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

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

(1.1)

When $x=0$, $H_n^{(\alpha )}(\lambda )=H_n^{(\alpha )}(0|\lambda )$ 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

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

(1.2)

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

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

(1.3)

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

$$L{i}_k(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n^k}(k\in \mathbb{Z})(\text{see [7]}).$$

(1.4)

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

$$\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(k)}(x)\frac{t^n}{n!}(\mathit{\text{cf. }}\text{[7]}).$$

(1.5)

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

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

(1.6)

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

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

(1.7)

defines a linear functional on ℙ by setting

$$〈f(t)|x^n〉=a_n,\text{for all }n\ge 0(\text{see [14, 15]}).$$

(1.8)

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

$$〈t^k|x^n〉=n!{\delta }_{n,k}(\text{see [14, 15]}),$$

(1.9)

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

Let us consider $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^n〉}{k!}{t}^k$. Then we see that $〈f_L(t)|x^n〉=〈L|x^n〉$, and so $L=f_L(t)$ as linear functionals. The map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth, ℱ will denote 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 (see [14]). We shall call ℱ the umbral algebra. The umbral calculus is the study of umbral algebra. The order $o(f(t))$ of a nonzero power series $f(t)$ is the smallest integer k, for which the coefficient of $t^k$ does not vanish. A series $f(t)$ is called a delta series if $o(f(t))=1$, and an invertible series if $o(f(t))=0$. Let $f(t),g(t)\in \mathcal{F}$. Then we have

$$〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉=〈g(t)|f(t)p(x)〉(\text{see [14]}).$$

(1.10)

For $f(t),g(t)\in \mathcal{F}$ with $o(f(t))=1$, $o(g(t))=0$, there exists a unique sequence $S_n(x)$ ($deg{S}_n(x)=n$) such that $〈g(t)f{(t)}^k|S_n(x)〉=n!{\delta }_{n,k}$ for $n,k\ge 0$. The sequence $S_n(x)$ is called the Sheffer sequence for $(g(t),f(t))$, which is denoted by $S_n(x)\sim (g(t),f(t))$ (see [14, 15]). Let $f(t)\in \mathcal{F}$ and $p(t)\in \mathbb{P}$. Then we have

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

(1.11)

From (1.11), we note that

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

(1.12)

By (1.12), we get

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}(\text{see [14, 15]}).$$

(1.13)

From (1.13), we easily derive the following equation

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

(1.14)

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

$$〈f(t)|xp(x)〉=〈{\partial }_{t}f(t)|p(x)〉=〈f^{\prime }(t)|p(x)〉(\text{see [14]}).$$

(1.15)

Let $S_n(x)\sim (g(t),f(t))$. Then we have

$$\frac{1}{g(\overline{f}(x))}{e}^{y\overline{f}(t)}=\sum _{n=0}^{\mathrm{\infty }}{S}_n(y)\frac{t^n}{n!}\text{for all }y\in \mathbb{C},$$

(1.16)

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

$$f(t)S_n(x)=n{S}_{n-1}(x)(\text{see [14, 15]}).$$

(1.17)

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

$${(e^t-1)}^m=m!\sum _{l=m}^{\mathrm{\infty }}{S}_2(l,m)\frac{t^m}{m!}(m\in {\mathbb{Z}}_{\ge 0}).$$

(1.18)

For $S_n(x)\sim (g(t),t)$, it is well known that

$$S_{n+1}(x)=(x-\frac{g^{\prime }(t)}{g(t)})S_n(x)(n\ge 0)(\text{see [14, 15]}).$$

(1.19)

Let $S_n(x)\sim (g(t),f(t))$, $r_n(x)\sim (h(t),l(t))$. Then we have

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

(1.20)

where

$$C_{n,m}=\frac{1}{m!}〈\frac{h(\overline{f}(t))}{g(\overline{f}(t))}l{(\overline{f}(t))}^m|x^n〉(\text{see [14, 15]}).$$

(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 $T_n^{(r,k)}(x|\lambda )$, called higher-order Frobenius-Euler and poly-Bernoulli mixed-type polynomials, as follows:

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

(2.1)

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

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

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

$$T_n^{(r,k)}(x|\lambda )\sim (g_{r,k}(t)={\left(\frac{e^t-\lambda }{1-\lambda }\right)}^r\frac{1-e^{-t}}{L{i}_k(1-e^{-t})},t).$$

(2.2)

By (1.17) and (2.2), we get

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

(2.3)

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

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

(2.4)

By (1.16) and (2.2), we get

$$T_n^{(r,k)}(x|\lambda )=\frac{1}{g_{r,k}(t)}{x}^n={\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{x}^n.$$

(2.5)

In [7], it is known that

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

(2.6)

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

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

(2.7)

By (1.1), we easily see that

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

(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

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

In [7], it is known that

$$\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{x}^n=\sum _{j=0}^n\{\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m)\}{x}^j.$$

(2.9)

By (2.5) and (2.9), we get

$$\begin{array}{rl}{T}_n^{(r,k)}(x|\lambda )& ={\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{x}^n\\ =\sum _{j=0}^n\{\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m)\}{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r{x}^j\\ =\sum _{j=0}^n\{\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m)\}{H}_j^{(r)}(x|\lambda ).\end{array}$$

(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

$$T_n^{(r,k)}(x|\lambda )=\sum _{l=0}^n\{\sum _{j=l}^n\sum _{m=0}^{n-j}{(-1)}^{n-m-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{j}{l}\right)\frac{m!}{{(m+1)}^k}{H}_{j-l}^{(r)}(\lambda )S_2(n-j,m)\}{x}^l.$$

From (1.19) and (2.2), we have

$$T_{n+1}^{(r,k)}(x|\lambda )=(x-\frac{g_{r,k}^{\mathrm{\prime }}(t)}{g_{r,k}(t)})T_n^{(r,k)}(x|\lambda ).$$

(2.11)

Now, we note that

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

(2.12)

By (2.11) and (2.12), we get

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

(2.13)

It is easy to show that

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

(2.14)

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

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

(2.15)

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

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

(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

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

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

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

(2.17)

Thus, by (2.17), we get $B_n^{(k)}(x)=T_n^{(0,k)}(x|\lambda )$.

From (2.4), we have

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

(2.18)

Applying t on both sides of Theorem 2.3, we get

$$\begin{array}{r}(n+1)T_n^{(r,k)}(x|\lambda )\\ =nx{T}_{n-1}^{(r,k)}(x|\lambda )+T_n^{(r,k)}(x|\lambda )-rn{T}_{n-1}^{(r,k)}(x|\lambda )-\frac{rn\lambda }{1-\lambda }{T}_{n-1}^{(r+1,k)}(x|\lambda )\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0ex}{}{n+1}{l}\right)B_l\{(n+1-l)T_{n-l}^{(r,k)}(x|\lambda )-(n+1-l)T_{n-l}^{(r,k-1)}(x|\lambda )\}.\end{array}$$

(2.19)

Thus, by (2.19), we have

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

(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

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

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

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

(2.21)

By (1.15) and (2.21), we get

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

(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

$$\begin{array}{rl}{T}_n^{(r,k)}(x|\lambda )=& (x-r)T_{n-1}^{(r,k)}(x|\lambda )-\frac{r\lambda }{1-\lambda }{T}_{n-1}^{(r+1,k)}(x|\lambda )\\ +\sum _{l=0}^{n-1}\{{(-1)}^{n-1-l}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\sum _{m=0}^{n-1-l}{(-1)}^m\frac{(m+1)!}{{(m+2)}^k}{S}_2(n-1-l,m)\}{H}_l^{(r)}(x-1|\lambda ).\end{array}$$

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

On the one hand,

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

(2.23)

On the other hand, we get

$$\begin{array}{r}〈{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^{r}L{i}_k(1-e^{-t})|x^{n+1}〉\\ =〈L{i}_k(1-e^{-t})|{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r{x}^{n+1}〉\\ =〈{\int }_0^t{\left(L{i}_k(1-e^{-s})\right)}^{\prime }ds|H_{n+1}^{(r)}(x|\lambda )〉\\ =〈{\int }_0^t{e}^{-s}\frac{L{i}_k(1-e^{-s})}{(1-e^{-s})}ds|H_{n+1}^{(r)}(x|\lambda )〉\\ =\sum _{l=0}^n\left(\sum _{m=0}^l\left(\genfrac{}{}{0ex}{}{l}{m}\right){(-1)}^{l-m}{B}_m^{(k-1)}\right)\frac{1}{l!}〈{\int }_0^t{s}^{l}ds|H_{n+1}^{(r)}(x|\lambda )〉\\ =\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0ex}{}{l}{m}\right){(-1)}^{l-m}\frac{B_m^{(k-1)}}{(l+1)!}〈t^{l+1}|H_{n+1}^{(r)}(x|\lambda )〉\\ =\sum _{l=0}^n\sum _{m=0}^l\left(\genfrac{}{}{0ex}{}{l}{m}\right)\left(\genfrac{}{}{0ex}{}{n+1}{l+1}\right){(-1)}^{l-m}{B}_m^{(k-1)}{H}_{n-l}^{(r)}(\lambda ).\end{array}$$

(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

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

Now, we consider the following two Sheffer sequences:

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

(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

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

(2.26)

By (1.21) and (2.26), we get

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{e^t-1}{t}\right)}^s{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{t}^m|x^n〉\\ =& \frac{1}{m!}〈{\left(\frac{e^t-1}{t}\right)}^s{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}|t^m{x}^n〉\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)〈{\left(\frac{e^t-1}{t}\right)}^s{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}|x^{n-m}〉\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!}{(l+s)!}{S}_2(l+s,s)〈{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}|t^l{x}^{n-m}〉\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!l!}{(l+s)!}\frac{{(n-m)}_l}{l!}{S}_2(l+s,s)〈1|T_{n-m-l}^{(r,k)}(x|\lambda )〉\\ =& \left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{s+l}{l}\right)}{S}_2(l+s,s)T_{n-m-l}^{(r,k)}(\lambda ).\end{array}$$

(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

$$T_n^{(r,k)}(x|\lambda )=\sum _{m=0}^n\{\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{s+l}{l}\right)}{S}_2(l+s,s)T_{n-m-l}^{(r,k)}(\lambda )\}{\mathbb{B}}_m^{(s)}(x).$$

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

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

(2.28)

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

By the same method, we get

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

(2.29)

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

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

(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

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

(2.31)

By (1.21) and (2.31), we get

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

(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

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

It is known that

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

(2.33)

Let

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

(2.34)

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

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{(e^t-1)}^m|x^n〉\\ =& \sum _{l=0}^{\mathrm{\infty }}\frac{S_2(l+m,m)}{(l+m)!}〈{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}|t^{m+l}{x}^n〉\\ =& \sum _{l=0}^{n-m}\frac{S_2(l+m,m)}{(l+m)!}{(n)}_{m+l}〈1|{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{x}^{n-m-l}〉\\ =& \sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)S_2(l+m,m)T_{n-m-l}^{(r,k)}(\lambda ).\end{array}$$

(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

$$T_n^{(r,k)}(x|\lambda )=\sum _{m=0}^n\{\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)S_2(l+m,m)T_{n-m-l}^{(r,k)}(\lambda )\}{(x)}_m.$$

Finally, we consider the following two Sheffer sequences:

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

(2.36)

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

Let us assume that

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

(2.37)

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

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{(1-e^{-t})}^m|x^n〉\\ =& \sum _{l=0}^{\mathrm{\infty }}\frac{{(-1)}^l{S}_2(l+m,m)}{(l+m)!}〈{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}|t^{m+l}{x}^n〉\\ =& \sum _{l=0}^{n-m}\frac{{(-1)}^l{S}_2(l+m,m)}{(l+m)!}{(n)}_{m+l}〈1|{\left(\frac{1-\lambda }{e^t-\lambda }\right)}^r\frac{L{i}_k(1-e^{-t})}{1-e^{-t}}{x}^{n-m-l}〉\\ =& \sum _{l=0}^{n-m}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)S_2(l+m,m)T_{n-m-l}^{(r,k)}(\lambda ).\end{array}$$

(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

$$T_n^{(r,k)}(x|\lambda )=\sum _{m=0}^n\{\sum _{l=0}^{n-m}{(-1)}^l\left(\genfrac{}{}{0ex}{}{n}{l+m}\right)S_2(l+m,m)T_{n-m-l}^{(r,k)}(\lambda )\}{x}^{[m]}.$$