Theory and Modern Applications

# 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

$B{H}_{n}\left(x\right)=B{H}_{n}\left(x|a;b;\lambda ;\mu \right)=B{H}_{n}\left(x|{a}_{1},\dots ,{a}_{r};{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)$

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

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

$B{H}_{n}=B{H}_{n}\left(0\right)=B{H}_{n}\left(0;{a}_{1},\dots ,{a}_{r};{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{r};{\mu }_{1},\dots ,{\mu }_{r}\right)$

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

Recall that Barnes’ multiple Bernoulli polynomials, denoted by ${B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$, are given by the generating function as

$\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right){e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\frac{{t}^{n}}{n!},$
(2)

where ${a}_{1},\dots ,{a}_{r}\ne 0$ [1, 2]. In addition, the generalized Barnes’ multiple Frobenius-Euler polynomials, denoted by ${H}_{n}\left(x|b;\lambda ;\mu \right)={H}_{n}\left(x|{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)$, are given by the generating function as

$\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}\left(x|b;\lambda ;\mu \right)\frac{{t}^{n}}{n!}$
(3)

(see e.g. [37]). If ${\mu }_{1}=\cdots ={\mu }_{s}=1$, then

${H}_{n}\left(x|{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s}\right)={H}_{n}\left(x|{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};1,\dots ,1\right)$

are called Barnes-type Frobenius-Euler polynomials. If further ${\lambda }_{1}=\cdots ={\lambda }_{s}=\lambda$ and ${b}_{1}=\cdots ={b}_{s}=1$, then ${H}_{n}^{\left(s\right)}\left(x|\lambda \right)={H}_{n}\left(x|1,\dots ,1;\lambda ,\dots ,\lambda ;1,\dots ,1\right)$ are called Frobenius-Euler polynomials of order s (see e.g. [8, 9]). If ${\lambda }_{1}=\cdots ={\lambda }_{s}=-1$, then ${E}_{n}\left(x|{b}_{1},\dots ,{b}_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)={H}_{n}\left(x|{b}_{1},\dots ,{b}_{s};1,\dots ,1;{\mu }_{1},\dots ,{\mu }_{s}\right)$ 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 be a complex number field and let be the set of all formal power series in the variable t:

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{k!}{t}^{k}|{a}_{k}\in \mathbb{C}\right\}.$
(4)

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

$〈f\left(t\right)|{x}^{n}〉={a}_{n}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(5)

In particular,

$〈{t}^{k}|{x}^{n}〉=n!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(n,k\ge 0\right),$
(6)

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

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

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

$〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈f\left(t\right)|g\left(t\right)p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉,$
(7)

and

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)|{x}^{k}〉\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}|p\left(x\right)〉\frac{{x}^{k}}{k!},$
(8)

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

${t}^{k}p\left(x\right)=p\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right).$
(9)

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

Lemma 1 The sequence ${s}_{n}\left(x\right)$ is Sheffer for $\left(g\left(t\right),f\left(t\right)\right)$ if and only if

$\frac{1}{g\left(\overline{f}\left(t\right)\right)}{e}^{y\overline{f}\left(t\right)}=\sum _{k=0}^{\mathrm{\infty }}\frac{{s}_{k}\left(y\right)}{k!}{t}^{k}\phantom{\rule{1em}{0ex}}\left(y\in \mathbb{C}\right),$

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

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we have the following equations [[10], Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

$f\left(t\right){s}_{n}\left(x\right)=n{s}_{n-1}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right),$
(10)
${s}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^{j}|{x}^{n}〉{x}^{j},$
(11)
${s}_{n}\left(x+y\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){s}_{j}\left(x\right){p}_{n-j}\left(y\right),$
(12)

where ${p}_{n}\left(x\right)=g\left(t\right){s}_{n}\left(x\right)$.

Assume that ${p}_{n}\left(x\right)\sim \left(1,f\left(t\right)\right)$ and ${q}_{n}\left(x\right)\sim \left(1,g\left(t\right)\right)$. Then the transfer formula [[10], Corollary 3.8.2] is given by

${q}_{n}\left(x\right)=x{\left(\frac{f\left(t\right)}{g\left(t\right)}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 1\right).$

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ and ${r}_{n}\left(x\right)\sim \left(h\left(t\right),l\left(t\right)\right)$, assume that

${s}_{n}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{r}_{m}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$

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

${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}l{\left(\overline{f}\left(t\right)\right)}^{m}|{x}^{n}〉.$
(13)

## 3 Main results

From definitions (2), (3) and (1), ${B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)$, ${H}_{n}\left(x|{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)$ and $B{H}_{n}\left(x|{a}_{1},\dots ,{a}_{r};{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)$ are the Appell sequences for

$\prod _{i=1}^{r}\left(\frac{{e}^{{a}_{i}t}-1}{t}\right),\phantom{\rule{2em}{0ex}}\prod _{j=1}^{s}{\left(\frac{{e}^{{b}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)}^{{\mu }_{j}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\prod _{i=1}^{r}\left(\frac{{e}^{{a}_{i}t}-1}{t}\right)\prod _{j=1}^{s}{\left(\frac{{e}^{{b}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)}^{{\mu }_{j}},$

respectively. So,

${B}_{n}\left(x|{a}_{1},\dots ,{a}_{r}\right)\sim \left(\sum _{i=1}^{r}\left(\frac{{e}^{{a}_{i}t}-1}{t}\right),t\right),$
(14)
${H}_{n}\left(x|{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)\sim \left(\prod _{j=1}^{s}{\left(\frac{{e}^{{b}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)}^{{\mu }_{j}},t\right),$
(15)
$\begin{array}{r}B{H}_{n}\left(x|{a}_{1},\dots ,{a}_{r};{b}_{1},\dots ,{b}_{s};{\lambda }_{1},\dots ,{\lambda }_{s};{\mu }_{1},\dots ,{\mu }_{s}\right)\\ \phantom{\rule{1em}{0ex}}\sim \left(\prod _{i=1}^{r}\left(\frac{{e}^{{a}_{i}t}-1}{t}\right)\prod _{j=1}^{s}{\left(\frac{{e}^{{b}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)}^{{\mu }_{j}},t\right).\end{array}$
(16)

In particular,

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

where $a=\left({a}_{1},\dots ,{a}_{r}\right)$, $b=\left({b}_{1},\dots ,{b}_{s}\right)$, $\lambda =\left({\lambda }_{1},\dots ,{\lambda }_{s}\right)$ and $\mu =\left({\mu }_{1},\dots ,{\mu }_{s}\right)$.

### 3.1 Explicit expressions

Let ${\left(n\right)}_{j}=n\left(n-1\right)\cdots \left(n-j+1\right)$ ($j\ge 1$) with ${\left(n\right)}_{0}=1$. The (signed) Stirling numbers of the first kind ${S}_{1}\left(n,m\right)$ are defined by

${\left(x\right)}_{n}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right){x}^{m}.$

Notice that

${\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right).$
(17)

Theorem 1 We have

$B{H}_{n}\left(x|a;b;\lambda ;\mu \right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)B{H}_{n-j}{x}^{j}$
(18)
$=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}\left(a\right){H}_{l}\left(x|b;\lambda ;\mu \right)$
(19)
$=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}\left(b;\lambda ;\mu \right){B}_{l}\left(x|a\right).$
(20)

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

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

Thus, we obtain identity (18).

Next,

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

Thus, we obtain (19).

Finally, we obtain that

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

Thus, we get identity (20). □

### 3.2 The Sheffer identity

Theorem 2

$B{H}_{n}\left(x+y|a;b;\lambda ;\mu \right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)B{H}_{j}\left(x|a;b;\lambda ;\mu \right){y}^{n-j}.$
(21)

Proof By (16) with

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

using (12), we have (21). □

### 3.3 Recurrence

Theorem 3

$\begin{array}{r}\left(1-\frac{r}{n+1}\right)B{H}_{n+1}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=xB{H}_{n}\left(x|a;b;\lambda ;\mu \right)-\frac{1}{n+1}\sum _{i=1}^{r}{a}_{i}B{H}_{n+1}\left(x+{a}_{i}|a,{a}_{i};b;\lambda ;\mu \right)\\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n}\left(x+{b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right).\end{array}$
(22)

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

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

Proof By applying

${s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right)$
(23)

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

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

Since

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

and

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

is a series with order at least one, we have

$\begin{array}{r}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}B{H}_{n}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=\left(\frac{1}{t}\sum _{i=1}^{r}\left(\frac{{a}_{i}t{e}^{{a}_{i}t}}{{e}^{{a}_{i}t}-1}-1\right)+\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}{e}^{{b}_{j}t}}{1-{\lambda }_{j}}\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)B{H}_{n}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{t}\sum _{i=1}^{r}\left(\frac{{a}_{i}t{e}^{{a}_{i}t}}{{e}^{{a}_{i}t}-1}-1\right)B{H}_{n}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{2em}{0ex}}+\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}{e}^{{a}_{j}t}}{1-{\lambda }_{j}}\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{x}^{n}\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\left(\sum _{i=1}^{r}\frac{{a}_{i}t{e}^{{a}_{i}t}}{{e}^{{a}_{i}t}-1}-r\right)B{H}_{n+1}\left(x|a;b;\lambda ;\mu \right)+\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n}\left(x+{b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\sum _{i=1}^{r}{a}_{i}{e}^{{a}_{i}t}\frac{t}{{e}^{{a}_{i}t}-1}\prod _{\nu =1}^{r}\left(\frac{t}{{e}^{{a}_{\nu }t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{x}^{n+1}\\ \phantom{\rule{2em}{0ex}}-\frac{r}{n+1}B{H}_{n+1}\left(x|a;b;\lambda ;\mu \right)+\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n}\left(x+{b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\sum _{i=1}^{r}{a}_{i}B{H}_{n+1}\left(x+{a}_{i}|a,{a}_{i};b;\lambda ;\mu \right)-\frac{r}{n+1}B{H}_{n+1}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{2em}{0ex}}+\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n}\left(x+{b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right).\end{array}$

Here $\left(a,{a}_{i}\right)=\left({a}_{1},\dots ,{a}_{r},{a}_{i}\right)$ and ($i=1,2,\dots ,r$). Therefore, we obtain

$\begin{array}{r}\left(1-\frac{r}{n+1}\right)B{H}_{n+1}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=xB{H}_{n}\left(x|a;b;\lambda ;\mu \right)-\frac{1}{n+1}\sum _{i=1}^{r}{a}_{i}B{H}_{n+1}\left(x+{a}_{i}|a,{a}_{i};b;\lambda ;\mu \right)\\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n}\left(x+{b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right),\end{array}$

which is (22). □

### 3.4 More relations

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

$\begin{array}{rl}B{H}_{n}\left(x|a;b;\lambda ;\mu \right)=& xB{H}_{n-1}\left(x|a;b;\lambda ;\mu \right)-\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n-1}\left(x+{b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right)\\ +\sum _{m=1}^{n}\frac{{\left(-1\right)}^{m-1}\left(\genfrac{}{}{0}{}{n-1}{m-1}\right){B}_{m}}{m}B{H}_{n-m}\left(x|a;b;\lambda ;\mu \right)\sum _{i=1}^{r}{a}_{i}^{m}.\end{array}$
(24)

Proof For $n\ge 1$, we have

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

The third term is

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

Since

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

the first term is

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

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

$\begin{array}{r}-\sum _{\nu =1}^{s}\frac{{\mu }_{\nu }{b}_{\nu }}{1-{\lambda }_{\nu }}〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\frac{1-{\lambda }_{\nu }}{{e}^{{b}_{\nu }t}-{\lambda }_{\nu }}\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{e}^{\left(y+{b}_{\nu }\right)t}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{\nu =1}^{s}\frac{{\mu }_{\nu }{b}_{\nu }}{1-{\lambda }_{\nu }}B{H}_{n-1}\left(y+{b}_{\nu }|a;b;\lambda ;\mu +{e}_{\nu }\right).\end{array}$

Therefore, we obtain

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

which is identity (24). □

### 3.5 A relation including Bernoulli numbers

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

$\begin{array}{r}\left(n-m\right)B{H}_{n-m}\left(a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=-\left(n-m\right)\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n-m-1}\left({b}_{i}|a;b;\lambda ;\mu +{e}_{j}\right)\\ \phantom{\rule{2em}{0ex}}+\sum _{l=1}^{n-m}{\left(-1\right)}^{l-1}\left(\genfrac{}{}{0}{}{n-m}{l}\right){B}_{l}B{H}_{n-m-l}\left(a;b;\lambda ;\mu \right)\sum _{i=1}^{r}{a}_{i}^{l}.\end{array}$
(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

$\begin{array}{r}〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|{t}^{m}{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}={\left(n\right)}_{m}〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|{x}^{n-m}〉\\ \phantom{\rule{1em}{0ex}}={\left(n\right)}_{m}B{H}_{n-m}\left(a;b;\lambda ;\mu \right).\end{array}$

On the other hand, it is equal to

$\begin{array}{r}〈{\partial }_{t}\left(\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{t}^{m}\right)|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈\left({\partial }_{t}\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{t}^{m}|{x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\left({\partial }_{t}\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}\right){t}^{m}|{x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}\left({\partial }_{t}{t}^{m}\right)|{x}^{n-1}〉.\end{array}$
(26)

The third term of (26) is equal to

$\begin{array}{r}m〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{t}^{m-1}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=m{\left(n-1\right)}_{m-1}B{H}_{n-m}\left(a;b;\lambda ;\mu \right).\end{array}$

The second term of (26) is equal to

$\begin{array}{r}-\sum _{l=1}^{s}\frac{{\mu }_{l}{b}_{l}}{1-{\lambda }_{l}}〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\frac{1-{\lambda }_{l}}{{e}^{{b}_{l}t}-{\lambda }_{l}}\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{e}^{{b}_{l}t}|{t}^{m}{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-{\left(n-1\right)}_{m}\sum _{l=1}^{s}\frac{{\mu }_{l}{b}_{l}}{1-{\lambda }_{l}}〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\frac{1-{\lambda }_{l}}{{e}^{{b}_{l}t}-{\lambda }_{l}}\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}{e}^{{b}_{l}t}|{x}^{n-m-1}〉\\ \phantom{\rule{1em}{0ex}}=-{\left(n-1\right)}_{m}\sum _{l=1}^{s}\frac{{\mu }_{l}{b}_{l}}{1-{\lambda }_{l}}B{H}_{n-m-1}\left({b}_{l}|a;b;\lambda ;\mu +{e}_{l}\right).\end{array}$

Since

${\left(n-1\right)}_{l-1}{\left(n-l\right)}_{m}={\left(n-1\right)}_{l+m-1}={\left(n-1\right)}_{m-1}{\left(n-m\right)}_{l},$

the first term of (26) is equal to

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

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

$\begin{array}{r}{\left(n\right)}_{m}B{H}_{n-m}\left(a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=m{\left(n-1\right)}_{m-1}B{H}_{n-m}\left(a;b;\lambda ;\mu \right)\\ \phantom{\rule{2em}{0ex}}-{\left(n-1\right)}_{m}\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n-m-1}\left({b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right)\\ \phantom{\rule{2em}{0ex}}+{\left(n-1\right)}_{m-1}\sum _{l=1}^{n-m}{\left(-1\right)}^{l-1}\left(\genfrac{}{}{0}{}{n-m}{l}\right){B}_{l}B{H}_{n-m-l}\left(a;b;\lambda ;\mu \right)\sum _{i=1}^{r}{a}_{i}^{l}.\end{array}$

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

$\begin{array}{r}\left(n-m\right)B{H}_{n-m}\left(a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=-\left(n-m\right)\sum _{j=1}^{s}\frac{{\mu }_{j}{b}_{j}}{1-{\lambda }_{j}}B{H}_{n-m-1}\left({b}_{j}|a;b;\lambda ;\mu +{e}_{j}\right)\\ \phantom{\rule{2em}{0ex}}+\sum _{l=1}^{n-m}{\left(-1\right)}^{l-1}\left(\genfrac{}{}{0}{}{n-m}{l}\right){B}_{l}B{H}_{n-m-l}\left(a;b;\lambda ;\mu \right)\sum _{i=1}^{r}{a}_{i}^{l}.\end{array}$

Thus, we get (25). □

### 3.6 A relation with Stirling numbers

The Stirling numbers of the second kind ${S}_{2}\left(n,m\right)$ are defined by

$\frac{{\left({e}^{t}-1\right)}^{m}}{m!}=\sum _{n=m}^{\mathrm{\infty }}{S}_{2}\left(n,m\right)\frac{{t}^{n}}{n!}.$

Then

${\varphi }_{n}\left(x\right):=\sum _{m=0}^{n}{S}_{2}\left(n,m\right){x}^{m}\sim \left(1,ln\left(1+t\right)\right).$
(27)

Theorem 6

$B{H}_{n}\left(x|a;b;\lambda ;\mu \right)=\sum _{m=0}^{n}\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(l,m\right)B{H}_{n-l}{\varphi }_{m}\left(x\right).$
(28)

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

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

Thus, we get identity (28). □

### 3.7 A relation with falling factorials

Theorem 7

$B{H}_{n}\left(x|a;b;\lambda ;\mu \right)=\sum _{m=0}^{n}\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right)B{H}_{n-l}{\left(x\right)}_{m}.$
(29)

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

$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{i=1}^{r}\left(\frac{{e}^{{a}_{i}t}-1}{t}\right){\prod }_{j=1}^{s}{\left(\frac{{e}^{{b}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)}^{{\mu }_{j}}}{\left({e}^{t}-1\right)}^{m}|{x}^{n}〉\\ =〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|\frac{1}{m!}{\left({e}^{t}-1\right)}^{m}{x}^{n}〉\\ =〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|\sum _{l=m}^{\mathrm{\infty }}{S}_{2}\left(l,m\right)\frac{{t}^{l}}{l!}{x}^{n}〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right)〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|{x}^{n-l}〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right)B{H}_{n-l}.\end{array}$

Thus, we get identity (29). □

### 3.8 A relation with higher-order Frobenius-Euler polynomials

Theorem 8

$\begin{array}{r}B{H}_{n}\left(x|a;b;\lambda ;\mu \right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{{\left(1-\alpha \right)}^{p}}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\sum _{l=0}^{p}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0}{}{p}{l}\right)\left(\genfrac{}{}{0}{}{n-m}{\nu }\right){\left(-\alpha \right)}^{p-l}{l}^{\nu }B{H}_{n-m-\nu }\right){H}_{m}^{\left(p\right)}\left(x|\alpha \right).\end{array}$
(30)

Proof For (16) and

${H}_{n}^{\left(p\right)}\left(x|\alpha \right)\sim \left({\left(\frac{{e}^{t}-\alpha }{1-\alpha }\right)}^{p},t\right),$
(31)

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

$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{\left(\frac{{e}^{t}-\alpha }{1-\alpha }\right)}^{p}}{{\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}^{m}|{x}^{n}〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{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}}|{\left({e}^{t}-\alpha \right)}^{p}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{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{}{}{0}{}{p}{l}\right){\left(-\alpha \right)}^{p-l}{e}^{lt}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{p}}\sum _{l=0}^{p}\left(\genfrac{}{}{0}{}{p}{l}\right){\left(-\alpha \right)}^{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{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{p}}\sum _{l=0}^{p}\left(\genfrac{}{}{0}{}{p}{l}\right){\left(-\alpha \right)}^{p-l}〈{e}^{lt}|\sum _{\nu =0}^{\mathrm{\infty }}B{H}_{\nu }\frac{{t}^{\nu }}{\nu !}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{p}}\sum _{l=0}^{p}\left(\genfrac{}{}{0}{}{p}{l}\right){\left(-\alpha \right)}^{p-l}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0}{}{n-m}{\nu }\right)B{H}_{\nu }〈{e}^{lt}|{x}^{n-m-\nu }〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{p}}\sum _{l=0}^{p}\left(\genfrac{}{}{0}{}{p}{l}\right){\left(-\alpha \right)}^{p-l}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0}{}{n-m}{\nu }\right)B{H}_{\nu }{l}^{n-m-\nu }\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\alpha \right)}^{p}}\sum _{l=0}^{p}\left(\genfrac{}{}{0}{}{p}{l}\right){\left(-\alpha \right)}^{p-l}\sum _{\nu =0}^{n-m}\left(\genfrac{}{}{0}{}{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 ${\mathfrak{B}}_{n}^{\left(p\right)}\left(x\right)$ of order p are defined by

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

(see e.g. [[10], Section 2.2]).

Theorem 9

$B{H}_{n}\left(x|a;b;\lambda ;\mu \right)=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{n-m-l+p}{p}\right)}{S}_{2}\left(n-m-l+p,p\right)B{H}_{l}\right){\mathfrak{B}}_{m}^{\left(p\right)}\left(x\right).$
(32)

Proof For (16) and

${\mathfrak{B}}_{n}^{\left(p\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{p},t\right),$
(33)

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

$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{\left(\frac{{e}^{t}-1}{t}\right)}^{p}}{{\prod }_{i=1}^{r}\left(\frac{{e}^{{a}_{i}t}-1}{t}\right){\prod }_{j=1}^{s}{\left(\frac{{e}^{{b}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)}^{{\mu }_{j}}}{t}^{m}|{x}^{n}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|{\left(\frac{{e}^{t}-1}{t}\right)}^{p}{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|p!\sum _{l=0}^{\mathrm{\infty }}{S}_{2}\left(l+p,p\right)\frac{{t}^{l}}{\left(l+p\right)!}{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)p!\sum _{l=0}^{n-m}\frac{{\left(n-m\right)}_{l}}{\left(l+p\right)!}{S}_{2}\left(l+p,p\right)〈\prod _{i=1}^{r}\left(\frac{t}{{e}^{{a}_{i}t}-1}\right)\prod _{j=1}^{s}{\left(\frac{1-{\lambda }_{j}}{{e}^{{b}_{j}t}-{\lambda }_{j}}\right)}^{{\mu }_{j}}|{x}^{n-m-l}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{l+p}{p}\right)}{S}_{2}\left(l+p,p\right)B{H}_{n-m-l}\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{n-m-l+p}{p}\right)}{S}_{2}\left(n-m-l+p,p\right)B{H}_{l}.\end{array}$

Thus, we get identity (32). □

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## Acknowledgements

The authors would like to thank the referees for their valuable comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786) and was partially supported by Kwangwoon University in 2014.

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Correspondence to Taekyun Kim.

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The authors declare that they have no competing interests.

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All authors contributed equally to this work. All authors read and approved the final manuscript.

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Kim, D.S., Kim, T., Komatsu, T. et al. Barnes’ multiple Bernoulli and generalized Barnes’ multiple Frobenius-Euler mixed-type polynomials. Adv Differ Equ 2014, 238 (2014). https://doi.org/10.1186/1687-1847-2014-238