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 $râˆˆ{\mathbb{Z}}_{â‰¥0}$, as is well known, the Bernoulli polynomials of order r are defined by the generating function to be

$\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathbb{B}}_{n}^{\left(r\right)}\left(x\right)}{n!}{t}^{n}={\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{r}{e}^{xt}\phantom{\rule{1em}{0ex}}\left(\text{see [1â€“16]}\right).$
(1.1)

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

${Li}_{k}\left(x\right)=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{x}^{n}}{{n}^{k}}.$
(1.2)

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

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

$\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [5, 8]}\right).$
(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 , the Hermite polynomials of order Î½ are given by the generating function to be

${e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [6, 12, 13]}\right).$
(1.4)

When $x=0$, ${H}_{n}^{\left(\mathrm{Î½}\right)}={H}_{n}^{\left(\mathrm{Î½}\right)}\left(0\right)$ are called the Hermite numbers of order Î½.

In this paper, we consider the Hermite and poly-Bernoulli mixed-type polynomials $H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)$ which are defined by the generating function to be

${e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)\frac{{t}^{n}}{n!},$
(1.5)

where $kâˆˆ\mathbb{Z}$ and .

When $x=0$, $H{B}_{n}^{\left(\mathrm{Î½},k\right)}=H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(0\right)$ are called the Hermite and poly-Bernoulli mixed-type numbers.

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

$\mathcal{F}=\left\{f\left(t\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{k}\frac{{t}^{k}}{k!}|{a}_{k}âˆˆ\mathbb{C}\right\}.$
(1.6)

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

$ã€ˆL|p\left(x\right)ã€‰$ denotes the action of the linear functional L on the polynomial $p\left(x\right)$, and we recall that the vector space operations on ${\mathbb{P}}^{âˆ—}$ are defined by $ã€ˆL+M|p\left(x\right)ã€‰=ã€ˆL|p\left(x\right)ã€‰+ã€ˆM|p\left(x\right)ã€‰$, $ã€ˆcL|p\left(x\right)ã€‰=cã€ˆL|p\left(x\right)ã€‰$, where c is a complex constant in â„‚. For $f\left(t\right)âˆˆ\mathcal{F}$, let us define the linear functional on â„™ by setting

$ã€ˆf\left(t\right)|{x}^{n}ã€‰={a}_{n}\phantom{\rule{1em}{0ex}}\left(nâ‰¥0\right).$
(1.7)

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

$ã€ˆ{t}^{k}|{x}^{n}ã€‰=n!{\mathrm{Î´}}_{n,k}\phantom{\rule{1em}{0ex}}\left(n,kâ‰¥0\right),$
(1.8)

where ${\mathrm{Î´}}_{n,k}$ is the Kronecker symbol.

For ${f}_{L}\left(t\right)={âˆ‘}_{k=0}^{\mathrm{âˆž}}\frac{ã€ˆL|{x}^{k}ã€‰}{k!}{t}^{k}$, we have $ã€ˆ{f}_{L}\left(t\right)|{x}^{n}ã€‰=ã€ˆL|{x}^{n}ã€‰$. That is, $L={f}_{L}\left(t\right)$. The map $Lâ†¦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{âˆ—}$ onto â„±. Henceforth, â„± denotes both the algebra of formal power series in t and the vector space of all linear functionals on â„™, and so an element $f\left(t\right)$ of â„± will be thought of as both a formal power series and a linear functional. We call â„± the umbral algebra and the umbral calculus is the study of umbral algebra. The order $O\left(f\right)$ of the power series 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 $f\left(t\right),g\left(t\right)âˆˆ\mathcal{F}$, we have

$ã€ˆf\left(t\right)g\left(t\right)|p\left(x\right)ã€‰=ã€ˆf\left(t\right)|g\left(t\right)p\left(x\right)ã€‰=ã€ˆg\left(t\right)|f\left(t\right)p\left(x\right)ã€‰.$
(1.9)

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

$f\left(t\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{ã€ˆf\left(t\right)|{x}^{k}ã€‰}{k!}{t}^{k},\phantom{\rule{2em}{0ex}}p\left(x\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{ã€ˆ{t}^{k}|p\left(x\right)ã€‰}{k!}{x}^{k}\phantom{\rule{1em}{0ex}}\left(\text{see [8, 9, 11, 13, 14]}\right).$
(1.10)

By (1.10), we get

${p}^{\left(k\right)}\left(0\right)=ã€ˆ{t}^{k}|p\left(x\right)ã€‰=ã€ˆ1|{p}^{\left(k\right)}\left(x\right)ã€‰,$
(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

${e}^{yt}p\left(x\right)=p\left(x+y\right),\phantom{\rule{2em}{0ex}}ã€ˆ{e}^{yt}|p\left(x\right)ã€‰=p\left(y\right).$
(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 $ã€ˆg\left(t\right)f{\left(t\right)}^{k}|{x}^{n}ã€‰=n!{\mathrm{Î´}}_{n,k}$ ($n,kâ‰¥0$).

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 ${s}_{n}\left(x\right)âˆ¼\left(g\left(t\right),f\left(t\right)\right)$.

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

$ã€ˆf\left(t\right)|xp\left(x\right)ã€‰=ã€ˆ{\mathrm{âˆ‚}}_{t}f\left(t\right)|p\left(x\right)ã€‰=ã€ˆ\frac{df\left(t\right)}{dt}|p\left(x\right)ã€‰.$
(1.14)

For ${s}_{n}\left(x\right)âˆ¼\left(g\left(t\right),f\left(t\right)\right)$, we have the following equations:

$h\left(t\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{ã€ˆh\left(t\right)|{s}_{k}\left(x\right)ã€‰}{k!}g\left(t\right)f{\left(t\right)}^{k},\phantom{\rule{2em}{0ex}}p\left(x\right)=\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{ã€ˆg\left(t\right)f{\left(t\right)}^{k}|p\left(x\right)ã€‰}{k!}{s}_{k}\left(x\right),$
(1.15)

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

$\frac{1}{g\left(\stackrel{Â¯}{f}\left(t\right)\right)}{e}^{y\stackrel{Â¯}{f}\left(t\right)}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{s}_{n}\left(y\right)\frac{{t}^{n}}{n!},$
(1.16)

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

(1.17)
$f\left(t\right){s}_{n}\left(x\right)=n{s}_{nâˆ’1}\left(x\right),\phantom{\rule{2em}{0ex}}{s}_{n+1}\left(x\right)=\left(xâˆ’\frac{{g}^{â€²}\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{â€²}\left(t\right)}{s}_{n}\left(x\right),$
(1.18)

and the conjugate representation is given by

${s}_{n}\left(x\right)=\underset{j=0}{\overset{n}{âˆ‘}}\frac{1}{j!}ã€ˆg{\left(\stackrel{Â¯}{f}\left(t\right)\right)}^{âˆ’1}\stackrel{Â¯}{f}{\left(t\right)}^{j}|{x}^{n}ã€‰{x}^{j}.$
(1.19)

For ${s}_{n}\left(x\right)âˆ¼\left(g\left(t\right),f\left(t\right)\right)$, ${r}_{n}\left(x\right)âˆ¼\left(h\left(t\right),l\left(t\right)\right)$, we have

${s}_{n}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}{C}_{n,m}{r}_{m}\left(x\right),$
(1.20)

where

${C}_{n,m}=\frac{1}{m!}ã€ˆ\frac{h\left(\stackrel{Â¯}{f}\left(t\right)\right)}{g\left(\stackrel{Â¯}{f}\left(t\right)\right)}l{\left(\stackrel{Â¯}{f}\left(t\right)\right)}^{m}|{x}^{n}ã€‰\phantom{\rule{1em}{0ex}}\left(\text{see [8, 9, 13]}\right).$
(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

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ¼\left({e}^{\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{1âˆ’{e}^{âˆ’t}}{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)},t\right),$
(2.1)

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

${B}_{n}^{\left(k\right)}\left(x\right)âˆ¼\left(\frac{1âˆ’{e}^{âˆ’t}}{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)},t\right),$
(2.2)
(2.3)

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

$t{B}_{n}^{\left(k\right)}\left(x\right)=n{B}_{nâˆ’1}^{\left(k\right)}\left(x\right),\phantom{\rule{2em}{0ex}}t{H}_{n}^{\left(\mathrm{Î½}\right)}\left(x\right)=n{H}_{nâˆ’1}^{\left(\mathrm{Î½}\right)}\left(x\right),\phantom{\rule{2em}{0ex}}tH{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=nH{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(x\right).$
(2.4)

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

$\begin{array}{rcl}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)& =& {e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{x}^{n}={e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{B}_{n}^{\left(k\right)}\left(x\right)\\ =& \underset{m=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\frac{1}{m!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{m}{\left(n\right)}_{2m}{B}_{nâˆ’2m}^{\left(k\right)}\left(x\right)\\ =& \underset{m=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{2m}\right)\frac{\left(2m\right)!}{m!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{m}{B}_{nâˆ’2m}^{\left(k\right)}\left(x\right).\end{array}$
(2.5)

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

Proposition 1 For $nâ‰¥0$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{m=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{2m}\right)\frac{\left(2m\right)!}{m!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{m}{B}_{nâˆ’2m}^{\left(k\right)}\left(x\right).$

From (1.5), we can also derive

$\begin{array}{rcl}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)& =& \frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{x}^{n}=\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(x\right)=\underset{m=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(1âˆ’{e}^{âˆ’t}\right)}^{m}}{{\left(m+1\right)}^{k}}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(x\right)\\ =& \underset{m=0}{\overset{n}{âˆ‘}}\frac{1}{{\left(m+1\right)}^{k}}\underset{j=0}{\overset{m}{âˆ‘}}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(âˆ’1\right)}^{j}{e}^{âˆ’jt}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(x\right)\\ =& \underset{m=0}{\overset{n}{âˆ‘}}\frac{1}{{\left(m+1\right)}^{k}}\underset{j=0}{\overset{m}{âˆ‘}}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(âˆ’1\right)}^{j}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(xâˆ’j\right).\end{array}$
(2.6)

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

Theorem 2 For $nâ‰¥0$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}\frac{1}{{\left(m+1\right)}^{k}}\underset{j=0}{\overset{m}{âˆ‘}}\left(\genfrac{}{}{0}{}{m}{j}\right){\left(âˆ’1\right)}^{j}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(xâˆ’j\right).$

By (1.5), we get

$\begin{array}{rcl}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)& =& {e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{B}_{n}^{\left(k\right)}\left(x\right)=\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}{t}^{2l}{B}_{n}^{\left(k\right)}\left(x\right)\\ =& \underset{l=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\frac{1}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}\underset{m=0}{\overset{n}{âˆ‘}}\frac{1}{{\left(m+1\right)}^{k}}\underset{j=0}{\overset{m}{âˆ‘}}{\left(âˆ’1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right){t}^{2l}{\left(xâˆ’j\right)}^{n}\\ =& \underset{l=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\underset{j=0}{\overset{n}{âˆ‘}}\left\{\underset{m=j}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}\frac{{\left(âˆ’1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)}{{\left(m+1\right)}^{k}}\right\}{\left(xâˆ’j\right)}^{nâˆ’2l}.\end{array}$
(2.7)

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

Theorem 3 For $nâ‰¥0$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{l=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\underset{j=0}{\overset{n}{âˆ‘}}\left\{\underset{m=j}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}\frac{{\left(âˆ’1\right)}^{j}\left(\genfrac{}{}{0}{}{m}{j}\right)}{{\left(m+1\right)}^{k}}\right\}{\left(xâˆ’j\right)}^{nâˆ’2l}.$

By (2.6), we get

$\begin{array}{rcl}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)& =& \underset{m=0}{\overset{n}{âˆ‘}}\frac{{\left(1âˆ’{e}^{âˆ’t}\right)}^{m}}{{\left(m+1\right)}^{k}}{H}_{n}^{\left(\mathrm{Î½}\right)}\left(x\right)\\ =& \underset{m=0}{\overset{n}{âˆ‘}}\frac{1}{{\left(m+1\right)}^{k}}\underset{a=0}{\overset{nâˆ’m}{âˆ‘}}\frac{m!}{\left(a+m\right)!}{\left(âˆ’1\right)}^{a}{S}_{2}\left(a+m,m\right){\left(n\right)}_{a+m}{H}_{nâˆ’aâˆ’m}^{\left(\mathrm{Î½}\right)}\left(x\right)\\ =& \underset{m=0}{\overset{n}{âˆ‘}}\underset{a=0}{\overset{nâˆ’m}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{nâˆ’aâˆ’m}m!}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{nâˆ’a}\right){S}_{2}\left(nâˆ’a,m\right){H}_{a}^{\left(\mathrm{Î½}\right)}\left(x\right)\\ =& {\left(âˆ’1\right)}^{n}\underset{a=0}{\overset{n}{âˆ‘}}\left\{\underset{m=0}{\overset{nâˆ’a}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m+a}m!}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{a}\right){S}_{2}\left(nâˆ’a,m\right)\right\}{H}_{a}^{\left(\mathrm{Î½}\right)}\left(x\right),\end{array}$
(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 $nâ‰¥0$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)={\left(âˆ’1\right)}^{n}\underset{a=0}{\overset{n}{âˆ‘}}\left\{\underset{m=0}{\overset{nâˆ’a}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{a+m}m!}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n}{a}\right){S}_{2}\left(nâˆ’a,m\right)\right\}{H}_{a}^{\left(\mathrm{Î½}\right)}\left(x\right).$

From (1.19) and (2.1), we have

$\begin{array}{rcl}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)& =& \underset{j=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}|{x}^{nâˆ’j}ã€‰{x}^{j}\\ =& \underset{j=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}|{B}_{nâˆ’j}^{\left(k\right)}\left(x\right)ã€‰{x}^{j}\\ =& \underset{j=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)\underset{l=0}{\overset{\left[\frac{nâˆ’j}{2}\right]}{âˆ‘}}\frac{{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}}{l!}{\left(nâˆ’j\right)}_{2l}ã€ˆ1|{B}_{nâˆ’jâˆ’2l}^{\left(k\right)}\left(x\right)ã€‰{x}^{j}\\ =& \underset{j=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)\underset{l=0}{\overset{\left[\frac{nâˆ’j}{2}\right]}{âˆ‘}}\frac{1}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}{\left(nâˆ’j\right)}_{2l}{B}_{nâˆ’jâˆ’2l}^{\left(k\right)}{x}^{j}\\ =& \underset{j=0}{\overset{n}{âˆ‘}}\left\{\underset{l=0}{\overset{\left[\frac{nâˆ’j}{2}\right]}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{nâˆ’j}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}{B}_{nâˆ’jâˆ’2l}^{\left(k\right)}\right\}{x}^{j}.\end{array}$
(2.9)

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

Theorem 5 For $nâ‰¥0$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{j=0}{\overset{n}{âˆ‘}}\left\{\underset{l=0}{\overset{\left[\frac{nâˆ’j}{2}\right]}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{nâˆ’j}{2l}\right)\frac{\left(2l\right)!}{l!}{\left(âˆ’\frac{\mathrm{Î½}}{2}\right)}^{l}{B}_{nâˆ’jâˆ’2l}^{\left(k\right)}\right\}{x}^{j}.$

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

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x+y\right)=\underset{j=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{j}\right)H{B}_{j}^{\left(\mathrm{Î½},k\right)}\left(x\right){y}^{nâˆ’j}.$
(2.10)

We note that

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ¼\left(g\left(t\right)={e}^{\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{1âˆ’{e}^{âˆ’t}}{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)},f\left(t\right)=t\right).$
(2.11)

From (1.18) and (2.11), we have

$H{B}_{n+1}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\left(xâˆ’\frac{{g}^{â€²}\left(t\right)}{g\left(t\right)}\right)H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right).$
(2.12)

Now, we observe that

$\begin{array}{rcl}\frac{{g}^{â€²}\left(t\right)}{g\left(t\right)}& =& {\left(log\left(g\left(t\right)\right)\right)}^{â€²}\\ =& {\left(log{e}^{\frac{\mathrm{Î½}{t}^{2}}{2}}+log\left(1âˆ’{e}^{âˆ’t}\right)âˆ’log\left({Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)\right)\right)}^{â€²}\\ =& \mathrm{Î½}t+\frac{{e}^{âˆ’t}}{1âˆ’{e}^{âˆ’t}}\left(1âˆ’\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)}{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}\right).\end{array}$
(2.13)

By (2.12) and (2.13), we get

$\begin{array}{c}H{B}_{n+1}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=xH{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’\frac{{g}^{â€²}\left(t\right)}{g\left(t\right)}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=xH{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’\mathrm{Î½}nH{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{t}{{e}^{t}âˆ’1}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)}{t\left(1âˆ’{e}^{âˆ’t}\right)}{x}^{n}.\hfill \end{array}$
(2.14)

It is easy to show that

$\begin{array}{rcl}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}& =& \underset{m=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\frac{1}{{m}^{k}}âˆ’\frac{1}{{m}^{kâˆ’1}}\right){\left(1âˆ’{e}^{âˆ’t}\right)}^{mâˆ’1}\\ =& \left(\frac{1}{{2}^{k}}âˆ’\frac{1}{{2}^{kâˆ’1}}\right)t+â‹¯.\end{array}$
(2.15)

Thus, by (2.15), we get

$\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)}{t\left(1âˆ’{e}^{âˆ’t}\right)}{x}^{n}=\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\frac{{x}^{n+1}}{n+1}.$
(2.16)

From (2.16), we can derive

$\begin{array}{c}{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{t}{{e}^{t}âˆ’1}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)}{t\left(1âˆ’{e}^{âˆ’t}\right)}{x}^{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\left(\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{B}_{l}}{l!}{t}^{l}\right)\left(H{B}_{n+1}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’H{B}_{n+1}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\underset{l=0}{\overset{n+1}{âˆ‘}}\frac{{B}_{l}}{l!}{t}^{l}\left(H{B}_{n+1}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’H{B}_{n+1}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n+1}\underset{l=0}{\overset{n+1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left(H{B}_{n+1âˆ’l}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’H{B}_{n+1âˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\right).\hfill \end{array}$
(2.17)

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

Theorem 6 For $nâ‰¥0$, we have

$\begin{array}{c}H{B}_{n+1}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=xH{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’\mathrm{Î½}nH{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\frac{1}{n+1}\underset{l=0}{\overset{n+1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left\{H{B}_{n+1âˆ’l}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’H{B}_{n+1âˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\right\}.\hfill \end{array}$
(2.18)

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

$\begin{array}{c}\left(n+1\right)H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(xt+1\right)H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’\mathrm{Î½}n\left(nâˆ’1\right)H{B}_{nâˆ’2}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\frac{1}{n+1}\underset{l=0}{\overset{n+1}{âˆ‘}}\left(\genfrac{}{}{0}{}{n+1}{l}\right)\left(n+1âˆ’l\right){B}_{l}\left\{H{B}_{nâˆ’l}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’H{B}_{nâˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=nxH{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(x\right)+H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’\mathrm{Î½}n\left(nâˆ’1\right)H{B}_{nâˆ’2}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}\left(H{B}_{nâˆ’l}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’H{B}_{nâˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\right),\hfill \end{array}$
(2.19)

where $nâ‰¥3$.

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

Theorem 7 For $nâ‰¥3$, we have

$\begin{array}{c}\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}H{B}_{nâˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(n+1\right)H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ’n\left(x+\frac{1}{2}\right)H{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}+n\left(nâˆ’1\right)\left(\mathrm{Î½}+\frac{1}{12}\right)H{B}_{nâˆ’2}^{\left(\mathrm{Î½},k\right)}\left(x\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\underset{l=0}{\overset{nâˆ’3}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{nâˆ’l}H{B}_{l}^{\left(\mathrm{Î½},k\right)}\left(x\right).\hfill \end{array}$

By (1.5) and (1.8), we get

$\begin{array}{c}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(y\right)\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|{x}^{n}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{\mathrm{âˆ‚}}_{t}\left({e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}\right)|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ\left({\mathrm{âˆ‚}}_{t}{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\right)\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{2em}{0ex}}+ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\left({\mathrm{âˆ‚}}_{t}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\right){e}^{yt}|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{2em}{0ex}}+ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\left({\mathrm{âˆ‚}}_{t}{e}^{yt}\right)|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î½}\left(nâˆ’1\right)ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|{x}^{nâˆ’2}ã€‰\hfill \\ \phantom{\rule{2em}{0ex}}+yã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{2em}{0ex}}+ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\left({\mathrm{âˆ‚}}_{t}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\right){e}^{yt}|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=âˆ’\mathrm{Î½}\left(nâˆ’1\right)H{B}_{nâˆ’2}^{\left(\mathrm{Î½},k\right)}\left(y\right)+yH{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(y\right)\hfill \\ \phantom{\rule{2em}{0ex}}+ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\left({\mathrm{âˆ‚}}_{t}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\right){e}^{yt}|{x}^{nâˆ’1}ã€‰.\hfill \end{array}$
(2.20)

Now, we observe that

${\mathrm{âˆ‚}}_{t}\left(\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\right)=\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{{\left(1âˆ’{e}^{âˆ’t}\right)}^{2}}{e}^{âˆ’t}.$
(2.21)

From (2.21), we have

$\begin{array}{c}ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\left({\mathrm{âˆ‚}}_{t}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\right){e}^{yt}|{x}^{nâˆ’1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\left(\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{{\left(1âˆ’{e}^{âˆ’t}\right)}^{2}}\right){e}^{âˆ’t}{e}^{yt}|\frac{1}{n}t{x}^{n}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|\frac{t}{{e}^{t}âˆ’1}{x}^{n}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|{B}_{n}\left(x\right)ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’t}\right)âˆ’{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{e}^{yt}|{x}^{nâˆ’l}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}\left\{H{B}_{nâˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(y\right)âˆ’H{B}_{nâˆ’l}^{\left(\mathrm{Î½},k\right)}\left(y\right)\right\},\hfill \end{array}$
(2.22)

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

$\frac{t}{{e}^{t}âˆ’1}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{B}_{n}}{n!}{t}^{n}.$

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

Theorem 8 For $nâ‰¥2$, we have

$\begin{array}{rcl}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)& =& âˆ’\mathrm{Î½}\left(nâˆ’1\right)H{B}_{nâˆ’2}^{\left(\mathrm{Î½},k\right)}\left(x\right)+xH{B}_{nâˆ’1}^{\left(\mathrm{Î½},k\right)}\left(x\right)\\ +\frac{1}{n}\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}\left(H{B}_{nâˆ’l}^{\left(\mathrm{Î½},kâˆ’1\right)}\left(x\right)âˆ’H{B}_{nâˆ’l}^{\left(\mathrm{Î½},k\right)}\left(x\right)\right).\end{array}$

Now, we compute

$ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)|{x}^{n+1}ã€‰$

in two different ways.

On the one hand,

$\begin{array}{c}ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)|{x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}\left(1âˆ’{e}^{âˆ’t}\right)|{x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}|\left(1âˆ’{e}^{âˆ’t}\right){x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}|{x}^{n+1}âˆ’{\left(xâˆ’1\right)}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{nâˆ’m}\left(\genfrac{}{}{0}{}{n+1}{m}\right)ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}|{x}^{m}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{nâˆ’m}\left(\genfrac{}{}{0}{}{n+1}{m}\right)H{B}_{m}^{\left(\mathrm{Î½},k\right)}.\hfill \end{array}$
(2.23)

On the other hand,

$\begin{array}{c}ã€ˆ{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)|{x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)|{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{âˆ«}_{0}^{t}{\left({Li}_{k}\left(1âˆ’{e}^{âˆ’s}\right)\right)}^{â€²}\phantom{\rule{0.2em}{0ex}}ds|{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ{âˆ«}_{0}^{t}{e}^{âˆ’s}\frac{{Li}_{kâˆ’1}\left(1âˆ’{e}^{âˆ’s}\right)}{1âˆ’{e}^{âˆ’s}}\phantom{\rule{0.2em}{0ex}}ds|{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}{x}^{n+1}ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=ã€ˆ\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{m=0}{\overset{l}{âˆ‘}}{\left(âˆ’1\right)}^{lâˆ’m}\left(\genfrac{}{}{0}{}{l}{m}\right){B}_{m}^{\left(kâˆ’1\right)}\frac{{t}^{l+1}}{\left(l+1\right)!}\right)|{H}_{n+1}^{\left(\mathrm{Î½}\right)}\left(x\right)ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{l}{âˆ‘}}{\left(âˆ’1\right)}^{lâˆ’m}\left(\genfrac{}{}{0}{}{l}{m}\right){B}_{m}^{\left(kâˆ’1\right)}\frac{1}{\left(l+1\right)!}ã€ˆ{t}^{l+1}|{H}_{n+1}^{\left(\mathrm{Î½}\right)}\left(x\right)ã€‰\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{l=0}{\overset{n}{âˆ‘}}\underset{m=0}{\overset{l}{âˆ‘}}{\left(âˆ’1\right)}^{lâˆ’m}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){B}_{m}^{\left(kâˆ’1\right)}{H}_{nâˆ’l}^{\left(\mathrm{Î½}\right)}.\hfill \end{array}$
(2.24)

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

Theorem 9 For $nâ‰¥0$, we have

$\begin{array}{c}\underset{m=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{nâˆ’m}\left(\genfrac{}{}{0}{}{n+1}{m}\right)H{B}_{m}^{\left(\mathrm{Î½},k\right)}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{m=0}{\overset{n}{âˆ‘}}\underset{l=m}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{lâˆ’m}\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{n+1}{l+1}\right){B}_{m}^{\left(kâˆ’1\right)}{H}_{nâˆ’l}^{\left(\mathrm{Î½}\right)}.\hfill \end{array}$

Let us consider the following two Sheffer sequences:

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)âˆ¼\left({e}^{\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{1âˆ’{e}^{âˆ’t}}{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)},t\right)$
(2.25)

and

${\mathbb{B}}_{n}^{\left(r\right)}\left(x\right)âˆ¼\left({\left(\frac{{e}^{t}âˆ’1}{t}\right)}^{r},t\right)\phantom{\rule{1em}{0ex}}\left(râˆˆ{\mathbb{Z}}_{â‰¥0}\right).$
(2.26)

Let us assume that

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}{C}_{n,m}{\mathbb{B}}_{m}^{\left(r\right)}\left(x\right).$
(2.27)

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

$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}ã€ˆ{\left(\frac{{e}^{t}âˆ’1}{t}\right)}^{r}{t}^{m}|{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{x}^{n}ã€‰\\ =& \frac{1}{m!}ã€ˆ{\left(\frac{{e}^{t}âˆ’1}{t}\right)}^{r}|{t}^{m}H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)ã€‰=\frac{1}{m!}{\left(n\right)}_{m}ã€ˆ{\left(\frac{{e}^{t}âˆ’1}{t}\right)}^{r}|H{B}_{nâˆ’m}^{\left(\mathrm{Î½},k\right)}\left(x\right)ã€‰\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\underset{l=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{r!}{\left(l+r\right)!}{S}_{2}\left(l+r,r\right)ã€ˆ{t}^{l}|H{B}_{nâˆ’m}^{\left(\mathrm{Î½},k\right)}\left(x\right)ã€‰\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}{\left(nâˆ’m\right)}_{l}\frac{r!}{\left(l+r\right)!}{S}_{2}\left(l+r,r\right)H{B}_{nâˆ’mâˆ’l}^{\left(\mathrm{Î½},k\right)}\\ =& \left(\genfrac{}{}{0}{}{n}{m}\right)\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{nâˆ’m}{l}\right)}{\left(\genfrac{}{}{0}{}{l+r}{r}\right)}{S}_{2}\left(l+r,r\right)H{B}_{nâˆ’mâˆ’l}^{\left(\mathrm{Î½},k\right)}.\end{array}$
(2.28)

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

Theorem 10 For $n,râˆˆ{\mathbb{Z}}_{â‰¥0}$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}\left\{\left(\genfrac{}{}{0}{}{n}{m}\right)\underset{l=0}{\overset{nâˆ’m}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{nâˆ’m}{l}\right)}{\left(\genfrac{}{}{0}{}{l+r}{r}\right)}{S}_{2}\left(l+r,r\right)H{B}_{nâˆ’mâˆ’l}^{\left(\mathrm{Î½},k\right)}\right\}{\mathbb{B}}_{m}^{\left(r\right)}\left(x\right).$

For , $râˆˆ{\mathbb{Z}}_{â‰¥0}$, the Frobenius-Euler polynomials of order r are defined by the generating function to be

${\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)}^{r}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1, 4, 7, 9, 10]}\right).$
(2.29)

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

${H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)âˆ¼\left({\left(\frac{{e}^{t}âˆ’\mathrm{Î»}}{1âˆ’\mathrm{Î»}}\right)}^{r},t\right).$
(2.30)

Let us assume that

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\underset{m=0}{\overset{n}{âˆ‘}}{C}_{n,m}{H}_{m}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$
(2.31)

By (1.21), we get

$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}ã€ˆ{\left(\frac{{e}^{t}âˆ’\mathrm{Î»}}{1âˆ’\mathrm{Î»}}\right)}^{r}{t}^{m}|{e}^{âˆ’\frac{\mathrm{Î½}{t}^{2}}{2}}\frac{{Li}_{k}\left(1âˆ’{e}^{âˆ’t}\right)}{1âˆ’{e}^{âˆ’t}}{x}^{n}ã€‰\\ =\frac{{\left(n\right)}_{m}}{m!{\left(1âˆ’\mathrm{Î»}\right)}^{r}}ã€ˆ\underset{l=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’l}{e}^{lt}|H{B}_{nâˆ’m}^{\left(\mathrm{Î½},k\right)}\left(x\right)ã€‰\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{l=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’l}ã€ˆ1|{e}^{lt}H{B}_{nâˆ’m}^{\left(\mathrm{Î½},k\right)}\left(x\right)ã€‰\\ =\frac{\left(\genfrac{}{}{0}{}{n}{m}\right)}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{l=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’l}H{B}_{nâˆ’m}^{\left(\mathrm{Î½},k\right)}\left(l\right).\end{array}$
(2.32)

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

Theorem 11 For $n,râˆˆ{\mathbb{Z}}_{â‰¥0}$, we have

$H{B}_{n}^{\left(\mathrm{Î½},k\right)}\left(x\right)=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{m=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{m}\right)\left\{\underset{l=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’l}H{B}_{nâˆ’m}^{\left(\mathrm{Î½},k\right)}\left(l\right)\right\}{H}_{m}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$

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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.

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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