Theory and Modern Applications

# 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 $UD\left[{\mathbb{Z}}_{p}\right]$ be the space of uniformly differentiable functions on ${\mathbb{Z}}_{p}$. For $f\in UD\left[{\mathbb{Z}}_{p}\right]$, the p-adic invariant integral on ${\mathbb{Z}}_{p}$ is defined by

$I\left(f\right){\int }_{{\mathbb{Z}}_{p}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{p}^{n}}\sum _{x=0}^{{p}^{n}-1}f\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{see [1, 2]}\right).$
(1)

Let ${f}_{1}$ be the translation of f with ${f}_{1}\left(x\right)=f\left(x+1\right)$. Then, by (1), we get

(2)

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

${\left(x\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l},$
(3)

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

${\left({e}^{t}-1\right)}^{m}=m!\sum _{l=m}^{\mathrm{\infty }}{S}_{2}\left(l,m\right)\frac{{t}^{l}}{l!}\phantom{\rule{1em}{0ex}}\left(\text{see [3–5]}\right).$
(4)

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

${\left(\frac{t}{{e}^{t}-1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [3, 6, 7]}\right).$
(5)

When $x=0$, ${B}_{n}^{\left(\alpha \right)}={B}_{n}^{\left(\alpha \right)}\left(0\right)$ 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 ${C}_{{p}^{n}}=\left\{\omega |{\omega }^{{p}^{n}}=1\right\}$ is the cyclic group of order ${p}^{n}$. It is well known that the twisted Bernoulli polynomials are defined as

$\frac{t}{\xi {e}^{t}-1}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!},\phantom{\rule{1em}{0ex}}\xi \in {T}_{p}\phantom{\rule{0.25em}{0ex}}\left(\text{see [8]}\right),$

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

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

$\left(\frac{log\left(1+t\right)}{t}\right){\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{D}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [9, 10]}\right).$
(6)

In the special case, $x=0$, ${D}_{n}={D}_{n}\left(0\right)$ 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

$\left(\frac{log\left(1+\xi t\right)}{\xi t}\right){\left(1+\xi t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}$
(7)

In the special case, $x=0$, ${D}_{n,\xi }={D}_{n,\xi }\left(0\right)$ 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:

(8)

By (8), we get

$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\frac{{t}^{n}}{n!}& =& {\int }_{{\mathbb{Z}}_{p}}\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}\left(\genfrac{}{}{0}{}{x}{n}\right){t}^{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{x}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right),\end{array}$
(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 $f\left(x\right)={\left(1+\xi t\right)}^{x}$. Then, from (2), we have

${\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{x}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)=\frac{log\left(1+\xi t\right)}{\xi t}.$
(10)

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

$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\frac{{t}^{n}}{n!}& =& \frac{log\left(1+\xi t\right)}{\xi t}\\ =& {\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{x}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& \sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\frac{{t}^{n}}{n!}.\end{array}$
(11)

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

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

${\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)={D}_{n,\xi }.$

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

${\left(\frac{log\left(1+t\right)}{t}\right)}^{n}{\left(1+t\right)}^{x-1}=\sum _{k=0}^{\mathrm{\infty }}{B}_{k}^{\left(k-n+1\right)}\left(x\right)\frac{{t}^{k}}{k!}\phantom{\rule{1em}{0ex}}\left(\text{see [3–5]}\right).$
(12)

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

${D}_{k,\xi }={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x\right)}_{k}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)={\xi }^{n}{B}_{k}^{\left(k+2\right)}\left(1\right)\phantom{\rule{1em}{0ex}}\left(k\ge 0\right),$
(13)

where ${B}_{k}^{\left(n\right)}\left(x\right)$ are the Bernoulli polynomials of order n.

In the special case, $x=0$, ${B}_{k}^{\left(n\right)}={B}_{k}^{\left(n\right)}\left(0\right)$ are called the n th Bernoulli numbers of order n.

From (11), we note that

$\begin{array}{rcl}{\left(1+\xi t\right)}^{x}{\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{y}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)& =& \left(\frac{log\left(1+\xi t\right)}{\xi t}\right){\left(1+\xi t\right)}^{x}\\ =& \sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}.\end{array}$
(14)

Thus, by (14), we get

${\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)={D}_{n,\xi }\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right),$
(15)

and, from (12), we have

${D}_{n,\xi }\left(x\right)={\xi }^{n}{B}_{n}^{\left(n+2\right)}\left(x+1\right).$
(16)

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

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

${D}_{n,\xi }\left(x\right)={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)$

and

${D}_{n,\xi }\left(x\right)={\xi }^{n}{B}_{n}^{\left(n+2\right)}\left(x+1\right).$

By Theorem 1, we easily see that

${D}_{n,\xi }={\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){B}_{l},$
(17)

where ${B}_{l}$ are the ordinary Bernoulli numbers.

From Theorem 2, we have

$\begin{array}{rcl}{D}_{n,\xi }\left(x\right)& =& {\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\\ =& {\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){B}_{l}\left(x\right),\end{array}$
(18)

where ${B}_{l}\left(x\right)$ are the Bernoulli polynomials defined by a generating function to be

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

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

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

${D}_{n,\xi }\left(x\right)={\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){B}_{l}\left(x\right).$

In (11), we have

$\frac{log\left(1+\xi t\right)}{\xi t}{\left(1+\xi t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}.$
(19)

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

$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\left(x\right)\frac{1}{n!}{\left({e}^{t}-\frac{1}{\xi }\right)}^{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{log\left(1+\xi \left({e}^{t}-\frac{1}{\xi }\right)\right)}{\xi \left({e}^{t}-\frac{1}{\xi }\right)}{\left(1+\xi \left({e}^{t}-\frac{1}{\xi }\right)\right)}^{x}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{t}{\xi {e}^{t}-1}{\left(\xi {e}^{t}\right)}^{x}\hfill \\ \phantom{\rule{1em}{0ex}}={\xi }^{x}\frac{t}{\xi {e}^{t}-1}{e}^{tx}\hfill \\ \phantom{\rule{1em}{0ex}}={\xi }^{x}\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}.\hfill \end{array}$
(20)

Therefore, we have

$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}& =& \sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\left(x\right)\frac{1}{n!}{\left({e}^{t}-\frac{1}{\xi }\right)}^{n}\\ =& \sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi }\left(x\right)\frac{1}{n!}{\xi }^{-n}n!\sum _{m=n}^{\mathrm{\infty }}{S}_{2}\left(m,n\right)\frac{{t}^{m}}{m!}\\ =& \sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{m}{D}_{n,\xi }\left(x\right){\xi }^{-n}{S}_{2}\left(m,n\right),\end{array}$
(21)

where ${S}_{2}\left(m,n\right)$ is the Stirling number of the second kind.

Hence,

${\xi }^{x}{B}_{n,\xi }\left(x\right)=\sum _{n=0}^{m}{D}_{n,\xi }\left(x\right){\xi }^{-n}{S}_{2}\left(m,n\right).$
(22)

Therefore, we have

${B}_{m,\xi }\left(x\right)=\sum _{n=0}^{m}{D}_{n,\xi }\left(x\right){\xi }^{-n-x}{S}_{2}\left(m,n\right).$
(23)

In particular,

${B}_{m,\xi }=\sum _{n=0}^{m}{D}_{n,\xi }{\xi }^{-n}{S}_{2}\left(m,n\right).$
(24)

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

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

${B}_{m,\xi }\left(x\right)=\sum _{n=0}^{m}{\xi }^{-n-x}{D}_{n,\xi }\left(x\right){S}_{2}\left(m,n\right).$

In particular,

${B}_{m,\xi }=\sum _{n=0}^{m}{\xi }^{-n}{D}_{n,\xi }{S}_{2}\left(m,n\right).$

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

${\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)={\xi }^{n}\sum _{n=0}^{m}{D}_{n}\left(x\right){S}_{2}\left(m,n\right).$

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

${x}^{\left(n\right)}=x\left(x+1\right)\cdots \left(x+n-1\right).$
(25)

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

${\stackrel{ˆ}{D}}_{n,\xi }={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(-x\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\in {\mathbb{Z}}_{n\ge 0}\right).$
(26)

By (26), we get

${x}^{\left(n\right)}={\left(-1\right)}^{n}{\left(-x\right)}_{n}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{n-l}{x}^{l}.$
(27)

From (26) and (27), we have

$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n,\xi }& =& {\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(-x\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{x}^{\left(n\right)}{\left(-1\right)}^{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}{B}_{l}.\end{array}$
(28)

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

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

${\stackrel{ˆ}{D}}_{n,\xi }={\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}{B}_{l}.$

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

$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\frac{{t}^{n}}{n!}& =& \sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(-x\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\frac{{t}^{n}}{n!}\\ =& {\int }_{{\mathbb{Z}}_{p}}\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}\left(\genfrac{}{}{0}{}{-x}{n}\right){t}^{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{-x}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right).\end{array}$
(29)

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

${\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{-x}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)=\frac{\left(1+\xi t\right)log\left(1+\xi t\right)}{\xi t},$
(30)

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

By (29) and (30), we get

$\begin{array}{rcl}\frac{1}{\xi t}\left(1+\xi t\right)log\left(1+\xi t\right)& =& {\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{-x}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& \sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\frac{{t}^{n}}{n!}.\end{array}$
(31)

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

$\frac{\left(1+\xi t\right)log\left(1+\xi t\right)}{\xi t}\frac{1}{{\left(1+\xi t\right)}^{x}}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}.$
(32)

Then, by (32), we get

${\int }_{{\mathbb{Z}}_{p}}{\left(1+\xi t\right)}^{-x-y}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}.$
(33)

From (33), we get

$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)& =& {\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(-x-y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right)\\ =& {\xi }^{n}\sum _{l=0}^{n}{\left(-1\right)}^{l}{S}_{1}\left(n,l\right){B}_{l}\left(x\right).\end{array}$
(34)

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

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

${\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(-x-y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)={\xi }^{n}\sum _{l=0}^{n}{\left(-1\right)}^{l}{S}_{1}\left(n,l\right){B}_{l}\left(x\right).$

From (32) and (33), we have

$\frac{log\left(1+\xi t\right)}{\xi t}{\left(1+\xi t\right)}^{1-x}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)\frac{{t}^{n}}{n!}.$
(35)

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

$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)\frac{1}{n!}{\left({e}^{t}-\frac{1}{\xi }\right)}^{n}& =& \frac{log\left(1+\xi \left({e}^{t}-\frac{1}{\xi }\right)\right)}{\xi \left({e}^{t}-\frac{1}{\xi }\right)}{\left(1+\xi \left({e}^{t}-\frac{1}{\xi }\right)\right)}^{1-x}\\ =& \frac{t}{\xi {e}^{t}-1}{\left(\xi {e}^{t}\right)}^{1-x}\\ =& {\xi }^{1-x}\frac{t}{\xi {e}^{t}-1}{e}^{t\left(1-x\right)}\\ =& {\xi }^{1-x}\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }\left(1-x\right)\frac{{t}^{n}}{n!}.\end{array}$
(36)

Therefore, we have

$\begin{array}{rcl}{\xi }^{1-x}\sum _{m=0}^{\mathrm{\infty }}{B}_{m,\xi }\left(1-x\right)\frac{{t}^{n}}{n!}& =& \sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)\frac{{\left({e}^{t}-\frac{1}{\xi }\right)}^{n}}{n!}\\ =& \sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right)\frac{1}{n!}{\xi }^{-n}n!\sum _{m=n}^{\mathrm{\infty }}{S}_{2}\left(m,n\right)\frac{{t}^{m}}{m!}\\ =& \sum _{m=0}^{\mathrm{\infty }}\left(\sum _{n=0}^{m}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right){\xi }^{-n}{S}_{2}\left(m,n\right)\right)\frac{{t}^{m}}{m!}.\end{array}$
(37)

Hence,

${\xi }^{1-x}{B}_{n,\xi }\left(1-x\right)=\sum _{n=0}^{m}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right){\xi }^{-n}{S}_{2}\left(m,n\right).$
(38)

Therefore, we have

${B}_{m,\xi }\left(1-x\right)=\sum _{n=0}^{m}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right){\xi }^{-n+x-1}{S}_{2}\left(m,n\right).$
(39)

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

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

${B}_{m,\xi }\left(1-x\right)=\sum _{n=0}^{m}{\xi }^{-m+x-1}{\stackrel{ˆ}{D}}_{n,\xi }\left(x\right){S}_{2}\left(m,n\right).$

From Theorem 1 and (26), we have

$\begin{array}{rcl}{\left(-1\right)}^{n}\frac{{D}_{n,\xi }}{n!}& =& {\left(-1\right)}^{n}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{x}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{-x+n-1}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\xi }^{n}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n-1}{n-m}\right){\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{-x}{m}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& \sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n-1}{n-m}\right){\xi }^{n-m}\frac{{\stackrel{ˆ}{D}}_{m,\xi }}{m!}\\ =& \sum _{m=1}^{n}\left(\genfrac{}{}{0}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{{\stackrel{ˆ}{D}}_{m,\xi }}{m!}\end{array}$
(40)

and

$\begin{array}{rcl}{\left(-1\right)}^{n}\frac{{\stackrel{ˆ}{D}}_{n,\xi }}{n!}& =& {\left(-1\right)}^{n}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{-x}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{x+n-1}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& {\xi }^{n}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n-1}{n-m}\right){\int }_{0}^{1}\left(\genfrac{}{}{0}{}{x}{m}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)\\ =& \sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{{D}_{m,\xi }}{m!}\\ =& \sum _{m=1}^{n}\left(\genfrac{}{}{0}{}{n-1}{m-1}\right){\xi }^{n-m}\frac{{D}_{m,\xi }}{m!}.\end{array}$
(41)

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

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

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

and

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

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## Acknowledgements

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

## Author information

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### Corresponding author

Correspondence to Jongkyum Kwon.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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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