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# Some explicit identities on Changhee-Genocchi polynomials and numbers

## Abstract

In this paper, we introduce a new family of functions, which is called the Changhee-Genocchi polynomials. We study some explicit identities on these polynomials, which are related to Genocchi polynomials and Changhee polynomials. Also, we represent Changhee-Genocchi polynomials by gamma and beta functions.

We also study some properties of higher-order Changhee-Genocchi polynomials related to Changhee polynomials and Daehee polynomials.

## Introduction

The Genocchi polynomials are defined by the generating function (see [1, 2])

$$\frac{2t}{e^{t}+1} e^{xt} = \sum _{n}^{\infty}G_{n}(x) \frac{t^{n}}{n!}.$$
(1)

When $$x=0$$, $$G_{n}=G_{n}(0)$$ are called the Genocchi numbers. From (1) we see that

\begin{aligned} \sum_{n=0}^{\infty}G_{n}(x) \frac{t^{n}}{n!} &= \biggl(\frac{2t}{e^{t}+1} \biggr) e^{xt} = \Biggl( \sum_{l=0}^{\infty}G_{l} \frac{t^{l}}{l!} \Biggr) \Biggl( \sum_{m=0}^{\infty}x^{m} \frac{t^{m}}{m!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{n} {n \choose l} G_{l} x^{n-l} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(2)

We consider Changhee-Genocchi polynomials defined by the generating function

$$\frac{2\log(1+t)}{2+t} (1+t)^{x} = \sum _{n=0}^{\infty}CG_{n}(x) \frac{t^{n}}{n!}.$$
(3)

When $$x=0$$, $$CG_{n} = CG_{n}(0)$$ are called the Changhee-Genocchi numbers.

The gamma and beta functions are defined by the following definite integrals:

$$\Gamma(\alpha) = \int_{0}^{\infty}e^{-t} t^{\alpha-1}\,dt,\quad \alpha>0,$$
(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,\quad \alpha>0,\beta>0. \end{aligned}
(5)

From (4) and (5) we have (see )

$$\Gamma(\alpha+1) = \alpha\Gamma(\alpha),\qquad B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha +\beta)}.$$
(6)

We recall that the classical Stirling numbers of the first kind $$S_{1}(n,k)$$ and $$S_{2}(n,k)$$ are defined by the relations (see )

\begin{aligned} &(x)_{n} = \sum_{k=0}^{n} S_{1}(n,k) x^{k} \quad\mbox{and}\\ &x^{n} = \sum_{k=0}^{n} S_{2}(n,k) (x)_{k}, \end{aligned}

respectively. Here $$(x)_{n} = x(x-1)\cdots(x-n+1)$$ denotes the falling factorial polynomial of order n. We also have

\begin{aligned} &\sum_{n=m}^{\infty}S_{2}(n,m) \frac{t^{n}}{n!} = \frac{(e^{t}-1)^{m}}{m!} \quad\mbox{and}\\ &\sum_{n=m}^{\infty}S_{1}(n,m) \frac{t^{n}}{n!} = \frac{(\log(1+t))^{m}}{m!}. \end{aligned}
(7)

In this paper, we introduce a new family of functions, which is called the Changhee-Genocchi polynomials.

We study some properties of these polynomials, which are related to Genocchi polynomials and Changhee polynomials. Also we represent Changhee-Genocchi polynomials by gamma and beta functions.

We also study higher-order Changhee-Genocchi polynomials related to Changhee polynomials and Daehee polynomials.

Most of the ideas in this paper come from Kim and Kim . Specifically, equations (14), (21), and (22) are related to the papers .

## Changhee-Genocchi polynomials

First, we relate our newly defined Changhee-Genocchi polynomials to Genocchi polynomials.

Replacing t by $$e^{t}-1$$ in (3) and applying (7), we get

\begin{aligned} \frac{2t}{e^{t}+1} e^{tx} &= \sum _{n=0}^{\infty}CG_{n}(x) \frac{1}{n!} \bigl(e^{t}-1\bigr)^{n} \\ &= \sum_{n=0}^{\infty}CG_{n}(x) \frac{1}{n!} n! \sum_{k=n}^{\infty}S_{2}(k,n) \frac{t^{k}}{k!} \\ &= \sum_{k=0}^{\infty}\Biggl( \sum _{n=0}^{k} CG_{n}(x) S_{2}(k,n) \Biggr)\frac{t^{k}}{k!}. \end{aligned}
(8)

The left-hand side of (8) is the generating function of the Genocchi polynomials.

Thus, by comparing the coefficients of (1) and (8) we have the following theorem.

### Theorem 1

For any nonnegative integer k, we have

$$G_{k}(x) = \sum_{n=0}^{k} CG_{n}(x) S_{2}(k,n).$$
(9)

On the other hand, if we replace t by $$\log(1+t)$$ in (1) and apply (7), then we get

\begin{aligned} \frac{2\log(1+t)}{2+t} (1+t)^{x} &= \sum_{n=0}^{\infty}G_{n}(x) \frac{1}{n!} \bigl( \log(1+t) \bigr)^{n} \\ &= \sum_{n=0}^{\infty}G_{n}(x) \frac{1}{n!} n! \sum_{k=n}^{\infty}S_{1}(k,n) \frac{t^{k}}{k!} \\ &= \sum_{k=0}^{\infty}\Biggl( \sum _{n=0}^{k} G_{n}(x) S_{1}(k,n) \Biggr) \frac{t^{k}}{k!}, \end{aligned}
(10)

where $$S_{1}(k,n)$$ are the Stirling numbers of the first kind.

By comparing the coefficients of both sides of (10), we get the following theorem.

### Theorem 2

For any nonnegative integer k, we have

$$CG_{k}(x) = \sum_{n=0}^{k} G_{n}(x) S_{1}(k,n).$$
(11)

### Remark

When $$x=0$$ in (11), we can see that Changhee-Genocchi numbers are integers.

We can consider equation (11) as the inversion formula for (9). From (3) we can consider the following identity:

\begin{aligned} \sum_{n=0}^{\infty}CG_{n}(x) \frac{t^{n}}{n!} &= \frac{2\log(1+t)}{2+t} (1+t)^{x} = \Biggl( \sum_{l=0}^{\infty}CG_{l} \frac{t^{l}}{l!} \Biggr) \Biggl(\sum_{m=0}^{\infty}(x)_{m} \frac{t^{m}}{m!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl(\sum _{l=0}^{n}{n \choose l}CG_{l}(x)_{n-l} \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(12)

Thus, by comparing the coefficients of both sides of (12) we have

\begin{aligned} CG_{n}(x) &= \sum _{l=0}^{n} {n \choose l} CG_{l} (x)_{n-l} = \sum_{l=0}^{n} {n \choose l} CG_{n-l} (x)_{l} \\ &= \sum_{l=0}^{n} \Biggl( \sum _{m=0}^{n-l}{n \choose l} CG_{l} S_{1}(n-l, m) x^{m} \Biggr). \end{aligned}
(13)

From (13) we can derive the following theorem.

### Theorem 3

For any nonnegative integer n, we have

$$\int_{0}^{1} CG_{n}(x)\,dx = \sum _{l=0}^{n}\sum_{m=0}^{n-l}{n \choose l} CG_{l} S_{1}(n-l, m) \frac{1}{m+1}.$$
(14)

In this paper, we define the λ-Changhee-Genocchi polynomials by a generating function as follows:

$$\frac{2\log(1+t)}{(1+t)^{\lambda}+ 1} (1+t)^{\lambda x} = \sum _{n=0}^{\infty}CG_{n,\lambda} (x) \frac{t^{n}}{n!}.$$
(15)

We recall that the λ-Changhee polynomials are defined in  by

$$\frac{2}{(1+t)^{\lambda}+ 1} (1+t)^{\lambda x} = \sum _{n=0}^{\infty}Ch_{n,\lambda}(x) \frac{t^{n}}{n!}.$$
(16)

When $$\lambda=1$$, Changhee-Genocchi polynomials are well-known Changhee polynomials, cf. . In order to establish a reflexive symmetry on the Changhee-Genocchi polynomials, we consider the following:

\begin{aligned} \sum_{n=0}^{\infty}CG_{n}(1-x)\frac{t^{n}}{n!} &= \frac{2\log(1+t)}{1+(1+t)}(1+t)^{1-x} = -\frac{2\log(1+t)}{(1+t)^{-1}+1}(1+t)^{-x} \\ &= \sum_{n=0}^{\infty}CG_{n,-1}(x) \frac{t^{n}}{n!}. \end{aligned}
(17)

By comparing the coefficients of (17) we have the following theorem.

### Theorem 4

For $$n\in\mathbb {N}$$, we have

$$CG_{n}(1-x) = CG_{n,-1}(x).$$
(18)

Thus, from (3) and (18) we have

\begin{aligned} \sum_{n=0}^{\infty}CG_{n}\bigl(-x+(1-y)\bigr)\frac{t^{n}}{n!} &= \frac{2\log(1+t)}{2+t}(1+t)^{-x+(1-y)} \\ &= \frac{2\log(1+t)}{2+t}(1+t)^{-x}(1+t)^{1-y} \\ &= \Biggl(\sum_{m=0}^{\infty}CG_{m}(-x)\frac{t^{m}}{m!} \Biggr) \Biggl(\sum _{l=0}^{\infty}(1-y)_{l}(-x)\frac{t^{l}}{l!} \Biggr) \\ &= \sum_{n=0}^{\infty}\Biggl( \sum _{m=0}^{n}{n\choose m} CG_{m}(-x) (1-y)_{n-m} \Biggr)\frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\sum _{m=0}^{n} \sum_{k=0}^{n-m} {n\choose m} CG_{m}(-x)S_{1}(n-m, k) (1-y)^{k}. \end{aligned}
(19)

By comparing the coefficients of (19) we have

$$CG_{n}\bigl(1-(x+y)\bigr) = \sum _{m=0}^{n}\sum_{k=0}^{n-m}{n \choose m}CG_{m}(-x) S_{1}(n-m, k) (1-y)^{k}.$$
(20)

On the other hand, by (5), (6), and (20) we have

\begin{aligned} &\int_{0}^{1} y^{n} CG_{n} \bigl(1-(x+y)\bigr)\,dy \\ &\quad= \sum_{m=0}^{n}\sum _{k=0}^{n-m}{n\choose m}CG_{m}(-x) S_{1}(n-m, k) B(n+1, k+1) \\ &\quad= \sum_{m=0}^{n}\sum _{k=0}^{n-m}{n\choose m}CG_{m}(-x)S_{1}(n-m,k) \frac{\Gamma(n+1)\Gamma(k+1)}{\Gamma(n+k+2)}. \end{aligned}
(21)

Thus, by (18) and (21) we have the following identities, which relate the λ-Changhee-Genocchi polynomials, the Stirling numbers, and the beta and gamma polynomials:

\begin{aligned} &\int_{0}^{1} y^{n} CG_{n,-1}(x+y)\,dy \\ &\quad= -\sum_{l=0}^{n}\sum _{m=0}^{n-l}{n\choose l}S_{1}(n-l,m)CG_{l} \int_{0}^{1} y^{n} \bigl(1-(x+y) \bigr)^{m} \,dy \\ &\quad= -\sum_{l=0}^{n}\sum _{m=0}^{n-l}\sum_{k=0}^{m}{n \choose l} {m\choose k}S_{1}(n-l,m) (-x)^{m-k} CG_{l} \int_{0}^{1} y^{n} (1-y)^{k} \,dy \\ &\quad= -\sum_{l=0}^{n}\sum _{m=0}^{n-l}\sum_{k=0}^{m}{n \choose l} {m\choose k}S_{1}(n-l,m) (-x)^{m-k} CG_{l} B(n+1, k+1) \\ &\quad= -\sum_{l=0}^{n}\sum _{m=0}^{n-l}\sum_{k=0}^{m}{n \choose l} {m\choose k}S_{1}(n-l,m) (-x)^{m-k} CG_{l} \frac{\Gamma(n+1)\Gamma(k+1)}{\Gamma(n+k+2)}. \end{aligned}
(22)

From (16) we consider

\begin{aligned} \sum_{n=0}^{\infty}CG_{n,\lambda}(1-x) \frac{t^{n}}{n!} &= \frac{2\log(1+t)}{(1+t)^{\lambda}+ 1}(1+t)^{\lambda(1-x)} = \frac{2\log(1+t)}{1+(1+t)^{-\lambda}}(1+t)^{-\lambda x} \\ &= \sum_{n=0}^{\infty}CG_{n,-\lambda}(x) \frac{t^{n}}{n!}. \end{aligned}
(23)

By comparing the coefficients of (23) we have the following theorem.

### Theorem 5

For any nonnegative integer n, we have

$$CG_{n,\lambda}(1-x) = CG_{n,-\lambda}(x).$$
(24)

### Remark

If we take $$\lambda=1$$ in Theorem 5, then we have the result in Theorem 4.

From the second line of (23) and from (16) we have

\begin{aligned} & \Biggl( \sum _{l=1}^{\infty}\frac{(-1)^{l-1} t^{l}}{l} \Biggr) \Biggl( \sum _{m=0}^{\infty}Ch_{m,\lambda}(x) \frac{t^{m}}{m!} \Biggr) \\ &\quad= \sum_{n=1}^{\infty} \Biggl( \sum _{l=1}^{n} \frac{(-1)^{l-1}}{l} \frac{Ch_{n-l,\lambda}(x)}{(n-l)!}n! \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(25)

By comparing the coefficients of (23) and (25) we have the following theorem.

### Theorem 6

For any positive integer n, we have

$$CG_{n,\lambda}(x) = \sum_{l=1}^{n} \frac{(-1)^{l-1}}{l} Ch_{n-l,\lambda }(x)\frac{n!}{(n-l)!}.$$

For $$r\in\mathbb {N}$$, we define the Changhee-Genocchi polynomials $$CG_{n}^{(r)}(x)$$ of order r by the generating function

$$\biggl( \frac{2\log(1+t)}{2+t} \biggr)^{r} (1+t)^{x} = \sum_{n=0}^{\infty}CG_{n}^{(r)}(x)\frac{t^{n}}{n!}.$$
(26)

From (26) we have the following relation between the Changhee-Genocchi polynomials of order r and the Changhee polynomials of order r:

\begin{aligned} &\bigl(\log(1+t) \bigr)^{r} \biggl(\frac{2}{2+t} \biggr)^{r} (1+t)^{x} \\ &\quad= \Biggl( r!\sum_{l=r}^{\infty}S_{2}(l,r)\frac{t^{l}}{l!} \Biggr) \Biggl( \sum _{m=0}^{\infty}Ch_{m}^{(r)}(x) \frac{t^{m}}{m!} \Biggr) \\ &\quad= \Biggl( \sum_{l=0}^{\infty}S_{2}(l+r,r)\frac{r! t^{l+r}}{(l+r)!} \Biggr) \Biggl( \sum _{m=0}^{\infty}Ch_{m}^{(r)}(x) \frac{t^{m}}{m!} \Biggr) \\ &\quad= \Biggl( \sum_{l=0}^{\infty}S_{2}(l+r,r) {l+r \choose r}^{-1} \frac {t^{l}}{l!} \Biggr) \Biggl( \sum_{m=0}^{\infty}Ch_{m}^{(r)}(x) \frac{t^{m}}{m!} \Biggr) t^{r} \\ &\quad= \sum_{n=0}^{\infty}\Biggl( \sum _{l=0}^{n} {n\choose l} S_{2}(l+r,r){l+r \choose r}^{-1} Ch_{n-l}^{(r)}(x) \Biggr) \frac{t^{n+r}}{n!}. \end{aligned}
(27)

By comparing the coefficients of (26) and (27) we have the following theorem.

### Theorem 7

For any nonnegative integer n, we have

$$CG_{n}^{(r)}(x) = \sum_{l=0}^{n}{n \choose l} {l+r\choose r}^{-1}S_{2}(l+r,r)Ch_{n-l}^{(r)}(x).$$

For $$d\in\mathbb {N}$$ with $$d\equiv1\ (\operatorname{mod}2)$$, we have the following identity:

$$\sum_{a=0}^{d-1}(-1)^{a}(1+t)^{a} = \frac{1+(1+t)^{d}}{2+t}.$$
(28)

So, for such $$d\equiv1\ (\operatorname{mod} 2)$$, from (28), (3), and (15) we see that

\begin{aligned} \sum_{n=0}^{\infty}CG_{n}(x)\frac{t^{n}}{n!} &= \frac{2\log (1+t)}{2+t}(1+t)^{x} \\ &= \sum_{a=0}^{d-1}(-1)^{a} \frac{2\log(1+t)}{(1+t)^{d}+1}(1+t)^{d (\frac{a+x}{d} )} \\ &= \sum_{a=0}^{d-1}(-1)^{a}\sum _{n=0}^{\infty}CG_{n,d} \biggl( \frac {a+x}{d} \biggr)\frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl(\sum _{a=0}^{d-1}(-1)^{a} CG_{n,d} \biggl(\frac {a+x}{d} \biggr) \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(29)

By comparing the coefficients in (29), for $$d\equiv1\ (\operatorname{mod} 2)$$, we have the following theorem.

### Theorem 8

For any nonnegative integer n and $$d\equiv1\ (\operatorname{mod} 2)$$, we have

$$CG_{n}(x) = \sum_{a=0}^{d-1} (-1)^{a} CG_{n,d} \biggl(\frac{a+x}{d} \biggr).$$
(30)

We remark that, for $$d\equiv1\ (\operatorname{mod} 2)$$, from (9) and (30) we have the inversion of Theorem 8.

### Theorem 9

For any nonnegative integer n and $$d\equiv1\ (\operatorname{mod} 2)$$, we have

\begin{aligned} G_{k}(x) &= \sum_{n=0}^{k} CG_{n}(x) S_{2}(k,n) \\ &= \sum_{n=0}^{k} \Biggl( \sum _{a=0}^{d-1} (-1)^{a} CG_{n,d} \biggl(\frac {a+x}{d} \biggr) S_{2}(k,n) \Biggr). \end{aligned}

From the generating function of the Changhee-Genocchi polynomials in (1), replacing t by $$\lambda\log(1+t)$$, we get

\begin{aligned} \frac{2\lambda\log(1+t)}{(1+t)^{\lambda}+1}(1+t)^{\lambda x} &= \sum_{n=0}^{\infty}G_{n}(x) \frac{1}{n!} \bigl( \lambda\log(1+t) \bigr)^{n} \\ &= \sum_{n=0}^{\infty}\lambda^{n} G_{n}(x) \Biggl( \sum_{k=n}^{\infty}S_{1}(k,n)\frac{t^{k}}{k!} \Biggr) \\ &= \sum_{k=0}^{\infty}\Biggl( \sum _{n=0}^{k}\lambda^{n} G_{n}(x) S_{1}(k,n) \Biggr)\frac{t^{k}}{k!}. \end{aligned}
(31)

Thus, the left-hand side of (31) can be represented by the λ-Changhee-Genocchi polynomials as follows:

$$\frac{2\lambda\log(1+t)}{(1+t)^{\lambda}+1} (1+t)^{\lambda x} = \lambda\sum _{k=0}^{\infty}CG_{k,\lambda}(x)\frac{t^{k}}{k!}.$$
(32)

By comparing the coefficients of (31) and (32) we have the following theorem.

### Theorem 10

For any nonnegative integer k, we have

$$CG_{k,\lambda}(x) = \sum_{n=0}^{k} \lambda^{n-1} G_{n}(x) S_{1}(k,n).$$

From the generating function of the Changhee-Genocchi numbers in (3) we want to see the recurrence relation for the Changhee-Genocchi numbers:

\begin{aligned} 2\log(1+t) &= \sum _{n=0}^{\infty}CG_{n} \frac{t^{n}}{n!}(t+2) \\ &= \sum_{n=1}^{\infty}CG_{n} \frac{t^{n+1}}{n!} + \sum_{n=0}^{\infty}2 CG_{n} \frac{t^{n}}{n!} \\ &= \sum_{n=2}^{\infty}n CG_{n-1} \frac{t^{n}}{n!} + 2\sum_{n=1}^{\infty}CG_{n} \frac{t^{n}}{n!} \\ &= 2CG_{1} t + \sum_{n=2}^{\infty}(n CG_{n-1} + 2CG_{n})\frac{t^{n}}{n!}. \end{aligned}
(33)

On the other hand, from the left-hand side of (33) we have

$$2\log(1+t) = \sum_{n=1}^{\infty}(-1)^{n-1} 2(n-1)! \frac{t^{n}}{n!}.$$
(34)

By comparing the coefficients of (33) and (34) we have the following recurrence relation for the Changhee-Genocchi numbers.

### Theorem 11

We have

\begin{aligned} & CG_{0} = 0,\\ & nCG_{n-1} + 2CG_{n} = 2(n-1)!(-1)^{n-1} \quad\textit{for } n\ge1. \end{aligned}

From the higher-order Changhee-Genocchi polynomials

$$\biggl( \frac{2\log(1+t)}{2+t} \biggr)^{r}(1+t)^{x} = \sum_{n=0}^{\infty}CG_{n}^{(r)}(x) \frac{t^{n}}{n!}$$
(35)

we can deduce

$$CG_{0}^{(r)}(x) = CG_{1}^{(r)}(x) = \cdots= CG_{r-1}^{(r)}(x) = 0.$$
(36)

Thus, from (36) we can rewrite (35) as follows:

$$\biggl( \frac{2\log(1+t)}{2+t} \biggr)^{r}(1+t)^{x} = \sum_{n=0}^{\infty}CG_{n+r}^{(r)}(x) \frac{t^{n+r}}{(n+r)!}.$$
(37)

We recall that the Dahee polynomials are defined by the generating function (see [9, 19])

$$\frac{\log(1+t)}{t} (1+t)^{x} = \sum_{n=0}^{\infty}D_{n}(x) \frac{t^{n}}{n!}.$$

When $$x=0$$, $$D_{n} = D_{n}(0)$$ are called the Dahee numbers.

For $$r\in\mathbb {N}$$, the higher-order Daehee numbers are given by the generating function (see [9, 19, 20])

$$\biggl(\frac{\log(1+t)}{t} \biggr)^{r} = \sum _{n=0}^{\infty}D_{n}^{(r)}(x) \frac{t^{n}}{n!}.$$

From (28) we have

\begin{aligned} 2\log(1+t)\sum _{a=0}^{d-1}(-1)^{a}(1+t)^{a} &= \frac{2\log(1+t)}{2+t} + \frac{2\log(1+t)}{t+2}(1+t)^{d} \\ &= \frac{2\log(1+t)}{t} \Biggl( \sum_{a=0}^{d-1}(-1)^{a}(1+t)^{a} \Biggr) \\ &= \sum_{n=0}^{\infty}CG_{n} \frac{t^{n-1}}{n!} + \sum_{n=0}^{\infty}CG_{n}(d)\frac{t^{n-1}}{n!} \\ &= \sum_{n=0}^{\infty}\Biggl( 2\sum _{a=0}^{d-1}(-1)^{a} D_{n}(a) \Biggr) \frac{t^{n}}{n!} \\ &= \sum_{n=0}^{\infty}\biggl( \frac{CG_{n+1}}{n+1} + \frac {CG_{n+1}(d)}{n+1} \biggr)\frac{t^{n}}{n!}. \end{aligned}
(38)

Thus, from (38) we have the following theorem.

### Theorem 12

For any nonnegative integer n and $$d\equiv1\ (\operatorname{mod} 2)$$, we have

$$2\sum_{a=0}^{d-1}(-1)^{a} D_{n}(a) = \frac{CG_{n+1}}{n+1} + \frac{CG_{n+1,d}}{n+1}.$$

## Changhee-Genocchi polynomials arising from differential equations

In this section, we give new identities on the Changhee-Genocchi numbers by using differential equations. We use the idea recently developed by Kwon et al. .

By equation (3) we can write the generating function for the Changhee-Genocchi numbers as follows:

$$F(t) = \frac{2\log(1+t)}{2+t} = \sum_{n=0}^{\infty}CG_{n}\frac{t^{n}}{n!}.$$
(39)

Let

\begin{aligned} G(t) = \log(1+t) \quad\mbox{and}\quad H(t) = \frac{2}{2+t}. \end{aligned}

Then

\begin{aligned} G^{(N)}(t) &= \biggl(\frac{d}{dt} \biggr)^{N} G(t) = (-1)^{N-1}(N-1)! e^{-N\cdot G(t)}, \quad\mbox{and}\\ H^{(N)}(t) &= \biggl(\frac{d}{dt} \biggr)^{N} H(t)\\ &= \biggl(-\frac{1}{2} \biggr)^{N} N! e^{-(N+1)\cdot K(t)},\quad \mbox{where } K(t)= \log(1+t/2). \end{aligned}

Thus,

\begin{aligned} F^{(N)}(t) ={}& \biggl( \frac{d}{dt} \biggr)^{N} F(t) = \sum _{k=0}^{N}{N\choose k}G^{(N-k)}H^{(k)} \\ ={}& \sum_{k=0}^{N} {N\choose k} (-1)^{N-k-1} (N-k-1)! e^{-(N-k)G(t)} \\ &{} \times \biggl(-\frac{1}{2} \biggr)^{k} k! e^{-(k+1)K(t)} \\ ={}& \sum_{k=0}^{N} {N\choose k} (-1)^{N-1} \biggl(\frac{1}{2} \biggr)^{k} k! (N-k-1)! e^{-(N-k)G(t)} e^{-(k+1)K(t)}. \end{aligned}
(40)

On the other hand,

\begin{aligned} e^{-(N-k)G} e^{-(k+1)K} ={}& \Biggl( \sum_{n=0}^{\infty}(-N+k)^{n} \frac {G^{n}}{n!} \Biggr) \Biggl( \sum_{l=0}^{\infty}\bigl(-(k+1)\bigr)^{l}\frac{K^{l}}{l!} \Biggr) \\ ={}& \Biggl( \sum_{n=0}^{\infty}(-N+k)^{n} \sum_{m=n}^{\infty}S_{1}(m,n) \frac {t^{m}}{m!} \Biggr) \\ &{} \times \Biggl( \sum_{l=0}^{\infty}(-k-1)^{l} \sum_{j=l}^{\infty}\frac{1}{2^{j}} S_{1}(j,l)\frac{t^{j}}{j!} \Biggr) \\ ={}& \sum_{m=0}^{\infty}\Biggl(\sum _{n=0}^{m}(-N+k)^{n} S_{1}(m,n) \Biggr)\frac {t^{m}}{m!} \\ &{} \times\sum_{j=0}^{\infty}\Biggl(\sum_{l=0}^{j}(-k-1)^{l} S_{1}(j,l)\frac{1}{2^{j}} \Biggr)\frac{t^{j}}{j!} \\ ={}& \sum_{s=0}^{\infty}\Biggl( \sum _{m=0}^{s}{s\choose m} \sum _{n=0}^{m}(-N+k)^{n} S_{1}(m,n) \\ &{}\times\sum_{l=0}^{s-m}(-k-1)^{l} S_{1}(s-m,l) \frac {1}{2^{s-m}} \Biggr)\frac{t^{s}}{s!}. \end{aligned}
(41)

From (39) we have

$$F^{(N)}(t) = \biggl(\frac{d}{dt} \biggr)^{N} F(t) = \sum_{m=0}^{\infty}CG_{N+m} \frac{t^{m}}{m!}.$$
(42)

By comparing the coefficients of (40), (41), and (42) we have new identities on the Changhee-Genocchi numbers as follows.

### Theorem 13

For any nonnegative integer s, we have

\begin{aligned} CG_{s+N} ={}& \sum_{m=0}^{s}{s \choose m} \Biggl\{ \Biggl( \sum_{n=0}^{m}(-N+k)^{n} S_{1}(m,n) \Biggr) \Biggl( \sum_{l=0}^{s-m}(-k-1)^{l} S_{1}(s-m, l)\frac{1}{2^{s-m}} \Biggr) \Biggr\} \\ &{} \times\sum_{k=0}^{N} {N \choose k} (-1)^{N-1} \biggl(\frac {1}{2} \biggr)^{k} k! (N-k-1)!. \end{aligned}

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The authors would like to express their sincere gratitude to the Editor, who gave us valuable comments to improve this paper.

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Correspondence to Seog-Hoon Rim.

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