Theory and Modern Applications

# On Appell-type Changhee polynomials and numbers

## 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.

## 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 ). 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  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 ). 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)

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## Acknowledgements

This paper was supported by Wonkwang University in 2015.

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Correspondence to Lee-Chae Jang.

### Competing interests

The authors declare that they have no competing interests.

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All authors contributed equally to this work. All authors read and approved the final manuscript.

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