Barnes’ multiple Bernoulli and generalized Barnes’ multiple Frobenius-Euler mixed-type polynomials

Abstract

In this paper, by considering Barnes’ multiple Bernoulli polynomials as well as generalized Barnes’ multiple Frobenius-Euler polynomials, we define and investigate the mixed-type polynomials of these polynomials. From the properties of Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

MSC:05A15, 05A40, 11B68, 11B75, 33E20, 65Q05.

1 Introduction

In this paper, we consider the polynomials

$$B{H}_n(x)=B{H}_n(x|a;b;\lambda ;\mu )=B{H}_n(x|a_1,\dots ,a_r;b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;{\mu }_1,\dots ,{\mu }_s)$$

called Barnes’ multiple Bernoulli and generalized Barnes’ multiple Frobenius-Euler mixed-type polynomials, whose generating function is given by

$$\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}B{H}_n(x)\frac{t^n}{n!},$$

(1)

where $a_1,\dots ,a_r,b_1,\dots ,b_s,{\lambda }_1,\dots ,{\lambda }_s,{\mu }_1,\dots ,{\mu }_s\in \mathbb{C}$ with $a_1,\dots ,a_r,b_1,\dots ,b_s\ne 0$, ${\lambda }_1,\dots ,{\lambda }_s\ne 1$. When $x=0$,

$$B{H}_n=B{H}_n(0)=B{H}_n(0;a_1,\dots ,a_r;b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_r;{\mu }_1,\dots ,{\mu }_r)$$

are called Barnes’ multiple Bernoulli and generalized Barnes’ multiple Frobenius-Euler mixed-type numbers.

Recall that Barnes’ multiple Bernoulli polynomials, denoted by $B_n(x|a_1,\dots ,a_r)$, are given by the generating function as

$$\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)e^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x|a_1,\dots ,a_r)\frac{t^n}{n!},$$

(2)

where $a_1,\dots ,a_r\ne 0$ [1, 2]. In addition, the generalized Barnes’ multiple Frobenius-Euler polynomials, denoted by $H_n(x|b;\lambda ;\mu )=H_n(x|b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;{\mu }_1,\dots ,{\mu }_s)$, are given by the generating function as

$$\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n(x|b;\lambda ;\mu )\frac{t^n}{n!}$$

(3)

(see e.g. [3–7]). If ${\mu }_1=\cdots ={\mu }_s=1$, then

$$H_n(x|b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s)=H_n(x|b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;1,\dots ,1)$$

are called Barnes-type Frobenius-Euler polynomials. If further ${\lambda }_1=\cdots ={\lambda }_s=\lambda$ and $b_1=\cdots =b_s=1$, then $H_n^{(s)}(x|\lambda )=H_n(x|1,\dots ,1;\lambda ,\dots ,\lambda ;1,\dots ,1)$ are called Frobenius-Euler polynomials of order s (see e.g. [8, 9]). If ${\lambda }_1=\cdots ={\lambda }_s=-1$, then $E_n(x|b_1,\dots ,b_s;{\mu }_1,\dots ,{\mu }_s)=H_n(x|b_1,\dots ,b_s;1,\dots ,1;{\mu }_1,\dots ,{\mu }_s)$ are called generalized Barnes-type Euler polynomials. These polynomials arise naturally in connection with the study of Barnes-type Peters polynomials. Peters polynomials were mentioned in [[10], p.128] and were investigated in e.g. [11].

In this paper, by considering Barnes’ multiple Bernoulli polynomials as well as generalized Barnes’ multiple Frobenius-Euler polynomials, we define and investigate the mixed-type polynomials of these polynomials. From the properties of Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

2 Umbral calculus

Let ℂ be a complex number field and let ℱ be the set of all formal power series in the variable t:

$$\mathcal{F}=\{f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{k!}{t}^k|a_k\in \mathbb{C}\}.$$

(4)

Let $\mathbb{P}=\mathbb{C}[x]$ and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on ℙ. $〈L|p(x)〉$ is the action of the linear functional L on the polynomial $p(x)$, and we recall that the vector space operations on ${\mathbb{P}}^{\ast }$ are defined by $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant in ℂ. For $f(t)\in \mathcal{F}$, let us define the linear functional on ℙ by setting

$$〈f(t)|x^n〉=a_n(n\ge 0).$$

(5)

In particular,

$$〈t^k|x^n〉=n!{\delta }_{n,k}(n,k\ge 0),$$

(6)

where ${\delta }_{n,k}$ is the Kronecker symbol.

For $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^k〉}{k!}{t}^k$, we have $〈f_L(t)|x^n〉=〈L|x^n〉$. That is, $L=f_L(t)$. The map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth, ℱ denotes both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and so an element $f(t)$ of ℱ will be thought of as both a formal power series and a linear functional. We call ℱ umbral algebra, and umbral calculus is the study of umbral algebra. The order $O(f(t))$ of a power series $f(t)$ (≠0) is the smallest integer k for which the coefficient of $t^k$ does not vanish. If $O(f(t))=1$, then $f(t)$ is called a delta series; if $O(f(t))=0$, then $f(t)$ is called an invertible series. For $f(t),g(t)\in \mathcal{F}$ with $O(f(t))=1$ and $O(g(t))=0$, there exists a unique sequence $s_n(x)$ ($deg{s}_n(x)=n$) such that $〈g(t)f{(t)}^k|s_n(x)〉=n!{\delta }_{n,k}$ for $n,k\ge 0$ [[10], Theorem 2.3.1]. Such a sequence $s_n(x)$ is called the Sheffer sequence for $(g(t),f(t))$ which is denoted by $s_n(x)\sim (g(t),f(t))$.

For $f(t),g(t)\in \mathcal{F}$ and $p(x)\in \mathbb{P}$, we have

$$〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉=〈g(t)|f(t)p(x)〉,$$

(7)

and

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}〈f(t)|x^k〉\frac{t^k}{k!},p(x)=\sum _{k=0}^{\mathrm{\infty }}〈t^k|p(x)〉\frac{x^k}{k!},$$

(8)

[[10], Theorem 2.2.5]. Thus, by (8), we get

$$t^{k}p(x)=p(x)=\frac{d^{k}p(x)}{d{x}^k}\text{and}{e}^{yt}p(x)=p(x+y).$$

(9)

Sheffer sequences are characterized in the generating function [[10], Theorem 2.3.4].

Lemma 1 The sequence $s_n(x)$ is Sheffer for $(g(t),f(t))$ if and only if

$$\frac{1}{g(\overline{f}(t))}{e}^{y\overline{f}(t)}=\sum _{k=0}^{\mathrm{\infty }}\frac{s_k(y)}{k!}{t}^k(y\in \mathbb{C}),$$

where $\overline{f}(t)$ is the compositional inverse of $f(t)$.

For $s_n(x)\sim (g(t),f(t))$, we have the following equations [[10], Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

$$f(t)s_n(x)=n{s}_{n-1}(x)(n\ge 0),$$

(10)

$$s_n(x)=\sum _{j=0}^n\frac{1}{j!}〈g{(\overline{f}(t))}^{-1}\overline{f}{(t)}^j|x^n〉x^j,$$

(11)

$$s_n(x+y)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)s_j(x)p_{n-j}(y),$$

(12)

where $p_n(x)=g(t)s_n(x)$.

Assume that $p_n(x)\sim (1,f(t))$ and $q_n(x)\sim (1,g(t))$. Then the transfer formula [[10], Corollary 3.8.2] is given by

$$q_n(x)=x{\left(\frac{f(t)}{g(t)}\right)}^n{x}^{-1}{p}_n(x)(n\ge 1).$$

For $s_n(x)\sim (g(t),f(t))$ and $r_n(x)\sim (h(t),l(t))$, assume that

$$s_n(x)=\sum _{m=0}^n{C}_{n,m}{r}_m(x)(n\ge 0).$$

Then we have [[10], p.132]

$$C_{n,m}=\frac{1}{m!}〈\frac{h(\overline{f}(t))}{g(\overline{f}(t))}l{(\overline{f}(t))}^m|x^n〉.$$

(13)

3 Main results

From definitions (2), (3) and (1), $B_n(x|a_1,\dots ,a_r)$, $H_n(x|b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;{\mu }_1,\dots ,{\mu }_s)$ and $B{H}_n(x|a_1,\dots ,a_r;b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;{\mu }_1,\dots ,{\mu }_s)$ are the Appell sequences for

$$\prod _{i=1}^r\left(\frac{e^{a_{i}t}-1}{t}\right),\prod _{j=1}^s{\left(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)}^{{\mu }_j}\text{and}\prod _{i=1}^r\left(\frac{e^{a_{i}t}-1}{t}\right)\prod _{j=1}^s{\left(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)}^{{\mu }_j},$$

respectively. So,

$$B_n(x|a_1,\dots ,a_r)\sim (\sum _{i=1}^r\left(\frac{e^{a_{i}t}-1}{t}\right),t),$$

(14)

$$H_n(x|b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;{\mu }_1,\dots ,{\mu }_s)\sim (\prod _{j=1}^s{\left(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)}^{{\mu }_j},t),$$

(15)

$$\begin{array}{r}B{H}_n(x|a_1,\dots ,a_r;b_1,\dots ,b_s;{\lambda }_1,\dots ,{\lambda }_s;{\mu }_1,\dots ,{\mu }_s)\\ \sim (\prod _{i=1}^r\left(\frac{e^{a_{i}t}-1}{t}\right)\prod _{j=1}^s{\left(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)}^{{\mu }_j},t).\end{array}$$

(16)

In particular,

$$\begin{array}{r}t{B}_n(x|a)=\frac{d}{dx}{B}_n(x|a)=n{B}_{n-1}(x|a),\\ t{H}_n(x|b;\lambda ;\mu )=\frac{d}{dx}{H}_n(x|b;\lambda ;\mu )=n{H}_{n-1}(x|b;\lambda ;\mu ),\\ tB{H}_n(x|a;b;\lambda ;\mu )=\frac{d}{dx}B{H}_n(x|a;b;\lambda ;\mu )=nB{H}_{n-1}(x|a;b;\lambda ;\mu ),\end{array}$$

where $a=(a_1,\dots ,a_r)$, $b=(b_1,\dots ,b_s)$, $\lambda =({\lambda }_1,\dots ,{\lambda }_s)$ and $\mu =({\mu }_1,\dots ,{\mu }_s)$.

3.1 Explicit expressions

Let ${(n)}_j=n(n-1)\cdots (n-j+1)$ ($j\ge 1$) with ${(n)}_0=1$. The (signed) Stirling numbers of the first kind $S_1(n,m)$ are defined by

$${(x)}_n=\sum _{m=0}^n{S}_1(n,m)x^m.$$

Notice that

$${(x)}_n\sim (1,e^t-1).$$

(17)

Theorem 1 We have

$$B{H}_n(x|a;b;\lambda ;\mu )=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)B{H}_{n-j}{x}^j$$

(18)

$$=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}(a)H_l(x|b;\lambda ;\mu )$$

(19)

$$=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_{n-l}(b;\lambda ;\mu )B_l(x|a).$$

(20)

Proof By (11) with (16), we get

$$\begin{array}{r}〈g{(\overline{f}(t))}^{-1}\overline{f}{(t)}^j|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^j|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|t^j{x}^n〉\\ ={(n)}_j〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|x^{n-j}〉\\ ={(n)}_j〈\sum _{i=0}^{\mathrm{\infty }}B{H}_i\frac{t^i}{i!}|x^{n-j}〉\\ ={(n)}_{j}B{H}_{n-j}.\end{array}$$

Thus, we obtain identity (18).

Next,

$$\begin{array}{rl}B{H}_n(y|a;b;\lambda ;\mu )& =〈\sum _{i=0}^{\mathrm{\infty }}B{H}_i(y|a;b;\lambda ;\mu )\frac{t^i}{i!}|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)|\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}{x}^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)|\sum _{l=0}^{\mathrm{\infty }}{H}_l(y|b;\lambda ;\mu )\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l(y|b;\lambda ;\mu )〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l(y|b;\lambda ;\mu )〈\sum _{i=0}^{\mathrm{\infty }}{B}_i(a)\frac{t^i}{i!}|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l(y|b;\lambda ;\mu )B_{n-l}(a).\end{array}$$

Thus, we obtain (19).

Finally, we obtain that

$$\begin{array}{rl}B{H}_n(y|a;b;\lambda ;\mu )& =〈\sum _{i=0}^{\mathrm{\infty }}B{H}_i(y|a;b;\lambda ;\mu )\frac{t^i}{i!}|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|x^n〉\\ =〈\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)e^{yt}{x}^n〉\\ =〈\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\sum _{l=0}^{\mathrm{\infty }}{B}_l(y|a)\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l(y|a)〈\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l(y|a)〈\sum _{i=0}^{\mathrm{\infty }}{H}_i(b;\lambda ;\mu )\frac{t^i}{i!}|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l(y|a)H_{n-l}(b;\lambda ;\mu ).\end{array}$$

Thus, we get identity (20). □

3.2 The Sheffer identity

Theorem 2

$$B{H}_n(x+y|a;b;\lambda ;\mu )=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)B{H}_j(x|a;b;\lambda ;\mu )y^{n-j}.$$

(21)

Proof By (16) with

$$\begin{array}{rl}{p}_n(x)& =\prod _{i=1}^r\left(\frac{e^{a_{i}t}-1}{t}\right)\prod _{j=1}^s{\left(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)}^{{\mu }_j}B{H}_n(x|a;b;\lambda ;\mu )\\ =x^n\sim (1,t),\end{array}$$

using (12), we have (21). □

3.3 Recurrence

Theorem 3

$$\begin{array}{r}(1-\frac{r}{n+1})B{H}_{n+1}(x|a;b;\lambda ;\mu )\\ =xB{H}_n(x|a;b;\lambda ;\mu )-\frac{1}{n+1}\sum _{i=1}^r{a}_{i}B{H}_{n+1}(x+a_i|a,a_i;b;\lambda ;\mu )\\ -\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_n(x+b_j|a;b;\lambda ;\mu +e_j).\end{array}$$

(22)

Remark When $n=r-1$, as the left-hand side of (22) is equal to 0, we have

$$\begin{array}{rl}xB{H}_{r-1}(x|a;b;\lambda ;\mu )=& \frac{1}{r}\sum _{i=1}^r{a}_{i}B{H}_r(x+a_i|a,a_i;b;\lambda ;\mu )\\ +\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_{r-1}(x+b_j|a;b;\lambda ;\mu +e_j).\end{array}$$

Proof By applying

$$s_{n+1}(x)=(x-\frac{g^{\prime }(t)}{g(t)})\frac{1}{f^{\prime }(t)}{s}_n(x)$$

(23)

[[10], Corollary 3.7.2] with (16), we get

$$B{H}_{n+1}(x|a;b;\lambda ;\mu )=xB{H}_n(x|a;b;\lambda ;\mu )-\frac{g^{\prime }(t)}{g(t)}B{H}_n(x|a;b;\lambda ;\mu ).$$

Since

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}& ={(lng(t))}^{\prime }={(\sum _{i=1}^{r}ln(e^{a_{i}t}-1)-rlnt+\sum _{j=1}^s{\mu }_{j}ln(e^{b_{j}t}-{\lambda }_j)-\sum _{j=1}^s{\mu }_{j}ln(1-{\lambda }_j))}^{\prime }\\ =\sum _{i=1}^r\frac{a_i{e}^{a_{i}t}}{e^{a_{i}t}-1}-\frac{r}{t}+\sum _{j=1}^s\frac{{\mu }_j{b}_j{e}^{b_{j}t}}{e^{b_{j}t}-{\lambda }_j}=\frac{1}{t}\sum _{i=1}^r(\frac{a_{i}t{e}^{a_{i}t}}{e^{a_{i}t}-1}-1)+\sum _{j=1}^s\frac{{\mu }_j{b}_j{e}^{b_{j}t}}{e^{b_{j}t}-{\lambda }_j}\end{array}$$

and

$$\sum _{i=1}^r(\frac{a_{i}t{e}^{a_{i}t}}{e^{a_{i}t}-1}-1)=\frac{1}{2}\left(\sum _{i=1}^r{a}_i\right)t+\cdots$$

is a series with order at least one, we have

$$\begin{array}{r}\frac{g^{\prime }(t)}{g(t)}B{H}_n(x|a;b;\lambda ;\mu )\\ =(\frac{1}{t}\sum _{i=1}^r(\frac{a_{i}t{e}^{a_{i}t}}{e^{a_{i}t}-1}-1)+\sum _{j=1}^s\frac{{\mu }_j{b}_j{e}^{b_{j}t}}{1-{\lambda }_j}\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j})B{H}_n(x|a;b;\lambda ;\mu )\\ =\frac{1}{t}\sum _{i=1}^r(\frac{a_{i}t{e}^{a_{i}t}}{e^{a_{i}t}-1}-1)B{H}_n(x|a;b;\lambda ;\mu )\\ +\sum _{j=1}^s\frac{{\mu }_j{b}_j{e}^{a_{j}t}}{1-{\lambda }_j}\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{x}^n\\ =\frac{1}{n+1}(\sum _{i=1}^r\frac{a_{i}t{e}^{a_{i}t}}{e^{a_{i}t}-1}-r)B{H}_{n+1}(x|a;b;\lambda ;\mu )+\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_n(x+b_j|a;b;\lambda ;\mu +e_j)\\ =\frac{1}{n+1}\sum _{i=1}^r{a}_i{e}^{a_{i}t}\frac{t}{e^{a_{i}t}-1}\prod _{\nu =1}^r\left(\frac{t}{e^{a_{\nu }t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{x}^{n+1}\\ -\frac{r}{n+1}B{H}_{n+1}(x|a;b;\lambda ;\mu )+\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_n(x+b_j|a;b;\lambda ;\mu +e_j)\\ =\frac{1}{n+1}\sum _{i=1}^r{a}_{i}B{H}_{n+1}(x+a_i|a,a_i;b;\lambda ;\mu )-\frac{r}{n+1}B{H}_{n+1}(x|a;b;\lambda ;\mu )\\ +\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_n(x+b_j|a;b;\lambda ;\mu +e_j).\end{array}$$

Here $(a,a_i)=(a_1,\dots ,a_r,a_i)$ and ($i=1,2,\dots ,r$). Therefore, we obtain

$$\begin{array}{r}(1-\frac{r}{n+1})B{H}_{n+1}(x|a;b;\lambda ;\mu )\\ =xB{H}_n(x|a;b;\lambda ;\mu )-\frac{1}{n+1}\sum _{i=1}^r{a}_{i}B{H}_{n+1}(x+a_i|a,a_i;b;\lambda ;\mu )\\ -\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_n(x+b_j|a;b;\lambda ;\mu +e_j),\end{array}$$

which is (22). □

3.4 More relations

Theorem 4 For $n\ge 1$, we have

$$\begin{array}{rl}B{H}_n(x|a;b;\lambda ;\mu )=& xB{H}_{n-1}(x|a;b;\lambda ;\mu )-\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_{n-1}(x+b_j|a;b;\lambda ;\mu +e_j)\\ +\sum _{m=1}^n\frac{{(-1)}^{m-1}\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right)B_m}{m}B{H}_{n-m}(x|a;b;\lambda ;\mu )\sum _{i=1}^r{a}_i^m.\end{array}$$

(24)

Proof For $n\ge 1$, we have

$$\begin{array}{rl}B{H}_n(y|a;b;\lambda ;\mu )=& 〈\sum _{l=0}^{\mathrm{\infty }}B{H}_l(y|a;b;\lambda ;\mu )\frac{t^l}{l!}|x^n〉\\ =& 〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|x^n〉\\ =& 〈{\partial }_t\left(\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}\right)|x^{n-1}〉\\ =& 〈\left({\partial }_t\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|x^{n-1}〉\\ +〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\left({\partial }_t\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}\right)e^{yt}|x^{n-1}〉\\ +〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}\left({\partial }_t{e}^{yt}\right)|x^{n-1}〉.\end{array}$$

The third term is

$$y〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|x^{n-1}〉=yB{H}_{n-1}(y|a;b;\lambda ;\mu ).$$

Since

$$\begin{array}{rl}{\partial }_t\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)& =\sum _{i=1}^r{\left(\frac{t}{e^{a_{i}t}-1}\right)}^{\prime }\prod _{\nu \ne i}\frac{t}{e^{a_{\nu }t}-1}\\ =\frac{t^{r-1}}{{\prod }_{\nu =1}^r(e^{a_{\nu }t}-1)}(r-\sum _{i=1}^r\frac{a_{i}t{e}^{a_{i}t}}{e^{a_{i}t}-1})\\ =\frac{t^{r-1}}{{\prod }_{\nu =1}^r(e^{a_{\nu }t}-1)}(r-\sum _{i=1}^r\frac{-a_{i}t}{e^{-a_{i}t}-1})\\ =\frac{t^{r-1}}{{\prod }_{\nu =1}^r(e^{a_{\nu }t}-1)}(r-\sum _{i=1}^r\sum _{m=0}^{\mathrm{\infty }}\frac{{(-a_i)}^m{B}_m{t}^m}{m!})\\ =\frac{t^{r-1}}{{\prod }_{\nu =1}^r(e^{a_{\nu }t}-1)}(r-\sum _{m=0}^{\mathrm{\infty }}\sum _{i=1}^r{(-a_i)}^m\frac{B_m{t}^m}{m!})\\ =\frac{t^r}{{\prod }_{\nu =1}^r(e^{a_{\nu }t}-1)}\sum _{m=1}^{\mathrm{\infty }}\sum _{i=1}^r{a}_i^m\frac{{(-1)}^{m-1}{B}_m}{m!}{t}^{m-1},\end{array}$$

the first term is

$$\begin{array}{r}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|\sum _{m=1}^{\mathrm{\infty }}\sum _{i=1}^r{a}_i^m\frac{{(-1)}^{m-1}{B}_m}{m!}{t}^{m-1}{x}^{n-1}〉\\ =\sum _{m=1}^n\frac{{(-1)}^{m-1}\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right)B_m}{m}\sum _{i=1}^r{a}_i^m〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{yt}|x^{n-m}〉\\ =\sum _{m=1}^n\frac{{(-1)}^{m-1}\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right)B_m}{m}B{H}_{n-m}(y|a;b;\lambda ;\mu )\sum _{i=1}^r{a}_i^m.\end{array}$$

Since

$$\begin{array}{rl}{\partial }_t\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}& =\sum _{j=1}^s{\left({\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}\right)}^{\prime }\prod _{\nu \ne j}{\left(\frac{1-{\lambda }_{\nu }}{e^{b_{\nu }t}-{\lambda }_{\nu }}\right)}^{{\mu }_{\nu }}\\ =-\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}\sum _{\nu =1}^s\frac{{\mu }_{\nu }{b}_{\nu }{e}^{b_{\nu }t}}{1-{\lambda }_{\nu }}\frac{1-{\lambda }_{\nu }}{e^{b_{\nu }t}-{\lambda }_{\nu }},\end{array}$$

the second term is

$$\begin{array}{r}-\sum _{\nu =1}^s\frac{{\mu }_{\nu }{b}_{\nu }}{1-{\lambda }_{\nu }}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\frac{1-{\lambda }_{\nu }}{e^{b_{\nu }t}-{\lambda }_{\nu }}\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{(y+b_{\nu })t}|x^{n-1}〉\\ =-\sum _{\nu =1}^s\frac{{\mu }_{\nu }{b}_{\nu }}{1-{\lambda }_{\nu }}B{H}_{n-1}(y+b_{\nu }|a;b;\lambda ;\mu +e_{\nu }).\end{array}$$

Therefore, we obtain

$$\begin{array}{rl}B{H}_n(x|a;b;\lambda ;\mu )=& xB{H}_{n-1}(x|a;b;\lambda ;\mu )-\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_{n-1}(x+b_j|a;b;\lambda ;\mu +e_j)\\ +\sum _{m=1}^n\frac{{(-1)}^{m-1}\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right)B_m}{m}B{H}_{n-m}(x|a;b;\lambda ;\mu )\sum _{i=1}^r{a}_i^m,\end{array}$$

which is identity (24). □

3.5 A relation including Bernoulli numbers

Theorem 5 For $n-1\ge m\ge 1$, we have

$$\begin{array}{r}(n-m)B{H}_{n-m}(a;b;\lambda ;\mu )\\ =-(n-m)\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_{n-m-1}(b_i|a;b;\lambda ;\mu +e_j)\\ +\sum _{l=1}^{n-m}{(-1)}^{l-1}\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_{l}B{H}_{n-m-l}(a;b;\lambda ;\mu )\sum _{i=1}^r{a}_i^l.\end{array}$$

(25)

Proof We shall compute

$$〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^m|x^n〉$$

in two different ways. On the one hand, it is equal to

$$\begin{array}{r}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|t^m{x}^n〉\\ ={(n)}_m〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|x^{n-m}〉\\ ={(n)}_{m}B{H}_{n-m}(a;b;\lambda ;\mu ).\end{array}$$

On the other hand, it is equal to

$$\begin{array}{r}〈{\partial }_t\left(\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^m\right)|x^{n-1}〉\\ =〈\left({\partial }_t\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^m|x^{n-1}〉\\ +〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\left({\partial }_t\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}\right)t^m|x^{n-1}〉\\ +〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}\left({\partial }_t{t}^m\right)|x^{n-1}〉.\end{array}$$

(26)

The third term of (26) is equal to

$$\begin{array}{r}m〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^{m-1}|x^{n-1}〉\\ =m{(n-1)}_{m-1}B{H}_{n-m}(a;b;\lambda ;\mu ).\end{array}$$

The second term of (26) is equal to

$$\begin{array}{r}-\sum _{l=1}^s\frac{{\mu }_l{b}_l}{1-{\lambda }_l}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\frac{1-{\lambda }_l}{e^{b_{l}t}-{\lambda }_l}\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{b_{l}t}|t^m{x}^{n-1}〉\\ =-{(n-1)}_m\sum _{l=1}^s\frac{{\mu }_l{b}_l}{1-{\lambda }_l}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\frac{1-{\lambda }_l}{e^{b_{l}t}-{\lambda }_l}\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{e}^{b_{l}t}|x^{n-m-1}〉\\ =-{(n-1)}_m\sum _{l=1}^s\frac{{\mu }_l{b}_l}{1-{\lambda }_l}B{H}_{n-m-1}(b_l|a;b;\lambda ;\mu +e_l).\end{array}$$

Since

$${(n-1)}_{l-1}{(n-l)}_m={(n-1)}_{l+m-1}={(n-1)}_{m-1}{(n-m)}_l,$$

the first term of (26) is equal to

$$\begin{array}{r}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^m|\sum _{l=1}^{\mathrm{\infty }}\left(\sum _{\nu =1}^r{a}_{\nu }^l\right)\frac{{(-1)}^{l-1}{B}_l}{l!}{t}^{l-1}{x}^{n-1}〉\\ =\sum _{l=1}^{n-m}\left(\sum _{\nu =1}^r{a}_{\nu }^l\right)\frac{{(-1)}^{l-1}{B}_l}{l!}{(n-1)}_{l-1}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{t}^m|x^{n-l}〉\\ =\sum _{l=1}^{n-m}\left(\sum _{\nu =1}^r{a}_{\nu }^l\right)\frac{{(-1)}^{l-1}{B}_l}{l!}{(n-1)}_{l-1}{(n-l)}_{m}B{H}_{n-m-l}(a;b;\lambda ;\mu )\\ ={(n-1)}_{m-1}\sum _{l=1}^{n-m}{(-1)}^{l-1}\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_{l}B{H}_{n-m-l}(a;b;\lambda ;\mu )\sum _{\nu =1}^r{a}_{\nu }^l.\end{array}$$

Therefore, we get, for $n-1\ge m\ge 1$,

$$\begin{array}{r}{(n)}_{m}B{H}_{n-m}(a;b;\lambda ;\mu )\\ =m{(n-1)}_{m-1}B{H}_{n-m}(a;b;\lambda ;\mu )\\ -{(n-1)}_m\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_{n-m-1}(b_j|a;b;\lambda ;\mu +e_j)\\ +{(n-1)}_{m-1}\sum _{l=1}^{n-m}{(-1)}^{l-1}\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_{l}B{H}_{n-m-l}(a;b;\lambda ;\mu )\sum _{i=1}^r{a}_i^l.\end{array}$$

Dividing both sides by ${(n-1)}_{m-1}$, we obtain, for $n-1\ge m\ge 1$,

$$\begin{array}{r}(n-m)B{H}_{n-m}(a;b;\lambda ;\mu )\\ =-(n-m)\sum _{j=1}^s\frac{{\mu }_j{b}_j}{1-{\lambda }_j}B{H}_{n-m-1}(b_j|a;b;\lambda ;\mu +e_j)\\ +\sum _{l=1}^{n-m}{(-1)}^{l-1}\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)B_{l}B{H}_{n-m-l}(a;b;\lambda ;\mu )\sum _{i=1}^r{a}_i^l.\end{array}$$

Thus, we get (25). □

3.6 A relation with Stirling numbers

The Stirling numbers of the second kind $S_2(n,m)$ are defined by

$$\frac{{(e^t-1)}^m}{m!}=\sum _{n=m}^{\mathrm{\infty }}{S}_2(n,m)\frac{t^n}{n!}.$$

Then

$${\varphi }_n(x):=\sum _{m=0}^n{S}_2(n,m)x^m\sim (1,ln(1+t)).$$

(27)

Theorem 6

$$B{H}_n(x|a;b;\lambda ;\mu )=\sum _{m=0}^n\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)B{H}_{n-l}{\varphi }_m(x).$$

(28)

Proof For (16) and (27), assume that $B{H}_n(x|a;b;\lambda ;\mu )={\sum }_{m=0}^n{C}_{n,m}{\varphi }_m(x)$. By (13), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{i=1}^r(\frac{e^{a_{i}t}-1}{t}){\prod }_{j=1}^s{(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j})}^{{\mu }_j}}{(ln(1+t))}^m|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\frac{1}{m!}{(ln(1+t))}^m{x}^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\sum _{l=m}^{\mathrm{\infty }}{S}_1(l,m)\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|x^{n-l}〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)B{H}_{n-l}.\end{array}$$

Thus, we get identity (28). □

3.7 A relation with falling factorials

Theorem 7

$$B{H}_n(x|a;b;\lambda ;\mu )=\sum _{m=0}^n\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)B{H}_{n-l}{(x)}_m.$$

(29)

Proof For (16) and (17), assume that $B{H}_n(x|a;b;\lambda ;\mu )={\sum }_{m=0}^n{C}_{n,m}{(x)}_m$. By (13), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{i=1}^r(\frac{e^{a_{i}t}-1}{t}){\prod }_{j=1}^s{(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j})}^{{\mu }_j}}{(e^t-1)}^m|x^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\frac{1}{m!}{(e^t-1)}^m{x}^n〉\\ =〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\sum _{l=m}^{\mathrm{\infty }}{S}_2(l,m)\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|x^{n-l}〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)B{H}_{n-l}.\end{array}$$

Thus, we get identity (29). □

3.8 A relation with higher-order Frobenius-Euler polynomials

Theorem 8

$$\begin{array}{r}B{H}_n(x|a;b;\lambda ;\mu )\\ =\frac{1}{{(1-\alpha )}^p}\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)\left(\sum _{l=0}^p\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0ex}{}{p}{l}\right)\left(\genfrac{}{}{0ex}{}{n-m}{\nu }\right){(-\alpha )}^{p-l}{l}^{\nu }B{H}_{n-m-\nu }\right)H_m^{(p)}(x|\alpha ).\end{array}$$

(30)

Proof For (16) and

$$H_n^{(p)}(x|\alpha )\sim ({\left(\frac{e^t-\alpha }{1-\alpha }\right)}^p,t),$$

(31)

assume that $B{H}_n(x|a;b;\lambda ;\mu )={\sum }_{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\alpha )$. By (13), similarly to the proof of (25), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{e^t-\alpha }{1-\alpha })}^p}{{\prod }_{i=1}^r(\frac{e^{a_{i}t}-1}{t}){\prod }_{j=1}^s{(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j})}^{{\mu }_j}}{t}^m|x^n〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|{(e^t-\alpha )}^p{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right){(-\alpha )}^{p-l}{e}^{lt}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right){(-\alpha )}^{p-l}〈e^{lt}|\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right){(-\alpha )}^{p-l}〈e^{lt}|\sum _{\nu =0}^{\mathrm{\infty }}B{H}_{\nu }\frac{t^{\nu }}{\nu !}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right){(-\alpha )}^{p-l}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-m}{\nu }\right)B{H}_{\nu }〈e^{lt}|x^{n-m-\nu }〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right){(-\alpha )}^{p-l}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-m}{\nu }\right)B{H}_{\nu }{l}^{n-m-\nu }\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\alpha )}^p}\sum _{l=0}^p\left(\genfrac{}{}{0ex}{}{p}{l}\right){(-\alpha )}^{p-l}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-m}{\nu }\right)B{H}_{n-m-\nu }{l}^{\nu }.\end{array}$$

Thus, we get identity (30). □

3.9 A relation with higher-order Bernoulli polynomials

Bernoulli polynomials ${\mathfrak{B}}_n^{(p)}(x)$ of order p are defined by

$${\left(\frac{t}{e^t-1}\right)}^p{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathfrak{B}}_n^{(p)}(x)}{n!}{t}^n$$

(see e.g. [[10], Section 2.2]).

Theorem 9

$$B{H}_n(x|a;b;\lambda ;\mu )=\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)(\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{n-m-l+p}{p}\right)}{S}_2(n-m-l+p,p)B{H}_l){\mathfrak{B}}_m^{(p)}(x).$$

(32)

Proof For (16) and

$${\mathfrak{B}}_n^{(p)}(x)\sim ({\left(\frac{e^t-1}{t}\right)}^p,t),$$

(33)

assume that $B{H}_n(x|a;b;\lambda ;\mu )={\sum }_{m=0}^n{C}_{n,m}{\mathfrak{B}}_m^{(p)}(x)$. By (13), similarly to the proof of (25), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{e^t-1}{t})}^p}{{\prod }_{i=1}^r(\frac{e^{a_{i}t}-1}{t}){\prod }_{j=1}^s{(\frac{e^{b_{j}t}-{\lambda }_j}{1-{\lambda }_j})}^{{\mu }_j}}{t}^m|x^n〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|{\left(\frac{e^t-1}{t}\right)}^p{x}^{n-m}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|p!\sum _{l=0}^{\mathrm{\infty }}{S}_2(l+p,p)\frac{t^l}{(l+p)!}{x}^{n-m}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)p!\sum _{l=0}^{n-m}\frac{{(n-m)}_l}{(l+p)!}{S}_2(l+p,p)〈\prod _{i=1}^r\left(\frac{t}{e^{a_{i}t}-1}\right)\prod _{j=1}^s{\left(\frac{1-{\lambda }_j}{e^{b_{j}t}-{\lambda }_j}\right)}^{{\mu }_j}|x^{n-m-l}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{l+p}{p}\right)}{S}_2(l+p,p)B{H}_{n-m-l}\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{n-m-l+p}{p}\right)}{S}_2(n-m-l+p,p)B{H}_l.\end{array}$$

Thus, we get identity (32). □