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.

1 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 [3])

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

(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 [4])

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

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 [5]. Specifically, equations (14), (21), and (22) are related to the papers [5–8].

2 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 [9] 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. [10–18]. 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

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

3 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. [21].

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

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