Some new formulas for the products of the Frobenius-Euler polynomials

Abstract

The main purpose of this paper is, using the generating function methods and summation transform techniques, to establish some new formulas for the products of an arbitrary number of the Frobenius-Euler polynomials and give some illustrative special cases.

1 Introduction

Let λ be a complex number with $\lambda\neq1$. Frobenius [1] introduced and studied the so-called Frobenius-Euler polynomials $H_{n}(x|\lambda)$, which are usually defined by the following exponential generating function:

$$ \frac{1-\lambda}{e^{t}-\lambda}e^{xt}=\sum _{n=0}^{\infty}H_{n}(x|\lambda) \frac{t^{n}}{n!}. $$

(1.1)

In particular, the case $x=0$ in (1.1) gives the Frobenius-Euler numbers $H_{n}(\lambda)=H_{n}(0|\lambda)$. It is interesting to point out that the Frobenius-Euler polynomials can be defined recursively by the Frobenius-Euler numbers as follows:

$$ H_{n}(x|\lambda)=\sum_{k=0}^{n} \binom{n}{k}H_{k}(\lambda)x^{n-k}\quad(n\geq0), $$

(1.2)

where, and in what follows, $\binom{a}{k}$ is the binomial coefficient defined for a complex number a and a non-negative integer k by

$$ \binom{a}{0}=1,\qquad\binom{a}{k}=\frac{a(a-1)(a-2)\cdots (a-k+1)}{k!} \quad(k\geq1), $$

(1.3)

and the Frobenius-Euler numbers satisfy the recurrence relation

$$ H_{0}(\lambda)=1,\qquad \bigl(H(\lambda)+1 \bigr)^{n}-H_{n}(\lambda)= \textstyle\begin{cases} 1-\lambda,& n=0,\\ 0,&n\geq1, \end{cases} $$

(1.4)

with the usual convention about replacing $H^{n}(\lambda)$ by $H_{n}(\lambda)$; see, for example, [2, 3]. For some interesting arithmetic properties on the Frobenius-Euler polynomials and numbers, one is referred to [4–11].

We now turn to the Bernoulli polynomials $B_{n}(x)$ and the Euler polynomials $E_{n}(x)$, which are usually defined by the exponential generating functions

$$ \frac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n}(x) \frac{t^{n}}{n!}\quad\text{and} \quad\frac{2e^{xt}}{e^{t}+1}=\sum _{n=0}^{\infty}E_{n}(x)\frac{t^{n}}{n!}. $$

(1.5)

The rational numbers $B_{n}$ and the integers $E_{n}$ given by

$$ B_{n}=B_{n}(0)\quad\text{and}\quad E_{n}=2^{n}E_{n} \biggl(\frac{1}{2} \biggr) $$

(1.6)

are called the Bernoulli numbers and the Euler numbers, respectively. It is easily seen from (1.1) and (1.5) that the Frobenius-Euler polynomials give the Euler polynomials when $\lambda =-1$ in (1.1), and the Bernoulli polynomials can be expressed by the Frobenius-Euler polynomials as follows:

$$ m^{n-1}\sum_{k=0}^{m-1} \lambda^{k}B_{n} \biggl(\frac{k}{m} \biggr)= \frac{n}{\lambda-1}H_{n-1} \biggl(\frac{1}{\lambda} \biggr)\quad(m,n \geq1). $$

(1.7)

It is well known that the Bernoulli and Euler polynomials and numbers play important roles in different areas of mathematics, and numerous interesting properties for them have been studied by many authors; see, for example, [12–14].

In the year 1963, Carlitz [15] explored some formulas of products of the Frobenius-Euler polynomials and obtained three expressions of products of the Frobenius-Euler polynomials to deduce Nielsen’s [16] formulas on the Bernoulli and Euler polynomials. For example, Carlitz [15] showed that for non-negative integers m, n,

$$\begin{aligned} &H_{m}(x|\lambda)H_{n}(x|\mu) \\ & \quad = \frac{\lambda(\mu-1)}{\lambda\mu-1}\sum _{k=0}^{m} \binom{m}{k}H_{k}( \lambda)H_{m+n-k}(x|\lambda\mu) \\ &\quad \quad{} +\frac{\mu(\lambda-1)}{\lambda\mu-1}\sum_{k=0}^{n} \binom{n}{k}H_{k}(\mu)H_{m+n-k}(x|\lambda\mu)- \frac{(\lambda-1)(\mu-1)}{\lambda\mu-1}H_{m+n}(x|\lambda\mu), \end{aligned}$$

(1.8)

when $\lambda\mu\neq1$. In the year 2012, Kim et al. [17] used a nice method called the Frobenius-Euler basis to establish the following new sums of products of two Frobenius-Euler polynomials:

$$\begin{aligned}& \frac{1}{n+1}\sum_{k=0}^{n}H_{k}(x| \lambda)H_{n-k}(x|\lambda) \\& \quad =-\lambda\sum_{k=0}^{n-1} \binom{n}{k} \sum_{l=k}^{n} \frac{ H_{l-k}(\lambda)H_{n-l}(\lambda)-2 H_{n-k}(\lambda )}{n+1-k}H_{k}(x| \lambda)+H_{n}(x| \lambda), \end{aligned}$$

(1.9)

where n is a positive integer. Following the work of Carlitz and Kim et al., He and Wang [18] extended Carlitz’s [15] three formulas of products of the Frobenius-Euler polynomials, by virtue of which some analogues to the summation formula (1.9) were obtained. In the year 2014, Agoh and Dilcher [19] used a generalization of the idea showed in [20] to establish the following higher-order convolution identity for the Euler polynomials:

$$\begin{aligned} \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}E_{j_{1}}(x)\cdots E_{j_{k}}(x)&=\sum_{r=1}^{k} \binom{k}{r}(-2)^{r-1} \sum_{\substack{l_{0}+l_{1}+\cdots+l_{k-r}=n\\ l_{0},l_{1},\ldots,l_{k-r}\geq0}} \binom{n+k-1}{l_{0}} \\ &\quad{} \times E_{l_{0}}(x)E_{l_{1}}(0)\cdots E_{l_{k-r}}(0), \end{aligned}$$

(1.10)

where n is a non-negative integer and k is a positive integer k with $2\nmid k$. In the year 2016, by using identities for difference operators, techniques of symbolic computation, and tools from the probability theory, Dilcher and Vignat [21] extended (1.10) and obtained that for a non-negative integer n, a positive integer k with $2\nmid k$, and arbitrary real numbers $a_{1},\ldots,a_{k}$,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{n}{j_{1},\ldots,j_{k}} \frac{(a_{1})_{j_{1}}\cdots(a_{k})_{j_{k}}}{(a_{1}+\cdots +a_{k})_{n}}E_{j_{1}}(x) \cdots E_{j_{k}}(x) \\& \quad =\sum_{r=1}^{k}\sum _{ \vert J \vert =r}(-2)^{r-1}\sum_{\substack{l_{0}+l_{1}+\cdots +l_{k-r}=n\\l_{0},l_{1},\ldots,l_{k-r}\geq0}} \binom{n}{l_{0},l_{1},\ldots,l_{k-r}} \frac{(a_{i_{r+1}})_{l_{1}}\cdots (a_{i_{k}})_{l_{k-r}}}{(a_{1}+\cdots+a_{k})_{n-l_{0}}} \\& \quad \quad{} \times E_{l_{0}}(x)E_{l_{1}}(0)\cdots E_{l_{k-r}}(0), \end{aligned}$$

(1.11)

where, and in what follows, $(a)_{k}$ is the rising factorial defined for a complex number a and a non-negative integer k by

$$ (a)_{0}=1\quad\text{and}\quad(a)_{k}=a(a+1) (a+2)\cdots(a+k-1)\quad(k\geq1), $$

(1.12)

$\binom{n}{r_{1},\ldots,r_{k}}$ is the multinomial coefficient defined for a positive integer k and non-negative integers n, $r_{1},\ldots ,r_{k}$ by

$$ \binom{n}{r_{1},\ldots,r_{k}}=\frac{n!}{r_{1}!\cdots r_{k}!}, $$

(1.13)

$\vert J \vert$ is the cardinality of a subset $J\subseteq\{ 1,\ldots,k\}$ and $i_{r+1},\ldots,i_{k}\in\overline{J}=\{1,\ldots,k\}\setminus J$.

Motivated by the work of Dilcher and Vignat [21], in this paper we establish some new summation formulas for the products of an arbitrary number of the Frobenius-Euler polynomials by making use of the generating function methods and summation transform techniques developed in [22]. It turns out that some known formulas including (1.10) and (1.11) are deduced as special cases.

2 The statement of results

We first state the following formula for the products of an arbitrary number of the Frobenius-Euler polynomials and the rising factorials.

Theorem 2.1

Let $a_{1},\ldots,a_{k}$ be arbitrary complex numbers with k being a positive integer. Then, for a non-negative integer n,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{n}{j_{1},\ldots,j_{k}} \frac{(a_{1})_{j_{1}}\cdots(a_{k})_{j_{k}}}{(a_{1}+\cdots +a_{k})_{n}}H_{j_{1}}(x_{1}| \lambda_{1})\cdots H_{j_{k}}(x_{k}| \lambda_{k}) \\& \quad =\sum_{r=1}^{k}\frac{1-\lambda_{r}}{1-\lambda_{1}\cdots\lambda_{k}} \sum_{\substack{l_{1}+\cdots+l_{k}=n\\l_{1},\ldots,l_{k}\geq0}}\binom {n}{l_{1},\ldots,l_{k}}\frac{1}{(a_{1}+\cdots+a_{k})_{n-l_{r}}} \\& \quad\quad{} \times H_{l_{r}}(x_{r}|\lambda_{1} \cdots \lambda_{k})\prod_{i=1}^{r-1}(a_{i})_{l_{i}} H_{l_{i}}(x_{i}-x_{r}+1|\lambda_{i}) \\& \quad\quad{} \times\prod_{i=r+1}^{k} \lambda_{i}(a_{i})_{l_{i}} H_{l_{i}}(x_{i}-x_{r}| \lambda_{i})\quad( \lambda_{1}\cdots\lambda_{k} \neq1). \end{aligned}$$

(2.1)

We next discuss some special cases of Theorem 2.1. By taking $a_{1}=\cdots=a_{k}=1$ in Theorem 2.1, in light of (1.12) and (1.13), we get the following result.

Corollary 2.2

Let k be a positive integer. Then, for a non-negative integer n,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}H_{j_{1}}(x_{1}| \lambda_{1})\cdots H_{j_{k}}(x_{k}| \lambda_{k}) \\& \quad =\sum_{r=1}^{k}\frac{1-\lambda_{r}}{1-\lambda_{1}\cdots\lambda_{k}} \sum_{\substack{l_{1}+\cdots+l_{k}=n\\l_{1},\ldots,l_{k}\geq0}}\binom {n+k-1}{l_{r}}H_{l_{r}}(x_{r}| \lambda_{1}\cdots\lambda_{k})\qquad\qquad \\& \quad\quad{} \times\prod_{i=1}^{r-1}H_{l_{i}}(x_{i}-x_{r}+1| \lambda_{i})\prod_{i=r+1}^{k} \lambda_{i} H_{l_{i}}(x_{i}-x_{r}| \lambda_{i})\quad(\lambda_{1}\cdots\lambda_{k} \neq1). \end{aligned}$$

(2.2)

The above Corollary 2.2 can be also found in [23] where it was established by using the generalized beta integral technique. In fact, Corollary 2.2 can be used to give a different expression for the new sums of products of two Frobenius-Euler polynomials appearing in (1.9). For example, taking $k=2$ and then substituting x for $x_{1}$, y for $x_{2}$, λ for $\lambda _{1}$, and μ for $\lambda_{2}$ in Corollary 2.2 gives

$$\begin{aligned} &\sum_{k=0}^{n}H_{k}(x| \lambda)H_{n-k}(y|\mu) \\ &\quad = \frac{\mu(1-\lambda)}{1-\lambda\mu}\sum _{k=0}^{n}\binom{n+1}{k}H_{k}(x|\lambda \mu)H_{n-k}(y-x|\mu) \\ &\quad \quad{} +\frac{1-\mu}{1-\lambda\mu}\sum_{k=0}^{n} \binom{n+1}{k}H_{k}(y|\lambda\mu)H_{n-k}(x-y+1|\lambda) \quad(\lambda\mu\neq1). \end{aligned}$$

(2.3)

Since the Frobenius-Euler polynomials satisfy the following difference equation (see, e.g., [17]):

$$ H_{n}(x+1|\lambda)-\lambda H_{n}(x| \lambda)=(1-\lambda)x^{n}\quad(n\geq0), $$

(2.4)

so by applying (2.4) to (2.3), we get

$$\begin{aligned} &\sum_{k=0}^{n}H_{k}(x| \lambda)H_{n-k}(y|\mu) \\ &\quad =\frac{\mu(1-\lambda)}{1-\lambda\mu}\sum _{k=0}^{n}\binom{n+1}{k}H_{k}(x|\lambda \mu)H_{n-k}(y-x|\mu) \\ &\quad \quad{} +\frac{\lambda(1-\mu)}{1-\lambda\mu}\sum_{k=0}^{n} \binom{n+1}{k}H_{k}(y|\lambda\mu)H_{n-k}(x-y|\lambda) \\ &\quad \quad{} +\frac{(1-\lambda)(1-\mu)}{1-\lambda\mu}\sum_{k=0}^{n} \binom{n+1}{k}H_{k}(y|\lambda\mu) (x-y)^{n-k}\quad(\lambda \mu\neq1). \end{aligned}$$

(2.5)

It becomes obvious that the case $x=y$ and $\lambda=\mu$ in (2.5) gives

$$\begin{aligned} \sum_{k=0}^{n}H_{k}(x| \lambda)H_{n-k}(x|\lambda)&=\frac{2\lambda}{1+\lambda}\sum _{k=0}^{n}\binom{n+1}{k}H_{k} \bigl(x| \lambda^{2} \bigr)H_{n-k}(\lambda) \\ &\quad{} +\frac{1-\lambda}{1+\lambda}(n+1)H_{n} \bigl(x|\lambda^{2} \bigr) \quad(\lambda\neq-1), \end{aligned}$$

(2.6)

which can be regarded as an equivalent version of (1.9). For a different proof of (2.5), see [18] for details.

On the other hand, from (2.4) and the fact (see, e.g., [24])

$$ (x_{1}+y_{1}) (x_{2}+y_{2}) \cdots(x_{k}+y_{k})=\sum_{J\subseteq\{1,\ldots,k\}} \prod_{i\in J}x_{i}\prod _{i\in\overline{J}}y_{i}\quad(k\geq1), $$

(2.7)

we obtain that for a positive integer r,

$$\begin{aligned}& \prod_{i=1}^{r-1}H_{j_{i}}(x_{i}-x_{r}+1| \lambda_{i}) \\ & \quad =\sum_{J\subseteq\{1,\ldots,r-1\}}\prod_{i\in J} \lambda_{i}H_{j_{i}}(x_{i}-x_{r}| \lambda_{i})\prod_{i\in\overline{J}}(1- \lambda_{i}) (x_{i}-x_{r})^{j_{i}}. \end{aligned}$$

(2.8)

Thus, by applying (2.8) to Theorem 2.1 and then taking $x_{1}=\cdots=x_{k}=x$, we get the following result.

Corollary 2.3

Let $a_{1},\ldots,a_{k}$ be arbitrary complex numbers with k being a positive integer. Then, for a non-negative integer n,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{n}{j_{1},\ldots,j_{k}} \frac{(a_{1})_{j_{1}} \cdots(a_{k})_{j_{k}}}{(a_{1}+\cdots +a_{k})_{n}}H_{j_{1}}(x| \lambda_{1})\cdots H_{j_{k}}(x|\lambda_{k}) \\ & \quad =\sum_{r=1}^{k}\sum _{ \vert J \vert =r}\frac{\lambda_{J}}{1-\lambda_{1}\cdots\lambda _{k}} \sum_{\substack{l_{0}+l_{1}+\cdots+l_{k-r}=n\\l_{0},l_{1},\ldots ,l_{k-r}\geq0}} \binom{n}{l_{0},l_{1},\ldots,l_{k-r}} \frac{(a_{i_{r+1}})_{l_{1}}\cdots (a_{i_{k}})_{l_{k-r}}}{(a_{1}+\cdots+a_{k})_{n-l_{0}}} \\ & \quad\quad{} \times H_{l_{0}}(x|\lambda_{1}\cdots \lambda_{k}) \lambda_{i_{r+1}}H_{l_{1}}( \lambda_{i_{r+1}})\cdots \lambda_{i_{k}}H_{l_{k-r}}( \lambda_{i_{k}}), \end{aligned}$$

(2.9)

where $\lambda_{1}\cdots\lambda_{k}\neq1$, $\lambda_{J}=\prod_{j\in J}(1-\lambda_{j})$ and $i_{r+1},\ldots,i_{k}\in\overline{J}$.

In particular, if we take $\lambda_{1}=\cdots=\lambda_{k}=-1$ with $2\nmid k$ and let $a_{1},\ldots,a_{k}$ be real numbers in Corollary 2.3, we get Dilcher and Vignat’s identity (1.11) immediately. If we take $a_{1}=\cdots=a_{k}=1$ in Corollary 2.3, we obtain the following result.

Corollary 2.4

Let n be a non-negative integer. Then, for a positive integer k,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}H_{j_{1}}(x| \lambda_{1}) \cdots H_{j_{k}}(x|\lambda_{k}) \\& \quad =\sum_{r=1}^{k}\sum _{ \vert J \vert =r}\frac{\lambda_{J}}{1-\lambda_{1}\cdots\lambda _{k}} \sum_{\substack{l_{0}+l_{1}+\cdots+l_{k-r}=n\\l_{0},l_{1},\ldots ,l_{k-r}\geq0}} \binom{n+k-1}{l_{0}}H_{l_{0}}(x|\lambda_{1}\cdots \lambda_{k}) \\& \quad\quad{} \times\lambda_{i_{r+1}}H_{l_{1}}(\lambda_{i_{r+1}}) \cdots \lambda_{i_{k}}H_{l_{k-r}}(\lambda_{i_{k}}), \end{aligned}$$

(2.10)

where $\lambda_{1}\cdots\lambda_{k}\neq1$, $\lambda_{J}=\prod_{j\in J}(1-\lambda_{j})$ and $i_{r+1},\ldots,i_{k}\in\overline{J}$.

The above Corollary 2.4 can be also found in [23] where it was obtained by applying (2.8) to Corollary 2.2. If we take $\lambda_{1}=\cdots=\lambda_{k}=\lambda$ in Corollary 2.4, we obtain that for a non-negative integer n and a positive integer k,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}H_{j_{1}}(x|\lambda) \cdots H_{j_{k}}(x|\lambda) \\& \quad =\sum_{r=1}^{k}\binom{k}{r} \frac{\lambda^{k-r}(1-\lambda)^{r}}{1-\lambda^{k}} \sum_{\substack {l_{0}+l_{1}+\cdots+l_{k-r}=n\\l_{0},l_{1},\ldots,l_{k-r}\geq0}}\binom {n+k-1}{l_{0}}H_{l_{0}} \bigl(x|\lambda^{k} \bigr) \\& \quad\quad{} \times H_{l_{1}}(\lambda)\cdots H_{l_{k-r}}( \lambda). \end{aligned}$$

(2.11)

Obviously, the case $\lambda=-1$ and $2\nmid k$ in (2.11) gives Agoh and Dilcher’s identity (1.10). If we take $k=2$ in (2.11), we get that for a non-negative integer n,

$$\begin{aligned} \sum_{\substack{j_{1}+j_{2}=n\\j_{1},j_{2}\geq 0}}H_{j_{1}}(x|\lambda) H_{j_{2}}(x|\lambda)&=\frac{2\lambda(1-\lambda)}{1-\lambda^{2}} \sum _{\substack{l_{0}+l_{1}=n\\l_{0},l_{1}\geq0}} \binom{n+1}{l_{0}}H_{l_{0}} \bigl(x| \lambda^{2} \bigr)H_{l_{1}}(\lambda) \\ &\quad{} +\frac{(1-\lambda)^{2}}{1-\lambda^{2}}\binom{n+1}{n}H_{n} \bigl(x| \lambda^{2} \bigr)\quad(\lambda\neq-1), \end{aligned}$$

(2.12)

which gives formula (2.6) immediately.

3 The proof of Theorem 2.1

For convenience, we denote by $[t_{1}^{i_{1}}\cdots t_{k}^{i_{k}}]f(t_{1},\ldots, t_{k})$ the coefficients of $t_{1}^{i_{1}}\cdots t_{k}^{i_{k}}$ in the power series expansion of $f(t_{1},\ldots, t_{k})$. It is clear that for non-negative integers $i_{1},\ldots,i_{k}$, we have

$$ \biggl[\frac{t_{1}^{i_{1}}}{i_{1}!}\cdots\frac {t_{k}^{i_{k}}}{i_{k}!} \biggr]f(t_{1},\ldots, t_{k})=i_{1}!\cdots i_{k}!\cdot \bigl[t_{1}^{i_{1}}\cdots t_{k}^{i_{k}} \bigr]f(t_{1},\ldots, t_{k}). $$

(3.1)

We now recall the famous Euler’s pentagonal number theorem: for $\vert x \vert <1$,

$$ (1-x) \bigl(1-x^{2} \bigr) \bigl(1-x^{3} \bigr)\cdots=1+\sum_{n=1}^{\infty}(-1)^{n} \bigl\{ x^{\frac{1}{2}n(3n-1)}+x^{\frac{1}{2}n(3n+1)} \bigr\} , $$

(3.2)

which can be used effectively for the calculation of the number of partitions of n (see, e.g., [25]). In his original proof of (3.2), Euler used the following beautiful idea:

$$\begin{aligned}& (1+x_{1}) (1+x_{2}) (1+x_{3}) \cdots \\& \quad =(1+x_{1})+x_{2}(1+x_{1})+x_{3}(1+x_{1}) (1+x_{2})+\cdots. \end{aligned}$$

(3.3)

Obviously, the finite form of (3.3) can be expressed as (see, e.g., [26])

$$\begin{aligned}& (1+x_{1}) (1+x_{2})\cdots(1+x_{k}) \\& \quad =(1+x_{1})+x_{2}(1+x_{1})+ \cdots+x_{k}(1+x_{1}) (1+x_{2}) \cdots(1+x_{k-1}). \end{aligned}$$

(3.4)

If we replace $x_{r}$ by $x_{r}-1$ for $1\leq r \leq k$ in (3.4), then we have

$$ x_{1}\cdots x_{k}-1=\sum _{r=1}^{k}(x_{r}-1)x_{1}\cdots x_{r-1}, $$

(3.5)

where $x_{1}\cdots x_{r-1}$ is considered to be equal to 1 when $r=1$. By taking $x_{r}=\lambda_{r}e^{t_{r}}$ for $1\leq r \leq k$ in (3.5), we obtain that for a positive integer k,

$$ \lambda_{1}\cdots\lambda_{k}e^{t_{1}+\cdots+t_{k}}-1= \sum_{r=1}^{k} \bigl(\lambda_{r}e^{t_{r}}-1 \bigr)\prod_{i=1}^{r-1}\lambda_{i}e^{t_{i}}, $$

(3.6)

which implies

$$ \prod_{i=1}^{k} \frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda_{i}e^{t_{i}}-1}=\sum_{r=1}^{k} \frac{\lambda_{r}e^{t_{r}}-1}{\lambda_{1}\cdots\lambda _{k}e^{t_{1}+\cdots+t_{k}}-1}\prod_{i=1}^{r-1} \lambda_{i}e^{t_{i}}\prod_{i=1}^{k} \frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda_{i}e^{t_{i}}-1}. $$

(3.7)

Observe that

$$\begin{aligned}& \bigl(\lambda_{r}e^{t_{r}}-1 \bigr)\prod _{i=1}^{r-1}\lambda_{i}e^{t_{i}}\prod _{i=1}^{k}\frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda_{i}e^{t_{i}}-1} \\& \quad =(\lambda_{r}-1)e^{x_{r}(t_{1}+\cdots+t_{k})}\prod _{i=1}^{r-1} \lambda_{i}\frac{(\lambda_{i}-1)e^{(x_{i}-x_{r}+1)t_{i}}}{\lambda _{i}e^{t_{i}}-1} \prod_{i=r+1}^{k} \frac{(\lambda_{i}-1)e^{(x_{i}-x_{r})t_{i}}}{ \lambda_{i}e^{t_{i}}-1}. \end{aligned}$$

(3.8)

It follows from (3.7) and (3.8) that

$$\begin{aligned} \prod_{i=1}^{k} \frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda_{i}e^{t_{i}}-1} &=\sum _{r=1}^{k} \frac{(\lambda_{r}-1)e^{x_{r}(t_{1}+\cdots+t_{k})}}{\lambda_{1}\cdots \lambda_{k}e^{t_{1}+\cdots+t_{k}}-1} \\ &\quad{} \times\prod_{i=1}^{r-1} \lambda_{i} \frac{(\lambda_{i}-1)e^{(x_{i}-x_{r}+1)t_{i}}}{\lambda_{i}e^{t_{i}}-1} \prod_{i=r+1}^{k} \frac{(\lambda_{i}-1)e^{(x_{i}-x_{r})t_{i}}}{\lambda_{i}e^{t_{i}}-1}. \end{aligned}$$

(3.9)

It is obvious that substituting $1/\lambda$ for λ in (1.1) gives

$$ \frac{\lambda-1}{\lambda e^{t}-1}e^{xt}=\sum _{n=0}^{\infty}H_{n} \biggl(x \Big| \frac{1}{\lambda} \biggr)\frac{t^{n}}{n!}, $$

(3.10)

and from (1.3) and (1.12) we get that for a non-negative integer k and a complex number a,

$$ (a)_{k}=(-1)^{k}k!\cdot\binom{-a}{k}. $$

(3.11)

It follows from (1.13), (3.10) and (3.11) that for a non-negative integer n and complex numbers $a_{1},\ldots,a_{k}$,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{-a_{1}}{j_{1}}\cdots \binom{-a_{k}}{j_{k}} \biggl[\frac{t_{1}^{j_{1}}}{j_{1}!}\cdots\frac {t_{k}^{j_{k}}}{j_{k}!} \biggr] \Biggl(\prod_{i=1}^{k}\frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda _{i}e^{t_{i}}-1} \Biggr) \\& \quad =\sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots,j_{k}\geq0}}\binom {-a_{1}}{j_{1}}\cdots \binom{-a_{k}}{j_{k}}H_{j_{1}} \biggl(x_{1} \Big| \frac{1}{\lambda_{1}} \biggr)\cdots H_{j_{k}} \biggl(x_{k} \Big| \frac{1}{\lambda_{k}} \biggr) \\& \quad =\frac{(-1)^{n}}{n!} \sum_{\substack{j_{1}+\cdots+j_{k}=n\\ j_{1},\ldots,j_{k}\geq0}}\binom{n}{j_{1},\ldots,j_{k}} (a_{1})_{j_{1}}\cdots(a_{k})_{j_{k}} \\& \qquad{}\times H_{j_{1}} \biggl(x_{1} \Big| \frac{1}{\lambda_{1}} \biggr)\cdots H_{j_{k}} \biggl(x_{k} \Big| \frac{1}{\lambda_{k}} \biggr). \end{aligned}$$

(3.12)

On the other hand, since for a positive integer k and a non-negative integer N (see, e.g., [27]),

$$ (t_{1}+\cdots+t_{k})^{N}=\sum _{\substack{l_{1}+\cdots+l_{k}=N\\l_{1},\ldots,l_{k}\geq0}}\binom {N}{l_{1},\ldots,l_{k}}t_{1}^{l_{1}} \cdots t_{k}^{l_{k}}, $$

(3.13)

so by (3.10) and (3.13) we have

$$ \frac{(\lambda_{1}\cdots\lambda_{k}-1) e^{x_{r}(t_{1}+\cdots+t_{k})}}{\lambda_{1}\cdots\lambda _{k}e^{t_{1}+\cdots+t_{k}}-1} =\sum_{N=0}^{\infty}H_{N} \biggl(x_{r} \Big| \frac{1}{\lambda_{1}\cdots\lambda_{k}} \biggr) \sum _{\substack{l_{1}+\cdots+l_{k}=N\\l_{1},\ldots,l_{k}\geq0}}\frac {t_{1}^{l_{1}}}{l_{1}!}\cdots\frac{t_{k}^{l_{k}}}{l_{k}!}. $$

(3.14)

If we multiply both sides of (3.9) by $[t_{1}^{j_{1}}\cdots t_{k}^{j_{k}}]$, with the help of (3.10) and (3.14), we discover

$$\begin{aligned}& \bigl[t_{1}^{j_{1}}\cdots t_{k}^{j_{k}} \bigr] \Biggl(\prod_{i=1}^{k} \frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda_{i}e^{t_{i}}-1} \Biggr) \\& \quad =\sum_{r=1}^{k}\frac{\lambda_{r}-1}{\lambda_{1}\cdots\lambda_{k}-1}\sum _{\substack{l_{1},\ldots,l_{r-1},\\l_{r+1},\ldots,l_{k}\geq0}}\frac {H_{l_{1}+\cdots+l_{r-1}+j_{r}+l_{r+1}+\cdots+l_{k}} (x_{r}|\frac{1}{\lambda_{1}\cdots\lambda_{k}})}{l_{1}!\cdots l_{r-1}!\cdot j_{r}!\cdot l_{r+1}!\cdots l_{k}!} \\& \quad\quad{} \times\prod_{i=1}^{r-1} \lambda_{i} \frac{H_{j_{i}-l_{i}}(x_{i}-x_{r}+1|\frac{1}{\lambda _{i}})}{(j_{i}-l_{i})!}\prod_{i=r+1}^{k} \frac{H_{j_{i}-l_{i}}(x_{i}-x_{r}|\frac{1}{\lambda_{i}})}{(j_{i}-l_{i})!}. \end{aligned}$$

(3.15)

Hence, by replacing $l_{i}$ by $j_{i}-l_{i}$ for $i\neq r$ in (3.15), in light of (3.1), we obtain

$$\begin{aligned}& \biggl[\frac{t_{1}^{j_{1}}}{j_{1}!}\cdots\frac {t_{k}^{j_{k}}}{j_{k}!} \biggr] \Biggl( \prod _{i=1}^{k}\frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda _{i}e^{t_{i}}-1} \Biggr) \\ & \quad =\sum_{r=1}^{k}\frac{\lambda_{r}-1}{\lambda_{1}\cdots\lambda_{k}-1} \sum_{\substack{l_{1}+\cdots+l_{k}=j_{1}+\cdots+j_{k}\\l_{1},\ldots ,l_{k}\geq0}}H_{l_{r}} \biggl(x_{r} \Big| \frac{1}{\lambda_{1}\cdots\lambda_{k}} \biggr) \\ & \quad\quad{} \times\prod_{i=1}^{r-1} \lambda_{i} \binom{j_{i}}{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r}+1 \Big| \frac{1}{\lambda_{i}} \biggr) \prod_{i=r+1}^{k} \binom{j_{i}}{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r} \Big| \frac{1}{\lambda_{i}} \biggr). \end{aligned}$$

(3.16)

It follows from (3.16) that

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{-a_{1}}{j_{1}}\cdots \binom{-a_{k}}{j_{k}} \biggl[\frac{t_{1}^{j_{1}}}{j_{1}!}\cdots\frac {t_{k}^{j_{k}}}{j_{k}!} \biggr] \Biggl(\prod_{i=1}^{k}\frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda _{i}e^{t_{i}}-1} \Biggr) \\ & \quad =\sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots,j_{k}\geq0}}\sum_{r=1}^{k} \frac{\lambda_{r}-1}{\lambda_{1}\cdots\lambda_{k}-1} \sum_{\substack {l_{1}+\cdots+l_{k}=n\\l_{1},\ldots,l_{k}\geq0}}\binom{-a_{r}}{j_{r}}H_{l_{r}} \biggl(x_{r} \Big| \frac{1}{\lambda_{1}\cdots\lambda_{k}} \biggr) \\ & \quad\quad{} \times\prod_{i=1}^{r-1} \lambda_{i} \binom{-a_{i}}{j_{i}}\binom{j_{i}}{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r}+1 \Big| \frac{1}{\lambda_{i}} \biggr) \\ & \quad\quad{} \times\prod_{i=r+1}^{k} \binom{-a_{i}}{j_{i}} \binom{j_{i}}{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r} \Big| \frac{1}{\lambda_{i}} \biggr). \end{aligned}$$

(3.17)

It is clear from (1.3) that for non-negative integers k, n and a complex number a,

$$ \binom{a}{n}\binom{n}{k}=\binom{a}{k} \binom{a-k}{n-k}, $$

(3.18)

which together with the famous Chu-Vandermonde convolution identity showed in [28] yields that for non-negative integers $l_{1},\ldots,l_{k}$ with $l_{1}+\cdots+l_{k}=n$,

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{-a_{r}}{j_{r}}\prod _{\substack{i=1\\i\neq r}}^{k}\binom{-a_{i}}{j_{i}} \binom{j_{i}}{l_{i}} \\ & \quad =\prod_{\substack{i=1\\i\neq r}}^{k}\binom{-a_{i}}{l_{i}}\sum _{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots,j_{k}\geq0}}\binom {-a_{r}}{j_{r}}\prod _{\substack{i=1\\i\neq r}}^{k}\binom{-a_{i}-l_{i}}{j_{i}-l_{i}} \\ & \quad =\prod_{\substack{i=1\\i\neq r}}^{k}\binom{-a_{i}}{l_{i}} \binom{-(a_{1}+\cdots+a_{k})-(n-l_{r})}{n-(n-l_{r})}. \end{aligned}$$

(3.19)

By applying (3.19) to (3.17), in view of (3.11), we obtain

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{-a_{1}}{j_{1}}\cdots \binom{-a_{k}}{j_{k}} \biggl[\frac{t_{1}^{j_{1}}}{j_{1}!}\cdots\frac {t_{k}^{j_{k}}}{j_{k}!} \biggr] \Biggl(\prod_{i=1}^{k}\frac{(\lambda_{i}-1)e^{x_{i}t_{i}}}{\lambda _{i}e^{t_{i}}-1} \Biggr) \\& \quad =\frac{(-1)^{n}}{n!}\sum_{r=1}^{k} \frac{\lambda_{r}-1}{\lambda_{1}\cdots\lambda_{k}-1} \sum_{\substack {l_{1}+\cdots+l_{k}=n\\l_{1},\ldots,l_{k}\geq0}}\binom{n}{l_{1},\ldots ,l_{k}}(a_{1}+ \cdots+a_{k}+n-l_{r})_{l_{r}} \\& \quad\quad{} \times H_{l_{r}} \biggl(x_{r} \Big| \frac{1}{\lambda_{1}\cdots\lambda_{k}} \biggr) \prod_{i=1}^{r-1} \lambda_{i}(a_{i})_{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r}+1 \Big| \frac{1}{\lambda_{i}} \biggr) \\& \quad\quad{} \times\prod_{i=r+1}^{k}(a_{i})_{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r} \Big| \frac{1}{\lambda_{i}} \biggr). \end{aligned}$$

(3.20)

Observe that

$$ (a_{1}+\cdots+a_{k}+n-l_{r})_{l_{r}} \cdot(a_{1}+\cdots+a_{k})_{n-l_{r}}=(a_{1}+ \cdots+a_{k})_{n}. $$

(3.21)

By equating (3.12) and (3.20), in light of (3.21), we get

$$\begin{aligned}& \sum_{\substack{j_{1}+\cdots+j_{k}=n\\j_{1},\ldots ,j_{k}\geq0}}\binom{n}{j_{1},\ldots,j_{k}} \frac{(a_{1})_{j_{1}}\cdots(a_{k})_{j_{k}}}{(a_{1}+\cdots +a_{k})_{n}}H_{j_{1}} \biggl(x_{1} \Big| \frac{1}{\lambda_{1}} \biggr)\cdots H_{j_{k}} \biggl(x_{k} \Big| \frac{1}{\lambda_{k}} \biggr) \\& \quad =\sum_{r=1}^{k}\frac{\lambda_{r}-1}{\lambda_{1}\cdots\lambda_{k}-1} \sum_{\substack{l_{1}+\cdots+l_{k}=n\\l_{1},\ldots,l_{k}\geq0}}\binom {n}{l_{1},\ldots,l_{k}}\frac{1}{(a_{1}+\cdots+a_{k})_{n-l_{r}}} \\& \quad\quad{} \times H_{l_{r}} \biggl(x_{r} \Big| \frac{1}{\lambda_{1}\cdots\lambda_{k}} \biggr) \prod_{i=1}^{r-1} \lambda_{i}(a_{i})_{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r}+1 \Big| \frac{1}{\lambda_{i}} \biggr) \\& \quad\quad{} \times\prod_{i=r+1}^{k}(a_{i})_{l_{i}} H_{l_{i}} \biggl(x_{i}-x_{r} \Big| \frac{1}{\lambda_{i}} \biggr). \end{aligned}$$

(3.22)

Thus, by replacing $\lambda_{i}$ by $1/\lambda_{i}$ for $1\leq i\leq k$ in (3.22), the desired result follows immediately. This completes the proof of Theorem 2.1.