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 ${\mathbb{Z}}_p$, ${\mathbb{Q}}_p$ and ${\mathbb{C}}_p$ will, respectively, denote the rings of p-adic integers, the fields of p-adic numbers and the completion of algebraic closure of ${\mathbb{Q}}_p$. The p-adic norm $|\cdot {|}_p$ is normalized by $|p{|}_p=1/p$. Let $UD[{\mathbb{Z}}_p]$ be the space of uniformly differentiable functions on ${\mathbb{Z}}_p$. For $f\in UD[{\mathbb{Z}}_p]$, the p-adic invariant integral on ${\mathbb{Z}}_p$ is defined by

$$I(f){\int }_{{\mathbb{Z}}_p}f(x)d{\mu }_0(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{p^n}\sum _{x=0}^{p^n-1}f(x)(\text{see [1, 2]}).$$

(1)

Let $f_1$ be the translation of f with $f_1(x)=f(x+1)$. Then, by (1), we get

$$I(f_1)=I(f)+f^{\prime }(0),\text{where }{f}^{\prime }(0)=\frac{df(x)}{dx}{|}_{x=0}.$$

(2)

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

$${(x)}_n=x(x-1)\cdots (x-n+1)=\sum _{l=0}^n{S}_1(n,l)x^l,$$

(3)

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

$${(e^t-1)}^m=m!\sum _{l=m}^{\mathrm{\infty }}{S}_2(l,m)\frac{t^l}{l!}(\text{see [3–5]}).$$

(4)

For $\alpha \in \mathbb{Z}$, the Bernoulli polynomials of order α are defined by the generating function to be

$${\left(\frac{t}{e^t-1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(\alpha )}(x)\frac{t^n}{n!}(\text{see [3, 6, 7]}).$$

(5)

When $x=0$, $B_n^{(\alpha )}=B_n^{(\alpha )}(0)$ are called the Bernoulli numbers of order α.

For $n\in \mathbb{N}$, let $T_p$ be the p-adic locally constant space defined by

$$T_p=\bigcup _{n\ge 1}{C}_{p^n}=\underset{n\to \mathrm{\infty }}{lim}{C}_{p^n},$$

where $C_{p^n}=\{\omega |{\omega }^{p^n}=1\}$ is the cyclic group of order $p^n$. It is well known that the twisted Bernoulli polynomials are defined as

$$\frac{t}{\xi e^t-1}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }(x)\frac{t^n}{n!},\xi \in T_p(\text{see [8]}),$$

and the twisted Bernoulli numbers $B_{n,\xi }$ are defined as $B_{n,\xi }=B_{n,\xi }(0)$.

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

$$\left(\frac{log(1+t)}{t}\right){(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{D}_n(x)\frac{t^n}{n!}(\text{see [9, 10]}).$$

(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

$$\left(\frac{log(1+\xi t)}{\xi t}\right){(1+\xi t)}^x=\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }(x)\frac{t^n}{n!}$$

(7)

In the special case, $x=0$, $D_{n,\xi }=D_{n,\xi }(0)$ 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:

$${\int }_{{\mathbb{Z}}_p}{(x)}_{n}d{\mu }_0(x),\text{where }n\in {\mathbb{Z}}_{+}=\mathbb{N}\cup \{0\}(\text{see [10]}).$$

(8)

By (8), we get

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{\xi }^n{\int }_{{\mathbb{Z}}_p}{(x)}_{n}d{\mu }_0(x)\frac{t^n}{n!}& =& {\int }_{{\mathbb{Z}}_p}\sum _{n=0}^{\mathrm{\infty }}{\xi }^n\left(\genfrac{}{}{0ex}{}{x}{n}\right)t^{n}d{\mu }_0(x)\\ =& {\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{x}d{\mu }_0(x),\end{array}$$

(9)

where $t\in C_p$ with $|t{|}_p<-\frac{1}{p-1}$.

For $t\in C_p$ with $|t{|}_p<p^{-\frac{1}{p-1}}$, let us take $f(x)={(1+\xi t)}^x$. Then, from (2), we have

$${\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{x}d{\mu }_0(x)=\frac{log(1+\xi t)}{\xi t}.$$

(10)

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

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\frac{t^n}{n!}& =& \frac{log(1+\xi t)}{\xi t}\\ =& {\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{x}d{\mu }_0(x)\\ =& \sum _{n=0}^{\mathrm{\infty }}{\xi }^n{\int }_{{\mathbb{Z}}_p}{(x)}_{n}d{\mu }_0(x)\frac{t^n}{n!}.\end{array}$$

(11)

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

Theorem 1 For $n\ge 0$, we have

$${\xi }^n{\int }_{{\mathbb{Z}}_p}{(x)}_{n}d{\mu }_0(x)=D_{n,\xi }.$$

For $n\in \mathbb{Z}$, it is known that

$${\left(\frac{log(1+t)}{t}\right)}^n{(1+t)}^{x-1}=\sum _{k=0}^{\mathrm{\infty }}{B}_k^{(k-n+1)}(x)\frac{t^k}{k!}(\text{see [3–5]}).$$

(12)

Thus, replacing t by $e^{\xi t}-1$ in (12), we get

$$D_{k,\xi }={\xi }^n{\int }_{{\mathbb{Z}}_p}{(x)}_{k}d{\mu }_0(x)={\xi }^n{B}_k^{(k+2)}(1)(k\ge 0),$$

(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

$$\begin{array}{rcl}{(1+\xi t)}^x{\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{y}d{\mu }_0(y)& =& \left(\frac{log(1+\xi t)}{\xi t}\right){(1+\xi t)}^x\\ =& \sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }(x)\frac{t^n}{n!}.\end{array}$$

(14)

Thus, by (14), we get

$${\xi }^n{\int }_{{\mathbb{Z}}_p}{(x+y)}_{n}d{\mu }_0(y)=D_{n,\xi }(x)(n\ge 0),$$

(15)

and, from (12), we have

$$D_{n,\xi }(x)={\xi }^n{B}_n^{(n+2)}(x+1).$$

(16)

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

Theorem 2 For $n\ge 0$, we have

$$D_{n,\xi }(x)={\xi }^n{\int }_{{\mathbb{Z}}_p}{(x+y)}_{n}d{\mu }_0(y)$$

and

$$D_{n,\xi }(x)={\xi }^n{B}_n^{(n+2)}(x+1).$$

By Theorem 1, we easily see that

$$D_{n,\xi }={\xi }^n\sum _{l=0}^n{S}_1(n,l)B_l,$$

(17)

where $B_l$ are the ordinary Bernoulli numbers.

From Theorem 2, we have

$$\begin{array}{rcl}{D}_{n,\xi }(x)& =& {\xi }^n{\int }_{{\mathbb{Z}}_p}{(x+y)}_{n}d{\mu }_0(y)\\ =& {\xi }^n\sum _{l=0}^n{S}_1(n,l)B_l(x),\end{array}$$

(18)

where $B_l(x)$ are the Bernoulli polynomials defined by a generating function to be

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

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

Corollary 3 For $n\ge 0$, we have

$$D_{n,\xi }(x)={\xi }^n\sum _{l=0}^n{S}_1(n,l)B_l(x).$$

In (11), we have

$$\frac{log(1+\xi t)}{\xi t}{(1+\xi t)}^x=\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }(x)\frac{t^n}{n!}.$$

(19)

Replacing t by $e^t-\frac{1}{\xi }$, we put

$$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }(x)\frac{1}{n!}{(e^t-\frac{1}{\xi })}^n\hfill \\ =\frac{log(1+\xi (e^t-\frac{1}{\xi }))}{\xi (e^t-\frac{1}{\xi })}{(1+\xi (e^t-\frac{1}{\xi }))}^x\hfill \\ =\frac{t}{\xi e^t-1}{\left(\xi e^t\right)}^x\hfill \\ ={\xi }^x\frac{t}{\xi e^t-1}{e}^{tx}\hfill \\ ={\xi }^x\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }(x)\frac{t^n}{n!}.\hfill \end{array}$$

(20)

Therefore, we have

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }(x)\frac{t^n}{n!}& =& \sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }(x)\frac{1}{n!}{(e^t-\frac{1}{\xi })}^n\\ =& \sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }(x)\frac{1}{n!}{\xi }^{-n}n!\sum _{m=n}^{\mathrm{\infty }}{S}_2(m,n)\frac{t^m}{m!}\\ =& \sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^m{D}_{n,\xi }(x){\xi }^{-n}{S}_2(m,n),\end{array}$$

(21)

where $S_2(m,n)$ is the Stirling number of the second kind.

Hence,

$${\xi }^x{B}_{n,\xi }(x)=\sum _{n=0}^m{D}_{n,\xi }(x){\xi }^{-n}{S}_2(m,n).$$

(22)

Therefore, we have

$$B_{m,\xi }(x)=\sum _{n=0}^m{D}_{n,\xi }(x){\xi }^{-n-x}{S}_2(m,n).$$

(23)

In particular,

$$B_{m,\xi }=\sum _{n=0}^m{D}_{n,\xi }{\xi }^{-n}{S}_2(m,n).$$

(24)

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

Theorem 4 For $m\ge 0$, we have

$$B_{m,\xi }(x)=\sum _{n=0}^m{\xi }^{-n-x}{D}_{n,\xi }(x)S_2(m,n).$$

In particular,

$$B_{m,\xi }=\sum _{n=0}^m{\xi }^{-n}{D}_{n,\xi }{S}_2(m,n).$$

Remark For $m\ge 0$, by (18), we have

$${\xi }^n{\int }_{{\mathbb{Z}}_p}{(x+y)}^{m}d{\mu }_0(y)={\xi }^n\sum _{n=0}^m{D}_n(x)S_2(m,n).$$

For $n\in {\mathbb{Z}}_{n\ge 0}$, the rising factorial sequence is defined by

$$x^{(n)}=x(x+1)\cdots (x+n-1).$$

(25)

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

$${\hat{D}}_{n,\xi }={\xi }^n{\int }_{{\mathbb{Z}}_p}{(-x)}_{n}d{\mu }_0(x)(n\in {\mathbb{Z}}_{n\ge 0}).$$

(26)

By (26), we get

$$x^{(n)}={(-1)}^n{(-x)}_n=\sum _{l=0}^n{S}_1(n,l){(-1)}^{n-l}{x}^l.$$

(27)

From (26) and (27), we have

$$\begin{array}{rcl}{\hat{D}}_{n,\xi }& =& {\xi }^n{\int }_{{\mathbb{Z}}_p}{(-x)}_{n}d{\mu }_0(x)\\ =& {\xi }^n{\int }_{{\mathbb{Z}}_p}{x}^{(n)}{(-1)}^{n}d{\mu }_0(x)\\ =& {\xi }^n\sum _{l=0}^n{S}_1(n,l){(-1)}^l{B}_l.\end{array}$$

(28)

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

Theorem 5 For $n\ge 0$, we have

$${\hat{D}}_{n,\xi }={\xi }^n\sum _{l=0}^n{S}_1(n,l){(-1)}^l{B}_l.$$

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

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }\frac{t^n}{n!}& =& \sum _{n=0}^{\mathrm{\infty }}{\xi }^n{\int }_{{\mathbb{Z}}_p}{(-x)}_{n}d{\mu }_0(x)\frac{t^n}{n!}\\ =& {\int }_{{\mathbb{Z}}_p}\sum _{n=0}^{\mathrm{\infty }}{\xi }^n\left(\genfrac{}{}{0ex}{}{-x}{n}\right)t^{n}d{\mu }_0(x)\\ =& {\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{-x}d{\mu }_0(x).\end{array}$$

(29)

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

$${\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{-x}d{\mu }_0(x)=\frac{(1+\xi t)log(1+\xi t)}{\xi t},$$

(30)

where $|t{|}_p<p^{-\frac{1}{p}}$.

By (29) and (30), we get

$$\begin{array}{rcl}\frac{1}{\xi t}(1+\xi t)log(1+\xi t)& =& {\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{-x}d{\mu }_0(x)\\ =& \sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }\frac{t^n}{n!}.\end{array}$$

(31)

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

$$\frac{(1+\xi t)log(1+\xi t)}{\xi t}\frac{1}{{(1+\xi t)}^x}=\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }(x)\frac{t^n}{n!}.$$

(32)

Then, by (32), we get

$${\int }_{{\mathbb{Z}}_p}{(1+\xi t)}^{-x-y}d{\mu }_0(y)=\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }(x)\frac{t^n}{n!}.$$

(33)

From (33), we get

$$\begin{array}{rcl}{\hat{D}}_{n,\xi }(x)& =& {\xi }^n{\int }_{{\mathbb{Z}}_p}{(-x-y)}_{n}d{\mu }_0(y)(n\ge 0)\\ =& {\xi }^n\sum _{l=0}^n{(-1)}^l{S}_1(n,l)B_l(x).\end{array}$$

(34)

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

Theorem 6 For $n\ge 0$, we have

$${\hat{D}}_{n,\xi }(x)={\xi }^n{\int }_{{\mathbb{Z}}_p}{(-x-y)}_{n}d{\mu }_0(y)={\xi }^n\sum _{l=0}^n{(-1)}^l{S}_1(n,l)B_l(x).$$

From (32) and (33), we have

$$\frac{log(1+\xi t)}{\xi t}{(1+\xi t)}^{1-x}=\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }(x)\frac{t^n}{n!}.$$

(35)

Replacing t by $e^t-\frac{1}{\xi }$, we get

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }(x)\frac{1}{n!}{(e^t-\frac{1}{\xi })}^n& =& \frac{log(1+\xi (e^t-\frac{1}{\xi }))}{\xi (e^t-\frac{1}{\xi })}{(1+\xi (e^t-\frac{1}{\xi }))}^{1-x}\\ =& \frac{t}{\xi e^t-1}{\left(\xi e^t\right)}^{1-x}\\ =& {\xi }^{1-x}\frac{t}{\xi e^t-1}{e}^{t(1-x)}\\ =& {\xi }^{1-x}\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }(1-x)\frac{t^n}{n!}.\end{array}$$

(36)

Therefore, we have

$$\begin{array}{rcl}{\xi }^{1-x}\sum _{m=0}^{\mathrm{\infty }}{B}_{m,\xi }(1-x)\frac{t^n}{n!}& =& \sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }(x)\frac{{(e^t-\frac{1}{\xi })}^n}{n!}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_{n,\xi }(x)\frac{1}{n!}{\xi }^{-n}n!\sum _{m=n}^{\mathrm{\infty }}{S}_2(m,n)\frac{t^m}{m!}\\ =& \sum _{m=0}^{\mathrm{\infty }}(\sum _{n=0}^m{\hat{D}}_{n,\xi }(x){\xi }^{-n}{S}_2(m,n))\frac{t^m}{m!}.\end{array}$$

(37)

Hence,

$${\xi }^{1-x}{B}_{n,\xi }(1-x)=\sum _{n=0}^m{\hat{D}}_{n,\xi }(x){\xi }^{-n}{S}_2(m,n).$$

(38)

Therefore, we have

$$B_{m,\xi }(1-x)=\sum _{n=0}^m{\hat{D}}_{n,\xi }(x){\xi }^{-n+x-1}{S}_2(m,n).$$

(39)

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

Theorem 7 For $m\ge 0$, we have

$$B_{m,\xi }(1-x)=\sum _{n=0}^m{\xi }^{-m+x-1}{\hat{D}}_{n,\xi }(x)S_2(m,n).$$

From Theorem 1 and (26), we have

$$\begin{array}{rcl}{(-1)}^n\frac{D_{n,\xi }}{n!}& =& {(-1)}^n{\xi }^n{\int }_{{\mathbb{Z}}_p}\left(\genfrac{}{}{0ex}{}{x}{n}\right)d{\mu }_0(x)\\ =& {\xi }^n{\int }_{{\mathbb{Z}}_p}\left(\genfrac{}{}{0ex}{}{-x+n-1}{n}\right)d{\mu }_0(x)\\ =& {\xi }^n\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n-1}{n-m}\right){\int }_{{\mathbb{Z}}_p}\left(\genfrac{}{}{0ex}{}{-x}{m}\right)d{\mu }_0(x)\\ =& \sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n-1}{n-m}\right){\xi }^{n-m}\frac{{\hat{D}}_{m,\xi }}{m!}\\ =& \sum _{m=1}^n\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{{\hat{D}}_{m,\xi }}{m!}\end{array}$$

(40)

and

$$\begin{array}{rcl}{(-1)}^n\frac{{\hat{D}}_{n,\xi }}{n!}& =& {(-1)}^n{\xi }^n{\int }_{{\mathbb{Z}}_p}\left(\genfrac{}{}{0ex}{}{-x}{n}\right)d{\mu }_0(x)\\ =& {\xi }^n{\int }_{{\mathbb{Z}}_p}\left(\genfrac{}{}{0ex}{}{x+n-1}{n}\right)d{\mu }_0(x)\\ =& {\xi }^n\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n-1}{n-m}\right){\int }_0^1\left(\genfrac{}{}{0ex}{}{x}{m}\right)d{\mu }_0(x)\\ =& \sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{D_{m,\xi }}{m!}\\ =& \sum _{m=1}^n\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{D_{m,\xi }}{m!}.\end{array}$$

(41)

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

Theorem 8 For $n\in \mathbb{N}$, we have

$${(-1)}^n\frac{D_{n,\xi }}{n!}=\sum _{m=1}^n\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{{\hat{D}}_{m,\xi }}{m!}$$

and

$${(-1)}^n\frac{{\hat{D}}_{n,\xi }}{n!}=\sum _{m=1}^n\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{D_{m,\xi }}{m!}.$$