Theory and Modern Applications

# Higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials

## Abstract

In this paper, we consider higher-order Frobenius-Euler polynomials, associated with poly-Bernoulli polynomials, which are derived from polylogarithmic function. These polynomials are called higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials. The purpose of this paper is to give various identities of those polynomials arising from umbral calculus.

## 1 Introduction

For $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, the Frobenius-Euler polynomials of order α ($\alpha \in \mathbb{R}$) are defined by the generating function to be

${\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(\alpha \right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1–5]}\right).$
(1.1)

When $x=0$, ${H}_{n}^{\left(\alpha \right)}\left(\lambda \right)={H}_{n}^{\left(\alpha \right)}\left(0|\lambda \right)$ are called the Frobenius-Euler numbers of order α. As is well known, the Bernoulli polynomials of order α are defined by the generating function to be

${\left(\frac{t}{{e}^{t}-1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{\mathbb{B}}_{n}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [6–8]}\right).$
(1.2)

When $x=0$, ${\mathbb{B}}_{n}^{\left(\alpha \right)}={\mathbb{B}}_{n}^{\left(\alpha \right)}\left(x\right)$ is called the n th Bernoulli number of order α. In the special case, $\alpha =1$, ${\mathbb{B}}_{n}^{\left(1\right)}\left(x\right)={B}_{n}\left(x\right)$ is called the n th Bernoulli polynomial. When $x=0$, ${B}_{n}={B}_{n}\left(0\right)$ is called the n th ordinary Bernoulli number. Finally, we recall that the Euler polynomials of order α are given by

${\left(\frac{2}{{e}^{t}+1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{E}_{n}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [9–13]}\right).$
(1.3)

When $x=0$, ${E}_{n}^{\left(\alpha \right)}={E}_{n}^{\left(\alpha \right)}\left(0\right)$ is called the n th Euler number of order α. In the special case, $\alpha =1$, ${E}_{n}^{\left(1\right)}\left(x\right)={E}_{n}\left(x\right)$ is called the n th ordinary Euler polynomial. The classical polylogarithmic function $L{i}_{k}\left(x\right)$ is defined by

$L{i}_{k}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{{n}^{k}}\phantom{\rule{1em}{0ex}}\left(k\in \mathbb{Z}\right)\phantom{\rule{0.25em}{0ex}}\left(\text{see }\right).$
(1.4)

As is known, poly-Bernoulli polynomials are defined by the generating function to be

(1.5)

Let be the complex number field, and let be the set of all formal power series in the variable t over with

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{k!}{t}^{k}|{a}_{k}\in \mathbb{C}\right\}.$
(1.6)

Now, we use the notation $\mathbb{P}=\mathbb{C}\left[x\right]$. In this paper, ${\mathbb{P}}^{\ast }$ will be denoted by the vector space of all linear functionals on . Let us assume that $〈L|p\left(x\right)〉$ be the action of the linear functional L on the polynomial $p\left(x\right)$, and we remind 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 . The formal power series

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{k!}{t}^{k}\in \mathcal{F}$
(1.7)

defines a linear functional on by setting

(1.8)

From (1.7) and (1.8), we note that

$〈{t}^{k}|{x}^{n}〉=n!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(\text{see [14, 15]}\right),$
(1.9)

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

Let us consider ${f}_{L}\left(t\right)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|{x}^{n}〉}{k!}{t}^{k}$. Then we see that $〈{f}_{L}\left(t\right)|{x}^{n}〉=〈L|{x}^{n}〉$, and so $L={f}_{L}\left(t\right)$ as linear functionals. The map $L↦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto . Henceforth, will denote 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 (see ). We shall call the umbral algebra. The umbral calculus is the study of umbral algebra. The order $o\left(f\left(t\right)\right)$ of a nonzero power series $f\left(t\right)$ is the smallest integer k, for which the coefficient of ${t}^{k}$ does not vanish. A series $f\left(t\right)$ is called a delta series if $o\left(f\left(t\right)\right)=1$, and an invertible series if $o\left(f\left(t\right)\right)=0$. Let $f\left(t\right),g\left(t\right)\in \mathcal{F}$. Then we have

$〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈f\left(t\right)|g\left(t\right)p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉\phantom{\rule{1em}{0ex}}\left(\text{see }\right).$
(1.10)

For $f\left(t\right),g\left(t\right)\in \mathcal{F}$ with $o\left(f\left(t\right)\right)=1$, $o\left(g\left(t\right)\right)=0$, 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 $n,k\ge 0$. The 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 ${S}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ (see [14, 15]). Let $f\left(t\right)\in \mathcal{F}$ and $p\left(t\right)\in \mathbb{P}$. Then we have

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)|{x}^{k}〉\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}|p\left(x\right)〉\frac{{x}^{k}}{k!}.$
(1.11)

From (1.11), we note that

${p}^{\left(k\right)}\left(0\right)=〈{t}^{k}|p\left(x\right)〉=〈1|{p}^{\left(k\right)}\left(x\right)〉.$
(1.12)

By (1.12), we get

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\left(\text{see [14, 15]}\right).$
(1.13)

From (1.13), we easily derive the following equation

${e}^{yt}p\left(x\right)=p\left(x+y\right),\phantom{\rule{1em}{0ex}}〈{e}^{yt}|p\left(x\right)〉=p\left(y\right).$
(1.14)

For $p\left(x\right)\in \mathbb{P}$, $f\left(t\right)\in \mathcal{F}$, it is known that

$〈f\left(t\right)|xp\left(x\right)〉=〈{\partial }_{t}f\left(t\right)|p\left(x\right)〉=〈{f}^{\prime }\left(t\right)|p\left(x\right)〉\phantom{\rule{1em}{0ex}}\left(\text{see }\right).$
(1.15)

Let ${S}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$. Then we have

(1.16)

where $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$ with $\overline{f}\left(f\left(t\right)\right)=t$, and

$f\left(t\right){S}_{n}\left(x\right)=n{S}_{n-1}\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{see [14, 15]}\right).$
(1.17)

The Stirling number of the second kind is defined by the generating function to be

${\left({e}^{t}-1\right)}^{m}=m!\sum _{l=m}^{\mathrm{\infty }}{S}_{2}\left(l,m\right)\frac{{t}^{m}}{m!}\phantom{\rule{1em}{0ex}}\left(m\in {\mathbb{Z}}_{\ge 0}\right).$
(1.18)

For ${S}_{n}\left(x\right)\sim \left(g\left(t\right),t\right)$, it is well known that

${S}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right){S}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right)\phantom{\rule{0.25em}{0ex}}\left(\text{see [14, 15]}\right).$
(1.19)

Let ${S}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, ${r}_{n}\left(x\right)\sim \left(h\left(t\right),l\left(t\right)\right)$. Then we have

${S}_{n}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{r}_{m}\left(x\right),$
(1.20)

where

${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}l{\left(\overline{f}\left(t\right)\right)}^{m}|{x}^{n}〉\phantom{\rule{1em}{0ex}}\left(\text{see [14, 15]}\right).$
(1.21)

In this paper, we study higher-order Frobeniuns-Euler polynomials associated with poly-Bernoulli polynomials, which are called higher-order Frobenius-Euler and poly-Beroulli mixed-type polynomials. The purpose of this paper is to give various identities of those polynomials arising from umbral calculus.

## 2 Higher-order Frobenius-Euler polynomials, associated poly-Bernoulli polynomials

Let us consider the polynomials ${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)$, called higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials, as follows:

${\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!},$
(2.1)

where $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, $r,k\in \mathbb{Z}$.

When $x=0$, ${T}_{n}^{\left(r,k\right)}\left(\lambda \right)={T}_{n}^{\left(r,k\right)}\left(0|\lambda \right)$ is called the n th higher-order Frobenius-Euler and poly-Bernoulli mixed type number.

From (1.16) and (2.1), we note that

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\sim \left({g}_{r,k}\left(t\right)={\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}\frac{1-{e}^{-t}}{L{i}_{k}\left(1-{e}^{-t}\right)},t\right).$
(2.2)

By (1.17) and (2.2), we get

$t{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=n{T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right).$
(2.3)

From (2.1), we can easily derive the following equation

$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)& =\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right){B}_{l}^{\left(k\right)}\left(x\right)\\ =\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(x|\lambda \right){B}_{l}^{\left(k\right)}.\end{array}$
(2.4)

By (1.16) and (2.2), we get

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\frac{1}{{g}_{r,k}\left(t\right)}{x}^{n}={\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}.$
(2.5)

In , it is known that

$\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}=\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(x-j\right)}^{n}.$
(2.6)

Thus, by (2.5) and (2.6), we get

$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)& ={\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}\\ =\sum _{m=0}^{\mathrm{\infty }}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}{\left(x-j\right)}^{n}\\ =\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right){H}_{n}^{\left(r\right)}\left(x-j|\lambda \right).\end{array}$
(2.7)

By (1.1), we easily see that

${H}_{n}^{\left(r\right)}\left(x|\lambda \right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right){x}^{l}.$
(2.8)

Therefore, by (2.7) and (2.8), we obtain the following theorem.

Theorem 2.1 For $r,k\in \mathbb{Z}$, $n\ge 0$, we have

$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)& =\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right){\left(x-j\right)}^{l}\\ =\sum _{l=0}^{n}\left\{\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right)\sum _{m=0}^{\mathrm{\infty }}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)\right\}{\left(x-j\right)}^{l}.\end{array}$

In , it is known that

$\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}=\sum _{j=0}^{n}\left\{\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right\}{x}^{j}.$
(2.9)

By (2.5) and (2.9), we get

$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)& ={\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}\\ =\sum _{j=0}^{n}\left\{\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right\}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}{x}^{j}\\ =\sum _{j=0}^{n}\left\{\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{n-m-j}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{j}\right)m!{S}_{2}\left(n-j,m\right)\right\}{H}_{j}^{\left(r\right)}\left(x|\lambda \right).\end{array}$
(2.10)

Therefore, by (2.8) and (2.10), we obtain the following theorem.

Theorem 2.2 For $r,k\in \mathbb{Z}$, $n\in {\mathbb{Z}}_{\ge 0}$, we have

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{l=0}^{n}\left\{\sum _{j=l}^{n}\sum _{m=0}^{n-j}{\left(-1\right)}^{n-m-j}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{j}{l}\right)\frac{m!}{{\left(m+1\right)}^{k}}{H}_{j-l}^{\left(r\right)}\left(\lambda \right){S}_{2}\left(n-j,m\right)\right\}{x}^{l}.$

From (1.19) and (2.2), we have

${T}_{n+1}^{\left(r,k\right)}\left(x|\lambda \right)=\left(x-\frac{{g}_{r,k}^{\mathrm{\prime }}\left(t\right)}{{g}_{r,k}\left(t\right)}\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right).$
(2.11)

Now, we note that

$\begin{array}{rl}\frac{{g}_{r,k}^{\mathrm{\prime }}\left(t\right)}{{g}_{r,k}\left(t\right)}& ={\left(log{g}_{r,k}\left(t\right)\right)}^{\prime }\\ ={\left(rlog\left({e}^{t}-\lambda \right)-rlog\left(1-\lambda \right)+log\left(1-{e}^{-t}\right)-logL{i}_{k}\left(1-{e}^{t}\right)\right)}^{\prime }\\ =r+\frac{r\lambda }{{e}^{t}\lambda }+\left(\frac{t}{{e}^{t}-1}\right)\frac{L{i}_{k}\left(1-{e}^{-t}\right)-L{i}_{k-1}\left(1-{e}^{-t}\right)}{tL{i}_{k}\left(1-{e}^{-t}\right)}.\end{array}$
(2.12)

By (2.11) and (2.12), we get

$\begin{array}{rcl}{T}_{n+1}^{\left(r,k\right)}\left(x|\lambda \right)& =& x{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-r{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r+1}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}\\ -{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)-L{i}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}\left(\frac{t}{{e}^{t}-1}\right){x}^{n}\\ =& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)-L{i}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{x}^{l}.\end{array}$
(2.13)

It is easy to show that

$\begin{array}{rl}\frac{L{i}_{k}\left(1-{e}^{-t}\right)-L{i}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}& =\frac{1}{1-{e}^{-t}}\sum _{n=1}^{\mathrm{\infty }}\left\{\frac{{\left(1-{e}^{-t}\right)}^{n}}{{n}^{k}}-\frac{{\left(1-{e}^{-t}\right)}^{n}}{{n}^{k-1}}\right\}\\ =\left(\frac{1-{e}^{-t}}{{2}^{k}}-\frac{1-{e}^{-t}}{{2}^{k-1}}\right)+\cdots \\ =\left(\frac{1}{{2}^{k}}-\frac{1}{{2}^{k-1}}\right)t+\cdots .\end{array}$
(2.14)

For any delta series $f\left(t\right)$, we have

$\frac{f\left(t\right)}{t}{x}^{n}=f\left(t\right)\frac{1}{n+1}{x}^{n+1}.$
(2.15)

Thus, by (2.13), (2.14) and (2.15), we get

$\begin{array}{rcl}{T}_{n+1}^{\left(r,k\right)}\left(x|\lambda \right)& =& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}\frac{1}{l+1}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)-L{i}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{l+1}\\ =& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\sum _{l=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{l}\right)}{l+1}{B}_{n-l}\left\{{T}_{l+1}^{\left(r,k\right)}\left(x|\lambda \right)-{T}_{l+1}^{\left(r,k-1\right)}\left(x|\lambda \right)\right\}\\ =& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\frac{1}{n+1}\sum _{l=1}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{n+1-l}\left\{{T}_{l}^{\left(r,k\right)}\left(x|\lambda \right)-{T}_{l}^{\left(r,k-1\right)}\left(x|\lambda \right)\right\}\\ =& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{n+1-l}\left\{{T}_{l}^{\left(r,k\right)}\left(x|\lambda \right)-{T}_{l}^{\left(r,k-1\right)}\left(x|\lambda \right)\right\}\\ =& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left\{{T}_{n+1-l}^{\left(r,k\right)}\left(x|\lambda \right)-{T}_{n+1-l}^{\left(r,k-1\right)}\left(x|\lambda \right)\right\}.\end{array}$
(2.16)

Therefore, by (2.16), we obtain the following theorem.

Theorem 2.3 For $r,k\in \mathbb{Z}$, $n\in {\mathbb{Z}}_{\ge 0}$, we have

$\begin{array}{rl}{T}_{n+1}^{\left(r,k\right)}\left(x|\lambda \right)=& \left(x-r\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left\{{T}_{n+1-l}^{\left(r,k\right)}\left(x|\lambda \right)-{T}_{n+1-l}^{\left(r,k-1\right)}\left(x|\lambda \right)\right\}.\end{array}$

Remark 1 If $r=0$, then we have

$\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}=\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{\left(1-{e}^{-t}\right)}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{T}_{n}^{\left(0,k\right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!}.$
(2.17)

Thus, by (2.17), we get ${B}_{n}^{\left(k\right)}\left(x\right)={T}_{n}^{\left(0,k\right)}\left(x|\lambda \right)$.

From (2.4), we have

$\begin{array}{rcl}tx{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)& =& t\left(x\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right){B}_{l}^{\left(k\right)}\left(x\right)\right)\\ =& \sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right)\left\{lx{B}_{l-1}^{\left(k\right)}\left(x\right)+{B}_{l}^{\left(k\right)}\left(x\right)\right\}\\ =& nx\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{l}\right){H}_{n-1-l}^{\left(r\right)}\left(\lambda \right){B}_{l}^{\left(k\right)}\left(x\right)+\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{n-l}^{\left(r\right)}\left(\lambda \right){B}_{l}^{\left(k\right)}\left(x\right)\\ =& nx{T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right)+{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right).\end{array}$
(2.18)

Applying t on both sides of Theorem 2.3, we get

$\begin{array}{r}\left(n+1\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\\ \phantom{\rule{1em}{0ex}}=nx{T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right)+{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)-rn{T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{rn\lambda }{1-\lambda }{T}_{n-1}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ \phantom{\rule{2em}{0ex}}-\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left\{\left(n+1-l\right){T}_{n-l}^{\left(r,k\right)}\left(x|\lambda \right)-\left(n+1-l\right){T}_{n-l}^{\left(r,k-1\right)}\left(x|\lambda \right)\right\}.\end{array}$
(2.19)

Thus, by (2.19), we have

$\begin{array}{r}\left(n+1\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)+n\left(r-\frac{1}{2}-x\right){T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right)+\sum _{l=0}^{n-2}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{T}_{l}^{\left(r,k\right)}\left(x|\lambda \right)\\ \phantom{\rule{1em}{0ex}}=-\frac{r\lambda n}{1-\lambda }{T}_{n-1}^{\left(r+1,k\right)}\left(x|\lambda \right)+\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{T}_{l}^{\left(r,k-1\right)}\left(x|\lambda \right).\end{array}$
(2.20)

Therefore, by (2.20), we obtain the following theorem.

Theorem 2.4 For $r,k\in \mathbb{Z}$, $n\in \mathbb{Z}$ with $n\ge 2$, we have

$\begin{array}{r}\left(n+1\right){T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)+n\left(r-\frac{1}{2}-x\right){T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right)+\sum _{l=0}^{n-2}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{T}_{l}^{\left(r,k\right)}\left(x|\lambda \right)\\ \phantom{\rule{1em}{0ex}}=-\frac{r\lambda n}{1-\lambda }{T}_{n-1}^{\left(r+1,k\right)}\left(x|\lambda \right)+\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{T}_{l}^{\left(r,k-1\right)}\left(x|\lambda \right).\end{array}$

From (1.14) and (2.5), we note that

$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(y|\lambda \right)& =〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n}〉\\ =〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|x{x}^{n-1}〉.\end{array}$
(2.21)

By (1.15) and (2.21), we get

$\begin{array}{rcl}{T}_{n}^{\left(r,k\right)}\left(y|\lambda \right)& =& 〈{\partial }_{t}\left({\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}\right)|{x}^{n-1}〉\\ =& 〈\left({\partial }_{t}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\right)\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-1}〉\\ +〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\left({\partial }_{t}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉\\ +〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{\partial }_{t}{e}^{yt}|{x}^{n-1}〉.\end{array}$
(2.22)

Therefore, by (2.22), we obtain the following theorem.

Theorem 2.5 For $r,k\in \mathbb{Z}$, $n\ge 1$, we have

$\begin{array}{rl}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=& \left(x-r\right){T}_{n-1}^{\left(r,k\right)}\left(x|\lambda \right)-\frac{r\lambda }{1-\lambda }{T}_{n-1}^{\left(r+1,k\right)}\left(x|\lambda \right)\\ +\sum _{l=0}^{n-1}\left\{{\left(-1\right)}^{n-1-l}\left(\genfrac{}{}{0}{}{n-1}{l}\right)\sum _{m=0}^{n-1-l}{\left(-1\right)}^{m}\frac{\left(m+1\right)!}{{\left(m+2\right)}^{k}}{S}_{2}\left(n-1-l,m\right)\right\}{H}_{l}^{\left(r\right)}\left(x-1|\lambda \right).\end{array}$

Now, we compute $〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}L{i}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉$ in two different ways.

On the one hand,

$\begin{array}{r}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}L{i}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|\left(1-{e}^{-t}\right){x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n+1}-{\left(x-1\right)}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(-1\right)}^{n-m}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{m}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(-1\right)}^{n-m}〈1|{T}_{m}^{\left(r,k\right)}\left(x|\lambda \right)〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(-1\right)}^{n-m}{T}_{m}^{\left(r,k\right)}\left(\lambda \right).\end{array}$
(2.23)

On the other hand, we get

$\begin{array}{r}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}L{i}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈L{i}_{k}\left(1-{e}^{-t}\right)|{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}{x}^{n+1}〉\\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}{\left(L{i}_{k}\left(1-{e}^{-s}\right)\right)}^{\prime }\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|\lambda \right)〉\\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}{e}^{-s}\frac{L{i}_{k}\left(1-{e}^{-s}\right)}{\left(1-{e}^{-s}\right)}\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|\lambda \right)〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\sum _{m=0}^{l}\left(\genfrac{}{}{0}{}{l}{m}\right){\left(-1\right)}^{l-m}{B}_{m}^{\left(k-1\right)}\right)\frac{1}{l!}〈{\int }_{0}^{t}{s}^{l}\phantom{\rule{0.2em}{0ex}}ds|{H}_{n+1}^{\left(r\right)}\left(x|\lambda \right)〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}\left(\genfrac{}{}{0}{}{l}{m}\right){\left(-1\right)}^{l-m}\frac{{B}_{m}^{\left(k-1\right)}}{\left(l+1\right)!}〈{t}^{l+1}|{H}_{n+1}^{\left(r\right)}\left(x|\lambda \right)〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){\left(-1\right)}^{l-m}{B}_{m}^{\left(k-1\right)}{H}_{n-l}^{\left(r\right)}\left(\lambda \right).\end{array}$
(2.24)

Therefore, by (2.23) and (2.24), we obtain the following theorem.

Theorem 2.6 For $r,k\in \mathbb{Z}$, $n\in {\mathbb{Z}}_{\ge 0}$, we have

$\begin{array}{r}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{m}\right){\left(-1\right)}^{n-m}{T}_{m}^{\left(r,k\right)}\left(\lambda \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){B}_{m}^{\left(k-1\right)}{H}_{n-l}^{\left(r\right)}\left(\lambda \right).\end{array}$

Now, we consider the following two Sheffer sequences:

$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}\frac{1-{e}^{-t}}{L{i}_{k}\left(1-{e}^{-t}\right)},t\right),\\ {\mathbb{B}}^{\left(s\right)}\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{s},t\right),\end{array}$
(2.25)

where $s\in {\mathbb{Z}}_{\ge 0}$, $r,k\in \mathbb{Z}$ and $\lambda \in \mathbb{C}$ with $\lambda \ne 1$. Let us assume that

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}{C}_{n\cdot m}{\mathbb{B}}_{m}^{\left(s\right)}\left(x\right).$
(2.26)

By (1.21) and (2.26), we get

$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{{e}^{t}-1}{t}\right)}^{s}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{t}^{m}|{x}^{n}〉\\ =& \frac{1}{m!}〈{\left(\frac{{e}^{t}-1}{t}\right)}^{s}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{t}^{m}{x}^{n}〉\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)〈{\left(\frac{{e}^{t}-1}{t}\right)}^{s}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n-m}〉\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!}{\left(l+s\right)!}{S}_{2}\left(l+s,s\right)〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{t}^{l}{x}^{n-m}〉\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!l!}{\left(l+s\right)!}\frac{{\left(n-m\right)}_{l}}{l!}{S}_{2}\left(l+s,s\right)〈1|{T}_{n-m-l}^{\left(r,k\right)}\left(x|\lambda \right)〉\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{s+l}{l}\right)}{S}_{2}\left(l+s,s\right){T}_{n-m-l}^{\left(r,k\right)}\left(\lambda \right).\end{array}$
(2.27)

Therefore, by (2.26) and (2.27), we obtain the following theorem.

Theorem 2.7 For $r,k\in \mathbb{Z}$, $s\in {\mathbb{Z}}_{\ge 0}$, we have

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}\left\{\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{s+l}{l}\right)}{S}_{2}\left(l+s,s\right){T}_{n-m-l}^{\left(r,k\right)}\left(\lambda \right)\right\}{\mathbb{B}}_{m}^{\left(s\right)}\left(x\right).$

From (1.3) and (2.1), we note that

$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}\frac{1-{e}^{-t}}{L{i}_{k}\left(1-{e}^{-t}\right)},t\right),\\ {E}_{n}^{\left(r,s\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}+1}{2}\right)}^{s},t\right),\end{array}$
(2.28)

where $r,k\in \mathbb{Z}$, $s\in {\mathbb{Z}}_{\ge 0}$.

By the same method, we get

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\frac{1}{{2}^{s}}\sum _{m=0}^{n}\left\{\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{j=0}^{s}\left(\genfrac{}{}{0}{}{s}{j}\right){T}_{n-m}^{\left(r,k\right)}\left(j\right)\right\}{E}_{m}^{\left(s\right)}\left(x\right).$
(2.29)

From (1.1) and (2.1), we note that

$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}\frac{1-{e}^{-t}}{L{i}_{k}\left(1-{e}^{-t}\right)},t\right),\\ {H}_{n}^{\left(s\right)}\left(x|\mu \right)\sim \left({\left(\frac{{e}^{t}-\mu }{1-\mu }\right)}^{s},t\right),\end{array}$
(2.30)

where $r,k\in \mathbb{Z}$, and $\lambda ,\mu \in \mathbb{C}$ with $\lambda \ne 1$, $\mu \ne 1$, $s\in {\mathbb{Z}}_{\ge 0}$.

Let us assume that

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}{C}_{n,m}{H}_{m}^{\left(s\right)}\left(x|\mu \right).$
(2.31)

By (1.21) and (2.31), we get

$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{{e}^{t}-\mu }{1-\mu }\right)}^{s}{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{t}^{m}|{x}^{n}〉\\ =& \frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\mu \right)}^{s}}〈{\left({e}^{t}-\mu \right)}^{s}|{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n-m}〉\\ =& \frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\mu \right)}^{s}}\sum _{j=0}^{s}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(-\mu \right)}^{s-j}〈{e}^{jt}|{T}_{n-m}^{\left(r,k\right)}\left(x|\lambda \right)〉\\ =& \frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\mu \right)}^{s}}\sum _{j=0}^{s}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(-\mu \right)}^{s-j}{T}_{n-m}^{\left(r,k\right)}\left(j|\lambda \right).\end{array}$
(2.32)

Therefore, by (2.31) and (2.32), we obtain the following theorem.

Theorem 2.8 For $r,k\in \mathbb{Z}$, $s\in {\mathbb{Z}}_{\ge 0}$, we have

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\frac{1}{{\left(1-\mu \right)}^{s}}\sum _{m=0}^{n}\left\{\left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{j=0}^{s}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(-\mu \right)}^{s-j}{T}_{n-m}^{\left(r,k\right)}\left(j|\lambda \right)\right\}{H}_{m}^{\left(s\right)}\left(x|\mu \right).$

It is known that

$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}\frac{1-{e}^{-t}}{L{i}_{k}\left(1-{e}^{-t}\right)},t\right),\\ {\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right).\end{array}$
(2.33)

Let

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}{C}_{n,m}{\left(x\right)}_{m}.$
(2.34)

Then, by (1.21) and (2.34), we get

$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{\left({e}^{t}-1\right)}^{m}|{x}^{n}〉\\ =& \sum _{l=0}^{\mathrm{\infty }}\frac{{S}_{2}\left(l+m,m\right)}{\left(l+m\right)!}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{t}^{m+l}{x}^{n}〉\\ =& \sum _{l=0}^{n-m}\frac{{S}_{2}\left(l+m,m\right)}{\left(l+m\right)!}{\left(n\right)}_{m+l}〈1|{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n-m-l}〉\\ =& \sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{2}\left(l+m,m\right){T}_{n-m-l}^{\left(r,k\right)}\left(\lambda \right).\end{array}$
(2.35)

Therefore, by (2.34) and (2.35), we obtain the following theorem.

Theorem 2.9 For $r,k\in \mathbb{Z}$, we have

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}\left\{\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{2}\left(l+m,m\right){T}_{n-m-l}^{\left(r,k\right)}\left(\lambda \right)\right\}{\left(x\right)}_{m}.$

Finally, we consider the following two Sheffer sequences:

$\begin{array}{r}{T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}\frac{1-{e}^{-t}}{L{i}_{k}\left(1-{e}^{-t}\right)},t\right),\\ {x}^{\left[n\right]}\sim \left(1,1-{e}^{-t}\right),\end{array}$
(2.36)

where ${x}^{\left[n\right]}=x\left(x+1\right)\cdots \left(x+n-1\right)$.

Let us assume that

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}{C}_{n,m}{x}^{\left[m\right]}.$
(2.37)

Then, by (1.21) and (2.37), we get

$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{\left(1-{e}^{-t}\right)}^{m}|{x}^{n}〉\\ =& \sum _{l=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{l}{S}_{2}\left(l+m,m\right)}{\left(l+m\right)!}〈{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{t}^{m+l}{x}^{n}〉\\ =& \sum _{l=0}^{n-m}\frac{{\left(-1\right)}^{l}{S}_{2}\left(l+m,m\right)}{\left(l+m\right)!}{\left(n\right)}_{m+l}〈1|{\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}\frac{L{i}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n-m-l}〉\\ =& \sum _{l=0}^{n-m}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{2}\left(l+m,m\right){T}_{n-m-l}^{\left(r,k\right)}\left(\lambda \right).\end{array}$
(2.38)

Therefore, by (2.37) and (2.38), we obtain the following theorem.

Theorem 2.10 For $r,k\in \mathbb{Z}$, $n\ge 0$, we have

${T}_{n}^{\left(r,k\right)}\left(x|\lambda \right)=\sum _{m=0}^{n}\left\{\sum _{l=0}^{n-m}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{2}\left(l+m,m\right){T}_{n-m-l}^{\left(r,k\right)}\left(\lambda \right)\right\}{x}^{\left[m\right]}.$

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## Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant, funded by the Korea government (MOE) (No. 2012R1A1A2003786).

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, D.S., Kim, T. Higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials. Adv Differ Equ 2013, 251 (2013). https://doi.org/10.1186/1687-1847-2013-251

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• DOI: https://doi.org/10.1186/1687-1847-2013-251

### Keywords

• Formal Power Series
• Linear Functional
• Bernoulli Number
• Bernoulli Polynomial
• Euler Polynomial 