=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{x}^{n}}{{n}^{k}}\phantom{\rule{1em}{0ex}}\left(kâˆˆ\mathbb{Z}\right)\phantom{\rule{0.25em}{0ex}}\left(\text{see [8â€“10]}\right).$
(5)

For $k=1$, ${Li}_{1}\left(x\right)={âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{x}^{n}}{n}=âˆ’log\left(1âˆ’x\right)$. The Stirling number of the first kind is defined by

${\left(x\right)}_{n}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){x}^{l}\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right)\phantom{\rule{0.25em}{0ex}}\left(\text{see [16]}\right).$
(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 $kâˆˆ\mathbb{Z}$, we consider the poly-Bernoulli polynomials ${b}_{n}^{\left(k\right)}\left(x\right)$ of the second kind:

$\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}{\left(1+t\right)}^{x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}.$
(7)

When $x=0$, ${b}_{n}^{\left(k\right)}={b}_{n}^{\left(k\right)}\left(0\right)$ are called the poly-Bernoulli numbers of the second kind.

Indeed, for $k=1$, we have

$\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}{\left(1+t\right)}^{x}=\frac{t}{log\left(1+t\right)}{\left(1+t\right)}^{x}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}\left(x\right)\frac{{t}^{n}}{n!}.$
(8)

By (7) and (8), we get

${b}_{n}^{\left(1\right)}\left(x\right)={b}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right).$
(9)

It is well known that

$\frac{t{\left(1+t\right)}^{xâˆ’1}}{log\left(1+t\right)}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(n\right)}\left(x\right)\frac{{t}^{n}}{n!},$
(10)

where ${B}_{n}^{\left(\mathrm{Î±}\right)}\left(x\right)$ are the Bernoulli polynomials of order Î± which are given by the generating function to be

${\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{\mathrm{Î±}}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(\mathrm{Î±}\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1â€“18]}\right).$

By (1) and (10), we get

${b}_{n}\left(x\right)={B}_{n}^{\left(n\right)}\left(x+1\right)\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right).$

Now, we observe that

(11)

Thus, by (11), we get

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}^{\left(2\right)}\left(x\right)\frac{{t}^{n}}{n!}& =\frac{{\left(1+t\right)}^{x}}{log\left(1+t\right)}{âˆ«}_{0}^{t}\frac{x}{{e}^{x}âˆ’1}\phantom{\rule{0.2em}{0ex}}dx\\ =\frac{{\left(1+t\right)}^{x}}{log\left(1+t\right)}\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{B}_{l}}{l!}{âˆ«}_{0}^{t}{x}^{l}\phantom{\rule{0.2em}{0ex}}dx\\ =\left(\frac{t}{log\left(1+t\right)}\right){\left(1+t\right)}^{x}\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{B}_{l}}{\left(l+1\right)}\frac{{t}^{l}}{l!}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left\{\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\frac{{B}_{l}{b}_{nâˆ’l}\left(x\right)}{l+1}\right\}\frac{{t}^{n}}{n!}.\end{array}$
(12)

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

Theorem 2.1 For $nâ‰¥0$ we have

${b}_{n}^{\left(2\right)}\left(x\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\frac{{B}_{l}{b}_{nâˆ’l}\left(x\right)}{l+1}.$

From (11), we have

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}& =\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}{\left(1+t\right)}^{x}\\ =\frac{t}{log\left(1+t\right)}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{t}{\left(1+t\right)}^{x}.\end{array}$
(13)

We observe that

$\begin{array}{rl}\frac{1}{t}{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)& =\frac{1}{t}\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{{n}^{k}}{\left(1âˆ’{e}^{âˆ’t}\right)}^{n}\\ =\frac{1}{t}\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{n}}{{n}^{k}}n!\underset{l=n}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{2}\left(l,n\right)\frac{{\left(âˆ’t\right)}^{l}}{l!}\\ =\frac{1}{t}\underset{l=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{n=1}{\overset{l}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{n+l}}{{n}^{k}}n!{S}_{2}\left(l,n\right)\frac{{t}^{l}}{l!}\\ =\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{n=1}{\overset{l+1}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{n+l+1}}{{n}^{k}}n!\frac{{S}_{2}\left(l+1,n\right)}{l+1}\frac{{t}^{l}}{l!}.\end{array}$
(14)

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

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}& =\left(\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{m}\left(x\right)\frac{{t}^{m}}{m!}\right)\left\{\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{p=1}{\overset{l+1}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{p+l+1}}{{p}^{k}}p!\frac{{S}_{2}\left(l+1,p\right)}{l+1}\right)\frac{{t}^{l}}{l!}\right\}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left\{\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\underset{p=1}{\overset{l+1}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{p+l+1}p!}{{p}^{k}}\frac{{S}_{2}\left(l+1,p\right)}{l+1}\right){b}_{nâˆ’l}\left(x\right)\right\}\frac{{t}^{n}}{n!}.\end{array}$
(15)

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

Theorem 2.2 For $nâ‰¥0$, we have

${b}_{n}^{\left(k\right)}\left(x\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\underset{p=1}{\overset{l+1}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{p+l+1}}{{p}^{k}}p!\frac{{S}_{2}\left(l+1,p\right)}{l+1}\right){b}_{nâˆ’l}\left(x\right).$

By (7), we get

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({b}_{n}^{\left(k\right)}\left(x+1\right)âˆ’{b}_{n}^{\left(k\right)}\left(x\right)\right)\frac{{t}^{n}}{n!}& =\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}{\left(1+t\right)}^{x+1}âˆ’\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}{\left(1+t\right)}^{x}\\ =\frac{t{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}{\left(1+t\right)}^{x}\\ =\left(\frac{t}{log\left(1+t\right)}{\left(1+t\right)}^{x}\right){Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)\\ =\left(\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{b}_{l}\left(x\right)}{l!}{t}^{l}\right)\left\{\underset{p=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{m=1}{\overset{p}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m+p}m!}{{m}^{k}}{S}_{2}\left(p,m\right)\right)\right\}\frac{{t}^{p}}{p!}\end{array}$
(16)
$\begin{array}{r}=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{p=1}{\overset{n}{âˆ‘}}\underset{m=1}{\overset{p}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m+p}}{{m}^{k}}m!{S}_{2}\left(p,m\right)\frac{{b}_{nâˆ’p}\left(x\right)n!}{\left(nâˆ’p\right)!p!}\right)\frac{{t}^{n}}{n!}\\ =\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left\{\underset{p=1}{\overset{n}{âˆ‘}}\underset{m=1}{\overset{p}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{p}\right)\frac{{\left(âˆ’1\right)}^{m+p}m!}{{m}^{k}}{S}_{2}\left(p,m\right){b}_{nâˆ’p}\left(x\right)\right\}\frac{{t}^{n}}{n!}.\end{array}$
(17)

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

Theorem 2.3 For $nâ‰¥1$, we have

${b}_{n}^{\left(k\right)}\left(x+1\right)âˆ’{b}_{n}^{\left(k\right)}\left(x\right)=\underset{p=1}{\overset{n}{âˆ‘}}\underset{m=1}{\overset{p}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{p}\right)\frac{{\left(âˆ’1\right)}^{m+p}m!}{{m}^{k}}{S}_{2}\left(p,m\right){b}_{nâˆ’p}\left(x\right).$
(18)

From (13), we have

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}^{\left(k\right)}\left(x+y\right)\frac{{t}^{n}}{n!}& ={\left(\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}\right)}^{k}{\left(1+t\right)}^{x+y}\\ ={\left(\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{log\left(1+t\right)}\right)}^{k}{\left(1+t\right)}^{x}{\left(1+t\right)}^{y}\\ =\left(\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{l}^{\left(k\right)}\left(x\right)\frac{{t}^{l}}{l!}\right)\left(\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\left(y\right)}_{m}\frac{{t}^{m}}{m!}\right)\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{l=0}{\overset{n}{âˆ‘}}{\left(y\right)}_{l}{b}_{nâˆ’l}^{\left(k\right)}\left(x\right)\frac{n!}{\left(nâˆ’l\right)!l!}\right)\frac{{t}^{n}}{n!}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){b}_{nâˆ’l}^{\left(k\right)}\left(x\right){\left(y\right)}_{l}\right)\frac{{t}^{n}}{n!}.\end{array}$
(19)

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

Theorem 2.4 For $nâ‰¥0$, we have

${b}_{n}^{\left(k\right)}\left(x+y\right)=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){b}_{nâˆ’l}^{\left(k\right)}\left(x\right){\left(y\right)}_{l}.$

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## Acknowledgements

The present research has been conducted by the Research Grant of Kwangwoon University in 2014.

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

All authors contributed equally to this work. All authors read and approved the final manuscript.

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Kim, T., Kwon, H.I., Lee, S.H. et al. A note on poly-Bernoulli numbers and polynomials of the second kind. Adv Differ Equ 2014, 219 (2014). https://doi.org/10.1186/1687-1847-2014-219