# 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

$t log ( 1 + t ) ( 1 + t ) x = ∑ n = 0 ∞ b n (x) t n n ! (see [1–3]).$
(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 $b 0 =1$, $b 1 =1/2$, $b 2 =−1/12$, $b 3 =1/24$, $b 4 =−19/720$, $b 5 =3/160,…$ .

From (1), we have

$b n (x)= ∑ l = 0 n ( n l ) b l ( x ) n − l ,$
(2)

where $( x ) n =x(x−1)⋯(x−n+1)$ ($n≧0$). The Stirling number of the second kind is defined by

$x n = ∑ l = 0 n S 2 (n,l) ( x ) l (n≧0).$
(3)

The ordinary Bernoulli polynomials are given by

$t e t − 1 e x t = ∑ n = 0 ∞ B n (x) t n n ! (see [1–18]).$
(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

$Li k (x)= ∑ n = 1 ∞ x n n k (k∈Z)(see [8–10]).$
(5)

For $k=1$, $Li 1 (x)= ∑ n = 1 ∞ x n n =−log(1−x)$. The Stirling number of the first kind is defined by

$( x ) n = ∑ l = 0 n S 1 (n,l) x l (n≥0)(see [16]).$
(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 $k∈Z$, we consider the poly-Bernoulli polynomials $b n ( k ) (x)$ of the second kind:

$Li k ( 1 − e − t ) log ( 1 + t ) ( 1 + t ) x = ∑ n = 0 ∞ b n ( k ) (x) t n n ! .$
(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

$Li k ( 1 − e − t ) log ( 1 + t ) ( 1 + t ) x = t log ( 1 + t ) ( 1 + t ) x = ∑ n = 0 ∞ b n (x) t n n ! .$
(8)

By (7) and (8), we get

$b n ( 1 ) (x)= b n (x)(n≥0).$
(9)

It is well known that

$t ( 1 + t ) x − 1 log ( 1 + t ) = ∑ n = 0 ∞ B n ( n ) (x) t n n ! ,$
(10)

where $B n ( α ) (x)$ are the Bernoulli polynomials of order α which are given by the generating function to be

$( t e t − 1 ) α e x t = ∑ n = 0 ∞ B n ( α ) (x) t n n ! (see [1–18]).$

By (1) and (10), we get

$b n (x)= B n ( n ) (x+1)(n≥0).$

Now, we observe that

(11)

Thus, by (11), we get

$∑ n = 0 ∞ b n ( 2 ) ( x ) t n n ! = ( 1 + t ) x log ( 1 + t ) ∫ 0 t x e x − 1 d x = ( 1 + t ) x log ( 1 + t ) ∑ l = 0 ∞ B l l ! ∫ 0 t x l d x = ( t log ( 1 + t ) ) ( 1 + t ) x ∑ l = 0 ∞ B l ( l + 1 ) t l l ! = ∑ n = 0 ∞ { ∑ l = 0 n ( n l ) B l b n − l ( x ) l + 1 } t n n ! .$
(12)

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

Theorem 2.1 For $n≥0$ we have

$b n ( 2 ) (x)= ∑ l = 0 n ( n l ) B l b n − l ( x ) l + 1 .$

From (11), we have

$∑ n = 0 ∞ b n ( k ) ( x ) t n n ! = Li k ( 1 − e − t ) log ( 1 + t ) ( 1 + t ) x = t log ( 1 + t ) Li k ( 1 − e − t ) t ( 1 + t ) x .$
(13)

We observe that

$1 t Li k ( 1 − e − t ) = 1 t ∑ n = 1 ∞ 1 n k ( 1 − e − t ) n = 1 t ∑ n = 1 ∞ ( − 1 ) n n k n ! ∑ l = n ∞ S 2 ( l , n ) ( − t ) l l ! = 1 t ∑ l = 1 ∞ ∑ n = 1 l ( − 1 ) n + l n k n ! S 2 ( l , n ) t l l ! = ∑ l = 0 ∞ ∑ n = 1 l + 1 ( − 1 ) n + l + 1 n k n ! S 2 ( l + 1 , n ) l + 1 t l l ! .$
(14)

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

$∑ n = 0 ∞ b n ( k ) ( x ) t n n ! = ( ∑ m = 0 ∞ b m ( x ) t m m ! ) { ∑ l = 0 ∞ ( ∑ p = 1 l + 1 ( − 1 ) p + l + 1 p k p ! S 2 ( l + 1 , p ) l + 1 ) t l l ! } = ∑ n = 0 ∞ { ∑ l = 0 n ( n l ) ( ∑ p = 1 l + 1 ( − 1 ) p + l + 1 p ! p k S 2 ( l + 1 , p ) l + 1 ) b n − l ( x ) } t n n ! .$
(15)

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

Theorem 2.2 For $n≥0$, we have

$b n ( k ) (x)= ∑ l = 0 n ( n l ) ( ∑ p = 1 l + 1 ( − 1 ) p + l + 1 p k p ! S 2 ( l + 1 , p ) l + 1 ) b n − l (x).$

By (7), we get

$∑ n = 0 ∞ ( b n ( k ) ( x + 1 ) − b n ( k ) ( x ) ) t n n ! = Li k ( 1 − e − t ) log ( 1 + t ) ( 1 + t ) x + 1 − Li k ( 1 − e − t ) log ( 1 + t ) ( 1 + t ) x = t Li k ( 1 − e − t ) log ( 1 + t ) ( 1 + t ) x = ( t log ( 1 + t ) ( 1 + t ) x ) Li k ( 1 − e − t ) = ( ∑ l = 0 ∞ b l ( x ) l ! t l ) { ∑ p = 1 ∞ ( ∑ m = 1 p ( − 1 ) m + p m ! m k S 2 ( p , m ) ) } t p p !$
(16)
$= ∑ n = 1 ∞ ( ∑ p = 1 n ∑ m = 1 p ( − 1 ) m + p m k m ! S 2 ( p , m ) b n − p ( x ) n ! ( n − p ) ! p ! ) t n n ! = ∑ n = 1 ∞ { ∑ p = 1 n ∑ m = 1 p ( n p ) ( − 1 ) m + p m ! m k S 2 ( p , m ) b n − p ( x ) } t n n ! .$
(17)

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

Theorem 2.3 For $n≥1$, we have

$b n ( k ) (x+1)− b n ( k ) (x)= ∑ p = 1 n ∑ m = 1 p ( n p ) ( − 1 ) m + p m ! m k S 2 (p,m) b n − p (x).$
(18)

From (13), we have

$∑ n = 0 ∞ b n ( k ) ( x + y ) t n n ! = ( Li k ( 1 − e − t ) log ( 1 + t ) ) k ( 1 + t ) x + y = ( Li k ( 1 − e − t ) log ( 1 + t ) ) k ( 1 + t ) x ( 1 + t ) y = ( ∑ l = 0 ∞ b l ( k ) ( x ) t l l ! ) ( ∑ m = 0 ∞ ( y ) m t m m ! ) = ∑ n = 0 ∞ ( ∑ l = 0 n ( y ) l b n − l ( k ) ( x ) n ! ( n − l ) ! l ! ) t n n ! = ∑ n = 0 ∞ ( ∑ l = 0 n ( n l ) b n − l ( k ) ( x ) ( y ) l ) t n n ! .$
(19)

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

Theorem 2.4 For $n≥0$, we have

$b n ( k ) (x+y)= ∑ l = 0 n ( n l ) b n − l ( k ) (x) ( y ) l .$

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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.

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The authors declare that they have no competing interests.

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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