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The twisted Daehee numbers and polynomials

Abstract

We consider the Witt-type formula for the n th twisted Daehee numbers and polynomials and investigate some properties of those numbers and polynomials. In particular, the n th twisted Daehee numbers are closely related to higher-order Bernoulli numbers and Bernoulli numbers of the second kind.

1 Introduction

In this paper, we assume that Z p , Q p and C p will, respectively, denote the rings of p-adic integers, the fields of p-adic numbers and the completion of algebraic closure of Q p . The p-adic norm | | p is normalized by |p | p =1/p. Let UD[ Z p ] be the space of uniformly differentiable functions on Z p . For fUD[ Z p ], the p-adic invariant integral on Z p is defined by

I(f) Z p f(x)d μ 0 (x)= lim n 1 p n x = 0 p n 1 f(x)(see [1, 2]).
(1)

Let f 1 be the translation of f with f 1 (x)=f(x+1). Then, by (1), we get

I( f 1 )=I(f)+ f (0),where  f (0)= d f ( x ) d x | x = 0 .
(2)

As is known, the Stirling number of the first kind is defined by

( x ) n =x(x1)(xn+1)= l = 0 n S 1 (n,l) x l ,
(3)

and the Stirling number of the second kind is given by the generating function to be

( e t 1 ) m =m! l = m S 2 (l,m) t l l ! (see [3–5]).
(4)

For αZ, the Bernoulli polynomials of order α are defined by the generating function to be

( t e t 1 ) α e x t = n = 0 B n ( α ) (x) t n n ! (see [3, 6, 7]).
(5)

When x=0, B n ( α ) = B n ( α ) (0) are called the Bernoulli numbers of order α.

For nN, let T p be the p-adic locally constant space defined by

T p = n 1 C p n = lim n C p n ,

where C p n ={ω| ω p n =1} is the cyclic group of order p n . It is well known that the twisted Bernoulli polynomials are defined as

t ξ e t 1 e x t = n = 0 B n , ξ (x) t n n ! ,ξ T p (see [8]),

and the twisted Bernoulli numbers B n , ξ are defined as B n , ξ = B n , ξ (0).

Recently, Kim and Kim introduced the Daehee numbers and polynomials which are given by the generating function to be

( log ( 1 + t ) t ) ( 1 + t ) x = n = 0 D n (x) t n n ! (see [9, 10]).
(6)

In the special case, x=0, D n = D n (0) are called the n th Daehee numbers.

In the viewpoint of generalization of the Daehee numbers and polynomials, we consider the n th twisted Daehee polynomials defined by the generating function to be

( log ( 1 + ξ t ) ξ t ) ( 1 + ξ t ) x = n = 0 D n , ξ (x) t n n !
(7)

In the special case, x=0, D n , ξ = D n , ξ (0) are called the n th twisted Daehee numbers.

In this paper, we give a p-adic integral representation of the n th twisted Daehee numbers and polynomials, which are called the Witt-type formula for the n th twisted Daehee numbers and polynomials. We can derive some interesting properties related to the n th twisted Daehee numbers and polynomials. For this idea, we are indebted to papers [9, 10].

2 Witt-type formula for the n th twisted Daehee numbers and polynomials

First, we consider the following integral representation associated with falling factorial sequences:

Z p ( x ) n d μ 0 (x),where n Z + =N{0}(see [10]).
(8)

By (8), we get

n = 0 ξ n Z p ( x ) n d μ 0 ( x ) t n n ! = Z p n = 0 ξ n ( x n ) t n d μ 0 ( x ) = Z p ( 1 + ξ t ) x d μ 0 ( x ) ,
(9)

where t C p with |t | p < 1 p 1 .

For t C p with |t | p < p 1 p 1 , let us take f(x)= ( 1 + ξ t ) x . Then, from (2), we have

Z p ( 1 + ξ t ) x d μ 0 (x)= log ( 1 + ξ t ) ξ t .
(10)

By (9) and (10), we see that

n = 0 D n , ξ t n n ! = log ( 1 + ξ t ) ξ t = Z p ( 1 + ξ t ) x d μ 0 ( x ) = n = 0 ξ n Z p ( x ) n d μ 0 ( x ) t n n ! .
(11)

Therefore, by (11), we obtain the following theorem.

Theorem 1 For n0, we have

ξ n Z p ( x ) n d μ 0 (x)= D n , ξ .

For nZ, it is known that

( log ( 1 + t ) t ) n ( 1 + t ) x 1 = k = 0 B k ( k n + 1 ) (x) t k k ! (see [3–5]).
(12)

Thus, replacing t by e ξ t 1 in (12), we get

D k , ξ = ξ n Z p ( x ) k d μ 0 (x)= ξ n B k ( k + 2 ) (1)(k0),
(13)

where B k ( n ) (x) are the Bernoulli polynomials of order n.

In the special case, x=0, B k ( n ) = B k ( n ) (0) are called the n th Bernoulli numbers of order n.

From (11), we note that

( 1 + ξ t ) x Z p ( 1 + ξ t ) y d μ 0 ( y ) = ( log ( 1 + ξ t ) ξ t ) ( 1 + ξ t ) x = n = 0 D n , ξ ( x ) t n n ! .
(14)

Thus, by (14), we get

ξ n Z p ( x + y ) n d μ 0 (y)= D n , ξ (x)(n0),
(15)

and, from (12), we have

D n , ξ (x)= ξ n B n ( n + 2 ) (x+1).
(16)

Therefore, by (15) and (16), we obtain the following theorem.

Theorem 2 For n0, we have

D n , ξ (x)= ξ n Z p ( x + y ) n d μ 0 (y)

and

D n , ξ (x)= ξ n B n ( n + 2 ) (x+1).

By Theorem 1, we easily see that

D n , ξ = ξ n l = 0 n S 1 (n,l) B l ,
(17)

where B l are the ordinary Bernoulli numbers.

From Theorem 2, we have

D n , ξ ( x ) = ξ n Z p ( x + y ) n d μ 0 ( y ) = ξ n l = 0 n S 1 ( n , l ) B l ( x ) ,
(18)

where B l (x) are the Bernoulli polynomials defined by a generating function to be

t e t 1 e x t = n = 0 B n (x) t n n ! .

Therefore, by (17) and (18), we obtain the following corollary.

Corollary 3 For n0, we have

D n , ξ (x)= ξ n l = 0 n S 1 (n,l) B l (x).

In (11), we have

log ( 1 + ξ t ) ξ t ( 1 + ξ t ) x = n = 0 D n , ξ (x) t n n ! .
(19)

Replacing t by e t 1 ξ , we put

n = 0 D n , ξ ( x ) 1 n ! ( e t 1 ξ ) n = log ( 1 + ξ ( e t 1 ξ ) ) ξ ( e t 1 ξ ) ( 1 + ξ ( e t 1 ξ ) ) x = t ξ e t 1 ( ξ e t ) x = ξ x t ξ e t 1 e t x = ξ x n = 0 B n , ξ ( x ) t n n ! .
(20)

Therefore, we have

n = 0 B n , ξ ( x ) t n n ! = n = 0 D n , ξ ( x ) 1 n ! ( e t 1 ξ ) n = n = 0 D n , ξ ( x ) 1 n ! ξ n n ! m = n S 2 ( m , n ) t m m ! = m = 0 n = 0 m D n , ξ ( x ) ξ n S 2 ( m , n ) ,
(21)

where S 2 (m,n) is the Stirling number of the second kind.

Hence,

ξ x B n , ξ (x)= n = 0 m D n , ξ (x) ξ n S 2 (m,n).
(22)

Therefore, we have

B m , ξ (x)= n = 0 m D n , ξ (x) ξ n x S 2 (m,n).
(23)

In particular,

B m , ξ = n = 0 m D n , ξ ξ n S 2 (m,n).
(24)

Therefore, by (20) and (23), we obtain the following theorem.

Theorem 4 For m0, we have

B m , ξ (x)= n = 0 m ξ n x D n , ξ (x) S 2 (m,n).

In particular,

B m , ξ = n = 0 m ξ n D n , ξ S 2 (m,n).

Remark For m0, by (18), we have

ξ n Z p ( x + y ) m d μ 0 (y)= ξ n n = 0 m D n (x) S 2 (m,n).

For n Z n 0 , the rising factorial sequence is defined by

x ( n ) =x(x+1)(x+n1).
(25)

Let us define the n th twisted Daehee numbers of the second kind as follows:

D ˆ n , ξ = ξ n Z p ( x ) n d μ 0 (x)(n Z n 0 ).
(26)

By (26), we get

x ( n ) = ( 1 ) n ( x ) n = l = 0 n S 1 (n,l) ( 1 ) n l x l .
(27)

From (26) and (27), we have

D ˆ n , ξ = ξ n Z p ( x ) n d μ 0 ( x ) = ξ n Z p x ( n ) ( 1 ) n d μ 0 ( x ) = ξ n l = 0 n S 1 ( n , l ) ( 1 ) l B l .
(28)

Therefore, by (28), we obtain the following theorem.

Theorem 5 For n0, we have

D ˆ n , ξ = ξ n l = 0 n S 1 (n,l) ( 1 ) l B l .

Let us consider the generating function of the n th twisted Daehee numbers of the second kind as follows:

n = 0 D ˆ n , ξ t n n ! = n = 0 ξ n Z p ( x ) n d μ 0 ( x ) t n n ! = Z p n = 0 ξ n ( x n ) t n d μ 0 ( x ) = Z p ( 1 + ξ t ) x d μ 0 ( x ) .
(29)

From (2), we can derive the following equation:

Z p ( 1 + ξ t ) x d μ 0 (x)= ( 1 + ξ t ) log ( 1 + ξ t ) ξ t ,
(30)

where |t | p < p 1 p .

By (29) and (30), we get

1 ξ t ( 1 + ξ t ) log ( 1 + ξ t ) = Z p ( 1 + ξ t ) x d μ 0 ( x ) = n = 0 D ˆ n , ξ t n n ! .
(31)

Let us consider the n th twisted Daehee polynomials of the second kind as follows:

( 1 + ξ t ) log ( 1 + ξ t ) ξ t 1 ( 1 + ξ t ) x = n = 0 D ˆ n , ξ (x) t n n ! .
(32)

Then, by (32), we get

Z p ( 1 + ξ t ) x y d μ 0 (y)= n = 0 D ˆ n , ξ (x) t n n ! .
(33)

From (33), we get

D ˆ n , ξ ( x ) = ξ n Z p ( x y ) n d μ 0 ( y ) ( n 0 ) = ξ n l = 0 n ( 1 ) l S 1 ( n , l ) B l ( x ) .
(34)

Therefore, by (34), we obtain the following theorem.

Theorem 6 For n0, we have

D ˆ n , ξ (x)= ξ n Z p ( x y ) n d μ 0 (y)= ξ n l = 0 n ( 1 ) l S 1 (n,l) B l (x).

From (32) and (33), we have

log ( 1 + ξ t ) ξ t ( 1 + ξ t ) 1 x = n = 0 D ˆ n , ξ (x) t n n ! .
(35)

Replacing t by e t 1 ξ , we get

n = 0 D ˆ n , ξ ( x ) 1 n ! ( e t 1 ξ ) n = log ( 1 + ξ ( e t 1 ξ ) ) ξ ( e t 1 ξ ) ( 1 + ξ ( e t 1 ξ ) ) 1 x = t ξ e t 1 ( ξ e t ) 1 x = ξ 1 x t ξ e t 1 e t ( 1 x ) = ξ 1 x n = 0 B n , ξ ( 1 x ) t n n ! .
(36)

Therefore, we have

ξ 1 x m = 0 B m , ξ ( 1 x ) t n n ! = n = 0 D ˆ n , ξ ( x ) ( e t 1 ξ ) n n ! = n = 0 D ˆ n , ξ ( x ) 1 n ! ξ n n ! m = n S 2 ( m , n ) t m m ! = m = 0 ( n = 0 m D ˆ n , ξ ( x ) ξ n S 2 ( m , n ) ) t m m ! .
(37)

Hence,

ξ 1 x B n , ξ (1x)= n = 0 m D ˆ n , ξ (x) ξ n S 2 (m,n).
(38)

Therefore, we have

B m , ξ (1x)= n = 0 m D ˆ n , ξ (x) ξ n + x 1 S 2 (m,n).
(39)

Therefore, by (37) and (38), we obtain the following theorem.

Theorem 7 For m0, we have

B m , ξ (1x)= n = 0 m ξ m + x 1 D ˆ n , ξ (x) S 2 (m,n).

From Theorem 1 and (26), we have

( 1 ) n D n , ξ n ! = ( 1 ) n ξ n Z p ( x n ) d μ 0 ( x ) = ξ n Z p ( x + n 1 n ) d μ 0 ( x ) = ξ n m = 0 n ( n 1 n m ) Z p ( x m ) d μ 0 ( x ) = m = 0 n ( n 1 n m ) ξ n m D ˆ m , ξ m ! = m = 1 n ( n 1 m 1 ) ξ n m D ˆ m , ξ m !
(40)

and

( 1 ) n D ˆ n , ξ n ! = ( 1 ) n ξ n Z p ( x n ) d μ 0 ( x ) = ξ n Z p ( x + n 1 n ) d μ 0 ( x ) = ξ n m = 0 n ( n 1 n m ) 0 1 ( x m ) d μ 0 ( x ) = m = 0 n ( n 1 m 1 ) ξ n m D m , ξ m ! = m = 1 n ( n 1 m 1 ) ξ n m D m , ξ m ! .
(41)

Therefore, by (40) and (41), we obtain the following theorem.

Theorem 8 For nN, we have

( 1 ) n D n , ξ n ! = m = 1 n ( n 1 m 1 ) ξ n m D ˆ m , ξ m !

and

( 1 ) n D ˆ n , ξ n ! = m = 1 n ( n 1 m 1 ) ξ n m D m , ξ m ! .

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The authors are grateful for the valuable comments and suggestions of the referees.

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Correspondence to Jongkyum Kwon.

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Park, JW., Rim, SH. & Kwon, J. The twisted Daehee numbers and polynomials. Adv Differ Equ 2014, 1 (2014). https://doi.org/10.1186/1687-1847-2014-1

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Keywords

  • the n th twisted Daehee numbers and polynomials
  • Bernoulli numbers of the second kind
  • higher-order Bernoulli numbers