Advances in Difference Equations

, 2015:258 | Cite as

A note on degenerate poly-Bernoulli numbers and polynomials

Open Access
Research

Abstract

In this paper, we consider the degenerate poly-Bernoulli polynomials and present new and explicit formulas for computing them in terms of the degenerate Bernoulli polynomials and Stirling numbers of the second kind.

Keywords

degenerate poly-Bernoulli polynomial degenerate Bernoulli polynomial Stirling number of the second kind 

MSC

11B68 11B73 11B83 

1 Introduction

For \(\lambda\in\mathbb{C}\), Carlitz considered the degenerate Bernoulli polynomials given by the generating function
$$ \frac{t}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac{x}{\lambda}}=\sum _{n=0}^{\infty}\beta_{n} (x\mid \lambda ) \frac{t^{n}}{n!}\quad (\text{see [1--3]}). $$
(1.1)

When \(x=0\), \(\beta_{n} (\lambda )=\beta_{n} (0\mid\lambda )\) are called the degenerate Bernoulli numbers.

Thus, by (1.1), we get
$$ \beta_{n} (x\mid\lambda )=\sum_{l=0}^{n} \binom{n}{l}\beta _{l} (\lambda ) (x\mid\lambda )_{n-l}, $$
(1.2)
where \((x\mid\lambda )_{n}=x (x-\lambda ) (x-2\lambda )\cdots (x-\lambda (n-1 ) )\).
The classical polylogarithm function \(\operatorname{Li}_{k}\) is
$$ \operatorname{Li}_{k} (x )=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{k}}\quad (k\in\mathbb{Z} \text{; see [2, 4--11]}). $$
(1.3)
From (1.1), we note that
$$\begin{aligned} & \sum_{n=0}^{\infty}\lim_{\lambda\rightarrow0} \beta _{n} (x\mid\lambda )\frac{t^{n}}{n!} \\ &\quad =\lim_{\lambda\rightarrow0}\frac{t}{ (1+\lambda t )^{\frac {1}{\lambda}}-1} (1+\lambda t )^{\frac{x}{\lambda}} \\ & \quad =\frac{t}{e^{t}-1}e^{xt} \\ & \quad =\sum_{n=0}^{\infty}B_{n} (x ) \frac{t^{n}}{n!}, \end{aligned}$$
(1.4)
where \(B_{n} (x )\) are called the Bernoulli polynomials (see [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27]).
Thus, by (1.4), we get
$$ \lim_{\lambda\rightarrow0}\beta_{n} (x\mid\lambda )=B_{n} (x )\quad (n\ge0 ). $$
(1.5)
In [4, 14], the poly-Bernoulli polynomials are given by
$$ \frac{\operatorname{Li}_{k} (1-e^{-t} )}{e^{t}-1}e^{xt}=\sum_{n=0}^{\infty }B_{n}^{ (k )} (x )\frac{t^{n}}{n!}. $$
(1.6)
For \(k=1\), we have
$$\begin{aligned} \frac{\operatorname{Li}_{1} (1-e^{-t} )}{e^{t}-1}e^{xt}&=\frac {t}{e^{t}-1}e^{xt} \\ &=\sum _{n=0}^{\infty}B_{n} (x ) \frac {t^{n}}{n!}. \end{aligned}$$
(1.7)

By (1.4) and (1.7), we get \(B_{n}^{ (1 )} (x )=B_{n} (x )\).

The Stirling numbers of the second kind are given by
$$ x^{n}=\sum_{l=0}^{n}S_{2} (n,l ) (x )_{l} \quad (\text{see [1--27]}), $$
(1.8)
and the Stirling numbers of the first kind are defined by
$$ (x )_{n}=x (x-1 )\cdots (x-n+1 )=\sum_{l=0}^{n}S_{1} (n,l )x^{l}\quad (n\ge0 ). $$
(1.9)

The purpose of this paper is to construct the degenerate poly-Bernoulli polynomials and present new and explicit formulas for computing them in terms of the degenerate Bernoulli polynomials and Stirling numbers of the second kind.

2 Degenerate poly-Bernoulli numbers and polynomials

For \(\lambda\in\mathbb{C}\), \(k\in\mathbb{Z}\), we consider the degenerate poly-Bernoulli polynomials given by the generating function
$$ \frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac {1}{\lambda}}-1} (1+\lambda t )^{\frac{x}{\lambda}}=\sum _{n=0}^{\infty}\beta_{n}^{ (k )} (x\mid \lambda )\frac {t^{n}}{n!}. $$
(2.1)

When \(x=0\), \(\beta_{n}^{ (k )} (\lambda )=\beta _{n}^{ (k )} (0\mid\lambda )\) are called the degenerate poly-Bernoulli numbers. Note that \(\beta _{n}^{ (1 )} (x\mid\lambda )=\beta_{n} (x\mid \lambda )\) and \(\lim_{\lambda\rightarrow0}\beta_{n}^{ (k )} (x\mid \lambda )=B_{n}^{ (k )} (x )\).

From (2.1), we can derive the following equation:
$$\begin{aligned} \sum_{n=0}^{\infty}\beta_{n}^{ (k )} (x\mid\lambda )\frac{t^{n}}{n!} & = \biggl(\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} \biggr) (1+\lambda t )^{\frac{x}{\lambda}} \\ & = \Biggl(\sum_{l=0}^{\infty} \beta_{l}^{ (k )} (\lambda )\frac{t^{l}}{l!} \Biggr) \Biggl( \sum_{m=0}^{\infty} (x\mid \lambda )_{m}\frac{t^{m}}{m!} \Biggr) \\ & =\sum_{n=0}^{\infty} \Biggl(\sum _{l=0}^{n}\binom{n}{l}\beta_{l}^{ (k )} (\lambda ) (x\mid\lambda )_{n-l} \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.2)
Thus, by (2.2), we get
$$ \beta_{n}^{ (k )} (x\mid\lambda )=\sum _{l=0}^{n}\binom {n}{l}\beta_{l}^{ (k )} (\lambda ) (x\mid\lambda )_{n-l}. $$
(2.3)
Now, we observe that
$$\begin{aligned} & \frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+t )^{\frac{x}{\lambda }} \\ &\quad =\sum_{n=0}^{\infty}\beta_{n}^{ (k )} (x\mid\lambda )\frac{t^{n}}{n!} \\ &\quad =\frac{ (1+t )^{\frac{x}{\lambda}}}{ (1+\lambda t )^{\frac{1}{\lambda}}-1}\underset{ (k-2)\text{ times}}{\int_{0}^{t} \underbrace{\frac{1}{e^{y}-1}\int_{0}^{y} \frac{1}{e^{y}-1}\int_{0}^{y}\cdots \frac{1}{e^{y}-1}\int_{0}^{y}}} \frac{y}{e^{y}-1}\,dy\cdots dy. \end{aligned}$$
(2.4)
From (2.4), we have
$$\begin{aligned} & \sum_{n=0}^{\infty}\beta_{n}^{ (2 )} (x\mid\lambda )\frac{t^{n}}{n!} \\ & \quad =\frac{ (1+t )^{\frac{x}{\lambda}}}{ (1+\lambda t )^{\frac{1}{\lambda}}-1}\int_{0}^{t} \frac{y}{e^{y}-1}\,dy \\ &\quad =\frac{ (1+t )^{\frac{x}{\lambda}}}{ (1+\lambda t )^{\frac{1}{\lambda}}-1}\sum_{l=0}^{\infty} \frac{B_{l}}{l!}\int_{0}^{t}y^{l}\,dy \\ &\quad = \biggl(\frac{t}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac{x}{\lambda}} \biggr) \Biggl(\sum _{l=0}^{\infty }\frac{B_{l}}{l+1} \frac{t^{l}}{l!} \Biggr) \\ & \quad =\sum_{n=0}^{\infty} \Biggl\{ \sum _{l=0}^{n}\binom{n}{l}\frac {B_{l}}{l+1} \beta_{n-l} (x\mid\lambda ) \Biggr\} \frac {t^{n}}{n!}, \end{aligned}$$
(2.5)
where \(B_{n}=B_{n} (0 )\) are the Bernoulli numbers.

By comparing the coefficients on both sides of (2.5), we obtain the following theorem.

Theorem 2.1

For \(n\ge0\), we have
$$\begin{aligned} \begin{aligned} \beta_{n}^{ (2 )} (x\mid\lambda ) & =\sum _{l=0}^{n}\binom{n}{l}\frac{B_{l}}{l+1} \beta_{n-l} (x\mid\lambda ) \\ & =\beta_{n} (x\mid\lambda )-\frac{n}{4}\beta_{n-1} (x \mid \lambda )+\sum_{l=2}^{n} \binom{n}{l}\frac{B_{l}}{l+1}\beta _{n-l} (x\mid\lambda ). \end{aligned} \end{aligned}$$
Moreover,
$$\beta_{n}^{ (k )} (x\mid\lambda )=\sum _{l=0}^{n}\binom {n}{l}\beta_{l}^{ (k )} (\lambda ) (x\mid\lambda )_{n-l}. $$
By (2.4), we easily get
$$\begin{aligned} & \sum_{n=0}^{\infty}\beta_{n}^{ (k )} (x\mid\lambda )\frac{t^{n}}{n!} \\ & \quad =\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac {1}{\lambda}}-1} (1+t )^{\frac{x}{\lambda}} \\ &\quad =\frac{t}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+t )^{\frac{x}{\lambda}}\frac{\operatorname{Li}_{k} (1-e^{-t} )}{t}. \end{aligned}$$
(2.6)
We observe that
$$\begin{aligned} \frac{1}{t}\operatorname{Li}_{k} \bigl(1-e^{-t} \bigr) & = \frac{1}{t}\sum_{n=1}^{\infty } \frac{1}{n^{k}} \bigl(1-e^{-t} \bigr)^{n} \\ & =\frac{1}{t}\sum_{n=1}^{\infty} \frac{ (-1 )^{n}}{n^{k}}n!\sum_{l=n}^{\infty}S_{2} (l,n )\frac{ (-t )^{l}}{l!} \\ & =\frac{1}{t}\sum_{l=1}^{\infty}\sum _{n=1}^{l}\frac{ (-1 )^{n+l}}{n^{k}}n!S_{2} (l,n )\frac{t^{l}}{l!} \\ & =\sum_{l=0}^{\infty}\sum _{n=1}^{l+1}\frac{ (-1 )^{n+l+1}}{n^{k}}n!\frac{S_{2} (l+1,n )}{l+1} \frac {t^{l}}{l!}. \end{aligned}$$
(2.7)
From (2.6) and (2.7), we have
$$\begin{aligned} & \sum_{n=0}^{\infty}\beta_{n}^{ (k )} (x\mid\lambda )\frac{t^{n}}{n!} \\ & \quad = \Biggl(\sum_{m=0}^{\infty} \beta_{m} (x\mid\lambda )\frac {t^{m}}{m!} \Biggr) \Biggl(\sum _{l=0}^{\infty} \Biggl(\sum _{p=1}^{l+1}\frac { (-1 )^{p+l+1}}{p^{k}}p!\frac{S_{2} (l+1,p )}{l+1} \Biggr)\frac{t^{l}}{l!} \Biggr) \\ & \quad =\sum_{n=0}^{\infty} \Biggl\{ \sum _{l=0}^{n}\binom{n}{l} \Biggl(\sum _{p=1}^{l+1}\frac{ (-1 )^{p+l+1}p!}{p^{k}}\frac{S_{2} (l+1,p )}{l+1} \Biggr)\beta_{n-l} (x\mid\lambda ) \Biggr\} \frac{t^{n}}{n!}. \end{aligned}$$
(2.8)

By comparing the coefficients on both sides of (2.8), we obtain the following theorem.

Theorem 2.2

For \(n\ge0\), we have
$$\beta_{n}^{ (k )} (x\mid\lambda )=\sum _{l=0}^{n}\binom {n}{l} \Biggl(\sum _{p=1}^{l+1}\frac{ (-1 )^{p+l+1}p!}{p^{k}}\frac {S_{2} (l+1,p )}{l+1} \Biggr)\beta_{n-l} (x\mid\lambda ). $$
It is easy to show that
$$\begin{aligned} & \frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x+1}{\lambda}}-\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x}{\lambda}} \\ & \quad = (1+\lambda t )^{\frac{x}{\lambda}} \operatorname{Li}_{k} \bigl(1-e^{-t} \bigr) \\ &\quad = \Biggl(\sum_{l=0}^{\infty} (x\mid\lambda )_{l}\frac {t^{l}}{l!} \Biggr) \Biggl(\sum _{m=1}^{\infty}\frac{ (1-e^{-t} )^{m}}{m^{k}} \Biggr) \\ &\quad = \Biggl(\sum_{l=0}^{\infty} (x\mid\lambda )_{l}\frac {t^{l}}{l!} \Biggr) \Biggl(\sum _{m=0}^{\infty}\frac{ (1-e^{-t} )^{m+1}}{ (m+1 )^{k}} \Biggr) \\ &\quad = \Biggl(\sum_{l=0}^{\infty} (x\mid\lambda )_{l}\frac {t^{l}}{l!} \Biggr) \Biggl(\sum _{p=1}^{\infty} \Biggl(\sum_{m=0}^{p-1} \frac { (-1 )^{m+p+1}}{ (m+1 )^{k}} (m+1 )!S_{2} (p,m+1 ) \Biggr)\frac{t^{p}}{p!} \Biggr) \\ & \quad =\sum_{n=1}^{\infty} \Biggl\{ \sum _{p=1}^{n}\sum_{m=0}^{p-1} \frac{ (-1 )^{m+p+1}}{ (m+1 )^{k}} (m+1 )!S_{2} (p,m+1 )\binom{n}{p} (x\mid\lambda )_{n-p} \Biggr\} \frac {t^{n}}{n!}. \end{aligned}$$
(2.9)
On the other hand,
$$\begin{aligned} & \frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x+1}{\lambda}}-\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x}{\lambda}} \\ &\quad =\sum_{n=0}^{\infty} \bigl\{ \beta_{n}^{ (k )} (x+1\mid \lambda )-\beta_{n}^{ (k )} (x\mid\lambda ) \bigr\} \frac{t^{n}}{n!}. \end{aligned}$$
(2.10)

Therefore, by (2.9) and (2.10), we obtain the following theorem.

Theorem 2.3

For \(n\ge1\), we have
$$\begin{aligned} & \beta_{n}^{ (k )} (x+1\mid\lambda )-\beta_{n}^{ (k )} (x\mid\lambda ) \\ & \quad =\sum_{p=1}^{n} \Biggl(\sum _{m=0}^{p-1}\frac{ (-1 )^{m+k+1}}{ (m+1 )^{k}} (m+1 )!S_{2} (k+m+1 ) \Biggr)\binom{n}{p} (x\mid\lambda )_{n-p}. \end{aligned}$$
Now, we note that
$$\begin{aligned} & \frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x}{\lambda}} \\ & \quad =\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac {d}{\lambda}}-1}\sum_{a=0}^{d-1} (1+ \lambda t )^{\frac {l+x}{\lambda}} \\ &\quad = \biggl(\frac{\operatorname{Li}_{k} (1-e^{-t} )}{t} \biggr)\frac{1}{d}\sum _{a=0}^{d-1}\frac{dt}{ (1+\lambda t )^{\frac{d}{\lambda }}-1} (1+\lambda t )^{\frac{l+x}{\lambda}} \\ & \quad =\sum_{l=0}^{\infty} \Biggl(\sum _{p=1}^{l+1}\frac{ (-1 )^{p+l+1}}{p^{k}}p!\frac{S_{2} (l+1,p )}{l+1} \Biggr)\frac {t^{l}}{l!} \\ & \qquad {}\times\sum_{a=0}^{d-1}\sum _{m=0}^{\infty}\beta _{m} \biggl( \frac{l+x}{d}\,\Big|\,\frac{\lambda}{d} \biggr)d^{m-1} \frac {t^{m}}{m!} \\ &\quad =\sum_{a=0}^{d-1} \Biggl(\sum _{n=0}^{\infty} \Biggl(\sum_{l=0}^{n} \sum_{p=1}^{l+1}\binom{n}{l} \frac{ (-1 )^{p+l+1}}{p^{k}}p!\frac {S_{2} (l+1,p )}{l+1}\beta_{n-l} \biggl( \frac{l+x}{d}\,\Big|\, \frac{\lambda}{d} \biggr)d^{n-l-1} \Biggr) \frac{t^{n}}{n!} \Biggr) \\ &\quad =\sum_{n=0}^{\infty} \Biggl\{ \sum _{a=0}^{d-1}\sum_{l=0}^{n} \sum_{p=1}^{l+1}\binom{n}{l} \frac{ (-1 )^{p+l+1}}{p^{k}}p!\frac {S_{2} (l+1,p )}{l+1}\beta_{n-l} \biggl( \frac{l+x}{d}\,\Big|\, \frac{\lambda}{d} \biggr)d^{n-l-1} \Biggr\} \frac{t^{n}}{n!}, \end{aligned}$$
(2.11)
where d is a fixed positive integer.
On the other hand,
$$\begin{aligned} & \frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+\lambda t )^{\frac {x}{\lambda}} \\ & \quad =\sum_{n=0}^{\infty}\beta_{n}^{ (k )} (x\mid\lambda )\frac{t^{n}}{n!}. \end{aligned}$$
(2.12)

Therefore, by (2.11) and (2.12), we obtain the following theorem.

Theorem 2.4

For \(n\ge0\), \(d\in\mathbb{N}\) and \(k\in\mathbb{Z}\), we have
$$\begin{aligned} & \beta_{n}^{ (k )} (x\mid\lambda ) \\ &\quad =\sum_{a=0}^{d-1}\sum _{l=0}^{n}\sum_{p=1}^{l+1} \binom{n}{l}\frac { (-1 )^{p+l+1}}{p^{k}}p!\frac{S_{2} (l+1,p )}{l+1}\beta_{n-l} \biggl(\frac{l+x}{d}\,\Big|\, \frac{\lambda }{d} \biggr)d^{n-l-1}. \end{aligned}$$
From (2.4), we can derive the following equation:
$$\begin{aligned} & \sum_{n=0}^{\infty}\beta_{n}^{ (k )} (x+y\mid\lambda )\frac{t^{n}}{n!} \\ & \quad =\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac {1}{\lambda}}-1} (1+\lambda t )^{\frac{x+y}{\lambda}} \\ &\quad = \biggl(\frac{\operatorname{Li}_{k} (1-e^{-t} )}{ (1+\lambda t )^{\frac{1}{\lambda}}-1} (1+t\lambda )^{\frac{x}{\lambda }} \biggr) (1+\lambda t )^{\frac{y}{\lambda}} \\ & \quad = \Biggl(\sum_{l=0}^{\infty} \beta_{l}^{ (k )} (x\mid \lambda )\frac{t^{l}}{l!} \Biggr) \Biggl(\sum_{m=0}^{\infty} (y\mid\lambda )_{m}\frac{t^{m}}{m!} \Biggr) \\ & \quad =\sum_{n=0}^{\infty} \Biggl(\sum _{l=0}^{n}\binom{n}{l}\beta_{l}^{ (k )} (x\mid\lambda ) (y\mid\lambda )_{n-l} \Biggr)\frac{t^{n}}{n!}. \end{aligned}$$
(2.13)

Therefore, by (2.13), we obtain the following theorem.

Theorem 2.5

For \(n\ge0\), we have
$$\beta_{n}^{ (k )} (x+y\mid\lambda )=\sum _{l=0}^{n}\binom{n}{l}\beta_{l}^{ (k )} (x\mid\lambda ) (y\mid\lambda )_{n-l}. $$

Remark

$$\begin{aligned} & \frac{d}{dx}\beta_{n}^{ (k )} (x\mid \lambda ) \\ & \quad =\frac{d}{dx}\sum_{l=0}^{n} \binom{n}{l}\beta_{n-l}^{ (k )} (\lambda ) (x\mid\lambda )_{l} \\ & \quad =\sum_{l=0}^{n}\binom{n}{l} \beta_{n-l}^{ (k )} (\lambda )\sum_{j=0}^{l-1} \frac{1}{x-\lambda j}\prod_{i=0}^{l-1} (x-\lambda i ) \\ &\quad =\sum_{l=0}^{n}\binom{n}{l} \beta_{n-l}^{ (k )} (\lambda )\sum_{j=0}^{l-1} \prod_{\substack{i=0\\ i\neq j } }^{l-1} (x-\lambda i ). \end{aligned}$$

Notes

Acknowledgements

The authors would like to thank the referees for their valuable comments.

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© Kim and Kim 2015

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Authors and Affiliations

  1. 1.Department of MathematicsSogang UniversitySeoulRepublic of Korea
  2. 2.Department of MathematicsKwangwoon UniversitySeoulRepublic of Korea

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