Barnes-type Narumi polynomials

Abstract

In this paper, we study the Barnes-type Narumi polynomials with umbral calculus viewpoint. From our study, we derive various identities of the Barnes-type Narumi polynomials.

MSC:05A19, 05A40, 11B68.

1 Introduction

As is well known, the Narumi polynomials of order α are defined by the generating function to be

$${\left(\frac{t}{log(1+t)}\right)}^{\alpha }{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{N}_n^{(\alpha )}(x)\frac{t^n}{n!}(\text{see [1]}).$$

(1)

Let $r\in {\mathbb{Z}}_{>0}$. We consider the polynomials $N_n(x|a_1,\dots ,a_r)$ and ${\hat{N}}_n(x|a_1,\dots ,a_r)$, respectively, called the Barnes-type Narumi polynomials of the first kind and those of the second kind and respectively given by

$$\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{N}_n(x|a_1,\dots ,a_r)\frac{t^n}{n!}$$

(2)

and

$$\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t){(1+t)}^{a_j}}\right){(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{\hat{N}}_n(x|a_1,\dots ,a_r)\frac{t^n}{n!},$$

(3)

where $a_1,a_2,\dots ,a_r\ne 0$.

When $x=0$,

$$N_n(a_1,\dots ,a_r)=N_n(0|a_1,\dots ,a_r)$$

and

$${\hat{N}}_n(a_1,\dots ,a_r)={\hat{N}}_n(0|a_1,\dots ,a_r)$$

are respectively called the Barnes-type Narumi numbers of the first kind and those of the second kind.

Note that

$$\begin{array}{c}{N}_n(x|\underset{r}{\underbrace{1,\dots ,1}})=N_n^{(r)}(x),\hfill \\ {\hat{N}}_n(x|\underset{r}{\underbrace{1,\dots ,1}})={\hat{N}}_n^{(r)}(x)\hfill \end{array}$$

and

$${\hat{N}}_n(x|\underset{r}{\underbrace{1,\dots ,1}})=N_n^{(r)}(x-r).$$

In the previous paper [2], $N_n^{(\alpha )}(x)$ was denoted by $N_n^{(-\alpha )}$ and called the Narumi polynomial of order α.

The Bernoulli polynomials are defined by the generating function to be

$$\frac{t}{e^t-1}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x)\frac{t^n}{n!}(\text{see [3–6]}).$$

(4)

When $x=0$, $B_n=B_n(0)$ are called the Bernoulli numbers. In [7], it is known that the Cauchy numbers are given by

$$\frac{t}{log(1+t)}=\sum _{n=0}^{\mathrm{\infty }}{C}_n\frac{t^n}{n!}.$$

(5)

It is well known that the Stirling number of the first kind is given by

$${(x)}_n=x(x-1)\cdots (x-n+1)=\sum _{l=0}^{\mathrm{\infty }}{S}_1(n,l)x^l(n\ge 0)(\text{see [1, 2, 7–11]}).$$

(6)

From (6), we have

$${(log(1+t))}^n=n!\sum _{l=n}^{\mathrm{\infty }}{S}_1(l,n)\frac{t^l}{l!}(n\ge 0).$$

(7)

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

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

Let $\mathbb{P}=\mathbb{C}[x]$ and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on ℙ. $〈L|p(x)〉$ denotes the action of the linear functional L on $p(x)$ which satisfies $〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉$, and $〈cL|p(x)〉=c〈L|p(x)〉$, where c is a complex constant. The linear functional $〈f(t)|\cdot 〉$ on ℙ is defined by $〈f(t)|x^n〉=a_n$ ($n\ge 0$), where $f(t)\in \mathcal{F}$. Thus, we note that

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

(8)

where ${\delta }_{n,k}$ is the Kronecker symbol (see [12–18]).

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

For $f(t),g(t)\in \mathcal{F}$ and $p(x)\in \mathbb{P}$, we have

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

(9)

and

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

(10)

From (10), we can derive the following equation (11):

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k},e^{yt}p(x)=p(x+y)(\text{see [1]}).$$

(11)

Let $s_n(x)\sim (g(t),f(t))$. Then the following will be used:

$$\frac{d{s}_n(x)}{dx}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)〈\overline{f}(t)|x^{n-l}〉s_l(x),$$

(12)

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

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

(13)

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

(14)

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

(15)

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

(16)

and

$$s_{n+1}(x)=(x-\frac{g^{\prime }(t)}{g(t)})\frac{1}{f^{\prime }(t)}{s}_n(x)(n\ge 0)(\text{see [1, 19]}).$$

(17)

Let us assume that $s_n(x)\sim (g(t),f(t))$ and $r_n(x)\sim (h(t),l(t))$. Then we have

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

(18)

where

$$c_{n,m}=\frac{1}{m!}〈\frac{h(\overline{f}(t))}{g(\overline{f}(t))}l{(\overline{f}(t))}^m|x^n〉(\text{see [1, 5]}).$$

(19)

From (2), (3) and (13), we note that

$$N_n(x|a_1,\dots ,a_r)\sim (\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right),e^t-1)$$

(20)

and

$${\hat{N}}_n(x|a_1,\dots ,a_r)\sim (\prod _{j=1}^r\left(\frac{t{e}^{a_{j}t}}{e^{a_{j}t}-1}\right),e^t-1).$$

(21)

In this paper, we study the Barnes-type Narumi polynomials with umbral calculus viewpoint. From our study, we derive various identities of the Barnes-type Narumi polynomials.

2 Barnes-type Narumi polynomials

From (21), we note that

$$\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right)N_n(x|a_1,\dots ,a_r)\sim (1,e^t-1)$$

(22)

and

$${(x)}_n\sim (1,e^t-1).$$

(23)

Thus, by (22) and (23), we get

$$\begin{array}{rl}{N}_n(x|a_1,\dots ,a_r)& =\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t}\right){(x)}_n\\ =\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t}\right)x^m.\end{array}$$

(24)

Note that

$$\begin{array}{rl}\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t}\right)& =\left(\sum _{l_1=0}^{\mathrm{\infty }}\frac{a_1^{l_1+1}}{(l_1+1)!}{t}^{l_1}\right)\times \cdots \times \left(\sum _{l_r=0}^{\mathrm{\infty }}\frac{a_r^{l_r+1}}{(l_r+1)!}{t}^{l_r}\right)\\ =\sum _{l_1,\dots ,l_r=0}^{\mathrm{\infty }}\sum _{l_1+\cdots +l_r=i}\frac{a_1^{l_1+1}\cdots a_r^{l_r+1}}{(l_1+1)!\cdots (l_r+1)!}{t}^i.\end{array}$$

(25)

From (24) and (25), we have

$$\begin{array}{rcl}{N}_n(x|a_1,\dots ,a_r)& =& \sum _{m=0}^{\mathrm{\infty }}{S}_1(n,m)\sum _{i=0}^m\sum _{l_1+\cdots +l_r=i}\frac{a_1^{l_1+1}\cdots a_r^{l_r+1}}{(l_1+1)!\cdots (l_r+1)!}{t}^i{x}^m\\ =& \sum _{m=0}^n{S}_1(n,m)\sum _{i=0}^m\frac{i!}{(i+r)!}\sum _{l_1+\cdots +l_r=i}\left(\genfrac{}{}{0ex}{}{i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)\\ \times a_1^{l_1+1}\cdots a_r^{l_r+1}{x}^{m-i}\\ =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{S}_1(n,m)\frac{(m-i)!}{(m-i+r)!}\\ \times \left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{x}^i.\end{array}$$

(26)

By (21), we see that

$$\prod _{j=1}^r\left(\frac{t{e}^{a_{j}t}}{e^{a_{j}t}-1}\right){\hat{N}}_n(x|a_1,\dots ,a_r)\sim (1,e^t-1),$$

(27)

and we recall (23).

Thus, we have

$$\begin{array}{rcl}{\hat{N}}_n(x|a_1,\dots ,a_r)& =& \prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t{e}^{a_{j}t}}\right){(x)}_n=e^{-{\sum }_{j=1}^r{a}_{j}t}\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t}\right){(x)}_n\\ =& e^{-{\sum }_{j=1}^r{a}_{j}t}{N}_n(x|a_1,\dots ,a_r)\\ =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{S}_1(n,m)\frac{(m-i)!}{(m-i+r)!}\\ \times \left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{e}^{-{\sum }_{j=1}^r{a}_{j}t}{x}^i\\ =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{S}_1(n,m)\frac{(m-i)!}{(m-i+r)!}\\ \times \left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{(x-\sum _{j=1}^r{a}_j)}^i\end{array}$$

(28)

or

$$\begin{array}{rcl}{\hat{N}}_n(x|a_1,\dots ,a_r)& =& \prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t{e}^{a_{j}t}}\right){(x)}_n=\prod _{j=1}^r\left(\frac{e^{-a_{j}t}-1}{-t}\right){(x)}_n\\ =& \sum _{m=0}^n{S}_1(n,m)\sum _{i=0}^m{(-1)}^i\frac{i!}{(i+r)!}\\ \times \sum _{l_1+\cdots +l_r=i}\left(\genfrac{}{}{0ex}{}{i+r}{i_1+1,\dots ,i_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}{x}^{m-i}\\ =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{(-1)}^{m-i}{S}_1(n,m)\frac{(m-i)!}{(m-i+r)!}\\ \times \left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{x}^i.\end{array}$$

(29)

Therefore, by (26), (28) and (29), we obtain the following theorem.

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

$$\begin{array}{rcl}{N}_n(x|a_1,\dots ,a_r)& =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{S}_1(n,m)\frac{(m-i)!}{(m-i+r)!}\\ \times \left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{x}^i\end{array}$$

and

$$\begin{array}{rcl}{\hat{N}}_n(x|a_1,\dots ,a_r)& =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{S}_1(n,m)\frac{(m-i)!}{(m-i+r)!}\\ \times \left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{(x-\sum _{j=1}^r{a}_j)}^i\\ =& \sum _{i=0}^n\{\sum _{m=i}^n\sum _{l_1+\cdots +l_r=m-i}{(-1)}^{m-i}{S}_1(n,m)\\ \times \frac{(m-i)!}{(m-i+r)!}\left(\genfrac{}{}{0ex}{}{m-i+r}{l_1+1,\dots ,l_r+1}\right)\left(\genfrac{}{}{0ex}{}{m}{j}\right)a_1^{l_1+1}\cdots a_r^{l_r+1}\}{x}^i.\end{array}$$

From (14), we can derive the following equation (30):

$$N_n(x|a_1,\dots ,a_r)=\sum _{j=0}^n\frac{1}{j!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^j|x^n〉x^j,$$

(30)

where

$$\begin{array}{c}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^j|x^n〉\hfill \\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|{(log(1+t))}^j{x}^n〉\hfill \\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|j!\sum _{l=j}^{\mathrm{\infty }}{S}_1(l,j)\frac{t^l}{l!}{x}^n〉\hfill \\ =j!\sum _{l=j}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,j)N_{n-l}(a_1,\dots ,a_r).\hfill \end{array}$$

(31)

Thus, by (30) and (31), we obtain the following theorem.

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

$$N_n(x|a_1,\dots ,a_r)=\sum _{j=0}^n\{\sum _{l=j}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,j)N_{n-l}(a_1,\dots ,a_r)\}{x}^j.$$

By the same methods as in (28), (29) and (30), we get

$$\begin{array}{c}{\hat{N}}_n(x|a_1,\dots ,a_r)\hfill \\ =\sum _{j=0}^n\{\sum _{l=j}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,j)N_{n-l}(a_1,\dots ,a_r)\}{(x-\sum _{i=1}^r{a}_i)}^j\hfill \\ =\sum _{j=0}^n\{\sum _{l=j}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,j){\hat{N}}_{n-l}(a_1,\dots ,a_r)\}{x}^j.\hfill \end{array}$$

(32)

By (8), we get

$$\begin{array}{rl}{N}_n(y|a_1,\dots ,a_r)& =〈\sum _{i=0}^{\mathrm{\infty }}{N}_i(y|a_1,\dots ,a_r)\frac{t^i}{i!}|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(1+t)}^y|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|\sum _{m=0}^{\mathrm{\infty }}{(y)}_m\frac{t^m}{m!}{x}^m〉\\ =\sum _{m=0}^n{(y)}_m\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|x^{n-m}〉\\ =\sum _{m=0}^n{(y)}_m\left(\genfrac{}{}{0ex}{}{n}{m}\right)N_{n-m}(a_1,\dots ,a_r)\end{array}$$

(33)

and

$$\begin{array}{rl}{\hat{N}}_n(y|a_1,\dots ,a_r)& =〈\sum _{i=0}^{\mathrm{\infty }}{\hat{N}}_i(y|a_1,\dots ,a_r)\frac{t^i}{i!}|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t){(1+t)}^{a_j}}\right){(1+t)}^y|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t){(1+t)}^{a_j}}\right)|\sum _{m=0}^{\mathrm{\infty }}{(y)}_m\frac{t^m}{m!}{x}^n〉\\ =\sum _{m=0}^n{(y)}_m\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t){(1+t)}^{a_j}}\right)|x^{n-m}〉\\ =\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right){(y)}_m{\hat{N}}_{n-m}(a_1,\dots ,a_r).\end{array}$$

(34)

Therefore, by (33) and (34), we obtain the following theorem.

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

$$N_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)N_{n-m}(a_1,\dots ,a_r){(x)}_m$$

and

$${\hat{N}}_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right){\hat{N}}_{n-m}(a_1,\dots ,a_r){(x)}_m.$$

From (15), we note that

$$N_n(x+y|a_1,\dots ,a_r)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)N_j(x|a_1,\dots ,a_r){(y)}_{n-j}$$

(35)

and

$${\hat{N}}_n(x+y|a_1,\dots ,a_r)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right){\hat{N}}_j(x|a_1,\dots ,a_r){(y)}_{n-j}.$$

(36)

By (14), we get

$$(e^t-1)N_n(x|a_1,\dots ,a_r)=n{N}_{n-1}(x|a_1,\dots ,a_r)$$

(37)

and

$$\begin{array}{rl}(e^t-1)N_n(x|a_1,\dots ,a_r)& =e^t{N}_n(x|a_1,\dots ,a_r)-N_n(x|a_1,\dots ,a_r)\\ =N_n(x+1|a_1,\dots ,a_r)-N(x|a_1,\dots ,a_r).\end{array}$$

(38)

From (37) and (38), we have

$$N_n(x+1|a_1,\dots ,a_r)-N_n(x|a_1,\dots ,a_r)=n{N}_{n-1}(x|a_1,\dots ,a_r).$$

(39)

By the same method as (39), we get

$${\hat{N}}_n(x+1|a_1,\dots ,a_r)-{\hat{N}}_n(x|a_1,\dots ,a_r)=n{\hat{N}}_{n-1}(x|a_1,\dots ,a_r).$$

(40)

Recall that $N_n(x|a_1,\dots ,a_r)\sim ({\prod }_{j=1}^r(\frac{t}{e^{a_{j}t}-1}),e^t-1)$.

From (17), we can derive the following equation (41):

$$N_{n+1}(x|a_1,\dots ,a_r)=x{N}_n(x-1|a_1,\dots ,a_r)-e^{-t}\frac{g^{\prime }(t)}{g(t)}{N}_n(x|a_1,\dots ,a_r).$$

(41)

Now, we observe that

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}& ={(logg(t))}^{\prime }={(rlogt-\sum _{j=1}^{r}log(e^{a_{j}t}-1))}^{\prime }=\frac{r}{t}-\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{e^{a_{j}t}-1}\\ =\frac{{\sum }_{j=1}^r{\prod }_{i\ne j}^r(e^{a_{i}t}-1)\{e^{a_{j}t}-1-t{a}_j{e}^{a_{j}t}\}}{t{\prod }_{j=1}^r(e^{a_{j}t}-1)},\end{array}$$

(42)

where

$$\begin{array}{rl}r-\sum _{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}& =\frac{{\sum }_{j=1}^r{\prod }_{i\ne j}(e^{a_{i}t}-1)\{e^{a_{j}t}-1-a_{j}t{e}^{a_{j}t}\}}{{\prod }_{j=1}^r(e^{a_{j}t}-1)}\\ =\frac{-\frac{1}{2}({\sum }_{j=1}^r{a}_1{a}_2\cdots a_{j-1}{a}_j^2{a}_{j+1}\cdots a_r)t^{r+1}+\cdots }{(a_1{a}_2\cdots a_r)t^r}\\ =-\frac{1}{2}\left(\sum _{j=1}^r{a}_j\right)t+\cdots \end{array}$$

(43)

has at least the order 1.

By (42) and (43), we get

$$\begin{array}{c}\frac{g^{\prime }(t)}{g(t)}{N}_n(x|a_1,\dots ,a_r)\hfill \\ =\frac{r-{\sum }_{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}}{t}\left(\sum _{i=0}^n\{\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)N_{n-l}(a_1,\dots ,a_r)\}{x}^i\right)\hfill \\ =\sum _{i=0}^n\{\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)N_{n-l}(a_1,\dots ,a_r)\}\frac{r-{\sum }_{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}}{t}{x}^i\hfill \\ =\sum _{i=0}^n\{\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)N_{n-l}(a_1,\dots ,a_r)\}(r-\sum _{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1})\frac{x^{i+1}}{i+1}\hfill \\ =\sum _{i=0}^n\frac{1}{i+1}\{\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)N_{n-l}(a_1,\dots ,a_r)\}(r{x}^{i+1}-\sum _{j=1}^r\sum _{m=0}^{\mathrm{\infty }}{B}_m\frac{{(-a_{j}t)}^m}{m!}{x}^{i+1})\hfill \\ =-\sum _{i=0}^n\frac{1}{i+1}\{\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)N_{n-l}(a_1,\dots ,a_r)\}\hfill \\ \times \sum _{j=1}^r\sum _{m=1}^{i+1}{(-1)}^m\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)B_m{a}_j^m{x}^{i+1-m}\hfill \\ =-\sum _{i=0}^n\frac{1}{i+1}\{\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)N_{n-l}(a_1,\dots ,a_r)\}\hfill \\ \times \sum _{j=1}^r\sum _{m=0}^i{(-1)}^{i+1-m}\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)a_j^{i+1-m}{B}_{i+1-m}{x}^m.\hfill \end{array}$$

(44)

Therefore, by (41) and (44), we obtain the following theorem.

Theorem 4 For $n\ge 0$, we have

$$\begin{array}{r}{N}_{n+1}(x|a_1,\dots ,a_r)\\ =x{N}_n(x-1|a_1,\dots ,a_r)+\sum _{m=0}^n\{\sum _{i=m}^n\sum _{l=i}^n\sum _{j=1}^r\frac{1}{i+1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)S_1(l,i)\\ \times B_{i+1-m}{(-a_j)}^{i+1-m}{N}_{n-l}(a_1,\dots ,a_r)\}{(x-1)}^m.\end{array}$$

By the same method as the proof of Theorem 4, we get

$$\begin{array}{rcl}{\hat{N}}_{n+1}(x|a_1,\dots ,a_r)& =& (x-\sum _{j=1}^r{a}_j){\hat{N}}_n(x-1|a_1,\dots ,a_r)\\ +\sum _{m=0}^n\{\sum _{i=m}^n\sum _{l=i}^n\sum _{j=1}^r\frac{1}{i+1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)S_1(l,i)\\ \times B_{i+1-m}{(-a_j)}^{i+1-m}{\hat{N}}_{n-l}(a_1,\dots ,a_r)\}{(x-1)}^m.\end{array}$$

(45)

From (12) and (20), we can derive the following equation (46):

$$〈\overline{f}(t)|x^{n-l}〉=〈log(1+t)|x^{n-l}〉=〈\sum _{m=1}^{\mathrm{\infty }}\frac{{(-1)}^{m-1}}{m}{t}^m|x^{n-l}〉={(-1)}^{n-l-1}(n-l-1)!.$$

(46)

Thus, by (46), we get

$$\begin{array}{rl}\frac{d}{dx}{N}_n(x|a_1,\dots ,a_r)& =\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right){(-1)}^{n-l-1}(n-l-1)!N_l(x|a_1,\dots ,a_r)\\ =n!\sum _{l=0}^{n-1}\frac{{(-1)}^{n-l-1}}{l!(n-l)}{N}_l(x|a_1,\dots ,a_r).\end{array}$$

(47)

By the same method as (47), we get

$$\frac{d}{dx}{\hat{N}}_n(x|a_1,\dots ,a_r)=n!\sum _{l=0}^{n-1}\frac{{(-1)}^{n-l-1}}{l!(n-l)}{\hat{N}}_l(x|a_1,\dots ,a_r).$$

(48)

From (8), we note that, for $n\ge 1$,

$$\begin{array}{c}{N}_n(y|a_1,\dots ,a_r)\hfill \\ =〈\sum _{i=0}^{\mathrm{\infty }}{N}_i(y|a_1,\dots ,a_r)\frac{t^i}{i!}|x^n〉\hfill \\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(1+t)}^y|x^n〉\hfill \\ =〈{\partial }_t\left(\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(1+t)}^y\right)|x^{n-1}〉\hfill \\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\left({\partial }_t{(1+t)}^y\right)|x^{n-1}〉\hfill \\ +〈\left({\partial }_t\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\right){(1+t)}^y|x^{n-1}〉\hfill \\ =y{N}_{n-1}(y-1|a_1,\dots ,a_r)+〈\left({\partial }_t\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\right){(1+t)}^y|x^{n-1}〉.\hfill \end{array}$$

(49)

Now, we observe that

$$\begin{array}{c}{\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\hfill \\ =\sum _{j=1}^r\prod _{i\ne j}\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right)\frac{a_j{(1+t)}^{a_j-1}log(1+t)-({(1+t)}^{a_j}-1)\frac{1}{1+t}}{{(log(1+t))}^2}\hfill \\ =\frac{1}{1+t}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right)\sum _{j=1}^r\{\frac{a_j{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{1}{log(1+t)}\}\hfill \\ =\frac{1}{1+t}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right)\frac{{\sum }_{j=1}^r\{\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{t}{log(1+t)}\}}{t},\hfill \end{array}$$

(50)

where

$$\sum _{j=1}^r\{\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{t}{log(1+t)}\}=\frac{1}{2}\left(\sum _{j=1}^r{a}_j\right)t+\cdots$$

(51)

is a series with order greater than or equal to 1.

By (50) and (51), we get

$$\begin{array}{c}〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\right){(1+t)}^y|x^{n-1}〉\hfill \\ =〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right)\frac{{(1+t)}^y}{1+t}|\frac{{\sum }_{j=1}^r\{\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{t}{log(1+t)}\}}{t}{x}^{n-1}〉\hfill \\ =\frac{1}{n}〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{y-1}|\sum _{j=1}^r\{\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{t}{log(1+t)}\}{x}^n〉\hfill \\ =\frac{1}{n}\{\sum _{j=1}^r{a}_j〈\frac{log(1+t)}{{(1+t)}^{a_j}-1}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{y+a_j-1}|\frac{t}{log(1+t)}{x}^n〉\hfill \\ -r〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{y-1}|\frac{t}{log(1+t)}{x}^n〉\}\hfill \\ =\frac{1}{n}\sum _{j=1}^r{a}_j\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)C_l〈\frac{log(1+t)}{{(1+t)}^{a_j}-1}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{y+a_j-1}|x^{n-l}〉\hfill \\ -\frac{r}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)C_l〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{y-1}|x^{n-l}〉\hfill \\ =\frac{1}{n}\sum _{j=1}^r\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)a_j{C}_l{N}_{n-l}(y+a_j-1|a_1,\dots ,{\hat{a}}_j,\dots ,a_r)\hfill \\ -\frac{r}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)C_l{N}_{n-l}(y-1|a_1,\dots ,a_r),\hfill \end{array}$$

(52)

where ${\hat{a}}_j$ means that $a_j$ is omitted.

Therefore, by (49) and (52), we obtain the following theorem.

Theorem 5 For $n\ge 1$, we have

$$\begin{array}{rcl}{N}_n(x|a_1,\dots ,a_r)& =& x{N}_{n-1}(x-1|a_1,\dots ,a_r)\\ +\frac{1}{n}\sum _{j=1}^r\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)a_j{C}_l{N}_{n-l}(x+a_j-1|a_1,\dots ,{\hat{a}}_j,\dots ,a_r)\\ -\frac{r}{n}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{n}{l}\right)C_l{N}_{n-l}(x-1|a_1,\dots ,a_r),\end{array}$$

where $C_n$ are the Cauchy numbers with the generating function given by

$$\frac{t}{log(1+t)}=\sum _{n=0}^{\mathrm{\infty }}{C}_n\frac{t^n}{n!}.$$

By the same method as the proof of Theorem 5, we get

$$\begin{array}{rcl}{\hat{N}}_n(x|a_1,\dots ,a_r)& =& (x-\sum _{j=1}^r{a}_j){\hat{N}}_{n-1}(x-1|a_1,\dots ,a_r)\\ -\frac{r}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)C_l{\hat{N}}_{n-l}(x-1|a_1,\dots ,a_r)\\ -\frac{1}{n}\sum _{j=1}^r\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)a_j{C}_l{\hat{N}}_{n-l}(x-1|a_1,\dots ,{\hat{a}}_j,\dots ,a_r).\end{array}$$

(53)

Now we compute the following formula (54) in two different ways:

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

(54)

On the one hand,

$$\begin{array}{c}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m|x^n〉\hfill \\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|m!\sum _{l=m}^{\mathrm{\infty }}{S}_1(l,m)\frac{t^l}{l!}{x}^n〉\hfill \\ =m!\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|x^{n-l}〉\hfill \\ =m!\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)N_{n-l}(a_1,\dots ,a_r)\hfill \\ =m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m)N_l(a_1,\dots ,a_r).\hfill \end{array}$$

(55)

On the other hand,

$$\begin{array}{c}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m|x^n〉\hfill \\ =〈{\partial }_t\left(\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m\right)|x^{n-1}〉\hfill \\ =〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\left({\partial }_t\left({(log(1+t))}^m\right)\right)|x^{n-1}〉\hfill \\ +〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\right){(log(1+t))}^m|x^{n-1}〉.\hfill \end{array}$$

(56)

Note that

$$\begin{array}{c}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\left({\partial }_t\left({(log(1+t))}^m\right)\right)|x^{n-1}〉\hfill \\ =m〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\frac{1}{1+t}|log{(1+t)}^{m-1}{x}^{n-1}〉\hfill \\ =m〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(1+t)}^{-1}|(m-1)!\sum _{l=m-1}^{\mathrm{\infty }}{S}_1(l,m-1)\frac{t^l}{l!}{x}^{n-1}〉\hfill \\ =m!\sum _{l=m-1}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(l,m-1)〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(1+t)}^{-1}|x^{n-1-l}〉\hfill \\ =m!\sum _{l=m-1}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(l,m-1)N_{n-1-l}(-1|a_1,\dots ,a_r)\hfill \\ =m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-1-l,m-1)N_l(-1|a_1,\dots ,a_r)\hfill \end{array}$$

(57)

and

$$\begin{array}{c}〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\right){(log(1+t))}^m|x^{n-1}〉\hfill \\ =〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)\right)|m!\sum _{l=m}^{\mathrm{\infty }}{S}_1(l,m)\frac{t^l}{l!}{x}^{n-1}〉\hfill \\ =m!\sum _{l=m}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(l,m)\hfill \\ \times 〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{-1}|\frac{{\sum }_{j=1}^r\{\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{t}{log(1+t)}\}}{t}{x}^{n-1-l}〉\hfill \\ =m!\sum _{l=m}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)\frac{S_1(l,m)}{n-l}\hfill \\ \times 〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{-1}|\sum _{j=1}^r\{\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1}-\frac{t}{log(1+t)}\}{x}^{n-l}〉\hfill \\ =\frac{m!}{n}\sum _{l=m}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)\hfill \\ \times \{\sum _{j=1}^r{a}_j〈\frac{log(1+t)}{{(1+t)}^{a_j}-1}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{a_j-1}|\frac{t}{log(1+t)}{x}^{n-l}〉\hfill \\ -r〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}-1}{log(1+t)}\right){(1+t)}^{-1}|\frac{t}{log(1+t)}{x}^{n-l}〉\}\hfill \\ =\frac{m!}{n}\sum _{l=m}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,m)\{\sum _{j=1}^r{a}_j\sum _{k=0}^{n-l}{C}_k\left(\genfrac{}{}{0ex}{}{n-l}{k}\right)N_{n-l-k}(a_j-1|a_1,\dots ,{\hat{a}}_j,\dots ,a_r)\hfill \\ -r\sum _{k=0}^{n-l}{C}_k\left(\genfrac{}{}{0ex}{}{n-l}{k}\right)N_{n-l-k}(-1|a_1,\dots ,a_r)\}.\hfill \end{array}$$

(58)

Therefore, by (55), (56), (57) and (58), we obtain the following theorem.

Theorem 6 For $n-1\ge m\ge 1$, we have

$$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m)N_l(a_1,\dots ,a_r)\hfill \\ =\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1)N_l(-1|a_1,\dots ,a_r)\hfill \\ +\frac{1}{n}\sum _{l=m}^{n-1}\sum _{k=0}^{n-l}\sum _{j=1}^r\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{k}\right)a_j{C}_{n-l-k}{S}_1(l,m)N_k(a_j-1|a_1,\dots ,{\hat{a}}_j,\dots ,a_r)\hfill \\ -\frac{r}{n}\sum _{l=m}^{n-1}\sum _{k=0}^{n-l}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{k}\right)C_{n-l-k}{S}_1(l,m)N_k(-1|a_1,\dots ,a_r).\hfill \end{array}$$

By the same method as the proof of Theorem 6, we get

$$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{N}}_l(a_1,\dots ,a_r)\hfill \\ =\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1){\hat{N}}_l(-1|a_1,\dots ,a_r)\hfill \\ +\frac{1}{n}\sum _{j=1}^r\sum _{l=m}^{n-1}\sum _{k=0}^{n-l}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{k}\right)a_j{C}_{n-l-k}{S}_1(l,m){\hat{N}}_k(-1|a_1,\dots ,{\hat{a}}_j,\dots ,a_r)\hfill \\ -\frac{r}{n}\sum _{l=m}^{n-1}\sum _{k=0}^{n-l}\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{n-l}{k}\right)C_{n-l-k}{S}_1(l,m){\hat{N}}_k(-1|a_1,\dots ,a_r)\hfill \\ -\sum _{j=1}^r{a}_j\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m){\hat{N}}_k(-1|a_1,\dots ,a_r),\hfill \end{array}$$

(59)

where $n-1\ge m\ge 1$.

Let us consider the following two Sheffer sequences:

$$N_n(x|a_1,\dots ,a_r)\sim (\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right),e^t-1)$$

(60)

and (23).

We let

$$N_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n{C}_{n,m}{(x)}_m.$$

(61)

From (18) and (19), we note that

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)t^m|x^n〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right)|x^{n-m}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)N_{n-m}(a_1,\dots ,a_r).\end{array}$$

(62)

Therefore, by (61) and (62), we obtain the following theorem.

Theorem 7 For $n\ge 0$, we have

$$N_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)N_{n-m}(a_1,\dots ,a_r){(x)}_m.$$

By the same method as the proof of Theorem 7, we get

$${\hat{N}}_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right){\hat{N}}_{n-m}(a_1,\dots ,a_r){(x)}_m.$$

(63)

For

$$N_n(x|a_1,\dots ,a_r)\sim (\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right),e^t-1)$$

and

$$H_n^{(s)}(x|\lambda )\sim ({\left(\frac{e^t-\lambda }{1-\lambda }\right)}^s,t),\lambda \in \mathbb{C}\text{ with }\lambda \ne 1,$$

let us assume that

$$N_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\lambda ),$$

(64)

where $H_m^{(s)}(x|\lambda )$ are the Frobenius-Euler polynomials of order s defined by the generating function as

$${\left(\frac{1-\lambda }{e^t-\lambda }\right)}^s{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(s)}(x|\lambda )\frac{t^n}{n!}.$$

From (18) and (19), we note that

$$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!{(1-\lambda )}^s}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m{(1-\lambda +t)}^s|x^n〉\\ =& \frac{1}{m!{(1-\lambda )}^s}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m|\sum _{j=0}^{min\{s,n\}}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\lambda )}^{s-j}{t}^j{x}^n〉\\ =& \frac{1}{m!{(1-\lambda )}^s}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\lambda )}^{s-j}{(n)}_j\\ \times 〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1_t)}\right){(log(1+t))}^m|x^{n-j}〉\\ =& \frac{1}{m!{(1-\lambda )}^s}\sum _{j=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{j}\right){(1-\lambda )}^{s-j}{(n)}_{j}m!\\ \times \sum _{l=0}^{n-j-m}\left(\genfrac{}{}{0ex}{}{n-j}{l}\right)S_1(n-j-l,m)N_l(a_1,\dots ,a_r)\\ =& \sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right)\\ \times {(n)}_j{(1-\lambda )}^{-j}{S}_1(n-j-l,m)N_l(a_1,\dots ,a_r).\end{array}$$

(65)

Therefore, by (64) and (65), we obtain the following theorem.

Theorem 8 For $n\ge 0$, we have

$$\begin{array}{c}{N}_n(x|a_1,\dots ,a_r)\hfill \\ =\sum _{m=0}^n\{\sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j{(1-\lambda )}^{-j}\hfill \\ \times S_1(n-j-l,m)N_l(a_1,\dots ,a_r)\}{H}_m^{(s)}(x|\lambda ).\hfill \end{array}$$

By the same method as the proof of Theorem 8, we get

$$\begin{array}{r}{\hat{N}}_n(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\{\sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j\\ \times {(1-\lambda )}^{-j}{S}_1(n-j-l,m){\hat{N}}_l(a_1,\dots ,a_r)\}{H}_m^{(s)}(x|\lambda ).\end{array}$$

(66)

Now, we consider the following two Sheffer sequences:

$$N_n(x|a_1,\dots ,a_r)\sim (\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right),e^t-1)$$

(67)

and

$$B_n^{(s)}(x)\sim ({\left(\frac{e^t-1}{t}\right)}^s,e^t-1),$$

where $B_n^{(s)}(x)$ are the Bernoulli polynomials of order s given by the generating function as

$${\left(\frac{t}{e^t-1}\right)}^s{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(s)}(x)\frac{t^n}{n!}.$$

Let us assume that

$$N_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n{C}_{n,m}{B}_m^{(s)}(x).$$

(68)

By (18) and (19), we get

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m{\left(\frac{t}{log(1+t)}\right)}^s|x^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m|{\left(\frac{t}{log(1+t)}\right)}^s{x}^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m|\sum _{k=0}^{\mathrm{\infty }}{\mathbb{C}}_k^{(s)}\frac{t^k}{k!}{x}^n〉\\ =\frac{1}{m!}\sum _{k=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathbb{C}}_k^{(s)}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}-1}{log(1+t)}\right){(log(1+t))}^m|x^{n-k}〉\\ =\frac{1}{m!}\sum _{k=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{k}\right){\mathbb{C}}_k^{(s)}m!\sum _{l=0}^{n-m-k}\left(\genfrac{}{}{0ex}{}{n-k}{l}\right)S_1(n-l-k,m)N_l(a_1,\dots ,a_r)\\ =\sum _{k=0}^{n-m}\sum _{l=0}^{n-m-k}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{n-k}{l}\right){\mathbb{C}}_k^{(s)}{S}_1(n-k-l,m)N_l(a_1,\dots ,a_r),\end{array}$$

(69)

where ${\mathbb{C}}_k^{(s)}$ are the Cauchy numbers of the first kind of order s defined by the generating function as

$${\left(\frac{t}{log(1+t)}\right)}^s=\sum _{n=0}^{\mathrm{\infty }}{\mathbb{C}}_n^{(s)}\frac{t^n}{n!}.$$

Therefore, by (68) and (69), we obtain the following theorem.

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

$$\begin{array}{rcl}{N}_n(x|a_1,\dots ,a_r)& =& \sum _{m=0}^n\{\sum _{k=0}^{n-m}\sum _{l=0}^{n-m-k}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{n-k}{l}\right){\mathbb{C}}_k^{(s)}\\ \times S_1(n-k-l,m)N_l(a_1,\dots ,a_r)\}{B}_m^{(s)}(x).\end{array}$$

By the same method as the proof of Theorem 9, we get

$$\begin{array}{rcl}{\hat{N}}_n(x|a_1,\dots ,a_r)& =& \sum _{m=0}^n\{\sum _{k=0}^{n-m}\sum _{l=0}^{n-m-k}\left(\genfrac{}{}{0ex}{}{n}{k}\right)\left(\genfrac{}{}{0ex}{}{n-k}{l}\right){\mathbb{C}}_k^{(s)}\\ \times S_1(n-k-l,m){\hat{N}}_l(a_1,\dots ,a_r)\}{B}_m^{(s)}(x).\end{array}$$

Recently, several authors have studied umbral calculus (see [1–5, 7–18, 20]).