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On Appell-type Changhee polynomials and numbers

  • Jeong Gon Lee1,
  • Lee-Chae Jang2Email author,
  • Jong-Jin Seo3,
  • Sang-Ki Choi4 and
  • Hyuck In Kwon5
Advances in Difference Equations20162016:160

https://doi.org/10.1186/s13662-016-0866-7

Received: 10 February 2016

Accepted: 17 May 2016

Published: 21 June 2016

Abstract

In this paper, we consider the Appell-type Changhee polynomials and derive some properties of these polynomials. Furthermore, we investigate certain identities for these polynomials.

Keywords

Changhee polynomialsAppell-type Changhee polynomialsdegenerate Bernoulli polynomialsbeta functions

MSC

05A1011B6811S8005A19

1 Introduction

Let p be a fixed odd prime number. Throughout this paper, we denote by \(\mathbb{Z}_{p}\), \(\mathbb{Q}_{p}\), and \(\mathbb{C}_{p}\) the ring of p-adic integers, the field of p-adic numbers, and the completion of algebraic closure of \(\mathbb{Q}_{p}\). The p-adic norm \(|\cdot|_{p}\) is normalized as \(|p|_{p}=\frac{1}{p}\). Let \(C(\mathbb{Z}_{p})\) be the space of continuous functions on \(\mathbb{Z}_{p}\). For \(f \in C(\mathbb{Z}_{p})\), the fermionic p-adic integral on \(\mathbb{Z}_{p}\) is defined by Kim to be
$$ I_{-1}(f) = \int_{\mathbb{Z}_{p}} f(x)\, d \mu_{-1}(x) = \lim _{N \rightarrow\infty} \sum_{x=0}^{p^{N}-1} f(x) (-1)^{x} $$
(1)
(see [119]). For \(f_{1}(x) = f(x+1)\), we have
$$ I_{-1}(f_{1}) + I_{-1}(f) = 2f(0). $$
(2)
As is well known, the Changhee polynomials are defined by the generating function
$$ \int_{\mathbb{Z}_{p}} (1+t)^{x+y}\, d \mu_{-1}(y)= \frac{2}{2+t}(1+t)^{x} = \sum_{n=0}^{\infty}\operatorname{Ch}_{n}(x) \frac{t^{n}}{n!}. $$
(3)
When \(x=0\), \(\operatorname{Ch}_{n} = \operatorname{Ch}_{n}(0)\) are called the Changhee numbers (see [17, 18, 20]). The gamma and beta functions are defined by the following definite integrals: for \(\alpha>0 \), \(\beta>0\),
$$ \Gamma(\alpha) = \int_{0}^{\infty}e^{-t}t^{\alpha-1}\, dt $$
(4)
and
$$\begin{aligned} B(\alpha,\beta) &= \int_{0}^{1} t^{\alpha-1}(1-t)^{\beta-1} \,dt \\ &= \int_{0}^{\infty}\frac{t^{\alpha-1}}{(1+t)^{\alpha+\beta}} \,dt \end{aligned}$$
(5)
(see[20, 21]). Thus, by (4) and (5) we have
$$ \Gamma(\alpha+1) = \alpha\Gamma(\alpha), \qquad B(\alpha,\beta) = \frac {\Gamma(\alpha)\Gamma(\beta) }{\Gamma(\alpha+\beta)}. $$
(6)
Stirling numbers of the first kind are defined by
$$ \bigl(\log(1+t)\bigr)^{n} = n! \sum _{m=n}^{\infty}S_{1} (m,n) {t^{m} \over m!}, $$
(7)
and the Stirling numbers of the second kind are defined by
$$ \bigl(e^{t}-1\bigr)^{n}= n! \sum _{l=n}^{\infty}S_{2} (n,l) \frac{t^{l}}{l!} \quad (n \ge0). $$
(8)

Recently, Lim and Qi [20] have derived integral identities for Appell-type λ-Changhee numbers from the fermionic integral equation. The degenerate Bernoulli polynomials, a degenerate version of the well-known family of polynomials, were introduced by Carlitz, and after that, many researchers have studied the degenerate special polynomials (see [13, 20, 2228]).

The goal of this paper is to consider the Appell-type Changhee polynomials, another version of the Changhee polynomials in (3), and derive some properties of these polynomials. Furthermore, we investigate certain identities for these polynomials.

2 Some identities for Appell-type Changhee polynomials

Now we define the Appell-type Changhee polynomials \(\operatorname{Ch}_{n}^{*}(x)\) by
$$ \frac{2}{2+t}e^{xt} = \sum _{n=0}^{\infty}\operatorname{Ch}_{n}^{*}(x) \frac{t^{n}}{n!}. $$
(9)
When \(x=0\), the Changhee numbers \(\operatorname{Ch}_{n}^{*}=\operatorname{Ch}_{n}^{*}(0)\) are equal to the Changhee numbers \(\operatorname{Ch}_{n}=\operatorname{Ch}_{n}(0)\). From (9) we have
$$\begin{aligned} \frac{2}{2+t}e^{xt} &= \Biggl( \sum _{m=0}^{\infty}\operatorname{Ch}_{m}^{*} \frac{t^{m}}{m!} \Biggr) \Biggl(\sum_{l=0}^{\infty}x^{l} \frac{t^{l}}{l!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{m=0}^{n} {n \choose m} \operatorname{Ch}_{m}^{*} x^{n-m} \Biggr) \frac{t^{n}}{n!}. \end{aligned}$$
(10)
By (10) we have the following theorem.

Theorem 1

For \(n \in\mathbb{N}\), we have
$$ \operatorname{Ch}_{n}^{*}(x) = \sum _{m=0}^{n} {n \choose m} \operatorname{Ch}_{m}^{*} x^{n-m}. $$
(11)
By (9), replacing t by \(e^{t}-1\), we get
$$ \frac{2}{2+e^{t}-1} e^{x(e^{t}-1)} = \sum _{n=0}^{\infty}\operatorname{Ch}_{n}^{*}(x) \frac {(e^{t}-1)^{n}}{n!}. $$
(12)
Then we have
$$\begin{aligned} \mathrm{RHS}&= \sum_{n=0}^{\infty}\operatorname{Ch}_{n}^{*}(x)\frac{(e^{t}-1)^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\operatorname{Ch}_{n}^{*}(x) \frac{1}{n!} n! \sum_{l=n}^{\infty}S_{2}(l,n) \frac{t^{l}}{l!} \\ &= \sum_{l=0}^{\infty}\sum _{n=0}^{l} \operatorname{Ch}_{n}^{*}(x) S_{2}(l,n) \frac{t^{l}}{l!}, \end{aligned}$$
(13)
where \(S_{2}(l,n)\) are the Stirling numbers of the second kind, and
$$\begin{aligned} \mathrm{LHS}&= \frac{2}{1+e^{t}} e^{x(e^{t}-1)} \\ &= \sum_{m=0}^{\infty}E_{m} \frac{t^{m}}{m!} \sum_{n=0}^{\infty}\operatorname{Bel}_{n} (x) \frac{t^{n}}{n!} \\ &= \sum_{l=0}^{\infty}\sum _{n=0}^{l}{l \choose n}E_{n} \operatorname{Bel}_{l-n}(x) \frac{t^{l}}{l!}. \end{aligned}$$
(14)
It is well known that the Bell polynomials are defined by the generating function
$$ e^{x(e^{t}-1)}= \sum_{n=0}^{\infty}\operatorname{Bel}_{n} (x) \frac{t^{n}}{n!} $$
(see [8]). By (13) and (14) we have the following theorem.

Theorem 2

For \(l \in\mathbb{N}\), we have
$$ \sum_{n=0}^{l} \operatorname{Ch}_{n}^{*}(x) S_{2}(l,n) = \sum _{n=0}^{l}{l \choose n}E_{n} \operatorname{Bel}_{l-n}(x). $$
(15)
By (11) we can derive the following equation:
$$\begin{aligned} \frac{d}{dx} \operatorname{Ch}_{n}^{*}(x)&= \sum _{m=0}^{n-1} {n \choose m} \operatorname{Ch}_{m}^{*} (n-m) x^{n-m-1} \\ &= n\operatorname{Ch}_{n-1}^{*}(x). \end{aligned}$$
(16)
From (16) we get
$$\begin{aligned} n \int_{0}^{x} \operatorname{Ch}_{n-1}^{*} (s) \,ds&= \int_{0}^{x} \frac{d}{ds}\operatorname{Ch}_{n}^{*} (s)\,ds \\ &= \operatorname{Ch}_{n}^{*}(s) |_{s=0}^{x} \\ &= \operatorname{Ch}_{n}^{*}(x) - \operatorname{Ch}_{n}^{*}. \end{aligned}$$
(17)
By (17) we can derive the following theorem.

Theorem 3

For \(n \in\mathbb{N}\), we have
$$ \frac{\operatorname{Ch}_{n+1}^{*}(x) - \operatorname{Ch}_{n+1}^{*}}{n+1} = \int_{0}^{x} \operatorname{Ch}_{n}^{*}(s) \,ds. $$
(18)
By (4) we note that
$$\begin{aligned} 2 &= \Biggl(\sum_{n=0}^{\infty}\operatorname{Ch}_{n}^{*} \frac{t^{n}}{n!} \Biggr) (2+t) \\ &= \Biggl( \sum_{n=0}^{\infty}2 \operatorname{Ch}_{n}^{*} \frac{t^{n}}{n!} \Biggr) + t\sum _{n=0}^{\infty}\operatorname{Ch}_{n}^{*} \frac{t^{n}}{n!} \\ &= \Biggl( \sum_{n=0}^{\infty}2 \operatorname{Ch}_{n}^{*} \frac{t^{n}}{n!} \Biggr) +\sum _{n=1}^{\infty}n\operatorname{Ch}_{n-1}^{*} \frac{t^{n}}{n!} \\ &= 2\operatorname{Ch}_{0}^{*} + \sum_{n=1}^{\infty}\bigl(2\operatorname{Ch}_{n}^{*} + n\operatorname{Ch}_{n-1}^{*} \bigr) \frac {t^{n}}{n!}. \end{aligned}$$
(19)
By (19) we have the following theorem.

Theorem 4

For \(n \in\mathbb{N}\), we have
$$ \operatorname{Ch}_{0}^{*}=1, \qquad 2 \operatorname{Ch}_{n}^{*}+n\operatorname{Ch}_{n-1}^{*}=0 \quad \textit{if } n\geq1. $$
(20)
Now we observe that
$$\begin{aligned} \sum_{n=0}^{\infty}\operatorname{Ch}_{n}^{*}(1-x) \frac{t^{n}}{n!} &= \frac{2}{2+t} e^{(1-x)t} \\ &= \frac{2}{2+t}e^{t} e^{-xt} \\ &= \Biggl( \sum_{l=0}^{\infty}\operatorname{Ch}_{l}^{*}(1) \frac{t^{l}}{l!} \Biggr) \Biggl( \sum _{m=0}^{\infty}(-x)^{m} \frac {t^{m}}{m!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{m=0}^{n} {n \choose m} \operatorname{Ch}_{n-m}^{*}(1) (-x)^{m} \Biggr) \frac{t^{n}}{n!}. \end{aligned}$$
(21)
From (21) we obtain the following theorem.

Theorem 5

For \(n \in\mathbb{N}\), we have
$$ \operatorname{Ch}_{n}^{*}(1-x) = \sum _{m=0}^{n} {n \choose m} \operatorname{Ch}_{n-m}^{*}(1) (-x)^{m}. $$
(22)
By (22) we get
$$\begin{aligned} \int_{0}^{1} \operatorname{Ch}_{n}^{*}(1-x)x^{n} \,dx &= \sum_{m=0}^{n} {n \choose m} \operatorname{Ch}_{n-m}^{*}(1) (-1)^{m} \int_{0}^{1} x^{n+m} \,dx \\ &= \sum_{m=0}^{n} {n \choose m} (-1)^{m} \frac{\operatorname{Ch}_{n-m}^{*}(1)}{n+m+1}. \end{aligned}$$
(23)
From (16) we note that
$$\begin{aligned}& \int_{0}^{1} y^{n} \operatorname{Ch}_{n}^{*}(x+y) \,dy \\& \quad = \frac{y^{n+1}}{n+1} \operatorname{Ch}_{n}^{*}(x+y) \bigg|_{y=0}^{1} - \frac{1}{n+1} \int _{0}^{1} y^{n+1} \frac{d}{dy} \operatorname{Ch}_{n}^{*}(x+y)\,dy \\& \quad = \frac{\operatorname{Ch}_{n}^{*}(x+1)}{n+1}- \frac{n}{n+1} \int_{0}^{1} y^{n+1} \operatorname{Ch}_{n-1}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n}^{*}(x+1)}{n+1}- \frac{n}{n+1} \biggl( \frac {\operatorname{Ch}_{n-1}^{*}(x+y)}{n+2} y^{n+2} \bigg|_{y=0}^{1} \biggr) \\& \qquad {}+(-1)^{2} \frac{n}{n+1} \frac{1}{n+2} (n-1) \int_{0}^{1} y^{n+2}\operatorname{Ch}_{n-2}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n}^{*}(x+1)}{n+1}- \frac{n}{n+1} \frac{\operatorname{Ch}_{n-1}^{*}(x+1)}{n+2} + (-1)^{2}\frac{n}{n+1}\frac{n-1}{n+2} \int_{0}^{1} y^{n+2}\operatorname{Ch}_{n-2}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n}^{*}(x+1)}{n+1}- \frac{n}{n+1} \frac{\operatorname{Ch}_{n-1}^{*}(x+1)}{n+2} +(-1)^{2} \frac{n}{n+1}\frac{n-1}{n+2}\frac{\operatorname{Ch}_{n-2}^{*}(x+1)}{n+3} \\& \qquad {}+ (-1)^{3} \frac{n}{n+1}\frac{n-1}{n+2} \frac{n-2}{n+3} \int_{0}^{1} y^{n+3}\operatorname{Ch}_{n-3}^{*}(x+y) \,dy. \end{aligned}$$
(24)
Also, we get
$$ \int_{0}^{1} y^{2n-1} \operatorname{Ch}_{1}^{*}(x+y) \,dy = \frac{\operatorname{Ch}_{1}^{*} (x+y)}{2n} y^{2n} \bigg|_{y=0}^{1} - \frac{1}{2n} \int_{0}^{1} y^{2n} \operatorname{Ch}_{0}^{*}(x+y) \,dy. $$
(25)
From (11) we get
$$ \operatorname{Ch}_{0}^{*}(x) = 1, $$
(26)
and hence
$$\begin{aligned} \int_{0}^{1} y^{2n-1}\operatorname{Ch}_{1}^{*}(x+y) \,dy &= \frac{\operatorname{Ch}_{1}^{*}(x)}{2n} - \frac{1}{2n} \int_{0}^{1} y^{2n} \,dy \\ &= \frac{\operatorname{Ch}_{1}^{*}(x)}{2n} - \frac{1}{2n(2n+1)}. \end{aligned}$$
(27)
By (27), continuing the process in (24), we have
$$\begin{aligned}& \int_{0}^{1} y^{n} \operatorname{Ch}_{n}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n}^{*}(x+1)}{n+1}+\sum_{m=1}^{n} (-1)^{m} \operatorname{Ch}_{n-m}^{*}(x+1) \frac{n(n-1)\cdots(n-m+1)}{(n+1)(n+2)\cdots(n+m+1)}. \end{aligned}$$
(28)
We note that
$$\begin{aligned} \operatorname{Ch}_{n}^{*}(x+y)&= \operatorname{Ch}_{n}^{*} (x+1+y-1) \\ &= \sum_{l=1}^{n} {n \choose l} \operatorname{Ch}_{l}^{*}(x+1) (-1)^{n-l}(1-y)^{n-l}. \end{aligned}$$
(29)
By (29) we get
$$\begin{aligned}& \int_{0}^{1} y^{n} \operatorname{Ch}_{n}^{*}(x+y) \,dy \\& \quad = \sum_{l=1}^{n} {n \choose l} \operatorname{Ch}_{l}^{*}(x+1) (-1)^{n-l} \int_{0}^{1} y^{n} (1-y)^{n-l} \,dy \\& \quad = \sum_{l=1}^{n} {n \choose l} \operatorname{Ch}_{l}^{*}(x+1) (-1)^{n-l}B(n+1, n-l+1) \\& \quad = \sum_{l=0}^{n} {n\choose l} \operatorname{Ch}_{l}^{*} (x+1) (-1)^{n-l} \frac{\Gamma (n+1)\Gamma(n-l+1)}{\Gamma(2n-l+2)} \\& \quad = \sum_{l=0}^{n} (-1)^{n-l}{n \choose l} \frac {n!(n-l)!}{(2n-l+1)!}\operatorname{Ch}_{l}^{*}(x+1) \\& \quad = \sum_{l=0}^{n} (-1)^{n-l} \frac{n{n\choose l}}{(2n-l+1){{2n-l} \choose n}}\operatorname{Ch}_{l}^{*}(x+1). \end{aligned}$$
(30)

By (28) and (30) we have the following theorem.

Theorem 6

For \(n \in\mathbb{N}\), we have
$$\begin{aligned}& \sum_{l=0}^{n} (-1)^{n-l}\frac{n{n\choose l}}{(2n-l+1){{2n-l} \choose n}}\operatorname{Ch}_{l}^{*}(x+1) \\& \quad = \frac{\operatorname{Ch}_{n}^{*}(x+1)}{n+1}+\sum_{m=1}^{n} (-1)^{m} \operatorname{Ch}_{n-m}^{*}(x+1) \frac {n(n-1)\cdots(n-m+1)}{(n+1)(n+2)\cdots(n+m+1)}. \end{aligned}$$
(31)
From (16) we note that
$$\begin{aligned}& \int_{0}^{1} y^{n} \operatorname{Ch}_{n}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(x+y)}{n+1}y^{n} \bigg|_{y=0}^{1} - \frac{1}{n+1}n \int _{0}^{1} y^{n-1} \operatorname{Ch}_{n+1}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(x+1)}{n+1} - \frac{n}{n+1} \int_{0}^{1} y^{n-1}\operatorname{Ch}_{n+1}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(x+1)}{n+1} - \frac{n}{n+1} \frac {\operatorname{Ch}_{n+2}^{*}(x+1)}{n+2} + \frac{n(n-1)}{(n+1)(n+2)} \int_{0}^{1} y^{n-2}\operatorname{Ch}_{n+2}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(x+1)}{n+1} - \frac{n}{n+1} \frac {\operatorname{Ch}_{n+2}^{*}(x+1)}{n+2} + \frac{n(n-1)}{(n+1)(n+2)}\frac{\operatorname{Ch}_{n+3}^{*}(x+1)}{n+3} \\& \qquad {}-\frac{n(n-1)(n-2)}{(n+1)(n+2)(n+3)} \int_{0}^{1} y^{n-3}\operatorname{Ch}_{n+3}^{*}(x+y) \,dy. \end{aligned}$$
(32)
Also, we have
$$\begin{aligned}& \int_{0}^{1} y \operatorname{Ch}_{2n-1}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{2n}^{*}(x+y)}{2n}y \bigg|_{y=0}^{1} - \frac{1}{2n} \int_{0}^{1} 1 \cdot \operatorname{Ch}_{2n}^{*}(x+y) \,dy \\& \quad = \frac{\operatorname{Ch}_{2n}^{*}(x+1)}{2n}-\frac{1}{2n} \frac{1}{2n+1} \operatorname{Ch}_{2n+1}^{*}(x+y) \bigg|_{y=0}^{1} \\& \quad = \frac{\operatorname{Ch}_{2n}^{*}(x+1)}{2n}-\frac{\operatorname{Ch}_{2n+1}^{*}(x+1)-\operatorname{Ch}_{2n+1}^{*}(x)}{2n(2n+1)}. \end{aligned}$$
(33)

By (30), continuing the process in (28), we obtain the following theorem.

Theorem 7

For \(n \in\mathbb{N}\), we have
$$\begin{aligned}& \sum_{l=0}^{n} (-1)^{n-l}\frac{n{n\choose l}}{(2n-l+1){{2n-l} \choose n}}\operatorname{Ch}_{l}^{*}(x+1) \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(x+1)}{n+1}+\sum_{m=1}^{n-1} (-1)^{m} \operatorname{Ch}_{n+m+1}^{*}(x+1) \frac{n(n-1)\cdots(n-m+1)}{(n+1)(n+2)\cdots(n+m+1)} \\& \qquad {}+ (-1)^{n} \frac{n!}{(2n+1)_{n+1}} \bigl( \operatorname{Ch}_{2n+1}^{*}(x+1)- \operatorname{Ch}_{2n+1}^{*}(1) \bigr). \end{aligned}$$
(34)
Now, we have
$$\begin{aligned}& \int_{0}^{1} \operatorname{Ch}_{n}^{*}(x) \operatorname{Ch}_{m}^{*}(x) \,dx \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(x)\operatorname{Ch}_{m}^{*}(x)}{n+1} \bigg|_{0}^{1} - \frac{1}{n+1}m \int _{0}^{1} \operatorname{Ch}_{n+1}^{*}(x) \operatorname{Ch}_{m-1}^{*}(x) \,dx \\& \quad = \frac{1}{n+1} \bigl( \operatorname{Ch}_{n+1}^{*}(1) \operatorname{Ch}_{m}^{*}(1)-\operatorname{Ch}_{n+1}^{*}(0) \operatorname{Ch}_{m}^{*}(0) \bigr) \\& \qquad {} - \frac{m}{n+1} \int_{0}^{1} \operatorname{Ch}_{n+1}^{*}(x) \operatorname{Ch}_{m-1}^{*}(x) \,dx \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(1)\operatorname{Ch}_{m}^{*}(1)-\operatorname{Ch}_{n+1}^{*}\operatorname{Ch}_{m}^{*}}{n+1} - \frac {m}{n+1} \frac{\operatorname{Ch}_{n+2}^{*}(1)\operatorname{Ch}_{m-1}^{*}(1)-\operatorname{Ch}_{n+2}^{*}\operatorname{Ch}_{m-1}^{*}}{n+2} \\& \qquad {}+ (-1)^{2} \frac{m}{n+1}\frac{m-1}{n+2} \int_{0}^{1} \operatorname{Ch}_{n+2}^{*}(x) \operatorname{Ch}_{m-2}^{*}(x) \,dx \end{aligned}$$
(35)
and
$$\begin{aligned}& \int_{0}^{1} \operatorname{Ch}_{n+m-1}^{*}(x) \operatorname{Ch}_{1}^{*}(x) \,dx \\& \quad = \frac{\operatorname{Ch}_{n+m}^{*}(1) \operatorname{Ch}_{1}^{*}(1) - \operatorname{Ch}_{n+m}^{*} \operatorname{Ch}_{1}^{*}}{n+m} - \frac {1}{n+m} \int_{0}^{1} \operatorname{Ch}_{n+m}^{*}(x) \operatorname{Ch}_{0}^{*}(x)\,dx \\& \quad = \frac{\operatorname{Ch}_{n+m}^{*}(1) \operatorname{Ch}_{1}^{*}(1) - \operatorname{Ch}_{n+m}^{*} \operatorname{Ch}_{1}^{*}}{n+m} - \frac {1}{n+m} \frac{\operatorname{Ch}_{n+m+1}^{*}(1) - \operatorname{Ch}_{n+m+1}^{*}}{n+m+1}. \end{aligned}$$
(36)
By (30) with \(x=0\) we get
$$\begin{aligned}& \int_{0}^{1} \operatorname{Ch}_{n}^{*}(x) \operatorname{Ch}_{m}^{*}(x) \,dx \\& \quad = \sum_{j=0}^{m} {m \choose j} \operatorname{Ch}_{j}^{*} \int_{0}^{1} x^{m-j}\operatorname{Ch}_{m}^{*}(x) \,dx \\& \quad = \sum_{j=0}^{m} {m \choose j} \operatorname{Ch}_{j}^{*} \sum_{l=0}^{m-j} (-1)^{m-j-l}\frac {(m-j){m-j\choose l}}{(2(m-j)-l+1){2(m-j)-l\choose m-j}} \operatorname{Ch}_{l}^{*}(1) \\& \quad = \sum_{j=0}^{m} \sum _{l=0}^{m-j}{m \choose j} (-1)^{m-j-l} \frac {(m-j){m-j\choose l}}{(2(m-j)-l+1){2(m-j)-l\choose m-j}} \operatorname{Ch}_{j}^{*} \operatorname{Ch}_{l}^{*}(1). \end{aligned}$$
(37)

By (37), continuing the process in (35), we obtain the following theorem.

Theorem 8

For \(n \in\mathbb{N}\), we have
$$\begin{aligned}& \sum_{j=0}^{m} \sum _{l=0}^{m-j}{m \choose j} (-1)^{m-j-l} \frac {(m-j){m-j\choose l}}{(2(m-j)-l+1){2(m-j)-l\choose m-j}} \operatorname{Ch}_{j}^{*} \operatorname{Ch}_{l}^{*}(1) \\& \quad = \frac{\operatorname{Ch}_{n+1}^{*}(1)\operatorname{Ch}_{m}^{*}(1)-\operatorname{Ch}_{n+1}^{*}\operatorname{Ch}_{m}^{*}}{n+1} \\& \qquad {}+ \sum_{k=1}^{m-1} (-1)^{k} \frac{m(m-1)\cdots(m-k+1)}{(n+1)(n+2)\cdots(n+k+1)} \\& \qquad {}\times \bigl(\operatorname{Ch}_{n+k+1}^{*}(1)\operatorname{Ch}_{m-k}^{*}(1) - \operatorname{Ch}_{n+k+1}^{*}\operatorname{Ch}_{m-k}^{*} \bigr) \\& \qquad {}+ (-1)^{m} \frac{m!}{(n+m+1)_{m+1}} \bigl( \operatorname{Ch}_{n+m+1}^{*}(1) - \operatorname{Ch}_{n+m+1}^{*} \bigr). \end{aligned}$$
(38)

3 Remarks

In this section, by using the fermionic p-adic integral on \(\mathbb {Z}_{p}\), we derive some identities for Changhee polynomials, Stirling numbers of the first kind, and Euler numbers. By (2) we note that
$$\begin{aligned} \frac{2}{2+t}e^{xt} &= \int_{\mathbb{Z}_{p}} (1+t)^{y} e^{xt}\, d \mu_{-1}(y) \\ &= \int_{\mathbb{Z}_{p}} e^{y\log(1+t)+xt}\, d\mu_{-1}(y) \end{aligned}$$
(39)
and
$$\begin{aligned} e^{xt}e^{y\log(1+t)} &= \Biggl( \sum _{m=0}^{\infty}x^{m} \frac{t^{m}}{m!} \Biggr) \Biggl(\sum_{l=0}^{\infty}\frac{y^{l} (log(1+t))^{l}}{l!} \Biggr) \\ &= \Biggl( \sum_{m=0}^{\infty}x^{m} \frac{t^{m}}{m!} \Biggr) \Biggl( \sum_{l=0}^{\infty}y^{l} \sum_{k=l}^{\infty}S_{1}(k,l)\frac{t^{k}}{k!} \Biggr) \\ &= \Biggl( \sum_{m=0}^{\infty}x^{m} \frac{t^{m}}{m!} \Biggr) \Biggl( \sum_{k=0}^{\infty}\sum_{l=0}^{k} y^{l} S_{1}(k,l) \frac{t^{k}}{k!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{k=0}^{n} \sum_{l=0}^{k} {n \choose k} x^{n-k} y^{l} S_{1} (k,l) \Biggr) \frac{t^{n}}{n!}. \end{aligned}$$
(40)
Thus, by (39) and (40) we have
$$\begin{aligned} \sum_{n=0}^{\infty}\operatorname{Ch}_{n}^{*}(x) \frac{t^{n}}{n!} &= \int_{\mathbb {Z}_{p}}e^{y\log(1+t)}e^{xt}\, d \mu_{-1}(y) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{k=0}^{n} \sum_{l=0}^{k} {n \choose k} x^{n-k} \int_{\mathbb{Z}_{p}} y^{l} \, d\mu_{-1}(y) S_{1} (k,l) \Biggr)\frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{k=0}^{n} \sum_{l=0}^{k} {n \choose k} x^{n-k} E_{l} S_{1} (k,l) \Biggr) \frac{t^{n}}{n!}. \end{aligned}$$
(41)

From (41) we have the following theorem.

Theorem 9

For \(n \in\mathbb{N}\), we have
$$ \operatorname{Ch}_{n}^{*}(x) = \sum _{k=0}^{n} \sum_{l=0}^{k} {n \choose k} x^{n-k} E_{l} S_{1} (k,l). $$
(42)

Declarations

Acknowledgements

This paper was supported by Wonkwang University in 2015.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Division of Mathematics and Informational Statistics, Nanoscale Science and Technology Institute, Wonkwang University, Iksan, Republic of Korea
(2)
Graduate School of Education, Konkuk University, Seoul, Republic of Korea
(3)
Department of Applied Mathematics, Pukyung National University, Busan, Republic of Korea
(4)
Department of Mathematics Education, Konkuk University, Seoul, Republic of Korea
(5)
Department of Mathematics, Kwangwoon University, Seoul, Republic of Korea

References

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