Theory and Modern Applications

# Hermite and poly-Bernoulli mixed-type polynomials

## Abstract

In this paper, we consider Hermite and poly-Bernoulli mixed-type polynomials and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give various identities associated with Stirling numbers, Bernoulli and Frobenius-Euler polynomials of higher order.

## 1 Introduction

For $r\in {\mathbb{Z}}_{\ge 0}$, as is well known, the Bernoulli polynomials of order r are defined by the generating function to be

$\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathbb{B}}_{n}^{\left(r\right)}\left(x\right)}{n!}{t}^{n}={\left(\frac{t}{{e}^{t}-1}\right)}^{r}{e}^{xt}\phantom{\rule{1em}{0ex}}\left(\text{see [1–16]}\right).$
(1.1)

For $k\in \mathbb{Z}$, the polylogarithm is defined by

${Li}_{k}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{{n}^{k}}.$
(1.2)

Note that ${Li}_{1}\left(x\right)=-log\left(1-x\right)$.

The poly-Bernoulli polynomials are defined by the generating function to be

$\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [5, 8]}\right).$
(1.3)

When $x=0$, ${B}_{n}^{\left(k\right)}={B}_{n}^{\left(k\right)}\left(0\right)$ are called the poly-Bernoulli numbers (of index k).

For $\nu \phantom{\rule{0.25em}{0ex}}\left(\ne 0\right)\in \mathbb{R}$, the Hermite polynomials of order ν are given by the generating function to be

${e}^{-\frac{\nu {t}^{2}}{2}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(\nu \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [6, 12, 13]}\right).$
(1.4)

When $x=0$, ${H}_{n}^{\left(\nu \right)}={H}_{n}^{\left(\nu \right)}\left(0\right)$ are called the Hermite numbers of order ν.

In this paper, we consider the Hermite and poly-Bernoulli mixed-type polynomials $H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)$ which are defined by the generating function to be

${e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)\frac{{t}^{n}}{n!},$
(1.5)

where $k\in \mathbb{Z}$ and $\nu \phantom{\rule{0.25em}{0ex}}\left(\ne 0\right)\in \mathbb{R}$.

When $x=0$, $H{B}_{n}^{\left(\nu ,k\right)}=H{B}_{n}^{\left(\nu ,k\right)}\left(0\right)$ are called the Hermite and poly-Bernoulli mixed-type numbers.

Let be the set of all formal power series in the variable t over as follows:

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

Let $\mathbb{P}=\mathbb{C}\left[x\right]$ and ${\mathbb{P}}^{\ast }$ denote the vector space of all linear functionals on .

$〈L|p\left(x\right)〉$ denotes 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

$〈f\left(t\right)|{x}^{n}〉={a}_{n}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(1.7)

Then, by (1.6) and (1.7), we get

$〈{t}^{k}|{x}^{n}〉=n!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(n,k\ge 0\right),$
(1.8)

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 $O\left(f\right)$ of the power series $f\left(t\right)\ne 0$ is the smallest integer for which ${a}_{k}$ does not vanish. If $O\left(f\right)=0$, then $f\left(t\right)$ is called an invertible series. If $O\left(f\right)=1$, then $f\left(t\right)$ is called a delta series. For $f\left(t\right),g\left(t\right)\in \mathcal{F}$, we have

$〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈f\left(t\right)|g\left(t\right)p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉.$
(1.9)

Let $f\left(t\right)\in \mathcal{F}$ and $p\left(x\right)\in \mathbb{P}$. Then we have

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f\left(t\right)|{x}^{k}〉}{k!}{t}^{k},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈{t}^{k}|p\left(x\right)〉}{k!}{x}^{k}\phantom{\rule{1em}{0ex}}\left(\text{see [8, 9, 11, 13, 14]}\right).$
(1.10)

By (1.10), we get

${p}^{\left(k\right)}\left(0\right)=〈{t}^{k}|p\left(x\right)〉=〈1|{p}^{\left(k\right)}\left(x\right)〉,$
(1.11)

where ${p}^{\left(k\right)}\left(0\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}{|}_{x=0}$.

From (1.11), we have

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\left(\text{see [8, 9, 13]}\right).$
(1.12)

By (1.12), we easily get

${e}^{yt}p\left(x\right)=p\left(x+y\right),\phantom{\rule{2em}{0ex}}〈{e}^{yt}|p\left(x\right)〉=p\left(y\right).$
(1.13)

For $O\left(f\left(t\right)\right)=1$, $O\left(g\left(t\right)\right)=0$, there exists a unique sequence ${s}_{n}\left(x\right)$ of polynomials such that $〈g\left(t\right)f{\left(t\right)}^{k}|{x}^{n}〉=n!{\delta }_{n,k}$ ($n,k\ge 0$).

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 ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$.

Let $p\left(x\right)\in \mathbb{P}$, $f\left(t\right)\in \mathcal{F}$. Then we see that

$〈f\left(t\right)|xp\left(x\right)〉=〈{\partial }_{t}f\left(t\right)|p\left(x\right)〉=〈\frac{df\left(t\right)}{dt}|p\left(x\right)〉.$
(1.14)

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we have the following equations:

$h\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈h\left(t\right)|{s}_{k}\left(x\right)〉}{k!}g\left(t\right)f{\left(t\right)}^{k},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈g\left(t\right)f{\left(t\right)}^{k}|p\left(x\right)〉}{k!}{s}_{k}\left(x\right),$
(1.15)

where $h\left(t\right)\in \mathcal{F}$, $p\left(x\right)\in \mathbb{P}$,

$\frac{1}{g\left(\overline{f}\left(t\right)\right)}{e}^{y\overline{f}\left(t\right)}=\sum _{n=0}^{\mathrm{\infty }}{s}_{n}\left(y\right)\frac{{t}^{n}}{n!},$
(1.16)

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

(1.17)
$f\left(t\right){s}_{n}\left(x\right)=n{s}_{n-1}\left(x\right),\phantom{\rule{2em}{0ex}}{s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right),$
(1.18)

and the conjugate representation is given by

${s}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^{j}|{x}^{n}〉{x}^{j}.$
(1.19)

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

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

where

${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}l{\left(\overline{f}\left(t\right)\right)}^{m}|{x}^{n}〉\phantom{\rule{1em}{0ex}}\left(\text{see [8, 9, 13]}\right).$
(1.21)

In this paper, we consider Hermite and poly-Bernoulli mixed-type polynomials and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give various identities associated with Bernoulli and Frobenius-Euler polynomials of higher order.

## 2 Hermite and poly-Bernoulli mixed-type polynomials

From (1.5) and (1.16), we note that

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)\sim \left({e}^{\frac{\nu {t}^{2}}{2}}\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},t\right),$
(2.1)

and, by (1.3), (1.4) and (1.16), we get

${B}_{n}^{\left(k\right)}\left(x\right)\sim \left(\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},t\right),$
(2.2)
(2.3)

From (1.18), (2.1), (2.2) and (2.3), we have

$t{B}_{n}^{\left(k\right)}\left(x\right)=n{B}_{n-1}^{\left(k\right)}\left(x\right),\phantom{\rule{2em}{0ex}}t{H}_{n}^{\left(\nu \right)}\left(x\right)=n{H}_{n-1}^{\left(\nu \right)}\left(x\right),\phantom{\rule{2em}{0ex}}tH{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=nH{B}_{n-1}^{\left(\nu ,k\right)}\left(x\right).$
(2.4)

By (1.5), (1.8) and (2.1), we get

$\begin{array}{rcl}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)& =& {e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}={e}^{-\frac{\nu {t}^{2}}{2}}{B}_{n}^{\left(k\right)}\left(x\right)\\ =& \sum _{m=0}^{\left[\frac{n}{2}\right]}\frac{1}{m!}{\left(-\frac{\nu }{2}\right)}^{m}{\left(n\right)}_{2m}{B}_{n-2m}^{\left(k\right)}\left(x\right)\\ =& \sum _{m=0}^{\left[\frac{n}{2}\right]}\left(\genfrac{}{}{0}{}{n}{2m}\right)\frac{\left(2m\right)!}{m!}{\left(-\frac{\nu }{2}\right)}^{m}{B}_{n-2m}^{\left(k\right)}\left(x\right).\end{array}$
(2.5)

Therefore, by (2.5), we obtain the following proposition.

Proposition 1 For $n\ge 0$, we have

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=\sum _{m=0}^{\left[\frac{n}{2}\right]}\left(\genfrac{}{}{0}{}{n}{2m}\right)\frac{\left(2m\right)!}{m!}{\left(-\frac{\nu }{2}\right)}^{m}{B}_{n-2m}^{\left(k\right)}\left(x\right).$

From (1.5), we can also derive

$\begin{array}{rcl}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)& =& \frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{-\frac{\nu {t}^{2}}{2}}{x}^{n}=\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{H}_{n}^{\left(\nu \right)}\left(x\right)=\sum _{m=0}^{\mathrm{\infty }}\frac{{\left(1-{e}^{-t}\right)}^{m}}{{\left(m+1\right)}^{k}}{H}_{n}^{\left(\nu \right)}\left(x\right)\\ =& \sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(-1\right)}^{j}{e}^{-jt}{H}_{n}^{\left(\nu \right)}\left(x\right)\\ =& \sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(-1\right)}^{j}{H}_{n}^{\left(\nu \right)}\left(x-j\right).\end{array}$
(2.6)

Therefore, by (2.6), we obtain the following theorem.

Theorem 2 For $n\ge 0$, we have

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(-1\right)}^{j}{H}_{n}^{\left(\nu \right)}\left(x-j\right).$

By (1.5), we get

$\begin{array}{rcl}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)& =& {e}^{-\frac{\nu {t}^{2}}{2}}{B}_{n}^{\left(k\right)}\left(x\right)=\sum _{l=0}^{\mathrm{\infty }}\frac{1}{l!}{\left(-\frac{\nu }{2}\right)}^{l}{t}^{2l}{B}_{n}^{\left(k\right)}\left(x\right)\\ =& \sum _{l=0}^{\left[\frac{n}{2}\right]}\frac{1}{l!}{\left(-\frac{\nu }{2}\right)}^{l}\sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{j=0}^{m}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right){t}^{2l}{\left(x-j\right)}^{n}\\ =& \sum _{l=0}^{\left[\frac{n}{2}\right]}\sum _{j=0}^{n}\left\{\sum _{m=j}^{n}\left(\genfrac{}{}{0}{}{n}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(-\frac{\nu }{2}\right)}^{l}\frac{{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)}{{\left(m+1\right)}^{k}}\right\}{\left(x-j\right)}^{n-2l}.\end{array}$
(2.7)

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

Theorem 3 For $n\ge 0$, we have

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=\sum _{l=0}^{\left[\frac{n}{2}\right]}\sum _{j=0}^{n}\left\{\sum _{m=j}^{n}\left(\genfrac{}{}{0}{}{n}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(-\frac{\nu }{2}\right)}^{l}\frac{{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)}{{\left(m+1\right)}^{k}}\right\}{\left(x-j\right)}^{n-2l}.$

By (2.6), we get

$\begin{array}{rcl}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)& =& \sum _{m=0}^{n}\frac{{\left(1-{e}^{-t}\right)}^{m}}{{\left(m+1\right)}^{k}}{H}_{n}^{\left(\nu \right)}\left(x\right)\\ =& \sum _{m=0}^{n}\frac{1}{{\left(m+1\right)}^{k}}\sum _{a=0}^{n-m}\frac{m!}{\left(a+m\right)!}{\left(-1\right)}^{a}{S}_{2}\left(a+m,m\right){\left(n\right)}_{a+m}{H}_{n-a-m}^{\left(\nu \right)}\left(x\right)\\ =& \sum _{m=0}^{n}\sum _{a=0}^{n-m}\frac{{\left(-1\right)}^{n-a-m}m!}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{n-a}\right){S}_{2}\left(n-a,m\right){H}_{a}^{\left(\nu \right)}\left(x\right)\\ =& {\left(-1\right)}^{n}\sum _{a=0}^{n}\left\{\sum _{m=0}^{n-a}\frac{{\left(-1\right)}^{m+a}m!}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{a}\right){S}_{2}\left(n-a,m\right)\right\}{H}_{a}^{\left(\nu \right)}\left(x\right),\end{array}$
(2.8)

where ${S}_{2}\left(n,m\right)$ is the Stirling number of the second kind.

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

Theorem 4 For $n\ge 0$, we have

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)={\left(-1\right)}^{n}\sum _{a=0}^{n}\left\{\sum _{m=0}^{n-a}\frac{{\left(-1\right)}^{a+m}m!}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{a}\right){S}_{2}\left(n-a,m\right)\right\}{H}_{a}^{\left(\nu \right)}\left(x\right).$

From (1.19) and (2.1), we have

$\begin{array}{rcl}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)& =& \sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n-j}〉{x}^{j}\\ =& \sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)〈{e}^{-\frac{\nu {t}^{2}}{2}}|{B}_{n-j}^{\left(k\right)}\left(x\right)〉{x}^{j}\\ =& \sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{l=0}^{\left[\frac{n-j}{2}\right]}\frac{{\left(-\frac{\nu }{2}\right)}^{l}}{l!}{\left(n-j\right)}_{2l}〈1|{B}_{n-j-2l}^{\left(k\right)}\left(x\right)〉{x}^{j}\\ =& \sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{l=0}^{\left[\frac{n-j}{2}\right]}\frac{1}{l!}{\left(-\frac{\nu }{2}\right)}^{l}{\left(n-j\right)}_{2l}{B}_{n-j-2l}^{\left(k\right)}{x}^{j}\\ =& \sum _{j=0}^{n}\left\{\sum _{l=0}^{\left[\frac{n-j}{2}\right]}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(-\frac{\nu }{2}\right)}^{l}{B}_{n-j-2l}^{\left(k\right)}\right\}{x}^{j}.\end{array}$
(2.9)

Therefore, by (2.9), we obtain the following theorem.

Theorem 5 For $n\ge 0$, we have

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=\sum _{j=0}^{n}\left\{\sum _{l=0}^{\left[\frac{n-j}{2}\right]}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(-\frac{\nu }{2}\right)}^{l}{B}_{n-j-2l}^{\left(k\right)}\right\}{x}^{j}.$

Remark By (1.17) and (2.1), we easily get

$H{B}_{n}^{\left(\nu ,k\right)}\left(x+y\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)H{B}_{j}^{\left(\nu ,k\right)}\left(x\right){y}^{n-j}.$
(2.10)

We note that

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)\sim \left(g\left(t\right)={e}^{\frac{\nu {t}^{2}}{2}}\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},f\left(t\right)=t\right).$
(2.11)

From (1.18) and (2.11), we have

$H{B}_{n+1}^{\left(\nu ,k\right)}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)H{B}_{n}^{\left(\nu ,k\right)}\left(x\right).$
(2.12)

Now, we observe that

$\begin{array}{rcl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}& =& {\left(log\left(g\left(t\right)\right)\right)}^{\prime }\\ =& {\left(log{e}^{\frac{\nu {t}^{2}}{2}}+log\left(1-{e}^{-t}\right)-log\left({Li}_{k}\left(1-{e}^{-t}\right)\right)\right)}^{\prime }\\ =& \nu t+\frac{{e}^{-t}}{1-{e}^{-t}}\left(1-\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)}{{Li}_{k}\left(1-{e}^{-t}\right)}\right).\end{array}$
(2.13)

By (2.12) and (2.13), we get

$\begin{array}{c}H{B}_{n+1}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=xH{B}_{n}^{\left(\nu ,k\right)}\left(x\right)-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=xH{B}_{n}^{\left(\nu ,k\right)}\left(x\right)-\nu nH{B}_{n-1}^{\left(\nu ,k\right)}\left(x\right)-{e}^{-\frac{\nu {t}^{2}}{2}}\frac{t}{{e}^{t}-1}\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{x}^{n}.\hfill \end{array}$
(2.14)

It is easy to show that

$\begin{array}{rcl}\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}& =& \sum _{m=2}^{\mathrm{\infty }}\left(\frac{1}{{m}^{k}}-\frac{1}{{m}^{k-1}}\right){\left(1-{e}^{-t}\right)}^{m-1}\\ =& \left(\frac{1}{{2}^{k}}-\frac{1}{{2}^{k-1}}\right)t+\cdots .\end{array}$
(2.15)

Thus, by (2.15), we get

$\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{x}^{n}=\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\frac{{x}^{n+1}}{n+1}.$
(2.16)

From (2.16), we can derive

$\begin{array}{c}{e}^{-\frac{\nu {t}^{2}}{2}}\frac{t}{{e}^{t}-1}\frac{{Li}_{k}\left(1-{e}^{-t}\right)-{Li}_{k-1}\left(1-{e}^{-t}\right)}{t\left(1-{e}^{-t}\right)}{x}^{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\left(\sum _{l=0}^{\mathrm{\infty }}\frac{{B}_{l}}{l!}{t}^{l}\right)\left(H{B}_{n+1}^{\left(\nu ,k\right)}\left(x\right)-H{B}_{n+1}^{\left(\nu ,k-1\right)}\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\sum _{l=0}^{n+1}\frac{{B}_{l}}{l!}{t}^{l}\left(H{B}_{n+1}^{\left(\nu ,k\right)}\left(x\right)-H{B}_{n+1}^{\left(\nu ,k-1\right)}\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left(H{B}_{n+1-l}^{\left(\nu ,k\right)}\left(x\right)-H{B}_{n+1-l}^{\left(\nu ,k-1\right)}\left(x\right)\right).\hfill \end{array}$
(2.17)

Therefore, by (2.14) and (2.17), we obtain the following theorem.

Theorem 6 For $n\ge 0$, we have

$\begin{array}{c}H{B}_{n+1}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=xH{B}_{n}^{\left(\nu ,k\right)}\left(x\right)-\nu nH{B}_{n-1}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left\{H{B}_{n+1-l}^{\left(\nu ,k\right)}\left(x\right)-H{B}_{n+1-l}^{\left(\nu ,k-1\right)}\left(x\right)\right\}.\hfill \end{array}$
(2.18)

Let us take t on the both sides of (2.18). Then we have

$\begin{array}{c}\left(n+1\right)H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(xt+1\right)H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)-\nu n\left(n-1\right)H{B}_{n-2}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0}{}{n+1}{l}\right)\left(n+1-l\right){B}_{l}\left\{H{B}_{n-l}^{\left(\nu ,k\right)}\left(x\right)-H{B}_{n-l}^{\left(\nu ,k-1\right)}\left(x\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=nxH{B}_{n-1}^{\left(\nu ,k\right)}\left(x\right)+H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)-\nu n\left(n-1\right)H{B}_{n-2}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}\left(H{B}_{n-l}^{\left(\nu ,k\right)}\left(x\right)-H{B}_{n-l}^{\left(\nu ,k-1\right)}\left(x\right)\right),\hfill \end{array}$
(2.19)

where $n\ge 3$.

Thus, by (2.19), we obtain the following theorem.

Theorem 7 For $n\ge 3$, we have

$\begin{array}{c}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}H{B}_{n-l}^{\left(\nu ,k-1\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(n+1\right)H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)-n\left(x+\frac{1}{2}\right)H{B}_{n-1}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}+n\left(n-1\right)\left(\nu +\frac{1}{12}\right)H{B}_{n-2}^{\left(\nu ,k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{l=0}^{n-3}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}H{B}_{l}^{\left(\nu ,k\right)}\left(x\right).\hfill \end{array}$

By (1.5) and (1.8), we get

$\begin{array}{c}H{B}_{n}^{\left(\nu ,k\right)}\left(y\right)\hfill \\ \phantom{\rule{1em}{0ex}}=〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\left({e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}\right)|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\left({\partial }_{t}{e}^{-\frac{\nu {t}^{2}}{2}}\right)\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-1}〉\hfill \\ \phantom{\rule{2em}{0ex}}+〈{e}^{-\frac{\nu {t}^{2}}{2}}\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉\hfill \\ \phantom{\rule{2em}{0ex}}+〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\left({\partial }_{t}{e}^{yt}\right)|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=-\nu \left(n-1\right)〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-2}〉\hfill \\ \phantom{\rule{2em}{0ex}}+y〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-1}〉\hfill \\ \phantom{\rule{2em}{0ex}}+〈{e}^{-\frac{\nu {t}^{2}}{2}}\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=-\nu \left(n-1\right)H{B}_{n-2}^{\left(\nu ,k\right)}\left(y\right)+yH{B}_{n-1}^{\left(\nu ,k\right)}\left(y\right)\hfill \\ \phantom{\rule{2em}{0ex}}+〈{e}^{-\frac{\nu {t}^{2}}{2}}\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉.\hfill \end{array}$
(2.20)

Now, we observe that

${\partial }_{t}\left(\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right)=\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{{\left(1-{e}^{-t}\right)}^{2}}{e}^{-t}.$
(2.21)

From (2.21), we have

$\begin{array}{c}〈{e}^{-\frac{\nu {t}^{2}}{2}}\left({\partial }_{t}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\right){e}^{yt}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{e}^{-\frac{\nu {t}^{2}}{2}}\left(\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{{\left(1-{e}^{-t}\right)}^{2}}\right){e}^{-t}{e}^{yt}|\frac{1}{n}t{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|\frac{t}{{e}^{t}-1}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{B}_{n}\left(x\right)〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k-1}\left(1-{e}^{-t}\right)-{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{e}^{yt}|{x}^{n-l}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}\left\{H{B}_{n-l}^{\left(\nu ,k-1\right)}\left(y\right)-H{B}_{n-l}^{\left(\nu ,k\right)}\left(y\right)\right\},\hfill \end{array}$
(2.22)

where ${B}_{n}$ are the ordinary Bernoulli numbers which are defined by the generating function to be

$\frac{t}{{e}^{t}-1}=\sum _{n=0}^{\mathrm{\infty }}\frac{{B}_{n}}{n!}{t}^{n}.$

Therefore, by (2.20) and (2.22), we obtain the following theorem.

Theorem 8 For $n\ge 2$, we have

$\begin{array}{rcl}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)& =& -\nu \left(n-1\right)H{B}_{n-2}^{\left(\nu ,k\right)}\left(x\right)+xH{B}_{n-1}^{\left(\nu ,k\right)}\left(x\right)\\ +\frac{1}{n}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}\left(H{B}_{n-l}^{\left(\nu ,k-1\right)}\left(x\right)-H{B}_{n-l}^{\left(\nu ,k\right)}\left(x\right)\right).\end{array}$

Now, we compute

$〈{e}^{-\frac{\nu {t}^{2}}{2}}{Li}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉$

in two different ways.

On the one hand,

$\begin{array}{c}〈{e}^{-\frac{\nu {t}^{2}}{2}}{Li}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|\left(1-{e}^{-t}\right){x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{n+1}-{\left(x-1\right)}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}{\left(-1\right)}^{n-m}\left(\genfrac{}{}{0}{}{n+1}{m}\right)〈{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}|{x}^{m}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}{\left(-1\right)}^{n-m}\left(\genfrac{}{}{0}{}{n+1}{m}\right)H{B}_{m}^{\left(\nu ,k\right)}.\hfill \end{array}$
(2.23)

On the other hand,

$\begin{array}{c}〈{e}^{-\frac{\nu {t}^{2}}{2}}{Li}_{k}\left(1-{e}^{-t}\right)|{x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{Li}_{k}\left(1-{e}^{-t}\right)|{e}^{-\frac{\nu {t}^{2}}{2}}{x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}{\left({Li}_{k}\left(1-{e}^{-s}\right)\right)}^{\prime }\phantom{\rule{0.2em}{0ex}}ds|{e}^{-\frac{\nu {t}^{2}}{2}}{x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\int }_{0}^{t}{e}^{-s}\frac{{Li}_{k-1}\left(1-{e}^{-s}\right)}{1-{e}^{-s}}\phantom{\rule{0.2em}{0ex}}ds|{e}^{-\frac{\nu {t}^{2}}{2}}{x}^{n+1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\sum _{l=0}^{\mathrm{\infty }}\left(\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right){B}_{m}^{\left(k-1\right)}\frac{{t}^{l+1}}{\left(l+1\right)!}\right)|{H}_{n+1}^{\left(\nu \right)}\left(x\right)〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right){B}_{m}^{\left(k-1\right)}\frac{1}{\left(l+1\right)!}〈{t}^{l+1}|{H}_{n+1}^{\left(\nu \right)}\left(x\right)〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}{\left(-1\right)}^{l-m}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){B}_{m}^{\left(k-1\right)}{H}_{n-l}^{\left(\nu \right)}.\hfill \end{array}$
(2.24)

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

Theorem 9 For $n\ge 0$, we have

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

Let us consider the following two Sheffer sequences:

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)\sim \left({e}^{\frac{\nu {t}^{2}}{2}}\frac{1-{e}^{-t}}{{Li}_{k}\left(1-{e}^{-t}\right)},t\right)$
(2.25)

and

${\mathbb{B}}_{n}^{\left(r\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{r},t\right)\phantom{\rule{1em}{0ex}}\left(r\in {\mathbb{Z}}_{\ge 0}\right).$
(2.26)

Let us assume that

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{\mathbb{B}}_{m}^{\left(r\right)}\left(x\right).$
(2.27)

Then, by (1.20) and (1.21), we get

$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{{e}^{t}-1}{t}\right)}^{r}{t}^{m}|{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}〉\\ =& \frac{1}{m!}〈{\left(\frac{{e}^{t}-1}{t}\right)}^{r}|{t}^{m}H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)〉=\frac{1}{m!}{\left(n\right)}_{m}〈{\left(\frac{{e}^{t}-1}{t}\right)}^{r}|H{B}_{n-m}^{\left(\nu ,k\right)}\left(x\right)〉\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{\mathrm{\infty }}\frac{r!}{\left(l+r\right)!}{S}_{2}\left(l+r,r\right)〈{t}^{l}|H{B}_{n-m}^{\left(\nu ,k\right)}\left(x\right)〉\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}{\left(n-m\right)}_{l}\frac{r!}{\left(l+r\right)!}{S}_{2}\left(l+r,r\right)H{B}_{n-m-l}^{\left(\nu ,k\right)}\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0}{}{l+r}{r}\right)}{S}_{2}\left(l+r,r\right)H{B}_{n-m-l}^{\left(\nu ,k\right)}.\end{array}$
(2.28)

Therefore, by (2.27) and (2.28), we obtain the following theorem.

Theorem 10 For $n,r\in {\mathbb{Z}}_{\ge 0}$, we have

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

For $\lambda \phantom{\rule{0.25em}{0ex}}\left(\ne 1\right)\in \mathbb{C}$, $r\in {\mathbb{Z}}_{\ge 0}$, the Frobenius-Euler polynomials of order r are defined by the generating function to be

${\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{r}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(r\right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1, 4, 7, 9, 10]}\right).$
(2.29)

From (1.16) and (2.29), we note that

${H}_{n}^{\left(r\right)}\left(x|\lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r},t\right).$
(2.30)

Let us assume that

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

By (1.21), we get

$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{r}{t}^{m}|{e}^{-\frac{\nu {t}^{2}}{2}}\frac{{Li}_{k}\left(1-{e}^{-t}\right)}{1-{e}^{-t}}{x}^{n}〉\\ =\frac{{\left(n\right)}_{m}}{m!{\left(1-\lambda \right)}^{r}}〈\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(-\lambda \right)}^{r-l}{e}^{lt}|H{B}_{n-m}^{\left(\nu ,k\right)}\left(x\right)〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\frac{1}{{\left(1-\lambda \right)}^{r}}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(-\lambda \right)}^{r-l}〈1|{e}^{lt}H{B}_{n-m}^{\left(\nu ,k\right)}\left(x\right)〉\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1-\lambda \right)}^{r}}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(-\lambda \right)}^{r-l}H{B}_{n-m}^{\left(\nu ,k\right)}\left(l\right).\end{array}$
(2.32)

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

Theorem 11 For $n,r\in {\mathbb{Z}}_{\ge 0}$, we have

$H{B}_{n}^{\left(\nu ,k\right)}\left(x\right)=\frac{1}{{\left(1-\lambda \right)}^{r}}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)\left\{\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(-\lambda \right)}^{r-l}H{B}_{n-m}^{\left(\nu ,k\right)}\left(l\right)\right\}{H}_{m}^{\left(r\right)}\left(x|\lambda \right).$

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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. Hermite and poly-Bernoulli mixed-type polynomials. Adv Differ Equ 2013, 343 (2013). https://doi.org/10.1186/1687-1847-2013-343

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• DOI: https://doi.org/10.1186/1687-1847-2013-343

### Keywords

• Vector Space
• Formal Power Series
• Linear Functional
• Hermite Polynomial
• Bernoulli Number 