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# Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials

https://doi.org/10.1186/1687-1847-2014-92

• Accepted: 6 March 2014
• Published:

## Abstract

In this paper, we consider Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type 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, 65Q05.

## Keywords

• Formal Power Series
• Bernoulli Number
• Bernoulli Polynomial
• Stirling Number
• Polylogarithm Function

## 1 Introduction

In this paper, we consider the polynomials ${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$ whose generating function is given by
$\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\frac{{t}^{n}}{n!},$
(1)
where $r\in {\mathbb{Z}}_{>\mathbb{⊬}}$, $k\in \mathbb{Z}$, ${a}_{1},\dots ,{a}_{r}\ne 0$, ${\lambda }_{1},\dots ,{\lambda }_{r}\ne 1$ and
${Li}_{k}\left(x\right)=\sum _{m=1}^{\mathrm{\infty }}\frac{{x}^{m}}{{m}^{k}}$

is the k th polylogarithm function. ${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$ will be called Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials. When $x=0$, ${T}_{n}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)={T}_{n}^{\left(r,k\right)}\left(0|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$ will be called Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type numbers.

Recall that, for every integer k, the poly-Bernoulli polynomials ${B}_{n}^{\left(k\right)}\left(x\right)$ are defined by the generating function as follows:
$\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}$
(2)
(, cf. ). Also, as a natural generalization of higher-order Frobenius-Euler polynomials, Barnes’ multiple Frobenius-Euler polynomials ${H}_{n}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$ are defined by the generating function as follows:
$\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\frac{{t}^{n}}{n!},$
(3)
where ${a}_{1},\dots ,{a}_{r}\ne 0$. Note that the Frobenius-Euler polynomials of order r, ${H}_{n}^{\left(r\right)}\left(x|\lambda \right)$ are defined by the generating function
${\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(r\right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!}$

(see, e.g., ).

In this paper, we consider Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type 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 the 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 the umbral algebra and the 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 $n,k\ge 0$. 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)
[, Theorem 2.2.5]. Thus, by (8), we get
${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right).$
(9)

Sheffer sequences are characterized in the generating function [, 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 [, 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 [, 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 [, 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

We now note that ${B}_{n}^{\left(k\right)}\left(x\right)$, ${H}_{n}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$ and ${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$ are the Appell sequences for
$\begin{array}{c}{g}_{k}\left(t\right)=\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},\phantom{\rule{2em}{0ex}}{g}_{r}\left(t\right)=\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right),\hfill \\ {g}_{r,k}\left(t\right)=\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)}.\hfill \end{array}$
So,
${B}_{n}^{\left(k\right)}\left(x\right)\sim \left(\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},t\right),$
(14)
${H}_{n}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\sim \left(\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right),t\right),$
(15)
${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\sim \left(\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},t\right).$
(16)
In particular, we have
$t{B}_{n}^{\left(k\right)}\left(x\right)=\frac{d}{dx}{B}_{n}^{\left(k\right)}\left(x\right)=n{B}_{n-1}^{\left(k\right)}\left(x\right),$
(17)
$\begin{array}{rl}t{H}_{n}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)& =\frac{d}{dx}{H}_{n}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ =n{H}_{n-1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right),\end{array}$
(18)
$\begin{array}{rl}t{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)& =\frac{d}{dx}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ =n{T}_{n-1}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$
(19)
Notice that
$\frac{d}{dx}{Li}_{k}\left(x\right)=\frac{1}{x}{Li}_{k-1}\left(x\right).$

### 3.1 Explicit expressions

Write ${H}_{n}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right):={H}_{n}^{\left(r\right)}\left(0|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$. Let ${\left(n\right)}_{j}=n\left(n-1\right)\cdots \left(n-j+1\right)$ ($j\ge 1$) with ${\left(n\right)}_{0}=1$.

Theorem 1

$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}^{\left(k\right)}\left(x\right){H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\end{array}$
(20)
$\phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}^{\left(k\right)}{H}_{l}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$
(21)
$\phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{n}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)\left(\genfrac{}{}{0}{}{n}{l}\right)\frac{1}{{\left(m+1\right)}^{k}}{H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right){\left(x-j\right)}^{l}$
(22)
$\begin{array}{r}\phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\sum _{j=l}^{n}\sum _{m=0}^{n-j}{\left(-1\right)}^{n-m-j}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{j}{l}\right)\\ \phantom{\rule{2em}{0ex}}×\frac{m!}{{\left(m+1\right)}^{k}}{S}_{2}\left(n-j,m\right){H}_{j-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right){x}^{l}\end{array}$
(23)
$\phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){T}_{n-j}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right){x}^{j}.$
(24)
Proof By (1), (2) and (3), we have
$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)& =〈\sum _{i=0}^{\mathrm{\infty }}{T}_{i}^{\left(r,k\right)}\left(y|{a}_{1},\dots ,{a}_{r},{\lambda }_{1},\dots ,{\lambda }_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)|\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}{x}^{n}〉\\ =〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)|\sum _{l=0}^{\mathrm{\infty }}{B}_{l}^{\left(k\right)}\left(y\right)\frac{{t}^{l}}{l!}{x}^{n}〉\\ =\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}^{\left(k\right)}\left(y\right)〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)|{x}^{n-l}〉\\ =\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}^{\left(k\right)}\left(y\right)〈\sum _{i=0}^{\mathrm{\infty }}{H}_{i}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n-l}〉\\ =\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}^{\left(k\right)}\left(y\right){H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

So, we get (20).

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

Thus, we get (21).

In  we obtained that
$\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}=\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(x-j\right)}^{n}.$
So,
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){\left(x-j\right)}^{n}\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right){\left(x-j\right)}^{l}\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{n}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)\left(\genfrac{}{}{0}{}{n}{l}\right)\frac{1}{{\left(m+1\right)}^{k}}{H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right){\left(x-j\right)}^{l},\end{array}$

which is identity (22).

In  we obtained that
$\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}=\sum _{j=0}^{n}\left(\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right){x}^{j},$
where ${S}_{2}\left(l,m\right)$ are the Stirling numbers of the second kind, defined by
${\left({e}^{t}-1\right)}^{m}=m!\sum _{l=m}^{\mathrm{\infty }}{S}_{2}\left(l,m\right)\frac{{t}^{l}}{l!}.$
Thus,
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\left(\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right)\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){x}^{j}\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\left(\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right){H}_{j}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\left(\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right)\sum _{l=0}^{j}\left(\genfrac{}{}{0}{}{j}{l}\right){H}_{j-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right){x}^{l}\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\sum _{j=l}^{n}\sum _{m=0}^{n-j}{\left(-1\right)}^{n-m-j}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{j}{l}\right)\frac{m!}{{\left(m+1\right)}^{k}}{S}_{2}\left(n-j,m\right){H}_{j-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right){x}^{l},\end{array}$

which is identity (23).

By (11) with (16), we have
$\begin{array}{rl}〈g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^{j}|{x}^{n}〉& =〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{t}^{j}|{x}^{n}〉\\ ={\left(n\right)}_{j}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n-j}〉\\ ={\left(n\right)}_{j}〈\sum _{i=0}^{\mathrm{\infty }}{T}_{i}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\frac{{t}^{i}}{i!}|{x}^{n-j}〉\\ ={\left(n\right)}_{j}{T}_{n-j}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

Thus, we get (24). □

### 3.2 Sheffer identity

Theorem 2
${T}_{n}^{\left(r,k\right)}\left(x+y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){T}_{j}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right){y}^{n-j}.$
(25)
Proof By (16) with
$\begin{array}{rl}{p}_{n}\left(x\right)& =\prod _{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ ={x}^{n}\sim \left(1,t\right),\end{array}$

using (12), we have (25). □

### 3.3 Recurrence

Theorem 3
$\begin{array}{rl}{T}_{n+1}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)=& x{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ -\sum _{j=1}^{r}\frac{{a}_{j}}{1-{\lambda }_{j}}{T}_{n}^{\left(r+1,k\right)}\left(x+{a}_{j}|{a}_{1},\dots ,{a}_{r},{a}_{j};{\lambda }_{1},\dots ,{\lambda }_{r},{\lambda }_{j}\right)\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{n+1-l}\\ ×\left({T}_{l}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ -{T}_{l}^{\left(r,k-1\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right),\end{array}$
(26)

where ${B}_{n}$ is the nth ordinary Bernoulli number.

Proof By applying
${s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right)$
[, Corollary 3.7.2] with (16), we get
${T}_{n+1}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)=\left(x-\frac{{g}_{r,k}^{\prime }\left(t\right)}{{g}_{r,k}\left(t\right)}\right){T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).$
Now,
$\begin{array}{rl}\frac{{g}_{r,k}^{\prime }\left(t\right)}{{g}_{r,k}\left(t\right)}& ={\left(ln{g}_{r,k}\left(t\right)\right)}^{\prime }\\ ={\left(\sum _{j=1}^{r}ln\left({e}^{{a}_{j}t}-{\lambda }_{j}\right)-\sum _{j=1}^{r}ln\left(1-{\lambda }_{j}\right)+ln\left(1-{e}^{-t}\right)-ln{Li}_{k}\left(1-{e}^{-t}\right)\right)}^{\prime }\\ =\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-{\lambda }_{j}}+\frac{{e}^{-t}}{1-{e}^{-t}}\left(1-\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)}{{Li}_{k}\left(1-{e}^{-t}\right)}\right)\\ =\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-{\lambda }_{j}}+\frac{t}{{e}^{t}-1}\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{t{Li}_{k}\left(1-{e}^{-t}\right)}.\end{array}$
Since
${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)=\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n},$
we have
$\begin{array}{r}{T}_{n+1}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=x{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{1-{\lambda }_{j}}\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}\\ \phantom{\rule{2em}{0ex}}-\frac{t}{{e}^{t}-1}\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{x}^{n}.\end{array}$
Since
$\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}=\left(\frac{1}{{2}^{k}}-\frac{1}{{2}^{k-1}}\right)t+\cdots$
is a delta series, we get
$\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{x}^{n}=\frac{1}{n+1}\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n+1}.$
Therefore, by
$\frac{t}{{e}^{t}-1}{x}^{n+1}={B}_{n+1}\left(x\right)=\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{n+1-l}{x}^{l},$
we obtain
$\begin{array}{r}{T}_{n+1}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=x{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)-\sum _{j=1}^{r}\frac{{a}_{j}}{1-{\lambda }_{j}}{T}_{n}^{\left(r+1,k\right)}\left(x+{a}_{j}|{a}_{1},\dots ,{a}_{r},{a}_{j};{\lambda }_{1},\dots ,{\lambda }_{r},{\lambda }_{j}\right)\\ \phantom{\rule{2em}{0ex}}-\frac{1}{n+1}\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\frac{t}{{e}^{t}-1}{x}^{n+1}\\ \phantom{\rule{1em}{0ex}}=x{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)-\sum _{j=1}^{r}\frac{{a}_{j}}{1-{\lambda }_{j}}{T}_{n}^{\left(r+1,k\right)}\left(x+{a}_{j}|{a}_{1},\dots ,{a}_{r},{a}_{j};{\lambda }_{1},\dots ,{\lambda }_{r},{\lambda }_{j}\right)\\ \phantom{\rule{2em}{0ex}}-\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{n+1-l}\\ \phantom{\rule{2em}{0ex}}×\left({T}_{l}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)-{T}_{l}^{\left(r,k-1\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right),\end{array}$

which is identity (26). □

### 3.4 A more relation

Theorem 4
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=x{T}_{n-1}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{2em}{0ex}}-\sum _{j=1}^{r}\frac{{a}_{j}}{1-{\lambda }_{j}}{T}_{n-1}^{\left(r+1,k\right)}\left(x+{a}_{j}|{a}_{1},\dots ,{a}_{r},{a}_{j};{\lambda }_{1},\dots ,{\lambda }_{r},{\lambda }_{j}\right)\\ \phantom{\rule{2em}{0ex}}-\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}\left({T}_{l}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{2em}{0ex}}-{T}_{l}^{\left(r,k-1\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right).\end{array}$
(27)
Proof For $n\ge 1$, we have
$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)=& 〈\sum _{l=0}^{\mathrm{\infty }}{T}_{l}^{\left(r,k\right)}\left(y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ =& 〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n}〉\\ =& 〈{\partial }_{t}\left(\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}\right)|{x}^{n-1}〉\\ =& 〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-1}〉\\ +〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉\\ +〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\left({\partial }_{t}{e}^{yt}\right)|{x}^{n-1}〉\\ =& y{T}_{n-1}^{\left(r,k\right)}\left(y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ +〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-1}〉\\ +〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉.\end{array}$
Observe that
$\begin{array}{rl}{\partial }_{t}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)& =\prod _{j=1}^{r}\left(1-{\lambda }_{j}\right){\partial }_{t}\left(\frac{1}{{\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-{\lambda }_{j}\right)}\right)\\ =\prod _{j=1}^{r}\left(1-{\lambda }_{j}\right)\frac{-{\left({\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-{\lambda }_{j}\right)\right)}^{\prime }}{{\left({\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-{\lambda }_{j}\right)\right)}^{2}}\\ =-\prod _{j=1}^{r}\left(1-{\lambda }_{j}\right)\frac{{\sum }_{j=1}^{r}{a}_{j}{e}^{{a}_{j}t}{\prod }_{i\ne j}\left({e}^{{a}_{i}t}-{\lambda }_{i}\right)}{{\left({\prod }_{j=1}^{r}\left({e}^{{a}_{j}t}-{\lambda }_{j}\right)\right)}^{2}}\\ =-\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right)\\ =-\sum _{j=1}^{r}\frac{{a}_{j}{e}^{{a}_{j}t}}{1-{\lambda }_{j}}\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right).\end{array}$
Thus,
$\begin{array}{r}〈\left({\partial }_{t}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{j=1}^{r}\frac{{a}_{j}}{1-{\lambda }_{j}}〈\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\prod _{i=1}^{r}\left(\frac{1-{\lambda }_{i}}{{e}^{{a}_{i}t}-{\lambda }_{i}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{\left(y+{a}_{j}\right)t}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{j=1}^{r}\frac{{a}_{j}}{1-{\lambda }_{j}}{T}_{n-1}^{\left(r+1,k\right)}\left(y+{a}_{j}|{a}_{1},\dots ,{a}_{r},{a}_{j};{\lambda }_{1},\dots ,{\lambda }_{r},{\lambda }_{j}\right).\end{array}$
Since
$\begin{array}{rl}{\partial }_{t}\left(\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right)& =\frac{{e}^{-t}\left({Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)\right)}{{\left(1-{e}^{-t}\right)}^{2}}\\ =\frac{t}{{e}^{t}-1}\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}\end{array}$
and the fact that
$\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}=\left(\frac{1}{{2}^{k-1}}-\frac{1}{{2}^{k}}\right)t+\cdots$
is a delta series, we have
$\begin{array}{r}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{t}{{e}^{t}-1}\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{e}^{yt}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{t}{{e}^{t}-1}\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}}{1-{e}^{-t}}{e}^{yt}|\frac{{x}^{n}}{n}〉\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|\frac{t}{{e}^{t}-1}{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{l}〉\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}\left({T}_{l}^{\left(r,k-1\right)}\left(y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)-{T}_{l}^{\left(r,k\right)}\left(y|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right).\end{array}$

Therefore, we obtain the desired result. □

Remark After n is replaced by $n+1$, identity (27) becomes the recurrence formula (26).

### 3.5 Relations with poly-Bernoulli numbers and Barnes’ multiple Bernoulli numbers

Theorem 5
$\begin{array}{r}\sum _{m=0}^{n}{\left(-1\right)}^{n-m}\left(\genfrac{}{}{0}{}{n+1}{m}\right){T}_{m}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){B}_{m}^{\left(k-1\right)}{H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$
(28)
Proof We shall compute
$〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){Li}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉$
in two different ways. On the one hand,
$\begin{array}{r}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){Li}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|\left(1-{e}^{-t}\right){x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n+1}-{\left(x-1\right)}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(-1\right)}^{n-m}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{m}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(-1\right)}^{n-m}{T}_{m}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$
On the other hand,
$\begin{array}{r}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){Li}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈{Li}_{k}\left(1-{e}^{-t}\right)|\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right){x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈{Li}_{k}\left(1-{e}^{-t}\right)|{H}_{n+1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}{\left({Li}_{k}\left(1-{e}^{-s}\right)\right)}^{\prime }\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}{e}^{-s}\frac{{Li}_{k-1}\left(1-{e}^{-s}\right)}{1-{e}^{-s}}\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}\left(\sum _{j=0}^{\mathrm{\infty }}\frac{{\left(-s\right)}^{j}}{j!}\right)\left(\sum _{m=0}^{\mathrm{\infty }}\frac{{B}_{m}^{\left(k-1\right)}}{m!}{s}^{m}\right)\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ \phantom{\rule{1em}{0ex}}=〈\sum _{l=0}^{\mathrm{\infty }}\left(\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right){B}_{m}^{\left(k-1\right)}\right)\frac{1}{l!}{\int }_{0}^{t}{s}^{l}\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right)\frac{{B}_{m}^{\left(k-1\right)}}{\left(l+1\right)!}〈{t}^{l+1}|{H}_{n+1}^{\left(r\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right)\frac{{B}_{m}^{\left(k-1\right)}}{\left(l+1\right)!}{\left(n+1\right)}_{l+1}{H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){B}_{m}^{\left(k-1\right)}{H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

Here, ${H}_{n-l}^{\left(r\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)={H}_{n-l}^{\left(r\right)}\left(0|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)$. Thus, we get (28). □

### 3.6 Relations with the Stirling numbers of the second kind and the falling factorials

Theorem 6
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right){T}_{n-l}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right){\left(x\right)}_{m}.\end{array}$
(29)
Proof For (16) and ${\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right)$, assume that
${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)=\sum _{m=0}^{n}{C}_{n,m}{\left(x\right)}_{m}.$
By (13), we have
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)}}{\left({e}^{t}-1\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{\left({e}^{t}-1\right)}^{m}{x}^{n}〉\\ =\frac{1}{m!}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|m!\sum _{l=m}^{n}{S}_{2}\left(l,m\right)\frac{{t}^{l}}{l!}{x}^{n}〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right)〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n-l}〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right){T}_{n-l}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

Thus, we get identity (29). □

### 3.7 Relations with the Stirling numbers of the second kind and the rising factorials

Theorem 7
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right){T}_{n-l}^{\left(r,k\right)}\left(-m|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right){\left(x\right)}^{\left(m\right)}.\end{array}$
(30)
Proof For (16) and ${\left(x\right)}^{\left(n\right)}=x\left(x+1\right)\cdots \left(x+n-1\right)\sim \left(1,1-{e}^{-t}\right)$, assume that ${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)={\sum }_{m=0}^{n}{C}_{n,m}{\left(x\right)}^{\left(m\right)}$. By (13), we have
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{j=1}^{r}\left(\frac{{e}^{{a}_{j}t}-{\lambda }_{j}}{1-{\lambda }_{j}}\right)\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)}}{\left(1-{e}^{-t}\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{-mt}|{\left({e}^{t}-1\right)}^{m}{x}^{n}〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right)〈{e}^{-mt}|\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n-l}〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right)〈{e}^{-mt}|{T}_{n-l}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)〉\\ =\sum _{l=m}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{2}\left(l,m\right){T}_{n-l}^{\left(r,k\right)}\left(-m|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

Thus, we get identity (30). □

### 3.8 Relations with higher-order Frobenius-Euler polynomials

Theorem 8
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\lambda \right)}^{s}}\sum _{l=0}^{s}\left(\genfrac{}{}{0}{}{s}{l}\right){\left(-\lambda \right)}^{s-l}{T}_{n-m}^{\left(r,k\right)}\left(l|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right){H}_{m}^{\left(s\right)}\left(x|\lambda \right).\end{array}$
(31)
Proof For (16) and
${H}_{n}^{\left(s\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{s},t\right),$
assume that ${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)={\sum }_{m=0}^{n}{C}_{n,m}{H}_{m}^{\left(s\right)}\left(x|\lambda \right)$. By (13), we have
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{s}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{t}^{m}|{x}^{n}〉\\ =\frac{1}{m!{\left(1-\lambda \right)}^{s}}\sum _{l=0}^{s}\left(\genfrac{}{}{0}{}{s}{l}\right){\left(-\lambda \right)}^{s-l}〈{e}^{lt}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{t}^{m}{x}^{n}〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\lambda \right)}^{s}}\sum _{l=0}^{s}\left(\genfrac{}{}{0}{}{s}{l}\right){\left(-\lambda \right)}^{s-l}〈{e}^{lt}|\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\lambda \right)}^{s}}\sum _{l=0}^{s}\left(\genfrac{}{}{0}{}{s}{l}\right){\left(-\lambda \right)}^{s-l}{T}_{n-m}^{\left(r,k\right)}\left(l|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

Thus, we get identity (31). □

### 3.9 Relations with higher-order Bernoulli polynomials

Bernoulli polynomials ${\mathfrak{B}}_{n}^{\left(r\right)}\left(x\right)$ of order r are defined by
${\left(\frac{t}{{e}^{t}-1}\right)}^{r}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathfrak{B}}_{n}^{\left(r\right)}\left(x\right)}{n!}{t}^{n}$

(see, e.g., [, Section 2.2]).

Theorem 9
$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{l+s}{l}\right)}{S}_{2}\left(l+s,s\right){T}_{n-m-l}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\right){\mathfrak{B}}_{m}^{\left(s\right)}\left(x\right).\end{array}$
(32)
Proof For (16) and
${\mathfrak{B}}_{n}^{\left(s\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{s},t\right),$
assume that ${T}_{n}^{\left(r,k\right)}\left(x|{a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)={\sum }_{m=0}^{n}{C}_{n,m}{\mathfrak{B}}_{m}^{\left(s\right)}\left(x\right)$. By (13), we have
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{t}-1}{t}\right)}^{s}\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{t}^{m}|{x}^{n}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{\left(\frac{{e}^{t}-1}{t}\right)}^{s}{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|\sum _{l=0}^{n-m}\frac{s!}{\left(l+s\right)!}{S}_{2}\left(l+s,s\right){t}^{l}{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!}{\left(l+s\right)!}{S}_{2}\left(l+s,s\right){\left(n-m\right)}_{l}〈\prod _{j=1}^{r}\left(\frac{1-{\lambda }_{j}}{{e}^{{a}_{j}t}-{\lambda }_{j}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n-m-l}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!}{\left(l+s\right)!}{S}_{2}\left(l+s,s\right){\left(n-m\right)}_{l}{T}_{n-m-l}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right)\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{l+s}{l}\right)}{S}_{2}\left(l+s,s\right){T}_{n-m-l}^{\left(r,k\right)}\left({a}_{1},\dots ,{a}_{r};{\lambda }_{1},\dots ,{\lambda }_{r}\right).\end{array}$

Thus, we get identity (32). □

## Declarations

### Acknowledgements

The authors would like to thank the referees for their valuable comments and suggestions. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786). The fourth author was supported in part by the Grant-in-Aid for Scientific research (C) (No. 22540005), the Japan Society for the Promotion of Science.

## Authors’ Affiliations

(1)
Department of Mathematics, Sogang University, Seoul, 121-741, Republic of Korea
(2)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, Republic of Korea
(3)
Department of Applied Mathematics, Pukyong National University, Pusan, 608-739, Republic of Korea
(4)
Graduate School of Science and Technology, Hirosaki University, Hirosaki 036-8561, Japan

## References 