Barnes-type Daehee polynomials

Abstract

In this paper, we consider Barnes-type Daehee polynomials of the first kind and of the second kind. From the properties of the Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

MSC:05A15, 05A40, 11B68, 11B75, 65Q05.

1 Introduction

In this paper, we consider the polynomials $D_n(x|a_1,\dots ,a_r)$ and ${\hat{D}}_n(x|a_1,\dots ,a_r)$ called the Barnes-type Daehee polynomials of the first kind and of the second kind, whose generating functions are given by

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

(1)

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

(2)

respectively, where $a_1,\dots ,a_r\ne 0$. When $x=0$, $D_n(a_1,\dots ,a_r)=D_n(0|a_1,\dots ,a_r)$ and ${\hat{D}}_n(a_1,\dots ,a_r)={\hat{D}}_n(0|a_1,\dots ,a_r)$ are called the Barnes-type Daehee numbers of the first kind and of the second kind, respectively.

Recall that the Daehee polynomials of the first kind and of the second kind of order r, denoted by $D_n^{(r)}(x)$ and ${\hat{D}}_n^{(r)}(x)$, respectively, are given by the generating functions to be

$$\begin{array}{c}{\left(\frac{ln(1+t)}{t}\right)}^r{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{D}_n^{(r)}(x)\frac{t^n}{n!},\hfill \\ {\left(\frac{(1+t)ln(1+t)}{t}\right)}^r{(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{\hat{D}}_n^{(r)}(x)\frac{t^n}{n!},\hfill \end{array}$$

respectively. If $a_1=\cdots =a_r=1$, then $D_n^{(r)}(x)=D_n(x|\underset{r}{\underbrace{1,\dots ,1}})$ and ${\hat{D}}_n^{(r)}(x)={\hat{D}}_n(x|\underset{r}{\underbrace{1,\dots ,1}})$. Daehee polynomials were defined by the second author [1] and have been investigated in [2–4].

In this paper, we consider Barnes-type Daehee polynomials of the first kind and of the second kind. From the properties of the Sheffer sequences of these polynomials arising from umbral calculus, we derive new and interesting identities.

2 Umbral calculus

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

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

(3)

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

$$〈f(t)|x^n〉=a_n(n\ge 0).$$

(4)

In particular,

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

(5)

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

For $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|x^k〉}{k!}{t}^k$, we have $〈f_L(t)|x^n〉=〈L|x^n〉$. That is, $L=f_L(t)$. The map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth, ℱ denotes both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and therefore an element $f(t)$ of ℱ will be thought of as both a formal power series and a linear functional. We call ℱ the umbral algebra and the umbral calculus is the study of umbral algebra. The order $O(f(t))$ of a power series $f(t)$ (≠0) is the smallest integer k 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 $f(t)$ is called an invertible series. For $f(t),g(t)\in \mathcal{F}$ with $O(f(t))=1$ and $O(g(t))=0$, there exists a unique sequence $s_n(x)$ ($deg{s}_n(x)=n$) such that $〈g(t)f{(t)}^k|s_n(x)〉=n!{\delta }_{n,k}$, for $n,k\ge 0$. Such a 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

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

(6)

and

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

(7)

[5], Theorem 2.2.5]. Thus, by (7), we get

$$t^{k}p(x)=p^{(k)}(x)=\frac{d^{k}p(x)}{d{x}^k}\text{and}{e}^{yt}p(x)=p(x+y).$$

(8)

Sheffer sequences are characterized by the generating function [5], Theorem 2.3.4].

Lemma 1 The sequence $s_n(x)$ is Sheffer for $(g(t),f(t))$ if and only if

$$\frac{1}{g(\overline{f}(t))}{e}^{y\overline{f}(t)}=\sum _{k=0}^{\mathrm{\infty }}\frac{s_k(y)}{k!}{t}^k(y\in \mathbb{C}),$$

where $\overline{f}(t)$ is the compositional inverse of $f(t)$.

For $s_n(x)\sim (g(t),f(t))$, we have the following equations [5], Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

$$f(t)s_n(x)=n{s}_{n-1}(x)(n\ge 0),$$

(9)

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

(10)

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

(11)

where $p_n(x)=g(t)s_n(x)$.

Assume that $p_n(x)\sim (1,f(t))$ and $q_n(x)\sim (1,g(t))$. Then the transfer formula [5], Corollary 3.8.2] is given by

$$q_n(x)=x{\left(\frac{f(t)}{g(t)}\right)}^n{x}^{-1}{p}_n(x)(n\ge 1).$$

For $s_n(x)\sim (g(t),f(t))$ and $r_n(x)\sim (h(t),l(t))$, assume that

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

Then we have [5], p.132]

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

(12)

3 Main results

We now note that $D_n(x|a_1,\dots ,a_r)$ is the Sheffer sequence for

$$g(t)=\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t}\right)\text{and}f(t)=e^t-1.$$

Therefore,

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

(13)

${\hat{D}}_n(x|a_1,\dots ,a_r)$ is the Sheffer sequence for

$$g(t)=\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t{e}^{a_{j}t}}\right)\text{and}f(t)=e^t-1.$$

So,

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

(14)

3.1 Explicit expressions

Recall that Barnes’ multiple Bernoulli polynomials $B_n(x|a_1,\dots ,a_r)$ are defined by the generating function as

$$\frac{t^r}{{\prod }_{j=1}^r(e^{a_{j}t}-1)}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n(x|a_1,\dots ,a_r)\frac{t^n}{n!},$$

(15)

where $a_1,\dots ,a_r\ne 0$ [6–8]. Let ${(n)}_j=n(n-1)\cdots (n-j+1)$ ($j\ge 1$) with ${(n)}_0=1$. The (signed) Stirling numbers of the first kind $S_1(n,m)$ are defined by

$${(x)}_n=\sum _{m=0}^n{S}_1(n,m)x^m.$$

Theorem 1

$$D_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n{S}_1(n,m)B_m(x|a_1,\dots ,a_r)$$

(16)

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

(17)

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

(18)

$${\hat{D}}_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n{S}_1(n,m)B_m(x+a_1+\cdots +a_r|a_1,\dots ,a_r)$$

(19)

$$\phantom{{\hat{D}}_n(x|a_1,\dots ,a_r)}=\sum _{j=0}^n(\sum _{l=j}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,j){\hat{D}}_{n-l}(a_1,\dots ,a_r))x^j$$

(20)

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

(21)

Proof Since

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

(22)

and

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

(23)

we have

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

So, we get (16).

Similarly, by

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

(24)

and (23), we have

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

Therefore, we get (19).

By (10) with (13), we get

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

Since

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

we obtain (17).

Similarly, by (10) with (14), we get

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

Since

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

we obtain (20).

Next, we obtain

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

Thus, we get the identity (18).

Similarly,

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

Thus, we get the identity (21). □

3.2 Sheffer identity

Theorem 2

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

(25)

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

(26)

Proof By (13) with

$$\begin{array}{rl}{p}_n(x)& =\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t}\right)D_n(x|a_1,\dots ,a_r)\\ ={(x)}_n\sim (1,e^t-1),\end{array}$$

using (11), we have (25).

By (14) with

$$\begin{array}{rl}{p}_n(x)& =\prod _{j=1}^r\left(\frac{e^{a_{j}t}-1}{t{e}^{a_{j}t}}\right){\hat{D}}_n(x|a_1,\dots ,a_r)\\ ={(x)}_n\sim (1,e^t-1),\end{array}$$

using (11), we have (26). □

3.3 Difference relations

Theorem 3

$$D_n(x+1|a_1,\dots ,a_r)-D_n(x|a_1,\dots ,a_r)=n{D}_{n-1}(x|a_1,\dots ,a_r),$$

(27)

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

(28)

Proof By (9) with (13), we get

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

By (8), we have (27).

Similarly, by (9) with (14), we get

$$(e^t-1){\hat{D}}_n(x|a_1,\dots ,a_r)=n{\hat{D}}_{n-1}(x|a_1,\dots ,a_r).$$

By (8), we have (28). □

3.4 Recurrence

Theorem 4

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

(29)

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

(30)

where $B_n$ is the nth ordinary Bernoulli number.

Proof By applying

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

(31)

[5], Corollary 3.7.2] with (13), we get

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

Now,

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}& ={(lng(t))}^{\prime }\\ ={(\sum _{j=1}^{r}ln(e^{a_{j}t}-1)-rlnt)}^{\prime }\\ =\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{e^{a_{j}t}-1}-\frac{r}{t}\\ =\frac{{\sum }_{j=1}^r{\prod }_{i\ne j}(e^{a_{i}t}-1)(a_{j}t{e}^{a_{j}t}-e^{a_{j}t}+1)}{t{\prod }_{j=1}^r(e^{a_{j}t}-1)}.\end{array}$$

Since

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

is a series with order ≥1, by (17) we have

$$\begin{array}{c}{D}_{n+1}(x|a_1,\dots ,a_r)\hfill \\ =x{D}_n(x-1|a_1,\dots ,a_r)-e^{-t}\frac{{\sum }_{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r}{t}{D}_n(x|a_1,\dots ,a_r)\hfill \\ =x{D}_n(x-1|a_1,\dots ,a_r)-e^{-t}\frac{{\sum }_{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r}{t}(\sum _{i=0}^n\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)D_{n-l}(a_1,\dots ,a_r)x^i)\hfill \\ =x{D}_n(x-1|a_1,\dots ,a_r)-\sum _{i=0}^n\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i)D_{n-l}(a_1,\dots ,a_r)e^{-t}(\sum _{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r)\frac{x^{i+1}}{i+1}.\hfill \end{array}$$

Since

$$\begin{array}{rl}{e}^{-t}(\sum _{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r)x^{i+1}& =e^{-t}(\sum _{j=1}^r\sum _{m=0}^{\mathrm{\infty }}\frac{{(-1)}^m{B}_m{a}_j^m}{m!}{t}^m-r)x^{i+1}\\ =e^{-t}(\sum _{j=1}^r\sum _{m=0}^{i+1}\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)B_m{(-a_j)}^m{x}^{i+1-m}-r{x}^{i+1})\\ =\sum _{j=1}^r\sum _{m=1}^{i+1}\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)B_m{(-a_j)}^m{(x-1)}^{i+1-m}\\ =\sum _{j=1}^r\sum _{m=0}^i\left(\genfrac{}{}{0ex}{}{i+1}{m}\right)B_{i+1-m}{(-a_j)}^{i+1-m}{(x-1)}^m,\end{array}$$

(32)

we have

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

which is the identity (29).

Next, by applying (31) with (14), we get

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

Now,

$$\begin{array}{rl}\frac{g^{\prime }(t)}{g(t)}& ={(lng(t))}^{\prime }\\ ={(\sum _{j=1}^{r}ln(e^{a_{j}t}-1)-rlnt-\left(\sum _{j=1}^r{a}_j\right)t)}^{\prime }\\ =\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{e^{a_{j}t}-1}-\frac{r}{t}-\sum _{j=1}^r{a}_j.\end{array}$$

By (20) we have

$$\begin{array}{c}{\hat{D}}_{n+1}(x|a_1,\dots ,a_r)\hfill \\ =(x+\sum _{j=1}^r{a}_j){\hat{D}}_n(x-1|a_1,\dots ,a_r)\hfill \\ -e^{-t}\frac{{\sum }_{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r}{t}{\hat{D}}_n(x|a_1,\dots ,a_r)\hfill \\ =(x+\sum _{j=1}^r{a}_j){\hat{D}}_n(x-1|a_1,\dots ,a_r)\hfill \\ -e^{-t}\frac{{\sum }_{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r}{t}(\sum _{i=0}^n\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i){\hat{D}}_{n-l}(a_1,\dots ,a_r)x^i)\hfill \\ =(x+\sum _{j=1}^r{a}_j){\hat{D}}_n(x-1|a_1,\dots ,a_r)\hfill \\ -\sum _{i=0}^n\sum _{l=i}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(l,i){\hat{D}}_{n-l}(a_1,\dots ,a_r)e^{-t}(\sum _{j=1}^r\frac{a_{j}t{e}^{a_{j}t}}{e^{a_{j}t}-1}-r)\frac{x^{i+1}}{i+1}.\hfill \end{array}$$

By (32), we have the identity (30). □

3.5 Differentiation

Theorem 5

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

(33)

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

(34)

Proof We shall use

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

(cf. [5], Theorem 2.3.12]). Since

$$\begin{array}{rl}〈\overline{f}(t)|x^{n-l}〉& =〈ln(1+t)|x^{n-l}〉=〈\sum _{m=1}^{\mathrm{\infty }}\frac{{(-1)}^{m-1}{t}^m}{m}|x^{n-l}〉\\ =\sum _{m=1}^{n-l}\frac{{(-1)}^{m-1}}{m}〈t^m|x^{n-l}〉\\ =\sum _{m=1}^{n-l}\frac{{(-1)}^{m-1}}{m}(n-l)!{\delta }_{m,n-l}\\ ={(-1)}^{n-l-1}(n-l-1)!,\end{array}$$

with (13), we have

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

which is the identity (33). Similarly, with (14), we have the identity (34). □

3.6 More relations

The classical Cauchy numbers $c_n$ are defined by

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

(see e.g. [9, 10]).

Theorem 6

$$\begin{array}{c}{D}_n(x|a_1,\dots ,a_r)=x{D}_{n-1}(x-1|a_1,\dots ,a_r)\hfill \\ \phantom{D_n(x|a_1,\dots ,a_r)=}+\frac{r}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)c_l{D}_{n-l}(x-1|a_1,\dots ,a_r)\hfill \\ \phantom{D_n(x|a_1,\dots ,a_r)=}-\frac{1}{n}\sum _{j=1}^r\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)a_j{c}_l{D}_{n-l}(x+a_j-1|a_1,\dots ,a_r,a_j),\hfill \end{array}$$

(35)

$$\begin{array}{c}{\hat{D}}_n(x|a_1,\dots ,a_r)=(x+\sum _{j=1}^r{a}_j){\hat{D}}_{n-1}(x-1|a_1,\dots ,a_r)\hfill \\ \phantom{{\hat{D}}_n(x|a_1,\dots ,a_r)=}+\frac{r}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)c_l{\hat{D}}_{n-l}(x-1|a_1,\dots ,a_r)\hfill \\ \phantom{{\hat{D}}_n(x|a_1,\dots ,a_r)=}-\frac{1}{n}\sum _{j=1}^r\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)a_j{c}_l{\hat{D}}_{n-l}(x-1|a_1,\dots ,a_r,a_j).\hfill \end{array}$$

(36)

Proof For $n\ge 1$, we have

$$\begin{array}{rcl}{D}_n(y|a_1,\dots ,a_r)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{D}_l(y|a_1,\dots ,a_r)\frac{t^l}{l!}|x^n〉\\ =& 〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(1+t)}^y|x^n〉\\ =& 〈{\partial }_t\left(\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(1+t)}^y\right)|x^{n-1}〉\\ =& 〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\left({\partial }_t{(1+t)}^y\right)|x^{n-1}〉\\ +〈\left({\partial }_t\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\right){(1+t)}^y|x^{n-1}〉\\ =& y{D}_{n-1}(y-1|a_1,\dots ,a_r)\\ +〈\left({\partial }_t\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\right){(1+t)}^y|x^{n-1}〉.\end{array}$$

Observe that

$$\begin{array}{rcl}{\partial }_t\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)& =& \sum _{j=1}^r\prod _{i\ne j}\left(\frac{ln(1+t)}{{(1+t)}^{a_i}-1}\right)\frac{\frac{1}{1+t}({(1+t)}^{a_j}-1)-ln(1+t)(a_j{(1+t)}^{a_j-1})}{{({(1+t)}^{a_j}-1)}^2}\\ =& \frac{1}{1+t}\prod _{i=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_i}-1}\right)\sum _{j=1}^r(\frac{1}{ln(1+t)}-\frac{a_j{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})\\ =& \frac{1}{1+t}\prod _{i=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_i}-1}\right)\frac{{\sum }_{j=1}^r(\frac{t}{ln(1+t)}-\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})}{t}.\end{array}$$

Since

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

is a series with order ≥1, we have

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

Therefore, we obtain

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

which is the identity (35).

Next, for $n\ge 1$ we have

$$\begin{array}{rcl}{\hat{D}}_n(y|a_1,\dots ,a_r)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{\hat{D}}_l(y|a_1,\dots ,a_r)\frac{t^l}{l!}|x^n〉\\ =& 〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(1+t)}^y|x^n〉\\ =& 〈{\partial }_t\left(\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(1+t)}^y\right)|x^{n-1}〉\\ =& 〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)\left({\partial }_t{(1+t)}^y\right)|x^{n-1}〉\\ +〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)\right){(1+t)}^y|x^{n-1}〉\\ =& y{\hat{D}}_{n-1}(y-1|a_1,\dots ,a_r)\\ +〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)\right){(1+t)}^y|x^{n-1}〉.\end{array}$$

Observe that

$$\begin{array}{c}{\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)\hfill \\ ={\partial }_t\left(\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\prod _{j=1}^r{(1+t)}^{a_j}\right)\hfill \\ =\left({\partial }_t\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\right)\prod _{j=1}^r{(1+t)}^{a_j}\hfill \\ +\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\left({\partial }_t\prod _{j=1}^r{(1+t)}^{a_j}\right)\hfill \\ =\frac{1}{1+t}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right)\frac{{\sum }_{j=1}^r(\frac{t}{ln(1+t)}-\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})}{t}\hfill \\ +\frac{1}{1+t}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right)\sum _{j=1}^r{a}_j.\hfill \end{array}$$

Thus, we have

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

Therefore, we obtain

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

which is the identity (36). □

3.7 Relations including the Stirling numbers of the first kind

Theorem 7 For $n-1\ge m\ge 1$, we have

$$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m)D_l(a_1,\dots ,a_r)\hfill \\ =\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1)D_l(-1|a_1,\dots ,a_r)\hfill \\ +\frac{1}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n}{l+1}\right)S_1(n-l-1,m)\hfill \\ \times (r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_i{D}_{l+1-i}(-1|a_1,\dots ,a_r)\hfill \\ -\sum _{j=1}^r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)a_j{c}_i{D}_{l+1-i}(a_j-1|a_1,\dots ,a_r,a_j)),\hfill \end{array}$$

(37)

$$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m){\hat{D}}_l(a_1,\dots ,a_r)\hfill \\ =\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1){\hat{D}}_l(-1|a_1,\dots ,a_r)\hfill \\ +\frac{1}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n}{l+1}\right)S_1(n-l-1,m)\hfill \\ \times (r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_i{\hat{D}}_{l+1-i}(-1|a_1,\dots ,a_r)\hfill \\ -\sum _{j=1}^r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)a_j{c}_i{\hat{D}}_{l+1-i}(-1|a_1,\dots ,a_r,a_j))\hfill \\ +\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\sum _{j=1}^r{a}_j{\hat{D}}_l(-1|a_1,\dots ,a_r).\hfill \end{array}$$

(38)

Proof We shall compute

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

in two different ways. On the one hand,

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

On the other hand,

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

(39)

The second term of (39) is equal to

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

The first term of (39) is equal to

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

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

$$\begin{array}{c}m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,m)D_l(a_1,\dots ,a_r)\hfill \\ =m!\sum _{l=0}^{n-m}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m-1)D_l(-1|a_1,\dots ,a_r)\hfill \\ +\frac{m!}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n}{l+1}\right)S_1(n-l-1,m)\hfill \\ \times (r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_{l+1-i}{D}_i(-1|a_1,\dots ,a_r)\hfill \\ -\sum _{j=1}^r\sum _{i=0}^{l+1}{a}_j\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_{l+1-i}{D}_i(a_j-1|a_1,\dots ,a_r,a_j)).\hfill \end{array}$$

Thus, we get (37).

Next, we shall compute

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

in two different ways. On the one hand,

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

On the other hand,

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

(40)

The second term of (40) is equal to

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

The first term of (40) is equal to

$$\begin{array}{c}〈\left({\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)\right){(ln(1+t))}^m|x^{n-1}〉\hfill \\ =〈{\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)|{(ln(1+t))}^m{x}^{n-1}〉\hfill \\ =〈{\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)|\sum _{l=0}^{n-m-1}\frac{m!}{(l+m)!}{S}_1(l+m,m)t^{l+m}{x}^{n-1}〉\hfill \\ =\sum _{l=0}^{n-m-1}\frac{m!}{(l+m)!}{S}_1(l+m,m){(n-1)}_{l+m}〈{\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)|x^{n-l-m-1}〉.\hfill \end{array}$$

From the proof of (36), we recall

$$\begin{array}{c}{\partial }_t\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)\hfill \\ =\frac{1}{1+t}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right)\frac{{\sum }_{j=1}^r(\frac{t}{ln(1+t)}-\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})}{t}\hfill \\ +\frac{1}{1+t}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right)\sum _{j=1}^r{a}_j.\hfill \end{array}$$

Hence, the first term of (39) is equal to

$$\begin{array}{c}\sum _{l=0}^{n-m-1}m!\left(\genfrac{}{}{0ex}{}{n-1}{l+m}\right)S_1(l+m,m)\hfill \\ \times (〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\frac{{\sum }_{j=1}^r(\frac{t}{ln(1+t)}-\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})}{t}{x}^{n-l-m-1}〉\hfill \\ +\left(\sum _{j=1}^r{a}_j\right)〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|x^{n-l-m-1}〉)\hfill \\ =m!\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\hfill \\ \times (〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\frac{{\sum }_{j=1}^r(\frac{t}{ln(1+t)}-\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})}{t}{x}^l〉\hfill \\ +\left(\sum _{j=1}^r{a}_j\right)〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|x^l〉)\hfill \\ =m!\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\hfill \\ \times (\frac{1}{l+1}〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\sum _{j=1}^r(\frac{t}{ln(1+t)}-\frac{a_{j}t{(1+t)}^{a_j}}{{(1+t)}^{a_j}-1})x^{l+1}〉\hfill \\ +\left(\sum _{j=1}^r{a}_j\right)〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|x^l〉)\hfill \\ =m!\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\hfill \\ \times (\frac{r}{l+1}〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\frac{t}{ln(1+t)}{x}^{l+1}〉\hfill \\ -\frac{1}{l+1}\left(\sum _{j=1}^r{a}_j\right)〈\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\frac{t}{ln(1+t)}{x}^{l+1}〉\hfill \\ +\left(\sum _{j=1}^r{a}_j\right)〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|x^l〉)\hfill \\ =m!\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\hfill \\ \times (\frac{r}{l+1}〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\sum _{i=0}^{l+1}{c}_i\frac{t^i}{i!}{x}^{l+1}〉\hfill \\ -\frac{1}{l+1}\left(\sum _{j=1}^r{a}_j\right)〈\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|\sum _{i=0}^{l+1}{c}_i\frac{t^i}{i!}{x}^{l+1}〉\hfill \\ +\left(\sum _{j=1}^r{a}_j\right)〈\prod _{i=1}^r\left(\frac{{(1+t)}^{a_i}ln(1+t)}{{(1+t)}^{a_i}-1}\right){(1+t)}^{-1}|x^l〉)\hfill \\ =m!\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\hfill \\ \times (\frac{r}{l+1}\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_i{\hat{D}}_{l+1-i}(-1|a_1,\dots ,a_r)\hfill \\ -\frac{1}{l+1}\sum _{j=1}^r{a}_j\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_i{\hat{D}}_{l+1-i}(-1|a_1,\dots ,a_r,a_j)+\sum _{j=1}^r{a}_j{\hat{D}}_l(-1|a_1,\dots ,a_r))\hfill \\ =\frac{m!}{n}\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n}{l+1}\right)S_1(n-l-1,m)\hfill \\ \times (r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)c_i{\hat{D}}_{l+1-i}(-1|a_1,\dots ,a_r)\hfill \\ -\sum _{j=1}^r\sum _{i=0}^{l+1}\left(\genfrac{}{}{0ex}{}{l+1}{i}\right)a_j{c}_i{\hat{D}}_{l+1-i}(-1|a_1,\dots ,a_r,a_j))\hfill \\ +m!\sum _{l=0}^{n-m-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)S_1(n-l-1,m)\sum _{j=1}^r{a}_j{\hat{D}}_l(-1|a_1,\dots ,a_r).\hfill \end{array}$$

Therefore, we get (38). □

3.8 Relations with the falling factorials

Theorem 8

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

(41)

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

(42)

Proof For (13) and (23), assume that $D_n(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{(x)}_m$. By (12), we have

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

Thus, we get the identity (41).

Similarly, for (13) and (23), assume that ${\hat{D}}_n(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{(x)}_m$. By (12), we have

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

Thus, we get the identity (42). □

3.9 Relations with higher-order Frobenius-Euler polynomials

For $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, the Frobenius-Euler polynomials of order r, $H_n^{(r)}(x|\lambda )$ are defined by the generating function

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

(see e.g. [11, 12]).

Theorem 9

$$\begin{array}{c}{D}_n(x|a_1,\dots ,a_r)=\sum _{m=0}^n(\sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0ex}{}{s}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{l}\right){(n)}_j\hfill \\ \phantom{D_n(x|a_1,\dots ,a_r)=}\times {(1-\lambda )}^{-j}{S}_1(n-j-l,m)D_l(a_1,\dots ,a_r))H_m^{(s)}(x|\lambda ),\hfill \end{array}$$

(43)

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

(44)

Proof For (13) and

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

(45)

assume that $D_n(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\lambda )$. By (12), similarly to the proof of (37), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{e^{ln(1+t)}-\lambda }{1-\lambda })}^s}{{\prod }_{j=1}^r(\frac{e^{a_{j}ln(1+t)}-1}{ln(1+t)})}{(ln(1+t))}^m|x^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m{(1-\lambda +t)}^s|x^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|\sum _{i=0}^{min\{s,n\}}\left(\genfrac{}{}{0ex}{}{s}{i}\right){(1-\lambda )}^{s-i}{t}^i{x}^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{i}\right){(1-\lambda )}^{s-i}{(n)}_i〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)|{(ln(1+t))}^m{x}^{n-i}〉\\ =\frac{1}{m!{(1-\lambda )}^s}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{i}\right){(1-\lambda )}^{s-i}{(n)}_i\sum _{l=0}^{n-m-i}m!\left(\genfrac{}{}{0ex}{}{n-i}{l}\right)S_1(n-i-l,m)D_l(a_1,\dots ,a_r)\\ =\sum _{i=0}^{n-m}\sum _{l=0}^{n-m-i}\left(\genfrac{}{}{0ex}{}{s}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{l}\right){(n)}_i{(1-\lambda )}^{-i}{S}_1(n-i-l,m)D_l(a_1,\dots ,a_r).\end{array}$$

Thus, we get the identity (43).

Next, for (14) and (45), assume that ${\hat{D}}_n(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\lambda )$. By (12), similarly to the proof of (38), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{e^{ln(1+t)}-\lambda }{1-\lambda })}^s}{{\prod }_{j=1}^r(\frac{e^{a_{j}ln(1+t)}-1}{e^{a_{j}ln(1+t)}ln(1+t)})}{(ln(1+t))}^m|x^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|{(1-\lambda +t)}^s{x}^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|\sum _{i=0}^{min\{s,n\}}\left(\genfrac{}{}{0ex}{}{s}{i}\right){(1-\lambda )}^{s-i}{t}^i{x}^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{i}\right){(1-\lambda )}^{s-i}{(n)}_i〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right)|{(ln(1+t))}^m{x}^{n-i}〉\\ =\frac{1}{m!{(1-\lambda )}^s}\sum _{i=0}^{n-m}\left(\genfrac{}{}{0ex}{}{s}{i}\right){(1-\lambda )}^{s-i}{(n)}_i\sum _{l=0}^{n-m-i}m!\left(\genfrac{}{}{0ex}{}{n-i}{l}\right)S_1(n-i-l,m){\hat{D}}_l(a_1,\dots ,a_r)\\ =\sum _{i=0}^{n-m}\sum _{l=0}^{n-m-i}\left(\genfrac{}{}{0ex}{}{s}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{l}\right){(n)}_i{(1-\lambda )}^{-i}{S}_1(n-i-l,m){\hat{D}}_l(a_1,\dots ,a_r).\end{array}$$

Thus, we get the identity (44). □

3.10 Relations with higher-order Bernoulli polynomials

Bernoulli polynomials ${\mathfrak{B}}_n^{(r)}(x)$ of order r are defined by

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

(see e.g. [5], Section 2.2]). In addition, Cauchy numbers of the first kind ${\mathfrak{C}}_n^{(r)}$ of order r are defined by

$${\left(\frac{t}{ln(1+t)}\right)}^r=\sum _{n=0}^{\mathrm{\infty }}\frac{{\mathfrak{C}}_n^{(r)}}{n!}{t}^n$$

(see e.g. [13], (2.1)], [14], (6)]).

Theorem 10

$$\begin{array}{c}{D}_n(x|a_1,\dots ,a_r)\hfill \\ =\sum _{m=0}^n(\sum _{i=0}^{n-m}\sum _{l=0}^{n-m-i}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{l}\right){\mathfrak{C}}_i^{(s)}{S}_1(n-i-l,m)D_l(a_1,\dots ,a_r)){\mathfrak{B}}_m^{(s)}(x),\hfill \end{array}$$

(46)

$$\begin{array}{c}{\hat{D}}_n(x|a_1,\dots ,a_r)\hfill \\ =\sum _{m=0}^n(\sum _{i=0}^{n-m}\sum _{l=0}^{n-m-i}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{l}\right){\mathfrak{C}}_i^{(s)}{S}_1(n-i-l,m){\hat{D}}_l(a_1,\dots ,a_r)){\mathfrak{B}}_m^{(s)}(x).\hfill \end{array}$$

(47)

Proof For (13) and

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

(48)

assume that $D_n(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{\mathfrak{B}}_m^{(s)}(x)$. By (12), similarly to the proof of (37), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{e^{ln(1+t)}-1}{ln(1+t)})}^s}{{\prod }_{j=1}^r(\frac{e^{a_{j}ln(1+t)}-1}{ln(1+t)})}{(ln(1+t))}^m|x^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|{\left(\frac{t}{ln(1+t)}\right)}^s{x}^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|\sum _{i=0}^{\mathrm{\infty }}{\mathfrak{C}}_i^{(s)}\frac{t^i}{i!}{x}^n〉\\ =\frac{1}{m!}\sum _{i=0}^{n-m}{\mathfrak{C}}_i^{(s)}\left(\genfrac{}{}{0ex}{}{n}{i}\right)〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|x^{n-i}〉\\ =\frac{1}{m!}\sum _{i=0}^{n-m}{\mathfrak{C}}_i^{(s)}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\sum _{l=0}^{n-m-i}m!\left(\genfrac{}{}{0ex}{}{n-i}{l}\right)S_1(n-i-l,m)D_l(a_1,\dots ,a_r)\\ =\sum _{i=0}^{n-m}\sum _{l=0}^{n-m-i}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{l}\right){\mathfrak{C}}_i^{(s)}{S}_1(n-i-l,m)D_l(a_1,\dots ,a_r).\end{array}$$

Thus, we get the identity (46).

Next, for (13) and (48), assume that ${\hat{D}}_n(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{\mathfrak{B}}_m^{(s)}(x)$. By (12), similarly to the proof of (38), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{(\frac{e^{ln(1+t)}-1}{ln(1+t)})}^s}{{\prod }_{j=1}^r(\frac{e^{a_{j}ln(1+t)}-1}{e^{a_{j}ln(1+t)}ln(1+t)})}{(ln(1+t))}^m|x^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|{\left(\frac{t}{ln(1+t)}\right)}^s{x}^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|\sum _{i=0}^{\mathrm{\infty }}{\mathfrak{C}}_i^{(s)}\frac{t^i}{i!}{x}^n〉\\ =\frac{1}{m!}\sum _{i=0}^{n-m}{\mathfrak{C}}_i^{(s)}\left(\genfrac{}{}{0ex}{}{n}{i}\right)〈\prod _{j=1}^r\left(\frac{{(1+t)}^{a_j}ln(1+t)}{{(1+t)}^{a_j}-1}\right){(ln(1+t))}^m|x^{n-i}〉\\ =\frac{1}{m!}\sum _{i=0}^{n-m}{\mathfrak{C}}_i^{(s)}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\sum _{l=0}^{n-m-i}m!\left(\genfrac{}{}{0ex}{}{n-i}{l}\right)S_1(n-i-l,m){\hat{D}}_l(a_1,\dots ,a_r)\\ =\sum _{i=0}^{n-m}\sum _{l=0}^{n-m-i}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n-i}{l}\right){\mathfrak{C}}_i^{(s)}{S}_1(n-i-l,m){\hat{D}}_l(a_1,\dots ,a_r).\end{array}$$

Thus, we get the identity (47). □