Theory and Modern Applications

# On the twisted Daehee polynomials with q-parameter

## Abstract

The n th twisted Daehee numbers with q-parameter are closely related to higher-order Bernoulli numbers and Bernoulli numbers of the second kind. In this paper, we give a p-adic integral representation of the twisted Daehee polynomials with q-parameter, and we derive some interesting properties related to the n th twisted Daehee polynomials with q-parameter.

## 1 Introduction

Let p be a fixed prime number. Throughout this paper, ${\mathbb{Z}}_{p}$, ${\mathbb{Q}}_{p}$, and ${\mathbb{C}}_{p}$ will, respectively, denote the ring of p-adic rational integers, the field of p-adic rational numbers, and the completions of algebraic closure of ${\mathbb{Q}}_{p}$. The p-adic norm is defined ${|p|}_{p}=\frac{1}{p}$.

When one talks of q-extension, q is variously considered as an indeterminate, a complex $q\in \mathbb{C}$, or a p-adic number $q\in {\mathbb{C}}_{p}$. If $q\in \mathbb{C}$, one normally assumes that $|q|<1$. If $q\in {\mathbb{C}}_{p}$, then we assume that ${|q-1|}_{p}<{p}^{-\frac{1}{p-1}}$ so that ${q}^{x}=exp\left(xlogq\right)$ for each $x\in {\mathbb{Z}}_{p}$. Throughout this paper, we use the notation

${\left[x\right]}_{q}=\frac{1-{q}^{x}}{1-q}.$

Note that ${lim}_{q\to 1}{\left[x\right]}_{q}=x$ for each $x\in {\mathbb{Z}}_{p}$.

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 Kim as follows:

$I\left(f\right)={\int }_{{\mathbb{Z}}_{p}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{p}^{n}}\sum _{x=0}^{{p}^{n}-1}f\left(x\right)\phantom{\rule{1em}{0ex}}\text{(see [1–3])}.$
(1.1)

Let ${f}_{1}$ be the translation of f with ${f}_{1}\left(x\right)=f\left(x+1\right)$. Then, by (1.1), we get

(1.2)

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

${\left(x\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l},$
(1.3)

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

${\left({e}^{t}-1\right)}^{m}=m!\sum _{l=m}^{\mathrm{\infty }}{S}_{2}\left(l,m\right)\frac{{t}^{l}}{l!}$
(1.4)

(see [46]).

Unsigned Stirling numbers of the first kind are given by

${x}^{\underline{n}}=x\left(x+1\right)\cdots \left(x+n-1\right)=\sum _{l=0}^{n}|{S}_{1}\left(n,l\right)|{x}^{l}.$
(1.5)

Note that if we replace x to −x in (1.3), then

$\begin{array}{rl}{\left(-x\right)}_{n}& ={\left(-1\right)}^{n}{x}^{\underline{n}}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}{x}^{l}\\ ={\left(-1\right)}^{n}\sum _{l=0}^{n}|{S}_{1}\left(n,l\right)|{x}^{l}.\end{array}$
(1.6)

Hence ${S}_{1}\left(n,l\right)=|{S}_{1}\left(n,l\right)|{\left(-1\right)}^{n-l}$.

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

${\left(\frac{t}{{e}^{t}-1}\right)}^{r}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [1, 4, 7–18])}.$
(1.7)

When $x=0$, ${B}_{n}^{\left(r\right)}={B}_{n}^{\left(r\right)}\left(0\right)$ are called the Bernoulli numbers of order r, and in the special case, $r=1$, ${B}_{n}^{\left(1\right)}\left(x\right)={B}_{n}\left(x\right)$ are called the ordinary Bernoulli polynomials.

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 ${p}^{n}$.

We assume that q is an indeterminate in ${\mathbb{C}}_{p}$ with ${|1-q|}_{p}<{p}^{-\frac{1}{p-1}}$. Then we define the q-analog of a falling factorial sequence as follows:

${\left(x\right)}_{n,q}=x\left(x-q\right)\left(x-2q\right)\cdots \left(x-\left(n-1\right)q\right)\phantom{\rule{1em}{0ex}}\left(n\ge 1\right),\phantom{\rule{2em}{0ex}}{\left(x\right)}_{0,q}=1.$

Note that

$\underset{q\to 1}{lim}{\left(x\right)}_{n,q}={\left(x\right)}_{n}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l}.$

Recently, DS Kim and T Kim introduced the Daehee polynomials as follows:

${D}_{n}\left(x\right)={\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right)\phantom{\rule{0.25em}{0ex}}\text{(see [2, 9, 19])}.$
(1.8)

When $x=0$, ${D}_{n}={D}_{n}\left(0\right)$ are called the nth Daehee numbers. From (1.8), we can derive the generating function to be

$\left(\frac{log\left(1+t\right)}{t}\right){\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{D}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [9])}.$
(1.9)

In addition, DS Kim et al. consider the Daehee polynomials with q-parameter, which are defined by the generating function to be

$\sum _{n=0}^{\mathrm{\infty }}{D}_{n,q}\frac{{t}^{n}}{n!}={\left(1+qt\right)}^{\frac{x}{q}}\frac{log\left(1+qt\right)}{q\left({\left(1+qt\right)}^{\frac{1}{q}}-1\right)}\phantom{\rule{1em}{0ex}}\text{(see [20, 21])}.$
(1.10)

When $x=0$, ${D}_{n,q}={D}_{n,q}\left(0\right)$ are called the Daehee numbers with q-parameter.

From the viewpoint of a generalization of the Daehee polynomials with q-parameter, we consider the twisted Daehee polynomials with q-parameter, defined to be

$\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi ,q}\frac{{t}^{n}}{n!}={\left(1+q\xi t\right)}^{\frac{x}{q}}\frac{log\left(1+q\xi t\right)}{q\left({\left(1+q\xi t\right)}^{\frac{1}{q}}-1\right)},$
(1.11)

where $t,q\in {\mathbb{C}}_{p}$ with ${|t|}_{p}<{|q|}_{p}{p}^{-\frac{1}{p-1}}$ and $\xi \in {T}_{p}$.

In this paper, we give a p-adic integral representation of the twisted Daehee polynomials with q-parameter, which is called the Witt-type formula for the twisted Daehee polynomials with q-parameter. We can derive some interesting properties related to the n th twisted Daehee polynomials with q-parameter.

## 2 Witt-type formula for the n th twisted Daehee polynomials with q-parameter

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

(2.1)

By (2.1),

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\frac{{t}^{n}}{n!}& =\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{q}^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(\frac{x+y}{q}\right)}_{n}d{\mu }_{0}\left(y\right)\frac{{t}^{n}}{n!}\\ ={\int }_{{\mathbb{Z}}_{p}}{\left(1+q\xi t\right)}^{\frac{x+y}{q}}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right),\end{array}$
(2.2)

where $t,q\in {\mathbb{C}}_{p}$ with ${|t|}_{p}<{|q|}_{p}{p}^{-\frac{1}{p-1}}$. For $t\in {\mathbb{C}}_{p}$ with ${|t|}_{p}<{|q|}_{p}{p}^{-\frac{1}{p-1}}$, put $f\left(x\right)={\left(1+q\xi t\right)}^{\frac{x+y}{q}}$. By (1.1), we get

$\begin{array}{rl}{\int }_{{\mathbb{Z}}_{p}}{\left(1+q\xi t\right)}^{\frac{x+y}{q}}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)& ={\left(1+q\xi t\right)}^{\frac{x}{q}}\frac{log\left(1+q\xi t\right)}{q\left({\left(1+q\xi t\right)}^{\frac{1}{q}}-1\right)}\\ =\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi ,q}\left(x\right)\frac{{t}^{n}}{n!}.\end{array}$
(2.3)

By (2.2) and (2.3), we obtain the following theorem.

Theorem 2.1 For $n\ge 0$, we have

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

In (2.3), by replacing t by $\frac{1}{\xi q}\left({e}^{\xi t}-1\right)$, we have

$\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi ,q}\left(x\right)\frac{1}{{\xi }^{n}{q}^{n}}\frac{{\left({e}^{\xi t}-1\right)}^{n}}{n!}={e}^{\frac{\xi tx}{q}}\frac{\frac{\xi t}{q}}{{e}^{\frac{\xi t}{q}}-1}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x\right)\frac{{\xi }^{n}}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.4)

and

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{D}_{n,\xi ,q}\left(x\right)}{{\xi }^{n}{q}^{n}}\frac{1}{n!}{\left({e}^{\xi t}-1\right)}^{n}& =\sum _{n=0}^{\mathrm{\infty }}\frac{{D}_{n,\xi ,q}\left(x\right)}{{\xi }^{n}{q}^{n}}\sum _{m=n}^{\mathrm{\infty }}{\xi }^{m}{S}_{2}\left(m,n\right)\frac{{t}^{m}}{m!}\\ =\sum _{m=0}^{\mathrm{\infty }}\sum _{n=0}^{m}\frac{{D}_{n,\xi ,q}\left(x\right)}{{\xi }^{n}{q}^{n}}{\xi }^{m}{S}_{2}\left(m,n\right)\frac{{t}^{m}}{m!}.\end{array}$
(2.5)

By (2.4) and (2.5), we obtain the following corollary.

Corollary 2.2 For $n\ge 0$, we have

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

By Theorem 2.1,

$\begin{array}{rl}{D}_{n,\xi ,q}\left(x\right)& ={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\\ ={\xi }^{n}{q}^{n}\sum _{l=0}^{n}\frac{1}{{q}^{l}}{S}_{1}\left(n,l\right){\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right).\end{array}$
(2.6)

By (1.2), we can derive easily that

$\begin{array}{rl}{\int }_{{\mathbb{Z}}_{p}}{e}^{\left(x+y\right)t}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)& =\frac{t}{{e}^{t}-1}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x\right)\frac{{t}^{n}}{n!}\\ =\sum _{l=0}^{\mathrm{\infty }}{\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\frac{{t}^{l}}{l!},\end{array}$
(2.7)

and so

${B}_{n}\left(x\right)={\int }_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right).$
(2.8)

By (1.6), (2.7), and (2.8), we obtain the following corollary.

Corollary 2.3 For $n\ge 0$, we have

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

From now on, we consider twisted Daehee polynomials of order $k\in \mathbb{N}$ with q-parameter. Twisted Daehee polynomials of order $k\in \mathbb{N}$ with q-parameter are defined by the multivariant p-adic invariant integral on ${\mathbb{Z}}_{p}$:

${D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left({x}_{1}+\cdots +{x}_{k}+x\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right),$
(2.9)

where n is a nonnegative integer and $k\in \mathbb{N}$. In the special case, $x=0$, ${D}_{n,\xi ,q}^{\left(k\right)}={D}_{n,\xi ,q}^{\left(k\right)}\left(0\right)$ are called the Daehee numbers of order k with q-parameter.

From (2.9), we can derive the generating function of ${D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)$ as follows:

$\begin{array}{r}\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}\\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{q}^{n}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{\frac{{x}_{1}+\cdots +{x}_{k}+x}{q}}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right){t}^{n}\\ \phantom{\rule{1em}{0ex}}={\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left(1+q\xi t\right)}^{\frac{{x}_{1}+\cdots +{x}_{k}+x}{q}}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\left(1+q\xi t\right)}^{\frac{x}{q}}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left(1+q\xi t\right)}^{\frac{{x}_{1}+\cdots +{x}_{k}}{q}}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\left(1+q\xi t\right)}^{\frac{x}{q}}{\left(\frac{log\left(1+q\xi t\right)}{q\left({\left(1+q\xi t\right)}^{\frac{1}{q}}-1\right)}\right)}^{k}.\end{array}$
(2.10)

Note that, by (2.9),

${D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)={\xi }^{n}{q}^{n}\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{{q}^{m}}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left({x}_{1}+\cdots +{x}_{k}+x\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right).$
(2.11)

Since

${\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{e}^{\left({x}_{1}+\cdots +{x}_{k}+x\right)t}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right)={\left(\frac{t}{{e}^{t}-1}\right)}^{k}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!},$

we can derive easily

${B}_{n}^{\left(k\right)}\left(x\right)={\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left({x}_{1}+\cdots +{x}_{k}+x\right)}^{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right).$
(2.12)

Thus, by (2.11) and (2.12), we have

$\begin{array}{rl}{D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)& ={\xi }^{n}{q}^{n}\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{{q}^{m}}{B}_{m}^{\left(k\right)}\left(x\right)\\ ={\xi }^{n}\sum _{m=0}^{n}{q}^{n-m}{S}_{1}\left(n,m\right){B}_{m}^{\left(k\right)}\left(x\right)\\ ={\xi }^{n}\sum _{m=0}^{n}|{S}_{1}\left(n,m\right)|{\left(-q\right)}^{n-m}{B}_{m}^{\left(k\right)}\left(x\right).\end{array}$
(2.13)

In (2.10), by replacing t by $\frac{1}{q\xi }\left({e}^{\xi t}-1\right)$, we get

$\sum _{n=0}^{\mathrm{\infty }}{D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)\frac{{\left({e}^{\xi t}-1\right)}^{n}}{{\xi }^{n}{q}^{n}n!}={e}^{\frac{\xi tx}{q}}{\left(\frac{\frac{\xi t}{q}}{{e}^{\frac{\xi t}{q}}-1}\right)}^{k}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\xi }^{n}{B}_{n}^{\left(k\right)}\left(x\right)}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.14)

and

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)}{{\xi }^{n}{q}^{n}}\frac{1}{n!}{\left({e}^{\xi t}-1\right)}^{n}& =\sum _{n=0}^{\mathrm{\infty }}\frac{{D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)}{{\xi }^{n}{q}^{n}}\sum _{l=n}^{\mathrm{\infty }}{S}_{2}\left(l,n\right){\xi }^{l}\frac{{t}^{l}}{l!}\\ =\sum _{m=0}^{\mathrm{\infty }}\left({\xi }^{m}\sum _{n=0}^{m}\frac{{D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)}{{\xi }^{n}{q}^{n}}{S}_{2}\left(m,n\right)\right)\frac{{t}^{m}}{m!}.\end{array}$
(2.15)

By (2.13), (2.14), and (2.15), we obtain the following theorem.

Theorem 2.4 For $n\ge 0$ and $k\in \mathbb{N}$, we have

${D}_{n,\xi ,q}^{\left(k\right)}\left(x\right)={\xi }^{n}\sum _{m=0}^{n}{q}^{n-m}{S}_{1}\left(n,m\right){B}_{m}^{\left(k\right)}\left(x\right)={\xi }^{n}\sum _{m=0}^{n}|{S}_{1}\left(n,m\right)|{\left(-q\right)}^{n-m}{B}_{m}^{\left(k\right)}\left(x\right)$

and

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

Now, we consider the twisted Daehee polynomials of the second kind with q-parameter as follows:

${\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(-y+x\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(2.16)

In the special case $x=0$, ${\stackrel{ˆ}{D}}_{n,\xi ,q}\left(0\right)={\stackrel{ˆ}{D}}_{n,\xi ,q}$ are called the twisted Daehee numbers of the second kind with q-parameter.

By (2.16), we have

${\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)={\xi }^{n}{q}^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(\frac{-y+x}{q}\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right),$
(2.17)

and so we can derive the generating function of ${\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)$ by (1.1) as follows:

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)\frac{{t}^{n}}{n!}=& \sum _{n=0}^{\mathrm{\infty }}{q}^{n}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(\frac{-y+x}{q}\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\frac{{t}^{n}}{n!}\\ =\sum _{n=0}^{\mathrm{\infty }}{q}^{n}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{\frac{-y+x}{q}}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right){t}^{n}\\ ={\int }_{{\mathbb{Z}}_{p}}{\left(1+q\xi t\right)}^{\frac{-y+x}{q}}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\\ ={\left(1+q\xi t\right)}^{\frac{x}{q}}\frac{log\left(1+q\xi t\right)}{q\left({\left(1+q\xi t\right)}^{\frac{1}{q}}-1\right)}{\left(1+q\xi t\right)}^{\frac{1}{q}}.\end{array}$
(2.18)

From (1.3), (1.6), and (2.17), we get

$\begin{array}{rl}{\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)& ={q}^{n}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}{\left(\frac{-y+x}{q}\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\\ ={q}^{n}{\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\sum _{l=0}^{n}\frac{{S}_{1}\left(n,l\right)}{{q}^{l}}{\left(-y+x\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right)\\ ={\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}{\int }_{{\mathbb{Z}}_{p}}{\left(y-x\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left(y\right){q}^{n-l}\\ ={\xi }^{n}\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}{B}_{l}\left(-x\right){q}^{n-l}\\ ={\left(-\xi \right)}^{n}\sum _{l=0}^{n}|{S}_{1}\left(n,l\right)|{B}_{l}\left(-x\right){q}^{n-l}.\end{array}$
(2.19)

By (1.10), it is easy to show that ${B}_{n}\left(-x\right)={\left(-1\right)}^{n}{B}_{n}\left(x+1\right)$. Thus, from (2.19), we have the following theorem.

Theorem 2.5 For $n\ge 0$, we have

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

By replacing t by $\frac{1}{q\xi }\left({e}^{\xi t}-1\right)$ in (2.18), we have

$\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)\frac{1}{{q}^{n}{\xi }^{n}}\frac{{\left({e}^{\xi t}-1\right)}^{n}}{n!}={e}^{\frac{\xi t}{q}\left(x+1\right)}\frac{\frac{\xi t}{q}}{{e}^{\frac{\xi t}{q}}-1}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\xi }^{n}{B}_{n}\left(x+1\right)}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.20)

and

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)\frac{1}{{q}^{n}{\xi }^{n}}\frac{{\left({e}^{\xi t}-1\right)}^{n}}{n!}& =\sum _{n=0}^{\mathrm{\infty }}\frac{{\stackrel{ˆ}{D}}_{n,\xi ,q}\left(x\right)}{{q}^{n}{\xi }^{n}}\sum _{m=n}^{\mathrm{\infty }}{S}_{2}\left(m,n\right)\frac{{\left(\xi t\right)}^{m}}{m!}\\ =\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{m=0}^{n}{\stackrel{ˆ}{D}}_{m,\xi ,q}\left(x\right){S}_{2}\left(n,m\right){q}^{-m}{\xi }^{n-m}\right)\frac{{t}^{n}}{n!}.\end{array}$
(2.21)

By (2.20) and (2.21), we obtain the following theorem.

Theorem 2.6 For $n\ge 0$, we have

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

Now, we consider higher-order twisted Daehee polynomials of the second kind with q-parameter. Higher-order twisted Daehee polynomials of the second kind with q-parameter are defined by the multivariant p-adic invariant integral on ${\mathbb{Z}}_{p}$:

${\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)={\xi }^{n}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left(-{x}_{1}-\cdots -{x}_{k}+x\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right),$
(2.22)

where n is a nonnegative integer and $k\in \mathbb{N}$. In the special case, $x=0$, ${\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}={\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(0\right)$ are called the higher-order twisted Daehee numbers of the second kind with q-parameter.

From (2.22), we can derive the generating function of ${\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)$ as follows:

$\begin{array}{r}\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}\\ \phantom{\rule{1em}{0ex}}=\sum _{n=0}^{\mathrm{\infty }}{\xi }^{n}{q}^{n}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{\frac{-{x}_{1}-\cdots -{x}_{k}+x}{q}}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right){t}^{n}\\ \phantom{\rule{1em}{0ex}}={\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left(1+q\xi t\right)}^{\frac{-{x}_{1}-\cdots -{x}_{k}+x}{q}}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\left(1+q\xi t\right)}^{\frac{x+k}{q}}{\left(\frac{log\left(1+q\xi t\right)}{q\left({\left(1+q\xi t\right)}^{\frac{1}{q}}-1\right)}\right)}^{k}.\end{array}$
(2.23)

By (2.22),

$\begin{array}{r}{\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)\\ \phantom{\rule{1em}{0ex}}={\xi }^{n}{q}^{n}\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{{q}^{m}}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left(-{x}_{1}-\cdots -{x}_{k}+x\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\xi }^{n}{q}^{n}\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{{\left(-q\right)}^{m}}{\int }_{{\mathbb{Z}}_{p}}\cdots {\int }_{{\mathbb{Z}}_{p}}{\left({x}_{1}+\cdots +{x}_{k}-x\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{1}\right)\cdots \phantom{\rule{0.2em}{0ex}}d{\mu }_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\xi }^{n}{q}^{n}\sum _{m=0}^{n}\frac{{S}_{1}\left(n,m\right)}{{\left(-q\right)}^{m}}{B}_{m}^{\left(k\right)}\left(-x\right)\\ \phantom{\rule{1em}{0ex}}={\xi }^{n}\sum _{m=0}^{n}{q}^{n-m}|{S}_{1}\left(n,m\right)|{B}_{m}^{\left(k\right)}\left(-x\right).\end{array}$
(2.24)

From (1.10), we know that ${B}_{n}^{\left(k\right)}\left(-x\right)={\left(-1\right)}^{n}{B}_{n}^{\left(k\right)}\left(k+x\right)$. Hence, by (2.24), we obtain the following theorem.

Theorem 2.7 For $n\ge 0$, we have

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

In (2.23), by replacing t by $\frac{1}{q\xi }\left({e}^{\xi t}-1\right)$, we get

$\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)\frac{{\left({e}^{\xi t}-1\right)}^{n}}{{\xi }^{n}{q}^{n}n!}={e}^{\frac{\xi t}{q}\left(x+k\right)}{\left(\frac{\frac{\xi t}{q}}{{e}^{\frac{\xi t}{q}}-1}\right)}^{k}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\xi }^{n}{B}_{n}^{\left(k\right)}\left(x+k\right)}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.25)

and

$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)}{{\xi }^{n}{q}^{n}}\frac{1}{n!}{\left({e}^{\xi t}-1\right)}^{n}& =\sum _{n=0}^{\mathrm{\infty }}\frac{{\stackrel{ˆ}{D}}_{n,\xi ,q}^{\left(k\right)}\left(x\right)}{{\xi }^{n}{q}^{n}}\sum _{l=n}^{\mathrm{\infty }}{S}_{2}\left(l,n\right){\xi }^{l}\frac{{t}^{l}}{l!}\\ =\sum _{n=0}^{\mathrm{\infty }}\left({\xi }^{n}\sum _{m=0}^{n}\frac{{\stackrel{ˆ}{D}}_{m,\xi ,q}^{\left(k\right)}\left(x\right)}{{\xi }^{m}{q}^{m}}{S}_{2}\left(n,m\right)\right)\frac{{t}^{n}}{n!}.\end{array}$
(2.26)

By (2.25) and (2.26), we obtain the following theorem.

Theorem 2.8 For $n\ge 0$ and $k\in \mathbb{N}$, we have

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

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## Acknowledgements

The author is grateful for the valuable comments and suggestions of the referees. This paper was supported by the Sehan University Research Fund in 2014.

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Correspondence to Jin-Woo Park.

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Park, JW. On the twisted Daehee polynomials with q-parameter. Adv Differ Equ 2014, 304 (2014). https://doi.org/10.1186/1687-1847-2014-304