Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials

Abstract

In this paper, we consider Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type 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 $T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$ whose generating function is given by

$$\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^n}{n!},$$

(1)

where $r\in {\mathbb{Z}}_{>\mathbb{⊬}}$, $k\in \mathbb{Z}$, $a_1,\dots ,a_r\ne 0$, ${\lambda }_1,\dots ,{\lambda }_r\ne 1$ and

$${Li}_k(x)=\sum _{m=1}^{\mathrm{\infty }}\frac{x^m}{m^k}$$

is the k th polylogarithm function. $T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$ will be called Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type polynomials. When $x=0$, $T_n^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=T_n^{(r,k)}(0|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$ will be called Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type numbers.

Recall that, for every integer k, the poly-Bernoulli polynomials $B_n^{(k)}(x)$ are defined by the generating function as follows:

$$\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_n^{(k)}(x)\frac{t^n}{n!}$$

(2)

([1], cf. [2]). Also, as a natural generalization of higher-order Frobenius-Euler polynomials, Barnes’ multiple Frobenius-Euler polynomials $H_n^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$ are defined by the generating function as follows:

$$\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)e^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_n^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^n}{n!},$$

(3)

where $a_1,\dots ,a_r\ne 0$. Note that 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., [3]).

In this paper, we consider Barnes’ multiple Frobenius-Euler and poly-Bernoulli mixed-type 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:

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

(4)

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

$$〈f(t)|x^n〉=a_n(n\ge 0).$$

(5)

In particular,

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

(6)

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 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

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

(7)

and

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

(8)

[[4], Theorem 2.2.5]. Thus, by (8), we get

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

(9)

Sheffer sequences are characterized in the generating function [[4], 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 [[4], Theorem 2.3.7, Theorem 2.3.5, Theorem 2.3.9]:

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

(10)

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

(11)

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

(12)

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 [[4], 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 [[4], p.132]

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

(13)

3 Main results

We now note that $B_n^{(k)}(x)$, $H_n^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$ and $T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$ are the Appell sequences for

$$\begin{array}{c}{g}_k(t)=\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},g_r(t)=\prod _{j=1}^r\left(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right),\hfill \\ g_{r,k}(t)=\prod _{j=1}^r\left(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)\frac{1-e^{-t}}{{Li}_k(1-e^{-t})}.\hfill \end{array}$$

So,

$$B_n^{(k)}(x)\sim (\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},t),$$

(14)

$$H_n^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\sim (\prod _{j=1}^r\left(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right),t),$$

(15)

$$T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\sim (\prod _{j=1}^r\left(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)\frac{1-e^{-t}}{{Li}_k(1-e^{-t})},t).$$

(16)

In particular, we have

$$t{B}_n^{(k)}(x)=\frac{d}{dx}{B}_n^{(k)}(x)=n{B}_{n-1}^{(k)}(x),$$

(17)

$$\begin{array}{rl}t{H}_n^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)& =\frac{d}{dx}{H}_n^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =n{H}_{n-1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r),\end{array}$$

(18)

$$\begin{array}{rl}t{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)& =\frac{d}{dx}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =n{T}_{n-1}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

(19)

Notice that

$$\frac{d}{dx}{Li}_k(x)=\frac{1}{x}{Li}_{k-1}(x).$$

3.1 Explicit expressions

Write $H_n^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r):=H_n^{(r)}(0|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$. Let ${(n)}_j=n(n-1)\cdots (n-j+1)$ ($j\ge 1$) with ${(n)}_0=1$.

Theorem 1

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l^{(k)}(x)H_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\end{array}$$

(20)

$$=\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}^{(k)}{H}_l^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$$

(21)

$$=\sum _{l=0}^n\sum _{m=0}^n\sum _{j=0}^m{(-1)}^j\left(\genfrac{}{}{0ex}{}{m}{j}\right)\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{1}{{(m+1)}^k}{H}_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r){(x-j)}^l$$

(22)

$$\begin{array}{r}=\sum _{l=0}^n(\sum _{j=l}^n\sum _{m=0}^{n-j}{(-1)}^{n-m-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{j}{l}\right)\\ \times \frac{m!}{{(m+1)}^k}{S}_2(n-j,m)H_{j-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r))x^l\end{array}$$

(23)

$$=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)T_{n-j}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)x^j.$$

(24)

Proof By (1), (2) and (3), we have

$$\begin{array}{rl}{T}_n^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)& =〈\sum _{i=0}^{\mathrm{\infty }}{T}_i^{(r,k)}(y|a_1,\dots ,a_r,{\lambda }_1,\dots ,{\lambda }_r)\frac{t^i}{i!}|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)|\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}{x}^n〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)|\sum _{l=0}^{\mathrm{\infty }}{B}_l^{(k)}(y)\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l^{(k)}(y)〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l^{(k)}(y)〈\sum _{i=0}^{\mathrm{\infty }}{H}_i^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^i}{i!}|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_l^{(k)}(y)H_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

So, we get (20).

We also have

$$\begin{array}{rl}{T}_n^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)& =〈\sum _{i=0}^{\mathrm{\infty }}{T}_i^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^i}{i!}|x^n〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^n〉\\ =〈\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)e^{yt}{x}^n〉\\ =〈\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|\sum _{l=0}^{\mathrm{\infty }}{H}_l^{(r)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l^{(r)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〈\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l^{(r)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〈\sum _{i=0}^{\mathrm{\infty }}{B}_i^{(k)}\frac{t^i}{i!}|x^{n-l}〉\\ =\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_l^{(r)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)B_{n-l}^{(k)}.\end{array}$$

Thus, we get (21).

In [5] we obtained that

$$\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^n=\sum _{m=0}^n\frac{1}{{(m+1)}^k}\sum _{j=0}^m{(-1)}^j\left(\genfrac{}{}{0ex}{}{m}{j}\right){(x-j)}^n.$$

So,

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^n\\ =\sum _{m=0}^n\frac{1}{{(m+1)}^k}\sum _{j=0}^m{(-1)}^j\left(\genfrac{}{}{0ex}{}{m}{j}\right)\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right){(x-j)}^n\\ =\sum _{m=0}^n\frac{1}{{(m+1)}^k}\sum _{j=0}^m{(-1)}^j\left(\genfrac{}{}{0ex}{}{m}{j}\right)\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)H_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r){(x-j)}^l\\ =\sum _{l=0}^n\sum _{m=0}^n\sum _{j=0}^m{(-1)}^j\left(\genfrac{}{}{0ex}{}{m}{j}\right)\left(\genfrac{}{}{0ex}{}{n}{l}\right)\frac{1}{{(m+1)}^k}{H}_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r){(x-j)}^l,\end{array}$$

which is identity (22).

In [5] we obtained that

$$\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^n=\sum _{j=0}^n(\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m))x^j,$$

where $S_2(l,m)$ are the Stirling numbers of the second kind, defined by

$${(e^t-1)}^m=m!\sum _{l=m}^{\mathrm{\infty }}{S}_2(l,m)\frac{t^l}{l!}.$$

Thus,

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{j=0}^n(\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m))\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)x^j\\ =\sum _{j=0}^n(\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m))H_j^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{j=0}^n(\sum _{m=0}^{n-j}\frac{{(-1)}^{n-m-j}}{{(m+1)}^k}\left(\genfrac{}{}{0ex}{}{n}{j}\right)m!S_2(n-j,m))\sum _{l=0}^j\left(\genfrac{}{}{0ex}{}{j}{l}\right)H_{j-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)x^l\\ =\sum _{l=0}^n(\sum _{j=l}^n\sum _{m=0}^{n-j}{(-1)}^{n-m-j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{j}{l}\right)\frac{m!}{{(m+1)}^k}{S}_2(n-j,m)H_{j-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r))x^l,\end{array}$$

which is identity (23).

By (11) with (16), we have

$$\begin{array}{rl}〈g{(\overline{f}(t))}^{-1}\overline{f}{(t)}^j|x^n〉& =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{t}^j|x^n〉\\ ={(n)}_j〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|x^{n-j}〉\\ ={(n)}_j〈\sum _{i=0}^{\mathrm{\infty }}{T}_i^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^i}{i!}|x^{n-j}〉\\ ={(n)}_j{T}_{n-j}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

Thus, we get (24). □

3.2 Sheffer identity

Theorem 2

$$T_n^{(r,k)}(x+y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)T_j^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)y^{n-j}.$$

(25)

Proof By (16) with

$$\begin{array}{rl}{p}_n(x)& =\prod _{j=1}^r\left(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j}\right)\frac{1-e^{-t}}{{Li}_k(1-e^{-t})}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =x^n\sim (1,t),\end{array}$$

using (12), we have (25). □

3.3 Recurrence

Theorem 3

$$\begin{array}{rl}{T}_{n+1}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=& x{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ -\sum _{j=1}^r\frac{a_j}{1-{\lambda }_j}{T}_n^{(r+1,k)}(x+a_j|a_1,\dots ,a_r,a_j;{\lambda }_1,\dots ,{\lambda }_r,{\lambda }_j)\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0ex}{}{n+1}{l}\right)B_{n+1-l}\\ \times (T_l^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ -T_l^{(r,k-1)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)),\end{array}$$

(26)

where $B_n$ is the nth ordinary Bernoulli number.

Proof By applying

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

[[4], Corollary 3.7.2] with (16), we get

$$T_{n+1}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=(x-\frac{g_{r,k}^{\prime }(t)}{g_{r,k}(t)})T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).$$

Now,

$$\begin{array}{rl}\frac{g_{r,k}^{\prime }(t)}{g_{r,k}(t)}& ={(ln{g}_{r,k}(t))}^{\prime }\\ ={(\sum _{j=1}^{r}ln(e^{a_{j}t}-{\lambda }_j)-\sum _{j=1}^{r}ln(1-{\lambda }_j)+ln(1-e^{-t})-ln{Li}_k(1-e^{-t}))}^{\prime }\\ =\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{e^{a_{j}t}-{\lambda }_j}+\frac{e^{-t}}{1-e^{-t}}(1-\frac{{Li}_{k-1}(1-e^{-t})}{{Li}_k(1-e^{-t})})\\ =\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{e^{a_{j}t}-{\lambda }_j}+\frac{t}{e^t-1}\frac{{Li}_k(1-e^{-t})-{Li}_{k-1}(1-e^{-t})}{t{Li}_k(1-e^{-t})}.\end{array}$$

Since

$$T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^n,$$

we have

$$\begin{array}{r}{T}_{n+1}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =x{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ -\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{1-{\lambda }_j}\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^n\\ -\frac{t}{e^t-1}\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right)\frac{{Li}_k(1-e^{-t})-{Li}_{k-1}(1-e^{-t})}{t(1-e^{-t})}{x}^n.\end{array}$$

Since

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

is a delta series, we get

$$\frac{{Li}_k(1-e^{-t})-{Li}_{k-1}(1-e^{-t})}{t(1-e^{-t})}{x}^n=\frac{1}{n+1}\frac{{Li}_k(1-e^{-t})-{Li}_{k-1}(1-e^{-t})}{1-e^{-t}}{x}^{n+1}.$$

Therefore, by

$$\frac{t}{e^t-1}{x}^{n+1}=B_{n+1}(x)=\sum _{l=0}^{n+1}\left(\genfrac{}{}{0ex}{}{n+1}{l}\right)B_{n+1-l}{x}^l,$$

we obtain

$$\begin{array}{r}{T}_{n+1}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =x{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)-\sum _{j=1}^r\frac{a_j}{1-{\lambda }_j}{T}_n^{(r+1,k)}(x+a_j|a_1,\dots ,a_r,a_j;{\lambda }_1,\dots ,{\lambda }_r,{\lambda }_j)\\ -\frac{1}{n+1}\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right)\frac{{Li}_k(1-e^{-t})-{Li}_{k-1}(1-e^{-t})}{1-e^{-t}}\frac{t}{e^t-1}{x}^{n+1}\\ =x{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)-\sum _{j=1}^r\frac{a_j}{1-{\lambda }_j}{T}_n^{(r+1,k)}(x+a_j|a_1,\dots ,a_r,a_j;{\lambda }_1,\dots ,{\lambda }_r,{\lambda }_j)\\ -\frac{1}{n+1}\sum _{l=0}^{n+1}\left(\genfrac{}{}{0ex}{}{n+1}{l}\right)B_{n+1-l}\\ \times (T_l^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)-T_l^{(r,k-1)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)),\end{array}$$

which is identity (26). □

3.4 A more relation

Theorem 4

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =x{T}_{n-1}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ -\sum _{j=1}^r\frac{a_j}{1-{\lambda }_j}{T}_{n-1}^{(r+1,k)}(x+a_j|a_1,\dots ,a_r,a_j;{\lambda }_1,\dots ,{\lambda }_r,{\lambda }_j)\\ -\frac{1}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}(T_l^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ -T_l^{(r,k-1)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)).\end{array}$$

(27)

Proof For $n\ge 1$, we have

$$\begin{array}{rl}{T}_n^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=& 〈\sum _{l=0}^{\mathrm{\infty }}{T}_l^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\frac{t^l}{l!}|x^n〉\\ =& 〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^n〉\\ =& 〈{\partial }_t\left(\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}\right)|x^{n-1}〉\\ =& 〈\left({\partial }_t\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^{n-1}〉\\ +〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\left({\partial }_t\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}\right)e^{yt}|x^{n-1}〉\\ +〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}\left({\partial }_t{e}^{yt}\right)|x^{n-1}〉\\ =& y{T}_{n-1}^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ +〈\left({\partial }_t\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^{n-1}〉\\ +〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\left({\partial }_t\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}\right)e^{yt}|x^{n-1}〉.\end{array}$$

Observe that

$$\begin{array}{rl}{\partial }_t\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)& =\prod _{j=1}^r(1-{\lambda }_j){\partial }_t\left(\frac{1}{{\prod }_{j=1}^r(e^{a_{j}t}-{\lambda }_j)}\right)\\ =\prod _{j=1}^r(1-{\lambda }_j)\frac{-{({\prod }_{j=1}^r(e^{a_{j}t}-{\lambda }_j))}^{\prime }}{{({\prod }_{j=1}^r(e^{a_{j}t}-{\lambda }_j))}^2}\\ =-\prod _{j=1}^r(1-{\lambda }_j)\frac{{\sum }_{j=1}^r{a}_j{e}^{a_{j}t}{\prod }_{i\ne j}(e^{a_{i}t}-{\lambda }_i)}{{({\prod }_{j=1}^r(e^{a_{j}t}-{\lambda }_j))}^2}\\ =-\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{e^{a_{j}t}-{\lambda }_j}\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right)\\ =-\sum _{j=1}^r\frac{a_j{e}^{a_{j}t}}{1-{\lambda }_j}\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right).\end{array}$$

Thus,

$$\begin{array}{r}〈\left({\partial }_t\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^{n-1}〉\\ =-\sum _{j=1}^r\frac{a_j}{1-{\lambda }_j}〈\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\prod _{i=1}^r\left(\frac{1-{\lambda }_i}{e^{a_{i}t}-{\lambda }_i}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{(y+a_j)t}|x^{n-1}〉\\ =-\sum _{j=1}^r\frac{a_j}{1-{\lambda }_j}{T}_{n-1}^{(r+1,k)}(y+a_j|a_1,\dots ,a_r,a_j;{\lambda }_1,\dots ,{\lambda }_r,{\lambda }_j).\end{array}$$

Since

$$\begin{array}{rl}{\partial }_t\left(\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}\right)& =\frac{e^{-t}({Li}_{k-1}(1-e^{-t})-{Li}_k(1-e^{-t}))}{{(1-e^{-t})}^2}\\ =\frac{t}{e^t-1}\frac{{Li}_{k-1}(1-e^{-t})-{Li}_k(1-e^{-t})}{t(1-e^{-t})}\end{array}$$

and the fact that

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

is a delta series, we have

$$\begin{array}{r}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\left({\partial }_t\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}\right)e^{yt}|x^{n-1}〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{t}{e^t-1}\frac{{Li}_{k-1}(1-e^{-t})-{Li}_k(1-e^{-t})}{t(1-e^{-t})}{e}^{yt}|x^{n-1}〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{t}{e^t-1}\frac{{Li}_{k-1}(1-e^{-t})-{Li}_k(1-e^{-t}}{1-e^{-t}}{e}^{yt}|\frac{x^n}{n}〉\\ =\frac{1}{n}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_{k-1}(1-e^{-t})-{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|\frac{t}{e^t-1}{x}^n〉\\ =\frac{1}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_{k-1}(1-e^{-t})-{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{yt}|x^l〉\\ =\frac{1}{n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)B_{n-l}(T_l^{(r,k-1)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)-T_l^{(r,k)}(y|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)).\end{array}$$

Therefore, we obtain the desired result. □

Remark After n is replaced by $n+1$, identity (27) becomes the recurrence formula (26).

3.5 Relations with poly-Bernoulli numbers and Barnes’ multiple Bernoulli numbers

Theorem 5

$$\begin{array}{r}\sum _{m=0}^n{(-1)}^{n-m}\left(\genfrac{}{}{0ex}{}{n+1}{m}\right)T_m^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{l=0}^n\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)\left(\genfrac{}{}{0ex}{}{n+1}{l+1}\right)B_m^{(k-1)}{H}_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

(28)

Proof We shall compute

$$〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right){Li}_k(1-e^{-t})|x^{n+1}〉$$

in two different ways. On the one hand,

$$\begin{array}{r}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right){Li}_k(1-e^{-t})|x^{n+1}〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|(1-e^{-t})x^{n+1}〉\\ =〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|x^{n+1}-{(x-1)}^{n+1}〉\\ =\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n+1}{m}\right){(-1)}^{n-m}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|x^m〉\\ =\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n+1}{m}\right){(-1)}^{n-m}{T}_m^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

On the other hand,

$$\begin{array}{r}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right){Li}_k(1-e^{-t})|x^{n+1}〉\\ =〈{Li}_k(1-e^{-t})|\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)x^{n+1}〉\\ =〈{Li}_k(1-e^{-t})|H_{n+1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =〈{\int }_0^t{\left({Li}_k(1-e^{-s})\right)}^{\prime }ds|H_{n+1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =〈{\int }_0^t{e}^{-s}\frac{{Li}_{k-1}(1-e^{-s})}{1-e^{-s}}ds|H_{n+1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =〈{\int }_0^t\left(\sum _{j=0}^{\mathrm{\infty }}\frac{{(-s)}^j}{j!}\right)\left(\sum _{m=0}^{\mathrm{\infty }}\frac{B_m^{(k-1)}}{m!}{s}^m\right)ds|H_{n+1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =〈\sum _{l=0}^{\mathrm{\infty }}\left(\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)B_m^{(k-1)}\right)\frac{1}{l!}{\int }_0^t{s}^{l}ds|H_{n+1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =\sum _{l=0}^n\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)\frac{B_m^{(k-1)}}{(l+1)!}〈t^{l+1}|H_{n+1}^{(r)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =\sum _{l=0}^n\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)\frac{B_m^{(k-1)}}{(l+1)!}{(n+1)}_{l+1}{H}_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{l=0}^n\sum _{m=0}^l{(-1)}^{l-m}\left(\genfrac{}{}{0ex}{}{l}{m}\right)\left(\genfrac{}{}{0ex}{}{n+1}{l+1}\right)B_m^{(k-1)}{H}_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

Here, $H_{n-l}^{(r)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=H_{n-l}^{(r)}(0|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)$. Thus, we get (28). □

3.6 Relations with the Stirling numbers of the second kind and the falling factorials

Theorem 6

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{m=0}^n(\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)T_{n-l}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)){(x)}_m.\end{array}$$

(29)

Proof For (16) and ${(x)}_n\sim (1,e^t-1)$, assume that

$$T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)=\sum _{m=0}^n{C}_{n,m}{(x)}_m.$$

By (13), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{j=1}^r(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j})\frac{1-e^{-t}}{{Li}_k(1-e^{-t})}}{(e^t-1)}^m|x^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|{(e^t-1)}^m{x}^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|m!\sum _{l=m}^n{S}_2(l,m)\frac{t^l}{l!}{x}^n〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|x^{n-l}〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)T_{n-l}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

Thus, we get identity (29). □

3.7 Relations with the Stirling numbers of the second kind and the rising factorials

Theorem 7

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{m=0}^n(\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)T_{n-l}^{(r,k)}(-m|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)){(x)}^{(m)}.\end{array}$$

(30)

Proof For (16) and ${(x)}^{(n)}=x(x+1)\cdots (x+n-1)\sim (1,1-e^{-t})$, assume that $T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)={\sum }_{m=0}^n{C}_{n,m}{(x)}^{(m)}$. By (13), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{1}{{\prod }_{j=1}^r(\frac{e^{a_{j}t}-{\lambda }_j}{1-{\lambda }_j})\frac{1-e^{-t}}{{Li}_k(1-e^{-t})}}{(1-e^{-t})}^m|x^n〉\\ =\frac{1}{m!}〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{e}^{-mt}|{(e^t-1)}^m{x}^n〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)〈e^{-mt}|\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^{n-l}〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)〈e^{-mt}|T_{n-l}^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)〉\\ =\sum _{l=m}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right)S_2(l,m)T_{n-l}^{(r,k)}(-m|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

Thus, we get identity (30). □

3.8 Relations with higher-order Frobenius-Euler polynomials

Theorem 8

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{m=0}^n(\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\lambda )}^s}\sum _{l=0}^s\left(\genfrac{}{}{0ex}{}{s}{l}\right){(-\lambda )}^{s-l}{T}_{n-m}^{(r,k)}(l|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r))H_m^{(s)}(x|\lambda ).\end{array}$$

(31)

Proof For (16) and

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

assume that $T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)={\sum }_{m=0}^n{C}_{n,m}{H}_m^{(s)}(x|\lambda )$. By (13), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{e^t-\lambda }{1-\lambda }\right)}^s\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{t}^m|x^n〉\\ =\frac{1}{m!{(1-\lambda )}^s}\sum _{l=0}^s\left(\genfrac{}{}{0ex}{}{s}{l}\right){(-\lambda )}^{s-l}〈e^{lt}\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|t^m{x}^n〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\lambda )}^s}\sum _{l=0}^s\left(\genfrac{}{}{0ex}{}{s}{l}\right){(-\lambda )}^{s-l}〈e^{lt}|\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{x}^{n-m}〉\\ =\frac{\left(\genfrac{}{}{0ex}{}{n}{m}\right)}{{(1-\lambda )}^s}\sum _{l=0}^s\left(\genfrac{}{}{0ex}{}{s}{l}\right){(-\lambda )}^{s-l}{T}_{n-m}^{(r,k)}(l|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

Thus, we get identity (31). □

3.9 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., [[4], Section 2.2]).

Theorem 9

$$\begin{array}{r}{T}_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\sum _{m=0}^n\left(\genfrac{}{}{0ex}{}{n}{m}\right)(\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{l+s}{l}\right)}{S}_2(l+s,s)T_{n-m-l}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)){\mathfrak{B}}_m^{(s)}(x).\end{array}$$

(32)

Proof For (16) and

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

assume that $T_n^{(r,k)}(x|a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)={\sum }_{m=0}^n{C}_{n,m}{\mathfrak{B}}_m^{(s)}(x)$. By (13), we have

$$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{e^t-1}{t}\right)}^s\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}{t}^m|x^n〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|{\left(\frac{e^t-1}{t}\right)}^s{x}^{n-m}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|\sum _{l=0}^{n-m}\frac{s!}{(l+s)!}{S}_2(l+s,s)t^l{x}^{n-m}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!}{(l+s)!}{S}_2(l+s,s){(n-m)}_l〈\prod _{j=1}^r\left(\frac{1-{\lambda }_j}{e^{a_{j}t}-{\lambda }_j}\right)\frac{{Li}_k(1-e^{-t})}{1-e^{-t}}|x^{n-m-l}〉\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{s!}{(l+s)!}{S}_2(l+s,s){(n-m)}_l{T}_{n-m-l}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r)\\ =\left(\genfrac{}{}{0ex}{}{n}{m}\right)\sum _{l=0}^{n-m}\frac{\left(\genfrac{}{}{0ex}{}{n-m}{l}\right)}{\left(\genfrac{}{}{0ex}{}{l+s}{l}\right)}{S}_2(l+s,s)T_{n-m-l}^{(r,k)}(a_1,\dots ,a_r;{\lambda }_1,\dots ,{\lambda }_r).\end{array}$$

Thus, we get identity (32). □