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On Appell-type Changhee polynomials and numbers

  • 1,
  • 2Email author,
  • 3,
  • 4 and
  • 5
Advances in Difference Equations20162016:160

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

  • Received: 10 February 2016
  • Accepted: 17 May 2016
  • Published:

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 polynomials
  • Appell-type Changhee polynomials
  • degenerate Bernoulli polynomials
  • beta functions

MSC

  • 05A10
  • 11B68
  • 11S80
  • 05A19

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, 570-749, Republic of Korea
(2)
Graduate School of Education, Konkuk University, Seoul, 143-701, Republic of Korea
(3)
Department of Applied Mathematics, Pukyung National University, Busan, 608-737, Republic of Korea
(4)
Department of Mathematics Education, Konkuk University, Seoul, 143-701, Republic of Korea
(5)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, Republic of Korea

References

  1. Bayad, A, Kim, T: Identities for the Bernoulli, the Euler and the Genocchi numbers and polynomials. Adv. Stud. Contemp. Math. (Kyungshang) 20(2), 247-253 (2010) MathSciNetMATHGoogle Scholar
  2. Bayad, A, Kim, T: Identities involving values of Bernstein, q-Bernoulli, and q-Euler polynomials. Russ. J. Math. Phys. 18(2), 133-143 (2011) MathSciNetView ArticleMATHGoogle Scholar
  3. Carlitz, L: Degenerate Stirling, Bernoulli and Eulerian numbers. Util. Math. 15, 51-88 (1979) MathSciNetMATHGoogle Scholar
  4. Kim, BM, Jang, LC: A note on the von Staudt-Clausen’s theorem for the weighted q-Genocchi numbers. Adv. Differ. Equ. 2015, 4 (2015) MathSciNetView ArticleGoogle Scholar
  5. Kim, DS, Kim, T: Some identities of degenerate Euler polynomials arising from p-adic fermionic integrals on \(\mathbb{Z}_{p}\). Integral Transforms Spec. Funct. 26(4), 295-302 (2015) MathSciNetView ArticleMATHGoogle Scholar
  6. Kim, DS, Kim, T: Some identities of degenerate special polynomials. Open Math. 13, 380-389 (2015) MathSciNetGoogle Scholar
  7. Kim, DS, Kim, T, Dolgy, DV: Degenerate q-Euler polynomials. Adv. Differ. Equ. 2015, 246 (2015) MathSciNetView ArticleGoogle Scholar
  8. Kim, DS, Kim, T: Some identities of Bell polynomials. Sci. China Math. 58(10), 1-10 (2015) MathSciNetView ArticleMATHGoogle Scholar
  9. Kim, T: Note on the Euler numbers and polynomials. Adv. Stud. Contemp. Math. (Kyungshang) 17(2), 131-136 (2008) MathSciNetMATHGoogle Scholar
  10. Kim, T: Some properties on the integral of the product of several Euler polynomials. Quaest. Math. 38(4), 553-562 (2015) MathSciNetView ArticleGoogle Scholar
  11. Kim, T: Degenerate Euler zeta function. Russ. J. Math. Phys. 22(4), 469-472 (2015) MathSciNetView ArticleMATHGoogle Scholar
  12. Kim, T: On the multiple q-Genocchi and Euler numbers. Russ. J. Math. Phys. 15(4), 481-486 (2008) MathSciNetView ArticleMATHGoogle Scholar
  13. Kim, T, Mansour, T: Umbral calculus associated with Frobenius-type Eulerian polynomials. Russ. J. Math. Phys. 21(4), 484-493 (2008) MathSciNetView ArticleMATHGoogle Scholar
  14. Kim, T: New approach to q-Euler, Genocchi numbers and their interpolation functions. Adv. Stud. Contemp. Math. (Kyungshang) 18(2), 105-112 (2009) MathSciNetMATHGoogle Scholar
  15. Kim, T: On Euler-Barnes multiple zeta functions. Russ. J. Math. Phys. 10(3), 261-267 (2003) MathSciNetMATHGoogle Scholar
  16. Kim, T: A study on the q-Euler numbers and the fermionic q-integral of the product of several type q-Bernstein polynomials on \(\mathbb{Z}_{p}\). Adv. Stud. Contemp. Math. (Kyungshang) 23(1), 5-11 (2013) MathSciNetMATHGoogle Scholar
  17. Kim, T, Kim, DS: A note on nonlinear Changhee differential equations. Russ. J. Math. Phys. 23(1), 1-5 (2016) MathSciNetView ArticleGoogle Scholar
  18. Wang, NL, Li, H: Some identities on the higher-order Daehee and Changhee numbers. Pure Appl. Math. J. 5, 33-37 (2015) Google Scholar
  19. Yilmaz Yasar, B, Özarslan, MA: Frobenius-Euler and Frobenius-Genocchi polynomials and their differential equations. New Trends Math. Sci. 3(2), 172-180 (2015) MathSciNetGoogle Scholar
  20. Lim, D, Qi, F: On the Appell type λ-Changhee polynomials. J. Nonlinear Sci. Appl. 9, 1872-1876 (2016) MathSciNetMATHGoogle Scholar
  21. Kim, T, Park, JW, Seo, JJ: A note on λ-zeta function. Glob. J. Pure Appl. Math. 11(5), 3501-3506 (2015) Google Scholar
  22. Adelberg, A: A finite difference approach to degenerate Bernoulli and Stirling polynomials. Discrete Math. 140(1-3), 1-21 (1995) MathSciNetView ArticleMATHGoogle Scholar
  23. Carlitz, L: A degenerate Staudt-Clausen theorem. Arch. Math. (Basel) 7, 28-33 (1956) MathSciNetView ArticleMATHGoogle Scholar
  24. Gaboury, S, Tremblay, R, Fugère, B-J: Some explicit formulas for certain new classes of Bernoulli, Euler and Genocchi polynomials. Proc. Jangjeon Math. Soc. 17(1), 115-123 (2014) MathSciNetMATHGoogle Scholar
  25. Howard, FT: Explicit formulas for degenerate Bernoulli numbers. Discrete Math. 162(1-3), 175-185 (1996) MathSciNetView ArticleMATHGoogle Scholar
  26. Liu, GD: Degenerate Bernoulli numbers and polynomials of higher order. J. Math. (Wuhan) 25(3), 283-288 (2005) (in Chinese) MathSciNetMATHGoogle Scholar
  27. Mahmudov, NI, Akkeles, A, Öneren, A: On two dimensional q-Bernoulli and q-Genocchi polynomials: properties and location of zeros. J. Comput. Anal. Appl. 18(5), 834-843 (2015) MathSciNetMATHGoogle Scholar
  28. Kwon, JK: A note on weighted Boole polynomials. Glob. J. Pure Appl. Math. 11(4), 2055-2063 (2015) Google Scholar

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