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.