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 [1–10]). 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, 13–18]).

Let $r\in {\mathbb{Z}}_{>0}$. Here we will consider the polynomials $S_n(x)=S_n(x|{\lambda }_1,\dots ,{\lambda }_r;{\mu }_1,\dots ,{\mu }_r)$ and ${\hat{S}}_n(x)={\hat{S}}_n(x|{\lambda }_1,\dots ,{\lambda }_r;{\mu }_1,\dots ,{\mu }_r)$, which are called Barnes-type Peters polynomials of the first kind and of the second kind, respectively, and are given by

$$\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^x=\sum _{n\ge 0}{S}_n(x)\frac{t^n}{n!},$$

(1.1)

$$\prod _{j=1}^r{\left(\frac{{(1+t)}^{{\lambda }_j}}{1+{(1+t)}^{{\lambda }_j}}\right)}^{{\mu }_j}{(1+t)}^x=\sum _{n\ge 0}{\hat{S}}_n(x)\frac{t^n}{n!},$$

(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 [19]. If ${\mu }_1=\cdots ={\mu }_r=1$, then $S_n(x|\mathit{\lambda })=S_n(x|{\lambda }_1,\dots ,{\lambda }_r)=S_n(x|{\lambda }_1,\dots ,{\lambda }_r;1,\dots ,1)$ and ${\hat{S}}_n(x|\mathit{\lambda })={\hat{S}}_n(x|{\lambda }_1,\dots ,{\lambda }_r)={\hat{S}}_n(x|{\lambda }_1,\dots ,{\lambda }_r;1,\dots ,1)$ are called Barnes-type Boole polynomials of the first kind and of the second kind. So,

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

We introduce the polynomials $E_n(x|\mathit{\lambda };\mathit{\mu })=E_n(x|{\lambda }_1,\dots ,{\lambda }_r;{\mu }_1,\dots ,{\mu }_r)$ with the generating function

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

These polynomials may be called generalized Barnes-type Euler polynomials. When ${\mu }_1=\cdots ={\mu }_r=1$, $E_n(x|\mathit{\lambda })=E_n(x|{\lambda }_1,\dots ,{\lambda }_r)=E_n(x|{\lambda }_1,\dots ,{\lambda }_r;1,\dots ,1)$ are called the Barnes-type Euler polynomials. If further ${\lambda }_1=\cdots ={\lambda }_r=1$, $E_n^{(r)}(x)=E_n(x|1,\dots ,1;1,\dots ,1)$ are called the Euler polynomials of order r. When $x=0$, $S_n=S_n(\mathit{\lambda };\mathit{\mu })=S_n(0|\mathit{\lambda };\mathit{\mu })$ and ${\hat{S}}_n={\hat{S}}_n(\mathit{\lambda };\mathit{\mu })={\hat{S}}_n(0|\mathit{\lambda };\mathit{\mu })$ 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 ℂ, 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 $p(x)$ by $〈L|p(x)〉$, and we define the vector space structure on ${\mathrm{\Pi }}^{\ast }$ by

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

where $c,c^{\prime }\in \mathbb{C}$ (see [19–22]). We define the algebra of formal power series in a single variable t to be

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

(1.3)

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

$$〈f(t)|x^n〉=a_n\text{for all }n\ge 0(\text{see [19–22]}).$$

(1.4)

By (1.3) and (1.4), we have

$$〈t^k|x^n〉=n!{\delta }_{n,k}\text{for all }n,k\ge 0(\text{see [19–22]}),$$

(1.5)

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

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

The order $O(f(t))$ of the non-zero power series $f(t)$ is the smallest integer k for which the coefficient of $t^k$ does not vanish (see [19–22]). If $O(f(t))=1$ (respectively, $O(f(t))=0$), then $f(t)$ is called a delta (respectively, an invertible) series. Suppose that $O(f(t))=1$ and $O(g(t))=0$, then there exists a unique sequence $s_n(x)$ of polynomials such that $〈g(t){(f(t))}^k|s_n(x)〉=n!{\delta }_{n,k}$, where $n,k\ge 0$ [[19], Theorem 2.3.1]. 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 [19–22]). For $f(t)\in \mathcal{H}$ and $p(x)\in \mathrm{\Pi }$, we have $〈e^{yt}|p(x)〉=p(y)$, $〈f(t)g(t)|p(x)〉=〈g(t)|f(t)p(x)〉$ and

$$f(t)=\sum _{n\ge 0}〈f(t)|x^n〉\frac{t^n}{n!},p(x)=\sum _{n\ge 0}〈t^n|p(x)〉\frac{x^n}{n!}$$

(1.6)

(see [19–22]). From (1.6), we obtain

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

(1.7)

where $p^{(k)}(0)$ denotes the k th derivative of $p(x)$ with respect to x at $x=0$. So, by (1.7), we get that $t^{k}p(x)=p^{(k)}(x)=\frac{d^k}{d{x}^k}p(x)$ for all $k\ge 0$ (see [19–22]).

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

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

(1.8)

for all $y\in \mathbb{C}$, where $\overline{f}(t)$ is the compositional inverse of $f(t)$ (see [19–22]). For $s_n(x)\sim (g(t),f(t))$ and $r_n(x)\sim (h(t),\ell (t))$, let

$$s_n(x)=\sum _{k=0}^n{c}_{n,k}{r}_k(x),$$

(1.9)

then we have

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

(1.10)

(see [19–22]).

It is immediate from (1.1)-(1.2), we see that $S_n(x)$ and ${\hat{S}}_n(x)$ are the Sheffer sequences for the pairs

$$S_n(x)\sim (\prod _{j=1}^r{(1+e^{{\lambda }_{j}t})}^{{\mu }_j},e^t-1),$$

(1.11)

$${\hat{S}}_n(x)\sim (\prod _{j=1}^r{\left(\frac{1+e^{{\lambda }_{j}t}}{e^{{\lambda }_{j}t}}\right)}^{{\mu }_j},e^t-1).$$

(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

$${(x)}_n=\sum _{m=0}^n{S}_1(n,m)x^m\sim (1,e^t-1),$$

(2.1)

where $S_1(n,m)$ is the Stirling number of the first kind. By (1.11) and (1.12) we have

$$\prod _{j=1}^r{(1+e^{{\lambda }_{j}t})}^{{\mu }_j}{S}_n(x)\sim (1,e^t-1)\text{and}\prod _{j=1}^r{\left(\frac{1+e^{{\lambda }_{j}t}}{e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{\hat{S}}_n(x)\sim (1,e^t-1).$$

(2.2)

So

$$\begin{array}{rl}{S}_n(x)& =\prod _{j=1}^r{(1+e^{{\lambda }_{j}t})}^{-{\mu }_j}{(x)}_n=\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r{(1+e^{{\lambda }_{j}t})}^{-{\mu }_j}{x}^m\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r{\left(\frac{2}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{x}^m\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)E_m(x|\mathit{\lambda };\mathit{\mu }),\end{array}$$

(2.3)

which implies

$$\begin{array}{rl}{\hat{S}}_n(x)& =\prod _{j=1}^r{\left(\frac{e^{{\lambda }_{j}t}}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{(x)}_n=e^{{\sum }_{j=1}^r{\lambda }_j{\mu }_{j}t}\prod _{j=1}^r{(1+e^{{\lambda }_{j}t})}^{-{\mu }_j}{(x)}_n\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)e^{{\sum }_{j=1}^r{\lambda }_j{\mu }_{j}t}\prod _{j=1}^r{\left(\frac{2}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{x}^m\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)e^{{\sum }_{j=1}^r{\lambda }_j{\mu }_{j}t}{E}_m(x|\mathit{\lambda };\mathit{\mu })\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)E_m(x+\sum _{j=1}^r{\lambda }_j{\mu }_j|\mathit{\lambda };\mathit{\mu }).\end{array}$$

(2.4)

Thus, we have the following result.

Theorem 1 For all $n\ge 0$,

$$\begin{array}{r}{S}_n(x)=2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)E_m(x|\mathit{\lambda };\mathit{\mu }),\\ {\hat{S}}_n(x)=2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)E_m(x+\sum _{j=1}^r{\lambda }_j{\mu }_j|\mathit{\lambda };\mathit{\mu }).\end{array}$$

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

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

where

$$\begin{array}{r}〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(log(1+t))}^j|x^n〉\\ =〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}|{(log(1+t))}^j{x}^n〉\\ =j!\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}|x^{n-\ell }〉\\ =j!\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }\end{array}$$

and

$$\begin{array}{r}〈\prod _{j=1}^r{\left(\frac{{(1+t)}^{{\lambda }_j}}{1+{(1+t)}^{{\lambda }_j}}\right)}^{{\mu }_j}{(log(1+t))}^j|x^n〉\\ =〈\prod _{j=1}^r{\left(\frac{{(1+t)}^{{\lambda }_j}}{1+{(1+t)}^{{\lambda }_j}}\right)}^{{\mu }_j}|{(log(1+t))}^j{x}^n〉\\ =j!\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)〈\prod _{j=1}^r{\left(\frac{{(1+t)}^{{\lambda }_j}}{1+{(1+t)}^{{\lambda }_j}}\right)}^{{\mu }_j}|x^{n-\ell }〉\\ =j!\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j){\hat{S}}_{n-\ell }.\end{array}$$

Hence, we can state the following formulas.

Theorem 2 For all $n\ge 0$,

$$S_n(x)=\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell })x^j\mathit{\text{and}}{\hat{S}}_n(x)=\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j){\hat{S}}_{n-\ell })x^j.$$

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

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

and

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

which implies the following formulas.

Theorem 3 For all $n\ge 0$,

$$S_n(x)=\sum _{j=0}^n{S}_{n-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right){(x)}_j\mathit{\text{and}}{\hat{S}}_n(x)=\sum _{j=0}^n{\hat{S}}_{n-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right){(x)}_j.$$

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

Corollary 1 For all $n\ge 0$,

$$S_n(x+y)=\sum _{j=0}^b{S}_j(x){(y)}_{n-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\mathit{\text{and}}{\hat{S}}_n(x+y)=\sum _{j=0}^b{\hat{S}}_j(x){(y)}_{n-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right).$$

3 Recurrence relations

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

Proposition 1 For all $n\ge 1$,

$$S_n(x+1)-S_n(x)=n{S}_{n-1}(x)\mathit{\text{and}}{\hat{S}}_n(x+1)-{\hat{S}}_n(x)=n{\hat{S}}_{n-1}(x).$$

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

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}{S}_n(x)& =\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{1+e^{{\lambda }_{i}t}}{S}_n(x)=\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{S}_n(x)\\ =\sum _{i=1}^r(\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{2}^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r{\left(\frac{2}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{x}^m)\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r\left(\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}\prod _{j=1}^r{\left(\frac{2}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{x}^m\right)\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i}{2}{E}_m(x+{\lambda }_i|\mathit{\lambda };\mathit{\mu }+e_i),\end{array}$$

where $e_i=(0,\dots ,0,1,0,\dots ,0)$ is a vector with 1 in the i th coordinate. Thus,

$$S_{n+1}(x)=x{S}_{n-1}(x)-2^{-1-{\sum }_{i=1}^r{\mu }_j}\sum _{m=0}^n\sum _{i=1}^r{S}_1(n,m){\lambda }_i{\mu }_i{E}_m(x+{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).$$

(3.1)

On the other hand, by Theorem 2, we have

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}{S}_n(x)& =\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{1+e^{{\lambda }_{i}t}}{S}_n(x)=\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{S}_n(x)\\ =\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell })x^j\\ =\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{x}^j)\\ =\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}{\lambda }_i^j{E}_j(x/{\lambda }_i))\\ =\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}{E}_j(1+x/{\lambda }_i))\end{array}$$

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

$$S_{n+1}(x)=x{S}_n(x-1)-\sum _{j=0}^n\sum _{\ell =j}^n\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }{E}_j(1+(x-1)/{\lambda }_i).$$

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

Theorem 4 For all $n\ge 0$,

$$\begin{array}{r}{S}_{n+1}(x)=x{S}_n(x-1)-2^{-1-{\sum }_{i=1}^r{\mu }_j}\sum _{m=0}^n\sum _{i=1}^r{S}_1(n,m){\lambda }_i{\mu }_i{E}_m(x+{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i),\\ S_{n+1}(x)=x{S}_n(x-1)-\sum _{j=0}^n\sum _{\ell =j}^n\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }{E}_j(1+(x-1)/{\lambda }_i).\end{array}$$

As a corollary, we get the following identity.

Corollary 2 For all $n\ge 0$,

$$\begin{array}{r}{2}^{-1-{\sum }_{i=1}^r{\mu }_j}\sum _{m=0}^n\sum _{i=1}^r{S}_1(n,m){\lambda }_i{\mu }_i{E}_m(x+{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i),\\ =\sum _{j=0}^n\sum _{\ell =j}^n\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j)S_{n-\ell }{E}_j(1+(x-1)/{\lambda }_i).\end{array}$$

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

$$\frac{g^{\prime }(t)}{g(t)}{S}_n(x)=\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{1+e^{{\lambda }_{i}t}}{\hat{S}}_n(x)-\mathit{\lambda }\mathit{\mu }{\hat{S}}_n(x),$$

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

$$\begin{array}{r}\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{1+e^{{\lambda }_{i}t}}{\hat{S}}_n(x)\\ =\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{\hat{S}}_n(x)\\ =\sum _{i=1}^r(\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{2}^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)e^{{\sum }_{j=1}^r{\lambda }_j{\mu }_{j}t}\prod _{j=1}^r{\left(\frac{2}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{x}^m)\\ =2^{-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r\left(\frac{{\lambda }_i{\mu }_i}{2}{e}^{{\lambda }_{i}t+{\sum }_{j=1}^r{\lambda }_j{\mu }_{j}t}\frac{2}{1+e^{{\lambda }_{i}t}}\prod _{j=1}^r{\left(\frac{2}{1+e^{{\lambda }_{j}t}}\right)}^{{\mu }_j}{x}^m\right)\\ =2^{-1-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r{\lambda }_i{\mu }_i{E}_n(x+\mathit{\lambda }(\mathit{\mu }+e_i)|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

So

$$\begin{array}{rl}{\hat{S}}_{n+1}(x)=& (x+\mathit{\lambda }\mathit{\mu }){\hat{S}}_n(x-1)\\ -2^{-1-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r{\lambda }_i{\mu }_i{E}_n(x+\mathit{\lambda }(\mathit{\mu }+e_i)-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

(3.2)

On the other hand, by Theorem 2, we have

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}{\hat{S}}_n(x)& =\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{1+e^{{\lambda }_{i}t}}{\hat{S}}_n(x)-\mathit{\lambda }\mathit{\mu }{\hat{S}}_n(x)=\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}{\hat{S}}_n(x)-\mathit{\lambda }\mathit{\mu }{\hat{S}}_n(x)\\ =\sum _{i=1}^r\frac{{\lambda }_i{\mu }_i{e}^{{\lambda }_{i}t}}{2}\frac{2}{1+e^{{\lambda }_{i}t}}\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j){\hat{S}}_{n-\ell })x^j-\mathit{\lambda }\mathit{\mu }{\hat{S}}_n(x)\\ =\sum _{j=0}^n(\sum _{\ell =j}^n\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j){\hat{S}}_{n-\ell }\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}{E}_j(1+x/{\lambda }_i))-\mathit{\lambda }\mathit{\mu }{\hat{S}}_n(x).\end{array}$$

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

Theorem 5 For all $n\ge 0$,

$$\begin{array}{c}\begin{array}{rl}{\hat{S}}_{n+1}(x)=& (x+\mathit{\lambda }\mathit{\mu }){\hat{S}}_n(x-1)\\ -2^{-1-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r{\lambda }_i{\mu }_i{E}_n(x+\mathit{\lambda }(\mathit{\mu }+e_i)-1|\mathit{\lambda };\mathit{\mu }+e_i),\end{array}\hfill \\ {\hat{S}}_{n+1}(x)=(x+\mathit{\lambda }\mathit{\mu }){\hat{S}}_n(x-1)-\sum _{j=0}^n\sum _{\ell =j}^n\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j){\hat{S}}_{n-\ell }{E}_j(1+(x-1)/{\lambda }_i).\hfill \end{array}$$

As a corollary, we get the following identity.

Corollary 3 For all $n\ge 0$,

$$\begin{array}{r}{2}^{-1-{\sum }_{j=1}^r{\mu }_j}\sum _{m=0}^n{S}_1(n,m)\sum _{i=1}^r{\lambda }_i{\mu }_i{E}_n(x+\mathit{\lambda }(\mathit{\mu }+e_i)-1|\mathit{\lambda };\mathit{\mu }+e_i),\\ =\sum _{j=0}^n\sum _{\ell =j}^n\sum _{i=1}^r\frac{{\lambda }_i^{j+1}{\mu }_i}{2}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(\ell ,j){\hat{S}}_{n-\ell }{E}_j(1+(x-1)/{\lambda }_i).\end{array}$$

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

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

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

Theorem 6 For all $n\ge 1$,

$$\frac{d}{dx}{S}_n(x)=n!\sum _{\ell =0}^{n-1}\frac{{(-1)}^{n-\ell -1}}{\ell !(n-\ell )}{S}_{\ell }(x)\mathit{\text{and}}\frac{d}{dx}{\hat{S}}_n(x)=n!\sum _{\ell =0}^{n-1}\frac{{(-1)}^{n-\ell -1}}{\ell !(n-\ell )}{\hat{S}}_{\ell }(x).$$

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

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

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

$$\begin{array}{r}〈\frac{d}{dt}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{{\mu }_j}{(1+t)}^y|x^{n-1}〉\\ =-\sum _{i=1}^r{\lambda }_i{\mu }_i〈{(1+{(1+t)}^{{\lambda }_i})}^{-1}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{y+{\lambda }_i-1}|x^{n-1}〉\\ =-\sum _{i=1}^r{\lambda }_i{\mu }_i{S}_{n-1}(y+{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

Hence,

$$S_n(x)=x{S}_{n-1}(x-1)-\sum _{i=1}^r{\lambda }_i{\mu }_i{S}_{n-1}(x+{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).$$

(3.3)

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

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

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

$$\begin{array}{r}〈\frac{d}{dt}\prod _{j=1}^r{\left(\frac{{(1+t)}^{{\lambda }_j}}{1+{(1+t)}^{{\lambda }_j}}\right)}^{{\mu }_j}{(1+t)}^y|x^{n-1}〉\\ =\sum _{i=1}^r{\lambda }_i{\mu }_i〈\frac{{(1+t)}^{{\lambda }_i}}{1+{(1+t)}^{{\lambda }_i}}\prod _{j=1}^r{\left(\frac{{(1+t)}^{{\lambda }_j}}{1+{(1+t)}^{{\lambda }_j}}\right)}^{{\mu }_j}{(1+t)}^{y-{\lambda }_i-1}|x^{n-1}〉\\ =\sum _{i=1}^r{\lambda }_i{\mu }_i{\hat{S}}_{n-1}(y-{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

Thus,

$${\hat{S}}_n(x)=x{\hat{S}}_{n-1}(x-1)+\sum _{i=1}^r{\lambda }_i{\mu }_i{\hat{S}}_{n-1}(x-{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).$$

(3.4)

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

Theorem 7 For $n\ge 1$,

$$\begin{array}{r}{S}_n(x)=x{S}_{n-1}(x-1)-\sum _{i=1}^r{\lambda }_i{\mu }_i{S}_{n-1}(x+{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i),\\ {\hat{S}}_n(x)=x{\hat{S}}_{n-1}(x-1)+\sum _{i=1}^r{\lambda }_i{\mu }_i{\hat{S}}_{n-1}(x-{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

Another result that can be obtained is the following.

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

$$\begin{array}{c}\begin{array}{rl}\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(n-\ell ,m)S_{\ell }=& \sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m-1)S_{\ell }(-1)\\ -\sum _{\ell =0}^{n-1-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m)\sum _{i=1}^r{\lambda }_i{\mu }_i{S}_{\ell }({\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i),\end{array}\hfill \\ \begin{array}{rl}\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(n-\ell ,m){\hat{S}}_{\ell }=& \sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m-1){\hat{S}}_{\ell }(-1)\\ +\sum _{\ell =0}^{n-1-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m)\sum _{i=1}^r{\lambda }_i{\mu }_i{\hat{S}}_{\ell }(-{\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}\hfill \end{array}$$

Proof Because of the similarity in the two cases $S_n(x)$ and ${\hat{S}}_n(x)$, we only give the proof of the first identity. In order to prove the first identity, we compute the following in two different ways:

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

On the one hand, it is equal to

$$\begin{array}{rl}A& =〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}|{(log(1+t))}^m{x}^n〉\\ =〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}|m!\sum _{\ell \ge m}{S}_1(\ell ,m)\frac{t^{\ell }}{\ell !}{x}^n〉\\ =m!\sum _{\ell =m}^n{S}_1(\ell ,m)\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}|x^{n-\ell }〉\\ =m!\sum _{\ell =m}^n{S}_1(\ell ,m)\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_{n-\ell }\\ =m!\sum _{\ell =0}^{n-m}{S}_1(n-\ell ,m)\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_{\ell }.\end{array}$$

(3.5)

On the other hand,

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

Here,

$$\begin{array}{r}〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}\frac{d}{dt}{(log(1+t))}^m|x^{n-1}〉\\ =m〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{-1}|{(log(1+t))}^{m-1}{x}^{n-1}〉\\ =m!\sum _{\ell =m-1}^{n-1}{S}_1(\ell ,m-1)〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{-1}|\frac{t^{\ell }}{\ell !}{x}^{n-1}〉\\ =m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m-1)〈\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{-1}|x^{\ell }〉\\ =m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m-1)S_{\ell }(-1)\end{array}$$

and

$$\begin{array}{r}〈\frac{d}{dt}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(log(1+t))}^m|x^{n-1}〉\\ =-\sum _{i=1}^r{\lambda }_i{\mu }_i〈{(1+{(1+t)}^{{\lambda }_i})}^{-1}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{{\lambda }_i-1}|{(log(1+t))}^m{x}^{n-1}〉\\ =-\sum _{i=1}^r{\lambda }_i{\mu }_i〈{(1+{(1+t)}^{{\lambda }_i})}^{-1}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{{\lambda }_i-1}|m!\sum _{\ell \ge m}{S}_1(\ell ,m)\frac{t^{\ell }}{\ell !}{x}^{n-1}〉\\ =-m!\sum _{i=1}^r\sum _{\ell =m}^{n-1}{\lambda }_i{\mu }_i\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(\ell ,m)\\ \times 〈{(1+{(1+t)}^{{\lambda }_i})}^{-1}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{{\lambda }_i-1}|x^{n-1-\ell }〉\\ =-m!\sum _{i=1}^r\sum _{\ell =0}^{n-1-m}{\lambda }_i{\mu }_i\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m)\\ \times 〈{(1+{(1+t)}^{{\lambda }_i})}^{-1}\prod _{j=1}^r{(1+{(1+t)}^{{\lambda }_j})}^{-{\mu }_j}{(1+t)}^{{\lambda }_i-1}|x^{\ell }〉\\ =-m!\sum _{i=1}^r\sum _{\ell =0}^{n-1-m}{\lambda }_i{\mu }_i\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m)S_{\ell }({\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

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

$$\begin{array}{r}m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)S_1(n-\ell ,m)S_{\ell }\\ =m!\sum _{\ell =0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m-1)S_{\ell }(-1)\\ -m!\sum _{i=1}^r\sum _{\ell =0}^{n-1-m}{\lambda }_i{\mu }_i\left(\genfrac{}{}{0ex}{}{n-1}{\ell }\right)S_1(n-1-\ell ,m)S_{\ell }({\lambda }_i-1|\mathit{\lambda };\mathit{\mu }+e_i).\end{array}$$

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

4 Identities

Let $S_n(x)={\sum }_{m=0}^n{c}_{n,m}{(x)}_m$ and ${\hat{S}}_n(x)={\sum }_{m=0}^n{\hat{c}}_{n,m}{(x)}_m$. By (1.9), (1.10) and (1.11), we obtain

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

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

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

Hence, we have the following identities.

Theorem 9 For all $n\ge 0$,

$$S_n(x)=\sum _{m=0}^n{S}_{n-m}\left(\genfrac{}{}{0ex}{}{n}{m}\right){(x)}_m\mathit{\text{and}}{\hat{S}}_n(x)=\sum _{m=0}^n{\hat{S}}_{n-m}\left(\genfrac{}{}{0ex}{}{n}{m}\right){(x)}_m.$$

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

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

and by Theorem 8, we have

$$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!{(1-\alpha )}^s}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\alpha )}^{s-j}{(n)}_j(m!\sum _{\ell =0}^{n-j-m}\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right)S_1(n-j-\ell ,m)S_{\ell })\\ =\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right){(1-\alpha )}^{-j}{(n)}_j{S}_1(n-j-\ell ,m)S_{\ell }.\end{array}$$

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

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

and by Theorem 8, we have

$$\begin{array}{rl}{\hat{c}}_{n,m}& =\frac{1}{m!{(1-\alpha )}^s}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\alpha )}^{s-j}{(n)}_j(m!\sum _{\ell =0}^{n-j-m}\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right)S_1(n-j-\ell ,m){\hat{S}}_{\ell })\\ =\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right){(1-\alpha )}^{-j}{(n)}_j{S}_1(n-j-\ell ,m){\hat{S}}_{\ell }.\end{array}$$

Therefore, we can state the following result.

Theorem 10 For all $n\ge 0$,

$$\begin{array}{r}{S}_n(x)=\sum _{m=0}^n(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right){(1-\alpha )}^{-j}{(n)}_j{S}_1(n-j-\ell ,m)S_{\ell })H_m^{(s)}(x|\alpha ),\\ {\hat{S}}_n(x)=\sum _{m=0}^n(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right){(1-\alpha )}^{-j}{(n)}_j{S}_1(n-j-\ell ,m){\hat{S}}_{\ell })H_m^{(s)}(x|\alpha ).\end{array}$$

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

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

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

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

and by Theorem 8, we obtain

$$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{i}\right)C_i^{(s)}(m!\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0ex}{}{n-i}{\ell }\right)S_1(n-i-\ell ,m)S_{\ell })\\ =\sum _{i=0}^{n-m}\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{\ell }\right)C_i^{(s)}{S}_1(n-i-\ell ,m)S_{\ell }.\end{array}$$

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

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

and by Theorem 8, we obtain

$$\begin{array}{rl}{c}_{n,m}& =\frac{1}{m!}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{i}\right)C_i^{(s)}(m!\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0ex}{}{n-i}{\ell }\right)S_1(n-i-\ell ,m){\hat{S}}_{\ell })\\ =\sum _{i=0}^{n-m}\sum _{\ell =0}^{n-i-m}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{\ell }\right)C_i^{(s)}{S}_1(n-i-\ell ,m){\hat{S}}_{\ell }.\end{array}$$

Hence, we have the following identities.

Theorem 11 For all $n\ge 0$,

$$\begin{array}{r}{S}_n(x)=\sum _{m=0}^n(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right)C_j^{(s)}{S}_1(n-j-\ell ,m)S_{\ell })B_m^{(s)}(x),\\ {\hat{S}}_n(x)=\sum _{m=0}^n(\sum _{j=0}^{n-m}\sum _{\ell =0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{\ell }\right)C_j^{(s)}{S}_1(n-j-\ell ,m){\hat{S}}_{\ell })B_m^{(s)}(x).\end{array}$$