# Barnes-type Peters polynomial with umbral calculus viewpoint

## Abstract

In this paper, we consider the Barnes-type Peters polynomials. We present several explicit formulas and recurrence relations for these polynomials. Also, we establish a connection between our polynomials and several known families of polynomials.

MSC:05A40, 11B83.

## 1 Introduction

The aim of this paper is to use umbral calculus to obtain several new and interesting identities of Barnes-type Peters polynomials. Umbral calculus has been used in numerous problems of mathematics (for example, see ). Umbral techniques have been used in different areas of physics; for example, it was used in group theory and quantum mechanics by Biedenharn et al. [11, 12] (for other examples, see [3, 10, 1318]).

Let $r\in {\mathbb{Z}}_{>0}$. Here we will consider the polynomials ${S}_{n}\left(x\right)={S}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r};{\mu }_{1},\dots ,{\mu }_{r}\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)={\stackrel{ˆ}{S}}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r};{\mu }_{1},\dots ,{\mu }_{r}\right)$, which are called Barnes-type Peters polynomials of the first kind and of the second kind, respectively, and are given by

$\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{x}=\sum _{n\ge 0}{S}_{n}\left(x\right)\frac{{t}^{n}}{n!},$
(1.1)
$\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{x}=\sum _{n\ge 0}{\stackrel{ˆ}{S}}_{n}\left(x\right)\frac{{t}^{n}}{n!},$
(1.2)

where ${\lambda }_{1},\dots ,{\lambda }_{r},{\mu }_{1},\dots ,{\mu }_{r}\in \mathbb{C}$ with ${\lambda }_{1},\dots ,{\lambda }_{r}\ne 0$. If $r=1$, then these polynomials are generalizations of Boole polynomials, see . If ${\mu }_{1}=\cdots ={\mu }_{r}=1$, then ${S}_{n}\left(x|\mathbit{\lambda }\right)={S}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r}\right)={S}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r};1,\dots ,1\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x|\mathbit{\lambda }\right)={\stackrel{ˆ}{S}}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r}\right)={\stackrel{ˆ}{S}}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r};1,\dots ,1\right)$ are called Barnes-type Boole polynomials of the first kind and of the second kind. So,

$\begin{array}{r}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-1}{\left(1+t\right)}^{x}=\sum _{n\ge 0}{S}_{n}\left(x|\mathbit{\lambda }\right)\frac{{t}^{n}}{n!},\\ \prod _{j=1}^{r}\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right){\left(1+t\right)}^{x}=\sum _{n\ge 0}{\stackrel{ˆ}{S}}_{n}\left(x|\mathbit{\lambda }\right)\frac{{t}^{n}}{n!}.\end{array}$

We introduce the polynomials ${E}_{n}\left(x|\mathbit{\lambda };\mathbit{\mu }\right)={E}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r};{\mu }_{1},\dots ,{\mu }_{r}\right)$ with the generating function

$\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{e}^{xt}=\sum _{n\ge 0}{E}_{n}\left(x|\mathbit{\lambda };\mathbit{\mu }\right)\frac{{t}^{n}}{n!}.$

These polynomials may be called generalized Barnes-type Euler polynomials. When ${\mu }_{1}=\cdots ={\mu }_{r}=1$, ${E}_{n}\left(x|\mathbit{\lambda }\right)={E}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r}\right)={E}_{n}\left(x|{\lambda }_{1},\dots ,{\lambda }_{r};1,\dots ,1\right)$ are called the Barnes-type Euler polynomials. If further ${\lambda }_{1}=\cdots ={\lambda }_{r}=1$, ${E}_{n}^{\left(r\right)}\left(x\right)={E}_{n}\left(x|1,\dots ,1;1,\dots ,1\right)$ are called the Euler polynomials of order r. When $x=0$, ${S}_{n}={S}_{n}\left(\mathbit{\lambda };\mathbit{\mu }\right)={S}_{n}\left(0|\mathbit{\lambda };\mathbit{\mu }\right)$ and ${\stackrel{ˆ}{S}}_{n}={\stackrel{ˆ}{S}}_{n}\left(\mathbit{\lambda };\mathbit{\mu }\right)={\stackrel{ˆ}{S}}_{n}\left(0|\mathbit{\lambda };\mathbit{\mu }\right)$ are called Barnes-type Peters numbers of the first kind and of the second kind, respectively.

Let Π be the algebra of polynomials in a single variable x over , and let ${\mathrm{\Pi }}^{\ast }$ be the vector space of all linear functionals on Π. We denote the action of a linear functional L on a polynomial $p\left(x\right)$ by $〈L|p\left(x\right)〉$, and we define the vector space structure on ${\mathrm{\Pi }}^{\ast }$ by

$〈cL+{c}^{\prime }{L}^{\prime }|p\left(x\right)〉=c〈L|p\left(x\right)〉+{c}^{\prime }〈{L}^{\prime }|p\left(x\right)〉,$

where $c,{c}^{\prime }\in \mathbb{C}$ (see ). We define the algebra of formal power series in a single variable t to be

$\mathcal{H}=\left\{f\left(t\right)=\sum _{k\ge 0}{a}_{k}\frac{{t}^{k}}{k!}|{a}_{k}\in \mathbb{C}\right\}.$
(1.3)

The formal power series in the variable t defines a linear functional on Π by setting

(1.4)

By (1.3) and (1.4), we have

(1.5)

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

Let ${f}_{L}\left(t\right)={\sum }_{n\ge 0}〈L|{x}^{n}〉\frac{{t}^{n}}{n!}$. From (1.5), we have $〈{f}_{L}\left(t\right)|{x}^{n}〉=〈L|{x}^{n}〉$. Thus, the map $L↦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathrm{\Pi }}^{\ast }$ onto . Therefore, is thought of as a set of both formal power series and linear functionals. We call umbral algebra. Umbral calculus is the study of umbral algebra.

The order $O\left(f\left(t\right)\right)$ of the non-zero power series $f\left(t\right)$ is the smallest integer k for which the coefficient of ${t}^{k}$ does not vanish (see ). If $O\left(f\left(t\right)\right)=1$ (respectively, $O\left(f\left(t\right)\right)=0$), then $f\left(t\right)$ is called a delta (respectively, an invertible) series. Suppose that $O\left(f\left(t\right)\right)=1$ and $O\left(g\left(t\right)\right)=0$, then there exists a unique sequence ${s}_{n}\left(x\right)$ of polynomials such that $〈g\left(t\right){\left(f\left(t\right)\right)}^{k}|{s}_{n}\left(x\right)〉=n!{\delta }_{n,k}$, where $n,k\ge 0$ [, Theorem 2.3.1]. The sequence ${s}_{n}\left(x\right)$ is called the Sheffer sequence for $\left(g\left(t\right),f\left(t\right)\right)$ which is denoted by ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ (see ). For $f\left(t\right)\in \mathcal{H}$ and $p\left(x\right)\in \mathrm{\Pi }$, we have $〈{e}^{yt}|p\left(x\right)〉=p\left(y\right)$, $〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉$ and

$f\left(t\right)=\sum _{n\ge 0}〈f\left(t\right)|{x}^{n}〉\frac{{t}^{n}}{n!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{n\ge 0}〈{t}^{n}|p\left(x\right)〉\frac{{x}^{n}}{n!}$
(1.6)

(see ). From (1.6), we obtain

$〈{t}^{k}|p\left(x\right)〉={p}^{\left(k\right)}\left(0\right),\phantom{\rule{2em}{0ex}}〈1|{p}^{\left(k\right)}\left(x\right)〉={p}^{\left(k\right)}\left(0\right),$
(1.7)

where ${p}^{\left(k\right)}\left(0\right)$ denotes the k th derivative of $p\left(x\right)$ with respect to x at $x=0$. So, by (1.7), we get that ${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}}{d{x}^{k}}p\left(x\right)$ for all $k\ge 0$ (see ).

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

$\frac{1}{g\left(\overline{f}\left(t\right)\right)}{e}^{y\overline{f}\left(t\right)}=\sum _{n\ge 0}{s}_{n}\left(y\right)\frac{{t}^{n}}{n!},$
(1.8)

for all $y\in \mathbb{C}$, where $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$ (see ). For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ and ${r}_{n}\left(x\right)\sim \left(h\left(t\right),\ell \left(t\right)\right)$, let

${s}_{n}\left(x\right)=\sum _{k=0}^{n}{c}_{n,k}{r}_{k}\left(x\right),$
(1.9)

then we have

${c}_{n,k}=\frac{1}{k!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}{\left(\ell \left(\overline{f}\left(t\right)\right)\right)}^{k}|{x}^{n}〉$
(1.10)

(see ).

It is immediate from (1.1)-(1.2), we see that ${S}_{n}\left(x\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)$ are the Sheffer sequences for the pairs

${S}_{n}\left(x\right)\sim \left(\prod _{j=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{{\mu }_{j}},{e}^{t}-1\right),$
(1.11)
${\stackrel{ˆ}{S}}_{n}\left(x\right)\sim \left(\prod _{j=1}^{r}{\left(\frac{1+{e}^{{\lambda }_{j}t}}{{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}},{e}^{t}-1\right).$
(1.12)

The aim of the present paper is to present several new identities for the Peters polynomials by the use of umbral calculus.

## 2 Explicit expressions

It is well known that

${\left(x\right)}_{n}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right){x}^{m}\sim \left(1,{e}^{t}-1\right),$
(2.1)

where ${S}_{1}\left(n,m\right)$ is the Stirling number of the first kind. By (1.11) and (1.12) we have

$\prod _{j=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{{\mu }_{j}}{S}_{n}\left(x\right)\sim \left(1,{e}^{t}-1\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\prod _{j=1}^{r}{\left(\frac{1+{e}^{{\lambda }_{j}t}}{{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{\stackrel{ˆ}{S}}_{n}\left(x\right)\sim \left(1,{e}^{t}-1\right).$
(2.2)

So

$\begin{array}{rl}{S}_{n}\left(x\right)& =\prod _{j=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{-{\mu }_{j}}{\left(x\right)}_{n}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\prod _{j=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{-{\mu }_{j}}{x}^{m}\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{x}^{m}\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){E}_{m}\left(x|\mathbit{\lambda };\mathbit{\mu }\right),\end{array}$
(2.3)

which implies

$\begin{array}{rl}{\stackrel{ˆ}{S}}_{n}\left(x\right)& =\prod _{j=1}^{r}{\left(\frac{{e}^{{\lambda }_{j}t}}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{\left(x\right)}_{n}={e}^{{\sum }_{j=1}^{r}{\lambda }_{j}{\mu }_{j}t}\prod _{j=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{-{\mu }_{j}}{\left(x\right)}_{n}\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){e}^{{\sum }_{j=1}^{r}{\lambda }_{j}{\mu }_{j}t}\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{x}^{m}\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){e}^{{\sum }_{j=1}^{r}{\lambda }_{j}{\mu }_{j}t}{E}_{m}\left(x|\mathbit{\lambda };\mathbit{\mu }\right)\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){E}_{m}\left(x+\sum _{j=1}^{r}{\lambda }_{j}{\mu }_{j}|\mathbit{\lambda };\mathbit{\mu }\right).\end{array}$
(2.4)

Thus, we have the following result.

Theorem 1 For all $n\ge 0$,

$\begin{array}{r}{S}_{n}\left(x\right)={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){E}_{m}\left(x|\mathbit{\lambda };\mathbit{\mu }\right),\\ {\stackrel{ˆ}{S}}_{n}\left(x\right)={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){E}_{m}\left(x+\sum _{j=1}^{r}{\lambda }_{j}{\mu }_{j}|\mathbit{\lambda };\mathbit{\mu }\right).\end{array}$

By (1.6), (1.8), (1.11) and (1.12), we have

$\begin{array}{r}{S}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{j}|{x}^{n}〉{x}^{j},\\ {\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{j}|{x}^{n}〉{x}^{j},\end{array}$

where

$\begin{array}{r}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{j}|{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{\left(log\left(1+t\right)\right)}^{j}{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=j!\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right)〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{x}^{n-\ell }〉\\ \phantom{\rule{1em}{0ex}}=j!\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }\end{array}$

and

$\begin{array}{r}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{j}|{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}|{\left(log\left(1+t\right)\right)}^{j}{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=j!\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right)〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}|{x}^{n-\ell }〉\\ \phantom{\rule{1em}{0ex}}=j!\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){\stackrel{ˆ}{S}}_{n-\ell }.\end{array}$

Hence, we can state the following formulas.

Theorem 2 For all $n\ge 0$,

${S}_{n}\left(x\right)=\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }\right){x}^{j}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){\stackrel{ˆ}{S}}_{n-\ell }\right){x}^{j}.$

Also, by the definitions, (2.1), (1.11) and (1.12), we have

$\begin{array}{rl}{S}_{n}\left(y\right)& =〈\sum _{i\ge 0}{S}_{i}\left(y\right)\frac{{t}^{i}}{i!}|{x}^{n}〉=〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{\left(1+t\right)}^{y}{x}^{n}〉\\ =\sum _{m=0}^{n}{\left(y\right)}_{m}\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{x}^{n-m}〉\\ =\sum _{m=0}^{n}{\left(y\right)}_{m}\left(\genfrac{}{}{0}{}{n}{m}\right){S}_{n-m}\end{array}$

and

$\begin{array}{rl}{\stackrel{ˆ}{S}}_{n}\left(y\right)& =〈\sum _{i\ge 0}{\stackrel{ˆ}{S}}_{i}\left(y\right)\frac{{t}^{i}}{i!}|{x}^{n}〉=〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n}〉\\ =〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}|{\left(1+t\right)}^{y}{x}^{n}〉\\ =\sum _{m=0}^{n}{\left(y\right)}_{m}\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}|{x}^{n-m}〉\\ =\sum _{m=0}^{n}{\left(y\right)}_{m}\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{S}}_{n-m},\end{array}$

which implies the following formulas.

Theorem 3 For all $n\ge 0$,

${S}_{n}\left(x\right)=\sum _{j=0}^{n}{S}_{n-j}\left(\genfrac{}{}{0}{}{n}{j}\right){\left(x\right)}_{j}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{j=0}^{n}{\stackrel{ˆ}{S}}_{n-j}\left(\genfrac{}{}{0}{}{n}{j}\right){\left(x\right)}_{j}.$

More generally, by (2.1) and (2.2) with ${p}_{n}\left(x\right)={\prod }_{j=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{{\mu }_{j}}{S}_{n}\left(x\right)={\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right)$, we obtain that ${S}_{n}\left(x+y\right)={\sum }_{j=0}^{b}{S}_{j}\left(x\right){\left(y\right)}_{n-j}\left(\genfrac{}{}{0}{}{n}{j}\right)$, and with ${p}_{n}\left(x\right)={\prod }_{j=1}^{r}{\left(\frac{1+{e}^{{\lambda }_{j}t}}{{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{\stackrel{ˆ}{S}}_{n}\left(x\right)={\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right)$, we obtain that ${\stackrel{ˆ}{S}}_{n}\left(x+y\right)={\sum }_{j=0}^{b}{\stackrel{ˆ}{S}}_{j}\left(x\right){\left(y\right)}_{n-j}\left(\genfrac{}{}{0}{}{n}{j}\right)$, which gives the following corollary.

Corollary 1 For all $n\ge 0$,

${S}_{n}\left(x+y\right)=\sum _{j=0}^{b}{S}_{j}\left(x\right){\left(y\right)}_{n-j}\left(\genfrac{}{}{0}{}{n}{j}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\stackrel{ˆ}{S}}_{n}\left(x+y\right)=\sum _{j=0}^{b}{\stackrel{ˆ}{S}}_{j}\left(x\right){\left(y\right)}_{n-j}\left(\genfrac{}{}{0}{}{n}{j}\right).$

## 3 Recurrence relations

Note that if ${a}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, then $f\left(t\right){a}_{n}\left(x\right)=n{a}_{n-1}\left(x\right)$, Thus, by (1.11) and (1.12), we have that ${S}_{n}\left(x+1\right)-{S}_{n}\left(x\right)=\left({e}^{t}-1\right){S}_{n}\left(x\right)=n{S}_{n-1}\left(x\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x+1\right)-{\stackrel{ˆ}{S}}_{n}\left(x\right)=\left({e}^{t}-1\right){\stackrel{ˆ}{S}}_{n}\left(x\right)=n{\stackrel{ˆ}{S}}_{n-1}\left(x\right)$, which give the following recurrences.

Proposition 1 For all $n\ge 1$,

${S}_{n}\left(x+1\right)-{S}_{n}\left(x\right)=n{S}_{n-1}\left(x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\stackrel{ˆ}{S}}_{n}\left(x+1\right)-{\stackrel{ˆ}{S}}_{n}\left(x\right)=n{\stackrel{ˆ}{S}}_{n-1}\left(x\right).$

Note that for ${a}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we have that ${a}_{n+1}\left(x\right)=\left(x-{g}^{\prime }\left(t\right)/g\left(t\right)\right)\frac{1}{{f}^{\prime }\left(t\right)}{a}_{n}\left(x\right)$. In the case (1.11), we obtain ${S}_{n+1}\left(x\right)=x{S}_{n}\left(x-1\right)-{e}^{-t}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{S}_{n}\left(x\right)$ with $g\left(t\right)={\prod }_{i=1}^{r}{\left(1+{e}^{{\lambda }_{j}t}\right)}^{{\mu }_{j}}$. Since $\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}={\sum }_{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}$ and by (2.3), we get

$\begin{array}{rl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{S}_{n}\left(x\right)& =\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}{S}_{n}\left(x\right)=\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{S}_{n}\left(x\right)\\ =\sum _{i=1}^{r}\left(\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{x}^{m}\right)\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}\left(\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{x}^{m}\right)\\ ={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}}{2}{E}_{m}\left(x+{\lambda }_{i}|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\end{array}$

where ${e}_{i}=\left(0,\dots ,0,1,0,\dots ,0\right)$ is a vector with 1 in the i th coordinate. Thus,

${S}_{n+1}\left(x\right)=x{S}_{n-1}\left(x\right)-{2}^{-1-{\sum }_{i=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}\sum _{i=1}^{r}{S}_{1}\left(n,m\right){\lambda }_{i}{\mu }_{i}{E}_{m}\left(x+{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).$
(3.1)

On the other hand, by Theorem 2, we have

$\begin{array}{rl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{S}_{n}\left(x\right)& =\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}{S}_{n}\left(x\right)=\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{S}_{n}\left(x\right)\\ =\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }\right){x}^{j}\\ =\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{x}^{j}\right)\\ =\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}{\lambda }_{i}^{j}{E}_{j}\left(x/{\lambda }_{i}\right)\right)\\ =\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}{E}_{j}\left(1+x/{\lambda }_{i}\right)\right)\end{array}$

(note that ${E}_{n}\left(x\right)=\frac{2}{1+{e}^{t}}{x}^{n}={\left(E+x\right)}^{n}={\sum }_{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){E}_{j}{x}^{n-j}$ and $\frac{2}{1+{e}^{{\lambda }_{j}t}}{x}^{j}={\lambda }_{i}^{j}{E}_{i}\left(x/{\lambda }_{i}\right)$), which implies

${S}_{n+1}\left(x\right)=x{S}_{n}\left(x-1\right)-\sum _{j=0}^{n}\sum _{\ell =j}^{n}\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }{E}_{j}\left(1+\left(x-1\right)/{\lambda }_{i}\right).$

Thus, by (3.1), we can state the following result.

Theorem 4 For all $n\ge 0$,

$\begin{array}{r}{S}_{n+1}\left(x\right)=x{S}_{n}\left(x-1\right)-{2}^{-1-{\sum }_{i=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}\sum _{i=1}^{r}{S}_{1}\left(n,m\right){\lambda }_{i}{\mu }_{i}{E}_{m}\left(x+{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\\ {S}_{n+1}\left(x\right)=x{S}_{n}\left(x-1\right)-\sum _{j=0}^{n}\sum _{\ell =j}^{n}\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }{E}_{j}\left(1+\left(x-1\right)/{\lambda }_{i}\right).\end{array}$

As a corollary, we get the following identity.

Corollary 2 For all $n\ge 0$,

$\begin{array}{r}{2}^{-1-{\sum }_{i=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}\sum _{i=1}^{r}{S}_{1}\left(n,m\right){\lambda }_{i}{\mu }_{i}{E}_{m}\left(x+{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\sum _{\ell =j}^{n}\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){S}_{n-\ell }{E}_{j}\left(1+\left(x-1\right)/{\lambda }_{i}\right).\end{array}$

In the case (1.12), we obtain ${\stackrel{ˆ}{S}}_{n+1}\left(x\right)=x{\stackrel{ˆ}{S}}_{n}\left(x-1\right)-{e}^{-t}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{\stackrel{ˆ}{S}}_{n}\left(x\right)$ with $g\left(t\right)={\prod }_{i=1}^{r}{\left(\frac{1+{e}^{{\lambda }_{j}t}}{{e}^{{\lambda }_{i}t}}\right)}^{{\mu }_{j}}$. Since $\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}={\sum }_{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}-{\sum }_{i=1}^{r}{\lambda }_{i}{\mu }_{i}$ and by (2.4), we get

$\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{S}_{n}\left(x\right)=\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}{\stackrel{ˆ}{S}}_{n}\left(x\right)-\mathbit{\lambda }\mathbit{\mu }{\stackrel{ˆ}{S}}_{n}\left(x\right),$

where $\mathbit{\lambda }\mathbit{\mu }={\sum }_{j=1}^{r}{\lambda }_{j}{\mu }_{j}$ and

$\begin{array}{r}\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}{\stackrel{ˆ}{S}}_{n}\left(x\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{\stackrel{ˆ}{S}}_{n}\left(x\right)\\ \phantom{\rule{1em}{0ex}}=\sum _{i=1}^{r}\left(\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right){e}^{{\sum }_{j=1}^{r}{\lambda }_{j}{\mu }_{j}t}\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{x}^{m}\right)\\ \phantom{\rule{1em}{0ex}}={2}^{-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}\left(\frac{{\lambda }_{i}{\mu }_{i}}{2}{e}^{{\lambda }_{i}t+{\sum }_{j=1}^{r}{\lambda }_{j}{\mu }_{j}t}\frac{2}{1+{e}^{{\lambda }_{i}t}}\prod _{j=1}^{r}{\left(\frac{2}{1+{e}^{{\lambda }_{j}t}}\right)}^{{\mu }_{j}}{x}^{m}\right)\\ \phantom{\rule{1em}{0ex}}={2}^{-1-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{E}_{n}\left(x+\mathbit{\lambda }\left(\mathbit{\mu }+{e}_{i}\right)|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$

So

$\begin{array}{rl}{\stackrel{ˆ}{S}}_{n+1}\left(x\right)=& \left(x+\mathbit{\lambda }\mathbit{\mu }\right){\stackrel{ˆ}{S}}_{n}\left(x-1\right)\\ -{2}^{-1-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{E}_{n}\left(x+\mathbit{\lambda }\left(\mathbit{\mu }+{e}_{i}\right)-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$
(3.2)

On the other hand, by Theorem 2, we have

$\begin{array}{rl}\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}{\stackrel{ˆ}{S}}_{n}\left(x\right)& =\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{1+{e}^{{\lambda }_{i}t}}{\stackrel{ˆ}{S}}_{n}\left(x\right)-\mathbit{\lambda }\mathbit{\mu }{\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}{\stackrel{ˆ}{S}}_{n}\left(x\right)-\mathbit{\lambda }\mathbit{\mu }{\stackrel{ˆ}{S}}_{n}\left(x\right)\\ =\sum _{i=1}^{r}\frac{{\lambda }_{i}{\mu }_{i}{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+{e}^{{\lambda }_{i}t}}\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){\stackrel{ˆ}{S}}_{n-\ell }\right){x}^{j}-\mathbit{\lambda }\mathbit{\mu }{\stackrel{ˆ}{S}}_{n}\left(x\right)\\ =\sum _{j=0}^{n}\left(\sum _{\ell =j}^{n}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){\stackrel{ˆ}{S}}_{n-\ell }\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}{E}_{j}\left(1+x/{\lambda }_{i}\right)\right)-\mathbit{\lambda }\mathbit{\mu }{\stackrel{ˆ}{S}}_{n}\left(x\right).\end{array}$

Therefore, by (3.2), we have the following result.

Theorem 5 For all $n\ge 0$,

$\begin{array}{c}\begin{array}{rl}{\stackrel{ˆ}{S}}_{n+1}\left(x\right)=& \left(x+\mathbit{\lambda }\mathbit{\mu }\right){\stackrel{ˆ}{S}}_{n}\left(x-1\right)\\ -{2}^{-1-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{E}_{n}\left(x+\mathbit{\lambda }\left(\mathbit{\mu }+{e}_{i}\right)-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\end{array}\hfill \\ {\stackrel{ˆ}{S}}_{n+1}\left(x\right)=\left(x+\mathbit{\lambda }\mathbit{\mu }\right){\stackrel{ˆ}{S}}_{n}\left(x-1\right)-\sum _{j=0}^{n}\sum _{\ell =j}^{n}\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){\stackrel{ˆ}{S}}_{n-\ell }{E}_{j}\left(1+\left(x-1\right)/{\lambda }_{i}\right).\hfill \end{array}$

As a corollary, we get the following identity.

Corollary 3 For all $n\ge 0$,

$\begin{array}{r}{2}^{-1-{\sum }_{j=1}^{r}{\mu }_{j}}\sum _{m=0}^{n}{S}_{1}\left(n,m\right)\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{E}_{n}\left(x+\mathbit{\lambda }\left(\mathbit{\mu }+{e}_{i}\right)-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\sum _{\ell =j}^{n}\sum _{i=1}^{r}\frac{{\lambda }_{i}^{j+1}{\mu }_{i}}{2}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(\ell ,j\right){\stackrel{ˆ}{S}}_{n-\ell }{E}_{j}\left(1+\left(x-1\right)/{\lambda }_{i}\right).\end{array}$

Recall that for ${a}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we have $\frac{d}{dx}{a}_{n}\left(x\right)={\sum }_{\ell =0}^{n-1}\left(\genfrac{}{}{0}{}{n}{\ell }\right)〈\overline{f}\left(t\right)|{x}^{n-\ell }〉{a}_{\ell }\left(x\right)$. Hence, in the case (1.11), namely ${a}_{n}\left(x\right)={S}_{n}\left(x\right)$, we have

$\begin{array}{rl}〈\overline{f}\left(t\right)|{x}^{n-\ell }〉& =〈log\left(1+t\right)|{x}^{n-\ell }〉\\ =〈\sum _{m\ge 1}\frac{{\left(-1\right)}^{m-1}{x}^{m}}{m}|{x}^{n-\ell }〉={\left(-1\right)}^{n-\ell -1}\left(n-\ell -1\right)!,\end{array}$

which implies $d/dx{S}_{n}\left(x\right)=n!{\sum }_{\ell =0}^{n-1}\frac{{\left(-1\right)}^{n-\ell -1}}{\ell !\left(n-\ell \right)}{S}_{\ell }\left(x\right)$. In the same way, we obtain the case ${\stackrel{ˆ}{S}}_{n}\left(x\right)$, which leads to the following result.

Theorem 6 For all $n\ge 1$,

$\frac{d}{dx}{S}_{n}\left(x\right)=n!\sum _{\ell =0}^{n-1}\frac{{\left(-1\right)}^{n-\ell -1}}{\ell !\left(n-\ell \right)}{S}_{\ell }\left(x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\frac{d}{dx}{\stackrel{ˆ}{S}}_{n}\left(x\right)=n!\sum _{\ell =0}^{n-1}\frac{{\left(-1\right)}^{n-\ell -1}}{\ell !\left(n-\ell \right)}{\stackrel{ˆ}{S}}_{\ell }\left(x\right).$

Now we find another recurrence relation by using the derivative operator. For $n\ge 1$, by (1.11) we have

$\begin{array}{rl}{S}_{n}\left(y\right)=& 〈\sum _{i\ge 0}{S}_{i}\left(y\right)\frac{{t}^{i}}{i!}|{x}^{n}〉=〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n}〉\\ =& 〈\frac{d}{dt}\left(\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{y}\right)|{x}^{n-1}〉\\ =& 〈\frac{d}{dt}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n-1}〉\\ +〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}\frac{d}{dt}{\left(1+t\right)}^{y}|{x}^{n-1}〉\\ =& 〈\frac{d}{dt}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n-1}〉+y{S}_{n-1}\left(y-1\right).\end{array}$

Observe that $\frac{d}{dt}{\prod }_{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}=-{\prod }_{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{{\mu }_{j}}{\sum }_{i=1}^{r}{\lambda }_{i}{\mu }_{i}\frac{{\left(1+t\right)}^{{\lambda }_{i}-1}}{1+{\left(1+t\right)}^{{\lambda }_{i}}}$. Thus,

$\begin{array}{r}〈\frac{d}{dt}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}〈{\left(1+{\left(1+t\right)}^{{\lambda }_{i}}\right)}^{-1}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{y+{\lambda }_{i}-1}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{S}_{n-1}\left(y+{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$

Hence,

${S}_{n}\left(x\right)=x{S}_{n-1}\left(x-1\right)-\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{S}_{n-1}\left(x+{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).$
(3.3)

Also, for $n\ge 1$, by (1.12) we have

$\begin{array}{rl}{\stackrel{ˆ}{S}}_{n}\left(y\right)=& 〈\sum _{i\ge 0}{\stackrel{ˆ}{S}}_{i}\left(y\right)\frac{{t}^{i}}{i!}|{x}^{n}〉\\ =& 〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n}〉\\ =& 〈\frac{d}{dt}\left[\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}\right]|{x}^{n-1}〉\\ =& 〈\frac{d}{dt}\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n-1}〉\\ +〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}\frac{d}{dt}{\left(1+t\right)}^{y}|{x}^{n-1}〉\\ =& 〈\frac{d}{dt}\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n-1}〉+y{\stackrel{ˆ}{S}}_{n-1}\left(y-1\right).\end{array}$

Observe that $\frac{d}{dt}{\prod }_{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}={\prod }_{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\sum }_{i=1}^{r}{\lambda }_{i}{\mu }_{i}{\left(1+t\right)}^{-{\lambda }_{i}-1}\frac{{\left(1+t\right)}^{{\lambda }_{i}}}{1+{\left(1+t\right)}^{{\lambda }_{i}}}$. So

$\begin{array}{r}〈\frac{d}{dt}\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}〈\frac{{\left(1+t\right)}^{{\lambda }_{i}}}{1+{\left(1+t\right)}^{{\lambda }_{i}}}\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(1+t\right)}^{y-{\lambda }_{i}-1}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{\stackrel{ˆ}{S}}_{n-1}\left(y-{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$

Thus,

${\stackrel{ˆ}{S}}_{n}\left(x\right)=x{\stackrel{ˆ}{S}}_{n-1}\left(x-1\right)+\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{\stackrel{ˆ}{S}}_{n-1}\left(x-{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).$
(3.4)

Hence, by (3.3) and (3.4), we obtain the following result.

Theorem 7 For $n\ge 1$,

$\begin{array}{r}{S}_{n}\left(x\right)=x{S}_{n-1}\left(x-1\right)-\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{S}_{n-1}\left(x+{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\\ {\stackrel{ˆ}{S}}_{n}\left(x\right)=x{\stackrel{ˆ}{S}}_{n-1}\left(x-1\right)+\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{\stackrel{ˆ}{S}}_{n-1}\left(x-{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$

Another result that can be obtained is the following.

Theorem 8 For $n-1\ge m\ge 1$,

$\begin{array}{c}\begin{array}{rl}\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(n-\ell ,m\right){S}_{\ell }=& \sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m-1\right){S}_{\ell }\left(-1\right)\\ -\sum _{\ell =0}^{n-1-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m\right)\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{S}_{\ell }\left({\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right),\end{array}\hfill \\ \begin{array}{rl}\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(n-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }=& \sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m-1\right){\stackrel{ˆ}{S}}_{\ell }\left(-1\right)\\ +\sum _{\ell =0}^{n-1-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m\right)\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}{\stackrel{ˆ}{S}}_{\ell }\left(-{\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}\hfill \end{array}$

Proof Because of the similarity in the two cases ${S}_{n}\left(x\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)$, we only give the proof of the first identity. In order to prove the first identity, we compute the following in two different ways:

$A=〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉.$

On the one hand, it is equal to

$\begin{array}{rl}A& =〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n}〉\\ =〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|m!\sum _{\ell \ge m}{S}_{1}\left(\ell ,m\right)\frac{{t}^{\ell }}{\ell !}{x}^{n}〉\\ =m!\sum _{\ell =m}^{n}{S}_{1}\left(\ell ,m\right)\left(\genfrac{}{}{0}{}{n}{\ell }\right)〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{x}^{n-\ell }〉\\ =m!\sum _{\ell =m}^{n}{S}_{1}\left(\ell ,m\right)\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{n-\ell }\\ =m!\sum _{\ell =0}^{n-m}{S}_{1}\left(n-\ell ,m\right)\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{\ell }.\end{array}$
(3.5)

On the other hand,

$\begin{array}{rl}A=& 〈\frac{d}{dt}\left[\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}\right]|{x}^{n-1}〉\\ =& 〈\frac{d}{dt}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\\ +〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}\frac{d}{dt}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-1}〉.\end{array}$

Here,

$\begin{array}{r}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}\frac{d}{dt}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=m〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{-1}|{\left(log\left(1+t\right)\right)}^{m-1}{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=m!\sum _{\ell =m-1}^{n-1}{S}_{1}\left(\ell ,m-1\right)〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{-1}|\frac{{t}^{\ell }}{\ell !}{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m-1\right)〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{-1}|{x}^{\ell }〉\\ \phantom{\rule{1em}{0ex}}=m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m-1\right){S}_{\ell }\left(-1\right)\end{array}$

and

$\begin{array}{r}〈\frac{d}{dt}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}〈{\left(1+{\left(1+t\right)}^{{\lambda }_{i}}\right)}^{-1}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{{\lambda }_{i}-1}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\sum _{i=1}^{r}{\lambda }_{i}{\mu }_{i}〈{\left(1+{\left(1+t\right)}^{{\lambda }_{i}}\right)}^{-1}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{{\lambda }_{i}-1}|m!\sum _{\ell \ge m}{S}_{1}\left(\ell ,m\right)\frac{{t}^{\ell }}{\ell !}{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-m!\sum _{i=1}^{r}\sum _{\ell =m}^{n-1}{\lambda }_{i}{\mu }_{i}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(\ell ,m\right)\\ \phantom{\rule{2em}{0ex}}×〈{\left(1+{\left(1+t\right)}^{{\lambda }_{i}}\right)}^{-1}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{{\lambda }_{i}-1}|{x}^{n-1-\ell }〉\\ \phantom{\rule{1em}{0ex}}=-m!\sum _{i=1}^{r}\sum _{\ell =0}^{n-1-m}{\lambda }_{i}{\mu }_{i}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m\right)\\ \phantom{\rule{2em}{0ex}}×〈{\left(1+{\left(1+t\right)}^{{\lambda }_{i}}\right)}^{-1}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(1+t\right)}^{{\lambda }_{i}-1}|{x}^{\ell }〉\\ \phantom{\rule{1em}{0ex}}=-m!\sum _{i=1}^{r}\sum _{\ell =0}^{n-1-m}{\lambda }_{i}{\mu }_{i}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m\right){S}_{\ell }\left({\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$

Altogether, we have, for $n-1\ge m\ge 1$,

$\begin{array}{r}m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n}{\ell }\right){S}_{1}\left(n-\ell ,m\right){S}_{\ell }\\ \phantom{\rule{1em}{0ex}}=m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m-1\right){S}_{\ell }\left(-1\right)\\ \phantom{\rule{2em}{0ex}}-m!\sum _{i=1}^{r}\sum _{\ell =0}^{n-1-m}{\lambda }_{i}{\mu }_{i}\left(\genfrac{}{}{0}{}{n-1}{\ell }\right){S}_{1}\left(n-1-\ell ,m\right){S}_{\ell }\left({\lambda }_{i}-1|\mathbit{\lambda };\mathbit{\mu }+{e}_{i}\right).\end{array}$

By dividing by m!, we complete the proof. □

## 4 Identities

Let ${S}_{n}\left(x\right)={\sum }_{m=0}^{n}{c}_{n,m}{\left(x\right)}_{m}$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)={\sum }_{m=0}^{n}{\stackrel{ˆ}{c}}_{n,m}{\left(x\right)}_{m}$. By (1.9), (1.10) and (1.11), we obtain

$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{t}^{m}{x}^{n}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}|{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right){S}_{n-m},\end{array}$

and by (1.9), (1.10) and (1.12), we obtain

$\begin{array}{rl}{\stackrel{ˆ}{c}}_{n,m}& =\frac{1}{m!}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}|{t}^{m}{x}^{n}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}|{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{S}}_{n-m}.\end{array}$

Hence, we have the following identities.

Theorem 9 For all $n\ge 0$,

${S}_{n}\left(x\right)=\sum _{m=0}^{n}{S}_{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right){\left(x\right)}_{m}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{m=0}^{n}{\stackrel{ˆ}{S}}_{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right){\left(x\right)}_{m}.$

Now, let ${S}_{n}\left(x\right)={\sum }_{m=0}^{n}{c}_{n,m}{H}_{m}^{\left(s\right)}\left(x|\alpha \right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)={\sum }_{m=0}^{n}{\stackrel{ˆ}{c}}_{n,m}{H}_{m}^{\left(s\right)}\left(x|\alpha \right)$, where ${H}_{n}^{\left(s\right)}\left(x|\alpha \right)\sim \left({\left(\frac{{e}^{t}-\alpha }{1-\alpha }\right)}^{s},t\right)$, with $\alpha \ne 1$. Then, by (1.9), (1.10) and (1.11), we obtain

$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{log\left(1+t\right)}-\alpha }{1-\alpha }\right)}^{s}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!{\left(1-\alpha \right)}^{s}}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}{\left(1-\alpha +t\right)}^{s}|{x}^{n}〉\\ =\frac{1}{m!{\left(1-\alpha \right)}^{s}}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|\sum _{j=0}^{min\left\{s,n\right\}}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(1-\alpha \right){t}^{j}{x}^{n}〉\\ =\frac{1}{m!{\left(1-\alpha \right)}^{s}}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(1-\alpha \right)}^{s-j}{\left(n\right)}_{j}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-j}〉,\end{array}$

and by Theorem 8, we have

$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!{\left(1-\alpha \right)}^{s}}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(1-\alpha \right)}^{s-j}{\left(n\right)}_{j}\left(m!\sum _{\ell =0}^{n-j-m}\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){S}_{1}\left(n-j-\ell ,m\right){S}_{\ell }\right)\\ =\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){\left(1-\alpha \right)}^{-j}{\left(n\right)}_{j}{S}_{1}\left(n-j-\ell ,m\right){S}_{\ell }.\end{array}$

By (1.9), (1.10) and (1.12), we obtain

$\begin{array}{rl}{\stackrel{ˆ}{c}}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{log\left(1+t\right)}-\alpha }{1-\alpha }\right)}^{s}\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!{\left(1-\alpha \right)}^{s}}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{\left(1-\alpha +t\right)}^{s}{x}^{n}〉\\ =\frac{1}{m!{\left(1-\alpha \right)}^{s}}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(1-\alpha \right)}^{s-j}{\left(n\right)}_{j}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-j}〉,\end{array}$

and by Theorem 8, we have

$\begin{array}{rl}{\stackrel{ˆ}{c}}_{n,m}& =\frac{1}{m!{\left(1-\alpha \right)}^{s}}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(1-\alpha \right)}^{s-j}{\left(n\right)}_{j}\left(m!\sum _{\ell =0}^{n-j-m}\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){S}_{1}\left(n-j-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }\right)\\ =\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){\left(1-\alpha \right)}^{-j}{\left(n\right)}_{j}{S}_{1}\left(n-j-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }.\end{array}$

Therefore, we can state the following result.

Theorem 10 For all $n\ge 0$,

$\begin{array}{r}{S}_{n}\left(x\right)=\sum _{m=0}^{n}\left(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){\left(1-\alpha \right)}^{-j}{\left(n\right)}_{j}{S}_{1}\left(n-j-\ell ,m\right){S}_{\ell }\right){H}_{m}^{\left(s\right)}\left(x|\alpha \right),\\ {\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{m=0}^{n}\left(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){\left(1-\alpha \right)}^{-j}{\left(n\right)}_{j}{S}_{1}\left(n-j-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }\right){H}_{m}^{\left(s\right)}\left(x|\alpha \right).\end{array}$

Finally, we express our polynomials ${S}_{n}\left(x\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)$ in terms of Bernoulli polynomials of order s. Let ${S}_{n}\left(x\right)={\sum }_{m=0}^{n}{c}_{n,m}{B}_{m}^{\left(s\right)}\left(x\right)$ and ${\stackrel{ˆ}{S}}_{n}\left(x\right)={\sum }_{m=0}^{n}{\stackrel{ˆ}{c}}_{n,m}{B}_{m}^{\left(s\right)}\left(x\right)$, where ${B}_{n}^{\left(s\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{s},t\right)$. Then, by (1.9), (1.10) and (1.11), we obtain

$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{log\left(1+t\right)}-1}{log\left(1+t\right)}\right)}^{s}\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{\left(\frac{t}{log\left(1+t\right)}\right)}^{s}{x}^{n}〉,\end{array}$

and by the fact that ${\left(\frac{t}{log\left(1+t\right)}\right)}^{s}={\sum }_{n\ge 0}{C}_{n}^{\left(s\right)}\frac{{t}^{n}}{n!}$, where ${C}_{n}^{\left(s\right)}$ is the Cauchy number of the first kind of order s, we derive

${c}_{n,m}=\frac{1}{m!}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{i}\right){C}_{i}^{\left(s\right)}〈\prod _{j=1}^{r}{\left(1+{\left(1+t\right)}^{{\lambda }_{j}}\right)}^{-{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-i}〉,$

and by Theorem 8, we obtain

$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{i}\right){C}_{i}^{\left(s\right)}\left(m!\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0}{}{n-i}{\ell }\right){S}_{1}\left(n-i-\ell ,m\right){S}_{\ell }\right)\\ =\sum _{i=0}^{n-m}\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(\genfrac{}{}{0}{}{n-i}{\ell }\right){C}_{i}^{\left(s\right)}{S}_{1}\left(n-i-\ell ,m\right){S}_{\ell }.\end{array}$

Also, by (1.9), (1.10) and (1.12), we obtain

$\begin{array}{rl}{\stackrel{ˆ}{c}}_{n,m}& =\frac{1}{m!}〈{\left(\frac{{e}^{log\left(1+t\right)}-1}{log\left(1+t\right)}\right)}^{s}\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{\left(\frac{t}{log\left(1+t\right)}\right)}^{s}{x}^{n}〉\\ =\frac{1}{m!}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{i}\right){C}_{i}^{\left(s\right)}〈\prod _{j=1}^{r}{\left(\frac{{\left(1+t\right)}^{{\lambda }_{j}}}{1+{\left(1+t\right)}^{{\lambda }_{j}}}\right)}^{{\mu }_{j}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-i}〉,\end{array}$

and by Theorem 8, we obtain

$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{i}\right){C}_{i}^{\left(s\right)}\left(m!\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0}{}{n-i}{\ell }\right){S}_{1}\left(n-i-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }\right)\\ =\sum _{i=0}^{n-m}\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(\genfrac{}{}{0}{}{n-i}{\ell }\right){C}_{i}^{\left(s\right)}{S}_{1}\left(n-i-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }.\end{array}$

Hence, we have the following identities.

Theorem 11 For all $n\ge 0$,

$\begin{array}{r}{S}_{n}\left(x\right)=\sum _{m=0}^{n}\left(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){C}_{j}^{\left(s\right)}{S}_{1}\left(n-j-\ell ,m\right){S}_{\ell }\right){B}_{m}^{\left(s\right)}\left(x\right),\\ {\stackrel{ˆ}{S}}_{n}\left(x\right)=\sum _{m=0}^{n}\left(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{\ell }\right){C}_{j}^{\left(s\right)}{S}_{1}\left(n-j-\ell ,m\right){\stackrel{ˆ}{S}}_{\ell }\right){B}_{m}^{\left(s\right)}\left(x\right).\end{array}$

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## Acknowledgements

The authors would like to thank the referees for their valuable comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786) and was partially supported by Kwangwoon University in 2014.

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