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

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

(1.1)

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

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

(1.2)

Note that ${Li}_1(x)=-log(1-x)$.

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

$$\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(k)}(x)\frac{t^n}{n!}(\text{see [5, 8]}).$$

(1.3)

When $x=0$, $B_n^{(k)}=B_n^{(k)}(0)$ are called the poly-Bernoulli numbers (of index k).

For $\nu (\ne 0)\in \mathbb{R}$, the Hermite polynomials of order ν are given by the generating function to be

$$e^{-\frac{\nu t^2}{2}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(\nu )}(x)\frac{t^n}{n!}(\text{see [6, 12, 13]}).$$

(1.4)

When $x=0$, $H_n^{(\nu )}=H_n^{(\nu )}(0)$ are called the Hermite numbers of order ν.

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

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

(1.5)

where $k\in \mathbb{Z}$ and $\nu (\ne 0)\in \mathbb{R}$.

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

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

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

(1.6)

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

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

$$〈f(t)|x^n〉=a_n(n\ge 0).$$

(1.7)

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

$$〈t^k|x^n〉=n!{\delta }_{n,k}(n,k\ge 0),$$

(1.8)

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

For $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^k〉}{k!}{t}^k$, we have $〈f_L(t)|x^n〉=〈L|x^n〉$. That is, $L=f_L(t)$. The map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth, ℱ denotes both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and so an element $f(t)$ of ℱ will be thought of as both a formal power series and a linear functional. We call ℱ the umbral algebra and the umbral calculus is the study of umbral algebra. The order $O(f)$ of the power series $f(t)\ne 0$ is the smallest integer for which $a_k$ does not vanish. If $O(f)=0$, then $f(t)$ is called an invertible series. If $O(f)=1$, then $f(t)$ is called a delta series. For $f(t),g(t)\in \mathcal{F}$, we have

$$〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉=〈g(t)|f(t)p(x)〉.$$

(1.9)

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

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈f(t)|x^k〉}{k!}{t}^k,p(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈t^k|p(x)〉}{k!}{x}^k(\text{see [8, 9, 11, 13, 14]}).$$

(1.10)

By (1.10), we get

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

(1.11)

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

From (1.11), we have

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}(\text{see [8, 9, 13]}).$$

(1.12)

By (1.12), we easily get

$$e^{yt}p(x)=p(x+y),〈e^{yt}|p(x)〉=p(y).$$

(1.13)

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

The sequence $s_n(x)$ is called the Sheffer sequence for $(g(t),f(t))$ which is denoted by $s_n(x)\sim (g(t),f(t))$.

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

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

(1.14)

For $s_n(x)\sim (g(t),f(t))$, we have the following equations:

$$h(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈h(t)|s_k(x)〉}{k!}g(t)f{(t)}^k,p(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈g(t)f{(t)}^k|p(x)〉}{k!}{s}_k(x),$$

(1.15)

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

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

(1.16)

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

$$s_n(x+y)=\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)s_k(y)p_{n-k}(x),\text{where }{p}_n(x)=g(t)s_n(x),$$

(1.17)

$$f(t)s_n(x)=n{s}_{n-1}(x),s_{n+1}(x)=(x-\frac{g^{\prime }(t)}{g(t)})\frac{1}{f^{\prime }(t)}{s}_n(x),$$

(1.18)

and the conjugate representation is given by

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

(1.19)

For $s_n(x)\sim (g(t),f(t))$, $r_n(x)\sim (h(t),l(t))$, we have

$$s_n(x)=\sum _{m=0}^n{C}_{n,m}{r}_m(x),$$

(1.20)

where

$$C_{n,m}=\frac{1}{m!}〈\frac{h(\overline{f}(t))}{g(\overline{f}(t))}l{(\overline{f}(t))}^m|x^n〉(\text{see [8, 9, 13]}).$$

(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

$$H{B}_n^{(\nu ,k)}(x)\sim (e^{\frac{\nu t^2}{2}}\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},t),$$

(2.1)

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

$$B_n^{(k)}(x)\sim (\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},t),$$

(2.2)

$$H_n^{(\nu )}(x)\sim (e^{\frac{\nu t^2}{2}},t),\text{where }n\ge 0.$$

(2.3)

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

$$t{B}_n^{(k)}(x)=n{B}_{n-1}^{(k)}(x),t{H}_n^{(\nu )}(x)=n{H}_{n-1}^{(\nu )}(x),tH{B}_n^{(\nu ,k)}(x)=nH{B}_{n-1}^{(\nu ,k)}(x).$$

(2.4)

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

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

(2.5)

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

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

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

From (1.5), we can also derive

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

(2.6)

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

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

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

By (1.5), we get

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

(2.7)

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

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

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

By (2.6), we get

$$\begin{array}{rcl}H{B}_n^{(\nu ,k)}(x)& =& \sum _{m=0}^n\frac{{(1-e^{-t})}^m}{{(m+1)}^k}{H}_n^{(\nu )}(x)\\ =& \sum _{m=0}^n\frac{1}{{(m+1)}^k}\sum _{a=0}^{n-m}\frac{m!}{(a+m)!}{(-1)}^a{S}_2(a+m,m){(n)}_{a+m}{H}_{n-a-m}^{(\nu )}(x)\\ =& \sum _{m=0}^n\sum _{a=0}^{n-m}\frac{{(-1)}^{n-a-m}m!}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{n-a}\right)S_2(n-a,m)H_a^{(\nu )}(x)\\ =& {(-1)}^n\sum _{a=0}^n\{\sum _{m=0}^{n-a}\frac{{(-1)}^{m+a}m!}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{a}\right)S_2(n-a,m)\}{H}_a^{(\nu )}(x),\end{array}$$

(2.8)

where $S_2(n,m)$ 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

$$H{B}_n^{(\nu ,k)}(x)={(-1)}^n\sum _{a=0}^n\{\sum _{m=0}^{n-a}\frac{{(-1)}^{a+m}m!}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{a}\right)S_2(n-a,m)\}{H}_a^{(\nu )}(x).$$

From (1.19) and (2.1), we have

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

(2.9)

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

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

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

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

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

(2.10)

We note that

$$H{B}_n^{(\nu ,k)}(x)\sim (g(t)=e^{\frac{\nu t^2}{2}}\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},f(t)=t).$$

(2.11)

From (1.18) and (2.11), we have

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

(2.12)

Now, we observe that

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

(2.13)

By (2.12) and (2.13), we get

$$\begin{array}{c}H{B}_{n+1}^{(\nu ,k)}(x)\hfill \\ =xH{B}_n^{(\nu ,k)}(x)-\frac{g^{\prime }(t)}{g(t)}H{B}_n^{(\nu ,k)}(x)\hfill \\ =xH{B}_n^{(\nu ,k)}(x)-\nu nH{B}_{n-1}^{(\nu ,k)}(x)-e^{-\frac{\nu t^2}{2}}\frac{t}{e^t-1}\frac{{Li}_k(1-e^{-t})-{Li}_{k-1}(1-e^{-t})}{t(1-e^{-t})}{x}^n.\hfill \end{array}$$

(2.14)

It is easy to show that

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

(2.15)

Thus, by (2.15), we get

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

(2.16)

From (2.16), we can derive

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

(2.17)

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

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

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

(2.18)

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

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

(2.19)

where $n\ge 3$.

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

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

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

By (1.5) and (1.8), we get

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

(2.20)

Now, we observe that

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

(2.21)

From (2.21), we have

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

(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

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

Now, we compute

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

in two different ways.

On the one hand,

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

(2.23)

On the other hand,

$$\begin{array}{c}〈e^{-\frac{\nu t^2}{2}}{Li}_k(1-e^{-t})|x^{n+1}〉\hfill \\ =〈{Li}_k(1-e^{-t})|e^{-\frac{\nu t^2}{2}}{x}^{n+1}〉\hfill \\ =〈{\int }_0^t{\left({Li}_k(1-e^{-s})\right)}^{\prime }ds|e^{-\frac{\nu t^2}{2}}{x}^{n+1}〉\hfill \\ =〈{\int }_0^t{e}^{-s}\frac{{Li}_{k-1}(1-e^{-s})}{1-e^{-s}}ds|e^{-\frac{\nu t^2}{2}}{x}^{n+1}〉\hfill \\ =〈\sum _{l=0}^{\mathrm{\infty }}\left(\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)B_m^{(k-1)}\frac{t^{l+1}}{(l+1)!}\right)|H_{n+1}^{(\nu )}(x)〉\hfill \\ =\sum _{l=0}^n\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)B_m^{(k-1)}\frac{1}{(l+1)!}〈t^{l+1}|H_{n+1}^{(\nu )}(x)〉\hfill \\ =\sum _{l=0}^n\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)\left(\genfrac{}{}{0ex}{}{n+1}{l+1}\right)B_m^{(k-1)}{H}_{n-l}^{(\nu )}.\hfill \end{array}$$

(2.24)

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

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

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

Let us consider the following two Sheffer sequences:

$$H{B}_n^{(\nu ,k)}(x)\sim (e^{\frac{\nu t^2}{2}}\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},t)$$

(2.25)

and

$${\mathbb{B}}_n^{(r)}(x)\sim ({\left(\frac{e^t-1}{t}\right)}^r,t)(r\in {\mathbb{Z}}_{\ge 0}).$$

(2.26)

Let us assume that

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

(2.27)

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

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{e^t-1}{t}\right)}^r{t}^m|e^{-\frac{\nu t^2}{2}}\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^n〉\\ =& \frac{1}{m!}〈{\left(\frac{e^t-1}{t}\right)}^r|t^{m}H{B}_n^{(\nu ,k)}(x)〉=\frac{1}{m!}{(n)}_m〈{\left(\frac{e^t-1}{t}\right)}^r|H{B}_{n-m}^{(\nu ,k)}(x)〉\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{\mathrm{\infty }}\frac{r!}{(l+r)!}{S}_2(l+r,r)〈t^l|H{B}_{n-m}^{(\nu ,k)}(x)〉\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}{(n-m)}_l\frac{r!}{(l+r)!}{S}_2(l+r,r)H{B}_{n-m-l}^{(\nu ,k)}\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{l+r}{r}\right)}{S}_2(l+r,r)H{B}_{n-m-l}^{(\nu ,k)}.\end{array}$$

(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

$$H{B}_n^{(\nu ,k)}(x)=\sum _{m=0}^n\{\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{l+r}{r}\right)}{S}_2(l+r,r)H{B}_{n-m-l}^{(\nu ,k)}\}{\mathbb{B}}_m^{(r)}(x).$$

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

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

(2.29)

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

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

(2.30)

Let us assume that

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

(2.31)

By (1.21), we get

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

(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

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