Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type polynomials

Abstract

In this paper, by considering Barnes-type Daehee polynomials of the first kind as well as poly-Cauchy polynomials of the first kind, we define and investigate the mixed-type polynomials of these polynomials. From the properties of 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 $D_n^{(k)}(x|a_1,\dots ,a_r)$ called the Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type polynomials, whose generating function is given by

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

(1)

where $a_1,\dots ,a_r\ne 0$. Here, ${Lif}_k(x)$ ($k\in \mathbb{Z}$) is the polyfactorial function [1] defined by

$${Lif}_k(x)=\sum _{m=0}^{\mathrm{\infty }}\frac{x^m}{m!{(m+1)}^k}.$$

When $x=0$, $D_n^{(k)}(a_1,\dots ,a_r)=D_n^{(k)}(0|a_1,\dots ,a_r)$ is called Barnes-type Daehee of the first kind and poly-Cauchy of the first kind mixed-type number.

Recall that the Barnes-type Daehee polynomials of the first kind, denoted by $D_n(x|a_1,\dots ,a_r)$, are given by the generating function

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

If $a_1=\cdots =a_r=1$, then $D_n^{(r)}(x)=D_n(x|\underset{r}{\underbrace{1,\dots ,1}})$ are the Daehee polynomials of the first kind of order r. Dahee polynomials were defined by the second author [2] and have been investigated in [3, 4].

The poly-Cauchy polynomials of the first kind, denoted by $c_n^{(k)}(x)$ [5, 6], are given by the generating function as

$${Lif}_k(ln(1+t)){(1+t)}^x=\sum _{n=0}^{\mathrm{\infty }}{c}_n^{(k)}(-x)\frac{t^n}{n!}.$$

In this paper, by considering Barnes-type Daehee polynomials of the first kind as well as poly-Cauchy polynomials of the first kind, we define and investigate the mixed-type polynomials of these polynomials. From the properties of 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:

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

(2)

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

$$〈f(t)|x^n〉=a_n(n\ge 0).$$

(3)

In particular,

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

(4)

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 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 and the umbral calculus is the study of the 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

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

(5)

and

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

(6)

[7], Theorem 2.2.5]. Thus, by (6), we get

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

(7)

Sheffer sequences are characterized by the generating function [7], 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 [7], Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

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

(8)

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

(9)

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

(10)

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 [7], 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 [7], p.132]

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

(11)

3 Main results

From the definition (1), $D_n^{(k)}(x|a_1,\dots ,a_r)$ is the Sheffer sequence for the pair

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

So,

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

(12)

3.1 Explicit expressions

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

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

(13)

where $a_1,\dots ,a_r\ne 0$ [8, 9]. 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

$$\begin{array}{r}{D}_n^{(k)}(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\sum _{l=0}^m{S}_1(n,m)\frac{\left(\genfrac{}{}{0ex}{}{m}{l}\right)}{{(m-l+1)}^k}{B}_l(x|a_1,\dots ,a_r)\end{array}$$

(14)

$$=\sum _{j=0}^n\sum _{l=0}^{n-j}\sum _{i=0}^l\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{i}\right)S_1(n-l,j)c_i^{(k)}{D}_{l-i}(a_1,\dots ,a_r)x^j$$

(15)

$$=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)D_{n-l}(a_1,\dots ,a_r)c_l^{(k)}(-x)$$

(16)

$$=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)c_{n-l}^{(k)}{D}_l(x|a_1,\dots ,a_r).$$

(17)

Proof Since

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

(18)

and

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

(19)

we have

$$\begin{array}{rl}{D}_n^{(k)}(x|a_1,\dots ,a_r)& =\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right){Lif}_k(t){(x)}_n\\ =\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right){Lif}_k(t)x^m\\ =\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right)\sum _{l=0}^m\frac{t^l}{l!{(l+1)}^k}{x}^m\\ =\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right)\sum _{l=0}^m\frac{{(m)}_l}{l!{(l+1)}^k}{x}^{m-l}\\ =\sum _{m=0}^n{S}_1(n,m)\sum _{l=0}^m\frac{{(m)}_l}{l!{(l+1)}^k}\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right)x^{m-l}\\ =\sum _{m=0}^n\sum _{l=0}^m{S}_1(n,m)\frac{\left(\genfrac{}{}{0ex}{}{m}{l}\right)}{{(l+1)}^k}{B}_{m-l}(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\sum _{l=0}^m{S}_1(n,m)\frac{\left(\genfrac{}{}{0ex}{}{m}{l}\right)}{{(m-l+1)}^k}{B}_l(x|a_1,\dots ,a_r).\end{array}$$

Thus, we get (14).

By (9) with (12), we get

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

On the other hand,

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

Thus, we obtain

$$\begin{array}{r}{D}_n^{(k)}(x|a_1,\dots ,a_r)\\ =\sum _{j=0}^n\sum _{l=0}^{n-j}\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_1(n-l,j)D_l^{(k)}(a_1,\dots ,a_r)x^j\\ =\sum _{j=0}^n\sum _{l=0}^{n-j}\sum _{i=0}^l\left(\genfrac{}{}{0ex}{}{n}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{i}\right)S_1(n-l,j)c_i^{(k)}{D}_{l-i}(a_1,\dots ,a_r)x^j,\end{array}$$

which is the identity (15).

Next,

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

Thus, we obtain (16).

Finally, we obtain

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

Thus, we get the identity (17). □

3.2 Sheffer identity

Theorem 2

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

(20)

Proof By (12) with

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

using (10), we have (20). □

3.3 Difference relations

Theorem 3

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

(21)

Proof By (8) with (12), we get

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

By (7), we have (21). □

3.4 Recurrence

Theorem 4

$$\begin{array}{rl}{D}_{n+1}^{(k)}(x|a_1,\dots ,a_r)=& x{D}_n^{(k)}(x-1|a_1,\dots ,a_r)\\ -\sum _{m=0}^n\sum _{j=1}^r\sum _{l=1}^{m+1}\sum _{i=0}^l\frac{\left(\genfrac{}{}{0ex}{}{m+1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{i}\right)}{(m+1){(l-i+1)}^k}{S}_1(n,m)\\ \times {(-a_j)}^{m+1-l}{B}_{m+1-l}{B}_i(x-1|a_1,\dots ,a_r)\\ +\sum _{m=0}^n\sum _{l=0}^m\frac{\left(\genfrac{}{}{0ex}{}{m}{l}\right)}{{(m+2-l)}^k}{S}_1(n,m)B_l(x-1|a_1,\dots ,a_r),\end{array}$$

(22)

where $B_n$ is the nth ordinary Bernoulli number.

Proof By applying

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

(23)

[7], Corollary 3.7.2] with (12), we get

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

Now,

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

Observe that

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

is a series with order ≥1. Since

$$D_n^{(k)}(x|a_1,\dots ,a_r)=\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right){Lif}_k(t){(x)}_n=\sum _{m=0}^n{S}_1(n,m)\prod _{j=1}^r\left(\frac{t}{e^{a_{j}t}-1}\right){Lif}_k(t)x^m,$$

we have

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

(24)

Since

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

the first term in (24) is

$$\begin{array}{r}\sum _{m=0}^n\frac{S_1(n,m)}{m+1}\sum _{j=1}^r\sum _{l=1}^{m+1}\left(\genfrac{}{}{0ex}{}{m+1}{l}\right){(-a_j)}^{m+1-l}{B}_{m+1-l}{Lif}_k(t)\left(\prod _{j=1}^r\frac{t}{e^{a_{j}t}-1}\right)x^l\\ =\sum _{m=0}^n\frac{S_1(n,m)}{m+1}\sum _{j=1}^r\sum _{l=1}^{m+1}\left(\genfrac{}{}{0ex}{}{m+1}{l}\right){(-a_j)}^{m+1-l}{B}_{m+1-l}\sum _{i=0}^l\frac{t^i}{i!{(i+1)}^k}{B}_l(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\frac{S_1(n,m)}{m+1}\sum _{j=1}^r\sum _{l=1}^{m+1}\left(\genfrac{}{}{0ex}{}{m+1}{l}\right){(-a_j)}^{m+1-l}{B}_{m+1-l}\sum _{i=0}^l\frac{\left(\genfrac{}{}{0ex}{}{l}{i}\right)}{{(i+1)}^k}{B}_{l-i}(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\sum _{j=1}^r\sum _{l=1}^{m+1}\sum _{i=0}^l\frac{\left(\genfrac{}{}{0ex}{}{m+1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{i}\right)}{(m+1){(l-i+1)}^k}{S}_1(n,m){(-a_j)}^{m+1-l}{B}_{m+1-l}{B}_i(x|a_1,\dots ,a_r).\end{array}$$

Since

$${Lif}_{k-1}(t)-{Lif}_k(t)=(\frac{1}{2^{k-1}}-\frac{1}{2^k})t+\cdots ,$$

(25)

the second term in (24) is

$$\begin{array}{r}\sum _{m=0}^n{S}_1(n,m)\frac{{Lif}_{k-1}(t)-{Lif}_k(t)}{t}{B}_m(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n{S}_1(n,m)({Lif}_{k-1}(t)-{Lif}_k(t))\frac{B_{m+1}(x|a_1,\dots ,a_r)}{m+1}\\ =\sum _{m=0}^n\frac{S_1(n,m)}{m+1}({Lif}_{k-1}(t)-{Lif}_k(t))B_{m+1}(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\frac{S_1(n,m)}{m+1}(\sum _{l=0}^{m+1}\frac{t^l}{l!{(l+1)}^{k-1}}{B}_{m+1}(x|a_1,\dots ,a_r)-\sum _{l=0}^{m+1}\frac{t^l}{l!{(l+1)}^k}{B}_{m+1}(x|a_1,\dots ,a_r))\\ =\sum _{m=0}^n\frac{S_1(n,m)}{m+1}(\sum _{l=0}^{m+1}\frac{\left(\genfrac{}{}{0ex}{}{m+1}{l}\right)}{{(l+1)}^{k-1}}{B}_{m+1-l}(x|a_1,\dots ,a_r)-\sum _{l=0}^{m+1}\frac{\left(\genfrac{}{}{0ex}{}{m+1}{l}\right)}{{(l+1)}^k}{B}_{m+1-l}(x|a_1,\dots ,a_r))\\ =\sum _{m=0}^n\frac{S_1(n,m)}{m+1}\sum _{l=1}^{m+1}\frac{\left(\genfrac{}{}{0ex}{}{m+1}{l}\right)l}{{(l+1)}^k}{B}_{m+1-l}(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\sum _{l=1}^{m+1}\left(\genfrac{}{}{0ex}{}{m}{l-1}\right)S_1(n,m)\frac{1}{{(l+1)}^k}{B}_{m+1-l}(x|a_1,\dots ,a_r)\\ =\sum _{m=0}^n\sum _{l=0}^m\frac{\left(\genfrac{}{}{0ex}{}{m}{l}\right)}{{(m+2-l)}^k}{S}_1(n,m)B_l(x|a_1,\dots ,a_r).\end{array}$$

Thus, we have

$$\begin{array}{rl}{D}_{n+1}^{(k)}(x|a_1,\dots ,a_r)=& x{D}_n^{(k)}(x-1|a_1,\dots ,a_r)\\ -\sum _{m=0}^n\sum _{j=1}^r\sum _{l=1}^{m+1}\sum _{i=0}^l\frac{\left(\genfrac{}{}{0ex}{}{m+1}{l}\right)\left(\genfrac{}{}{0ex}{}{l}{i}\right)}{(m+1){(l-i+1)}^k}{S}_1(n,m)\\ \times {(-a_j)}^{m+1-l}{B}_{m+1-l}{B}_i(x-1|a_1,\dots ,a_r)\\ +\sum _{m=0}^n\sum _{l=0}^m\frac{\left(\genfrac{}{}{0ex}{}{m}{l}\right)}{{(m+2-l)}^k}{S}_1(n,m)B_l(x-1|a_1,\dots ,a_r),\end{array}$$

which is the identity (22). □

3.5 Differentiation

Theorem 5

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

(26)

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. [7], 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 (12), we have

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

which is the identity (26). □

3.6 One more relation

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. [1, 10]).

Theorem 6

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

(27)

Proof For $n\ge 1$, we have

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

The third term is

$$\begin{array}{r}y〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){Lif}_k(ln(1+t)){(1+t)}^{y-1}|x^{n-1}〉\\ =y{D}_{n-1}^{(k)}(y-1|a_1,\dots ,a_r).\end{array}$$

By (25), the second term is

$$\begin{array}{r}〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\frac{{Lif}_{k-1}(ln(1+t))-{Lif}_k(ln(1+t))}{(1+t)ln(1+t)}{(1+t)}^y|x^{n-1}〉\\ =〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\frac{{Lif}_{k-1}(ln(1+t))-{Lif}_k(ln(1+t))}{t}{(1+t)}^{y-1}|\frac{t}{ln(1+t)}{x}^{n-1}〉\\ =〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\frac{{Lif}_{k-1}(ln(1+t))-{Lif}_k(ln(1+t))}{t}{(1+t)}^{y-1}|\sum _{l=0}^{\mathrm{\infty }}{c}_l\frac{t^l}{l!}{x}^{n-1}〉\\ =\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)c_l\\ \times 〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(1+t)}^{y-1}|\frac{{Lif}_{k-1}(ln(1+t))-{Lif}_k(ln(1+t))}{t}{x}^{n-1-l}〉\\ =\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n-1}{l}\right)c_l\\ \times 〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){(1+t)}^{y-1}|({Lif}_{k-1}(ln(1+t))-{Lif}_k(ln(1+t)))\frac{x^{n-l}}{n-l}〉\\ =\frac{1}{n}\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)c_l(〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){Lif}_{k-1}(ln(1+t)){(1+t)}^{y-1}|x^{n-l}〉\\ -〈\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right){Lif}_k(ln(1+t)){(1+t)}^{y-1}|x^{n-l}〉)\\ =\frac{1}{n}\sum _{l=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{l}\right)c_l(D_{n-l}^{(k-1)}(y-1|a_1,\dots ,a_r)-D_{n-l}^{(k)}(y-1|a_1,\dots ,a_r)).\end{array}$$

Since

$${\partial }_t\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)=\frac{1}{1+t}\prod _{j=1}^r\left(\frac{ln(1+t)}{{(1+t)}^{a_j}-1}\right)\frac{{\sum }_{i=1}^r(\frac{t}{ln(1+t)}-\frac{a_{i}t{(1+t)}^{a_i}}{{(1+t)}^{a_i}-1})}{t},$$

with

$$\sum _{i=1}^r(\frac{t}{ln(1+t)}-\frac{a_{i}t{(1+t)}^{a_i}}{{(1+t)}^{a_i}-1})=-\frac{1}{2}\left(\sum _{i=1}^r{a}_i\right)t+\cdots$$

a series with order (≥1), the first term is

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

Therefore, we obtain

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

which is the identity (27). □

3.7 A relation including the Stirling numbers of the first kind

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

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

(28)

Proof We shall compute

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

in two different ways. On the one hand,

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

On the other hand,

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

(29)

The third term of (29) is equal to

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

The second term of (29) is equal to

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

The first term of (29) is equal to

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

Therefore, we get for $n\ge m\ge 1$

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

Dividing both sides by $(m-1)!$, we obtain for $n\ge m\ge 1$

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

Thus, we get (28). □

3.8 A relation with the falling factorials

Theorem 8

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

(30)

Proof For (12) and (19), assume that $D_n^{(k)}(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{(x)}_m$. By (11), we have

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

Thus, we get the identity (30). □

3.9 A relation 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]).

Theorem 9

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

(31)

Proof For (12) and

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

(32)

assume that $D_n^{(k)}(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\lambda )$. By (11), similarly to the proof of (28), we have

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

Thus, we get the identity (31). □

3.10 A relation 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. [7], 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. [12], (2.1)], [13], (6)]).

Theorem 10

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

(33)

Proof For (12) and

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

(34)

assume that $D_n^{(k)}(x|a_1,\dots ,a_r)={\sum }_{m=0}^n{C}_{n,m}{\mathfrak{B}}_m^{(s)}(x)$. By (11), similarly to the proof of (28), we have

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

Thus, we get the identity (33). □