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 $qâˆˆ\mathbb{C}$, or a p-adic number $qâˆˆ{\mathbb{C}}_{p}$. If $qâˆˆ\mathbb{C}$, one normally assumes that $|q|<1$. If $qâˆˆ{\mathbb{C}}_{p}$, then we assume that ${|qâˆ’1|}_{p}<{p}^{âˆ’\frac{1}{pâˆ’1}}$ so that ${q}^{x}=exp\left(xlogq\right)$ for each $xâˆˆ{\mathbb{Z}}_{p}$. Throughout this paper, we use the notation

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

Note that ${lim}_{qâ†’1}{\left[x\right]}_{q}=x$ for each $xâˆˆ{\mathbb{Z}}_{p}$.

Let $UD\left({\mathbb{Z}}_{p}\right)$ be the space of uniformly differentiable functions on ${\mathbb{Z}}_{p}$. For $fâˆˆUD\left({\mathbb{Z}}_{p}\right)$, the p-adic invariant integral on ${\mathbb{Z}}_{p}$ is defined by Kim as follows:

$I\left(f\right)={âˆ«}_{{\mathbb{Z}}_{p}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(x\right)=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{p}^{n}}\underset{x=0}{\overset{{p}^{n}âˆ’1}{âˆ‘}}f\left(x\right)\phantom{\rule{1em}{0ex}}\text{(see [1â€“3])}.$
(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

${\left(x\right)}_{n}=x\left(xâˆ’1\right)â‹¯\left(xâˆ’n+1\right)=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){x}^{l},$
(1.3)

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

${\left({e}^{t}âˆ’1\right)}^{m}=m!\underset{l=m}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{2}\left(l,m\right)\frac{{t}^{l}}{l!}$
(1.4)

(see [4â€“6]).

Unsigned Stirling numbers of the first kind are given by

${x}^{\underset{Ì²}{n}}=x\left(x+1\right)â‹¯\left(x+nâˆ’1\right)=\underset{l=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,l\right)|{x}^{l}.$
(1.5)

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

$\begin{array}{rl}{\left(âˆ’x\right)}_{n}& ={\left(âˆ’1\right)}^{n}{x}^{\underset{Ì²}{n}}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}{x}^{l}\\ ={\left(âˆ’1\right)}^{n}\underset{l=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,l\right)|{x}^{l}.\end{array}$
(1.6)

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

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

${\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{r}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [1, 4, 7â€“18])}.$
(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 $nâˆˆ\mathbb{N}$, let ${T}_{p}$ be the p-adic locally constant space defined by

${T}_{p}=\underset{nâ‰¥1}{â‹ƒ}{C}_{{p}^{n}}=\underset{nâ†’\mathrm{âˆž}}{lim}{C}_{{p}^{n}},$

where ${C}_{{p}^{n}}=\left\{\mathrm{Ï‰}|{\mathrm{Ï‰}}^{{p}^{n}}=1\right\}$ is the cyclic group of order ${p}^{n}$.

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

${\left(x\right)}_{n,q}=x\left(xâˆ’q\right)\left(xâˆ’2q\right)â‹¯\left(xâˆ’\left(nâˆ’1\right)q\right)\phantom{\rule{1em}{0ex}}\left(nâ‰¥1\right),\phantom{\rule{2em}{0ex}}{\left(x\right)}_{0,q}=1.$

Note that

$\underset{qâ†’1}{lim}{\left(x\right)}_{n,q}={\left(x\right)}_{n}=\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){x}^{l}.$

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

${D}_{n}\left(x\right)={âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right)\phantom{\rule{0.25em}{0ex}}\text{(see [2, 9, 19])}.$
(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

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

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

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{D}_{n,q}\frac{{t}^{n}}{n!}={\left(1+qt\right)}^{\frac{x}{q}}\frac{log\left(1+qt\right)}{q\left({\left(1+qt\right)}^{\frac{1}{q}}âˆ’1\right)}\phantom{\rule{1em}{0ex}}\text{(see [20, 21])}.$
(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

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{D}_{n,\mathrm{Î¾},q}\frac{{t}^{n}}{n!}={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x}{q}}\frac{log\left(1+q\mathrm{Î¾}t\right)}{q\left({\left(1+q\mathrm{Î¾}t\right)}^{\frac{1}{q}}âˆ’1\right)},$
(1.11)

where $t,qâˆˆ{\mathbb{C}}_{p}$ with ${|t|}_{p}<{|q|}_{p}{p}^{âˆ’\frac{1}{pâˆ’1}}$ and $\mathrm{Î¾}âˆˆ{T}_{p}$.

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),

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\frac{{t}^{n}}{n!}& =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î¾}}^{n}{q}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(\frac{x+y}{q}\right)}_{n}d{\mathrm{Î¼}}_{0}\left(y\right)\frac{{t}^{n}}{n!}\\ ={âˆ«}_{{\mathbb{Z}}_{p}}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{x+y}{q}}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right),\end{array}$
(2.2)

where $t,qâˆˆ{\mathbb{C}}_{p}$ with ${|t|}_{p}<{|q|}_{p}{p}^{âˆ’\frac{1}{pâˆ’1}}$. For $tâˆˆ{\mathbb{C}}_{p}$ with ${|t|}_{p}<{|q|}_{p}{p}^{âˆ’\frac{1}{pâˆ’1}}$, put $f\left(x\right)={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x+y}{q}}$. By (1.1), we get

$\begin{array}{rl}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{x+y}{q}}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)& ={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x}{q}}\frac{log\left(1+q\mathrm{Î¾}t\right)}{q\left({\left(1+q\mathrm{Î¾}t\right)}^{\frac{1}{q}}âˆ’1\right)}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{D}_{n,\mathrm{Î¾},q}\left(x\right)\frac{{t}^{n}}{n!}.\end{array}$
(2.3)

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

Theorem 2.1 For $nâ‰¥0$, we have

${D}_{n,\mathrm{Î¾},q}\left(x\right)={\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right).$

In (2.3), by replacing t by $\frac{1}{\mathrm{Î¾}q}\left({e}^{\mathrm{Î¾}t}âˆ’1\right)$, we have

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{D}_{n,\mathrm{Î¾},q}\left(x\right)\frac{1}{{\mathrm{Î¾}}^{n}{q}^{n}}\frac{{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}}{n!}={e}^{\frac{\mathrm{Î¾}tx}{q}}\frac{\frac{\mathrm{Î¾}t}{q}}{{e}^{\frac{\mathrm{Î¾}t}{q}}âˆ’1}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}\left(x\right)\frac{{\mathrm{Î¾}}^{n}}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.4)

and

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{D}_{n,\mathrm{Î¾},q}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}\frac{1}{n!}{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}& =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{D}_{n,\mathrm{Î¾},q}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}\underset{m=n}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î¾}}^{m}{S}_{2}\left(m,n\right)\frac{{t}^{m}}{m!}\\ =\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\underset{n=0}{\overset{m}{âˆ‘}}\frac{{D}_{n,\mathrm{Î¾},q}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}{\mathrm{Î¾}}^{m}{S}_{2}\left(m,n\right)\frac{{t}^{m}}{m!}.\end{array}$
(2.5)

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

Corollary 2.2 For $nâ‰¥0$, we have

${B}_{n}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}{D}_{m,\mathrm{Î¾},q}\left(x\right){\mathrm{Î¾}}^{âˆ’m}{q}^{nâˆ’m}{S}_{2}\left(n,m\right).$

By Theorem 2.1,

$\begin{array}{rl}{D}_{n,\mathrm{Î¾},q}\left(x\right)& ={\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\\ ={\mathrm{Î¾}}^{n}{q}^{n}\underset{l=0}{\overset{n}{âˆ‘}}\frac{1}{{q}^{l}}{S}_{1}\left(n,l\right){âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right).\end{array}$
(2.6)

By (1.2), we can derive easily that

$\begin{array}{rl}{âˆ«}_{{\mathbb{Z}}_{p}}{e}^{\left(x+y\right)t}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)& =\frac{t}{{e}^{t}âˆ’1}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}\left(x\right)\frac{{t}^{n}}{n!}\\ =\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\frac{{t}^{l}}{l!},\end{array}$
(2.7)

and so

${B}_{n}\left(x\right)={âˆ«}_{{\mathbb{Z}}_{p}}{\left(x+y\right)}^{n}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right).$
(2.8)

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

Corollary 2.3 For $nâ‰¥0$, we have

${D}_{n,\mathrm{Î¾},q}\left(x\right)={\mathrm{Î¾}}^{n}\underset{l=0}{\overset{n}{âˆ‘}}{q}^{nâˆ’l}{S}_{1}\left(n,l\right){B}_{l}\left(x\right)={\mathrm{Î¾}}^{n}\underset{l=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,l\right)|{\left(âˆ’q\right)}^{nâˆ’l}{B}_{l}\left(x\right).$

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

${D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)={\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left({x}_{1}+â‹¯+{x}_{k}+x\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right),$
(2.9)

where n is a nonnegative integer and $kâˆˆ\mathbb{N}$. In the special case, $x=0$, ${D}_{n,\mathrm{Î¾},q}^{\left(k\right)}={D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(0\right)$ are called the Daehee numbers of order k with q-parameter.

From (2.9), we can derive the generating function of ${D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)$ as follows:

$\begin{array}{r}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}\\ \phantom{\rule{1em}{0ex}}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î¾}}^{n}{q}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{\frac{{x}_{1}+â‹¯+{x}_{k}+x}{q}}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right){t}^{n}\\ \phantom{\rule{1em}{0ex}}={âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{{x}_{1}+â‹¯+{x}_{k}+x}{q}}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x}{q}}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{{x}_{1}+â‹¯+{x}_{k}}{q}}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x}{q}}{\left(\frac{log\left(1+q\mathrm{Î¾}t\right)}{q\left({\left(1+q\mathrm{Î¾}t\right)}^{\frac{1}{q}}âˆ’1\right)}\right)}^{k}.\end{array}$
(2.10)

Note that, by (2.9),

${D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)={\mathrm{Î¾}}^{n}{q}^{n}\underset{m=0}{\overset{n}{âˆ‘}}\frac{{S}_{1}\left(n,m\right)}{{q}^{m}}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left({x}_{1}+â‹¯+{x}_{k}+x\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right).$
(2.11)

Since

${âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{e}^{\left({x}_{1}+â‹¯+{x}_{k}+x\right)t}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right)={\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{k}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!},$

we can derive easily

${B}_{n}^{\left(k\right)}\left(x\right)={âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left({x}_{1}+â‹¯+{x}_{k}+x\right)}^{n}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right).$
(2.12)

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

$\begin{array}{rl}{D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)& ={\mathrm{Î¾}}^{n}{q}^{n}\underset{m=0}{\overset{n}{âˆ‘}}\frac{{S}_{1}\left(n,m\right)}{{q}^{m}}{B}_{m}^{\left(k\right)}\left(x\right)\\ ={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}{q}^{nâˆ’m}{S}_{1}\left(n,m\right){B}_{m}^{\left(k\right)}\left(x\right)\\ ={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,m\right)|{\left(âˆ’q\right)}^{nâˆ’m}{B}_{m}^{\left(k\right)}\left(x\right).\end{array}$
(2.13)

In (2.10), by replacing t by $\frac{1}{q\mathrm{Î¾}}\left({e}^{\mathrm{Î¾}t}âˆ’1\right)$, we get

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)\frac{{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}}{{\mathrm{Î¾}}^{n}{q}^{n}n!}={e}^{\frac{\mathrm{Î¾}tx}{q}}{\left(\frac{\frac{\mathrm{Î¾}t}{q}}{{e}^{\frac{\mathrm{Î¾}t}{q}}âˆ’1}\right)}^{k}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¾}}^{n}{B}_{n}^{\left(k\right)}\left(x\right)}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.14)

and

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}\frac{1}{n!}{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}& =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}\underset{l=n}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{2}\left(l,n\right){\mathrm{Î¾}}^{l}\frac{{t}^{l}}{l!}\\ =\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({\mathrm{Î¾}}^{m}\underset{n=0}{\overset{m}{âˆ‘}}\frac{{D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}{S}_{2}\left(m,n\right)\right)\frac{{t}^{m}}{m!}.\end{array}$
(2.15)

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

Theorem 2.4 For $nâ‰¥0$ and $kâˆˆ\mathbb{N}$, we have

${D}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}{q}^{nâˆ’m}{S}_{1}\left(n,m\right){B}_{m}^{\left(k\right)}\left(x\right)={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,m\right)|{\left(âˆ’q\right)}^{nâˆ’m}{B}_{m}^{\left(k\right)}\left(x\right)$

and

${B}_{n}^{\left(k\right)}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}{D}_{m,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right){\mathrm{Î¾}}^{âˆ’m}{q}^{nâˆ’m}{S}_{2}\left(n,m\right).$

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

${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)={\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(âˆ’y+x\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right).$
(2.16)

In the special case $x=0$, ${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(0\right)={\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}$ are called the twisted Daehee numbers of the second kind with q-parameter.

By (2.16), we have

${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)={\mathrm{Î¾}}^{n}{q}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(\frac{âˆ’y+x}{q}\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right),$
(2.17)

and so we can derive the generating function of ${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)$ by (1.1) as follows:

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)\frac{{t}^{n}}{n!}=& \underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{q}^{n}{\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(\frac{âˆ’y+x}{q}\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\frac{{t}^{n}}{n!}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{q}^{n}{\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{\frac{âˆ’y+x}{q}}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right){t}^{n}\\ ={âˆ«}_{{\mathbb{Z}}_{p}}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{âˆ’y+x}{q}}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\\ ={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x}{q}}\frac{log\left(1+q\mathrm{Î¾}t\right)}{q\left({\left(1+q\mathrm{Î¾}t\right)}^{\frac{1}{q}}âˆ’1\right)}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{1}{q}}.\end{array}$
(2.18)

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

$\begin{array}{rl}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)& ={q}^{n}{\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(\frac{âˆ’y+x}{q}\right)}_{n}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\\ ={q}^{n}{\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}\underset{l=0}{\overset{n}{âˆ‘}}\frac{{S}_{1}\left(n,l\right)}{{q}^{l}}{\left(âˆ’y+x\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right)\\ ={\mathrm{Î¾}}^{n}\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}{âˆ«}_{{\mathbb{Z}}_{p}}{\left(yâˆ’x\right)}^{l}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left(y\right){q}^{nâˆ’l}\\ ={\mathrm{Î¾}}^{n}\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}{B}_{l}\left(âˆ’x\right){q}^{nâˆ’l}\\ ={\left(âˆ’\mathrm{Î¾}\right)}^{n}\underset{l=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,l\right)|{B}_{l}\left(âˆ’x\right){q}^{nâˆ’l}.\end{array}$
(2.19)

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

Theorem 2.5 For $nâ‰¥0$, we have

${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)={\mathrm{Î¾}}^{n}\underset{l=0}{\overset{n}{âˆ‘}}{S}_{1}\left(n,l\right){\left(âˆ’1\right)}^{l}{B}_{l}\left(âˆ’x\right){q}^{nâˆ’l}={\mathrm{Î¾}}^{n}\underset{l=0}{\overset{n}{âˆ‘}}|{S}_{1}\left(n,l\right)|{B}_{l}\left(x+1\right){\left(âˆ’q\right)}^{nâˆ’l}.$

By replacing t by $\frac{1}{q\mathrm{Î¾}}\left({e}^{\mathrm{Î¾}t}âˆ’1\right)$ in (2.18), we have

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)\frac{1}{{q}^{n}{\mathrm{Î¾}}^{n}}\frac{{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}}{n!}={e}^{\frac{\mathrm{Î¾}t}{q}\left(x+1\right)}\frac{\frac{\mathrm{Î¾}t}{q}}{{e}^{\frac{\mathrm{Î¾}t}{q}}âˆ’1}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¾}}^{n}{B}_{n}\left(x+1\right)}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.20)

and

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)\frac{1}{{q}^{n}{\mathrm{Î¾}}^{n}}\frac{{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}}{n!}& =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}\left(x\right)}{{q}^{n}{\mathrm{Î¾}}^{n}}\underset{m=n}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{2}\left(m,n\right)\frac{{\left(\mathrm{Î¾}t\right)}^{m}}{m!}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{m=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{D}}_{m,\mathrm{Î¾},q}\left(x\right){S}_{2}\left(n,m\right){q}^{âˆ’m}{\mathrm{Î¾}}^{nâˆ’m}\right)\frac{{t}^{n}}{n!}.\end{array}$
(2.21)

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

Theorem 2.6 For $nâ‰¥0$, we have

${B}_{n}\left(x+1\right)=\underset{m=0}{\overset{n}{âˆ‘}}{q}^{nâˆ’m}{\mathrm{Î¾}}^{âˆ’m}{\stackrel{Ë†}{D}}_{m,\mathrm{Î¾},q}\left(x\right){S}_{2}\left(n,m\right).$

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 ${\mathbb{Z}}_{p}$:

${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)={\mathrm{Î¾}}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left(âˆ’{x}_{1}âˆ’â‹¯âˆ’{x}_{k}+x\right)}_{n,q}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right),$
(2.22)

where n is a nonnegative integer and $kâˆˆ\mathbb{N}$. In the special case, $x=0$, ${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}={\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(0\right)$ 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 ${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)$ as follows:

$\begin{array}{r}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}\\ \phantom{\rule{1em}{0ex}}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î¾}}^{n}{q}^{n}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}\left(\genfrac{}{}{0}{}{\frac{âˆ’{x}_{1}âˆ’â‹¯âˆ’{x}_{k}+x}{q}}{n}\right)\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right){t}^{n}\\ \phantom{\rule{1em}{0ex}}={âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left(1+q\mathrm{Î¾}t\right)}^{\frac{âˆ’{x}_{1}âˆ’â‹¯âˆ’{x}_{k}+x}{q}}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\left(1+q\mathrm{Î¾}t\right)}^{\frac{x+k}{q}}{\left(\frac{log\left(1+q\mathrm{Î¾}t\right)}{q\left({\left(1+q\mathrm{Î¾}t\right)}^{\frac{1}{q}}âˆ’1\right)}\right)}^{k}.\end{array}$
(2.23)

By (2.22),

$\begin{array}{r}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)\\ \phantom{\rule{1em}{0ex}}={\mathrm{Î¾}}^{n}{q}^{n}\underset{m=0}{\overset{n}{âˆ‘}}\frac{{S}_{1}\left(n,m\right)}{{q}^{m}}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left(âˆ’{x}_{1}âˆ’â‹¯âˆ’{x}_{k}+x\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\mathrm{Î¾}}^{n}{q}^{n}\underset{m=0}{\overset{n}{âˆ‘}}\frac{{S}_{1}\left(n,m\right)}{{\left(âˆ’q\right)}^{m}}{âˆ«}_{{\mathbb{Z}}_{p}}â‹¯{âˆ«}_{{\mathbb{Z}}_{p}}{\left({x}_{1}+â‹¯+{x}_{k}âˆ’x\right)}^{m}\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{1}\right)â‹¯\phantom{\rule{0.2em}{0ex}}d{\mathrm{Î¼}}_{0}\left({x}_{k}\right)\\ \phantom{\rule{1em}{0ex}}={\mathrm{Î¾}}^{n}{q}^{n}\underset{m=0}{\overset{n}{âˆ‘}}\frac{{S}_{1}\left(n,m\right)}{{\left(âˆ’q\right)}^{m}}{B}_{m}^{\left(k\right)}\left(âˆ’x\right)\\ \phantom{\rule{1em}{0ex}}={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}{q}^{nâˆ’m}|{S}_{1}\left(n,m\right)|{B}_{m}^{\left(k\right)}\left(âˆ’x\right).\end{array}$
(2.24)

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

Theorem 2.7 For $nâ‰¥0$, we have

${\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{m}{q}^{nâˆ’m}{S}_{1}\left(n,m\right){B}_{m}^{\left(k\right)}\left(âˆ’x\right)={\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{m}{q}^{nâˆ’m}|{S}_{1}\left(n,m\right)|{B}_{m}^{\left(k\right)}\left(x+k\right).$

In (2.23), by replacing t by $\frac{1}{q\mathrm{Î¾}}\left({e}^{\mathrm{Î¾}t}âˆ’1\right)$, we get

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)\frac{{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}}{{\mathrm{Î¾}}^{n}{q}^{n}n!}={e}^{\frac{\mathrm{Î¾}t}{q}\left(x+k\right)}{\left(\frac{\frac{\mathrm{Î¾}t}{q}}{{e}^{\frac{\mathrm{Î¾}t}{q}}âˆ’1}\right)}^{k}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¾}}^{n}{B}_{n}^{\left(k\right)}\left(x+k\right)}{{q}^{n}}\frac{{t}^{n}}{n!}$
(2.25)

and

$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}\frac{1}{n!}{\left({e}^{\mathrm{Î¾}t}âˆ’1\right)}^{n}& =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\stackrel{Ë†}{D}}_{n,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)}{{\mathrm{Î¾}}^{n}{q}^{n}}\underset{l=n}{\overset{\mathrm{âˆž}}{âˆ‘}}{S}_{2}\left(l,n\right){\mathrm{Î¾}}^{l}\frac{{t}^{l}}{l!}\\ =\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left({\mathrm{Î¾}}^{n}\underset{m=0}{\overset{n}{âˆ‘}}\frac{{\stackrel{Ë†}{D}}_{m,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right)}{{\mathrm{Î¾}}^{m}{q}^{m}}{S}_{2}\left(n,m\right)\right)\frac{{t}^{n}}{n!}.\end{array}$
(2.26)

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

Theorem 2.8 For $nâ‰¥0$ and $kâˆˆ\mathbb{N}$, we have

${B}_{n}^{\left(k\right)}\left(x+k\right)=\underset{m=0}{\overset{n}{âˆ‘}}{\stackrel{Ë†}{D}}_{m,\mathrm{Î¾},q}^{\left(k\right)}\left(x\right){\mathrm{Î¾}}^{âˆ’m}{q}^{nâˆ’m}{S}_{2}\left(n,m\right).$

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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