Theory and Modern Applications

# Poly-Cauchy and Peters mixed-type polynomials

## Abstract

The Peters polynomials are a generalization of Boole polynomials. In this paper, we consider Peters and poly-Cauchy mixed-type polynomials and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give various identities of those polynomials associated with special polynomials.

## 1 Introduction

The Peters polynomials are defined by the generating function to be

$\sum _{n=0}^{\mathrm{\infty }}{S}_{n}\left(x;\lambda ,\mu \right)\frac{{t}^{n}}{n!}={\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{\left(1+t\right)}^{x}\phantom{\rule{1em}{0ex}}\left(\text{see [1]}\right).$
(1)

The first few of them are given by

${S}_{0}\left(x;\lambda ,\mu \right)={2}^{-\mu },\phantom{\rule{2em}{0ex}}{S}_{1}\left(x;\lambda ,\mu \right)={2}^{-\left(\mu +1\right)}\left(2x-\lambda \mu \right),\phantom{\rule{2em}{0ex}}\dots .$

If $\mu =1$, then ${S}_{n}\left(x;\lambda \right)={S}_{n}\left(x:\lambda ,1\right)$ are called Boole polynomials.

In particular, for $\mu =1$, $\lambda =1$, ${S}_{n}\left(x;1,1\right)={Ch}_{n}\left(x\right)$ are Changhee polynomials which are defined by

$\sum _{n=0}^{\mathrm{\infty }}{Ch}_{n}\left(x\right)\frac{{t}^{n}}{n!}=\frac{1}{t+2}{\left(1+t\right)}^{x}\phantom{\rule{1em}{0ex}}\left(\text{see [2]}\right).$

The generating functions for the poly-Cauchy polynomials of the first kind ${C}_{n}^{\left(k\right)}\left(x\right)$ and of the second kind ${\stackrel{ˆ}{C}}_{n}^{\left(k\right)}\left(x\right)$ are, respectively, given by

${Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-x}=\sum _{n=0}^{\mathrm{\infty }}{C}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}$
(2)

and

${Lif}_{k}\left(-log\left(1+t\right)\right){\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{C}}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!},$
(3)

where ${Lif}_{k}\left(t\right)={\sum }_{n=0}^{\mathrm{\infty }}\frac{{t}^{n}}{n!{\left(n+1\right)}^{k}}$ ($k\in \mathbb{Z}$) (see [3, 4]).

In this paper, we consider the poly-Cauchy of the first kind and Peters (respectively the poly-Cauchy of the second kind and Peters) mixed-type polynomials as follows:

${\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-x}=\sum _{n=0}^{\mathrm{\infty }}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\frac{{t}^{n}}{n!}$
(4)

and

${\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right){\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\frac{{t}^{n}}{n!}.$
(5)

For $\alpha \in {\mathbb{Z}}_{\ge 0}$, the Bernoulli polynomials of order α are defined by the generating function to be

${\left(\frac{t}{{e}^{t}-1}\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [2, 5–11]}\right).$
(6)

As is well known, the Frobenius-Euler polynomials of order α are given by

${\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{\alpha }{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(\alpha \right)}\left(x\mid \lambda \right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [2–14]}\right),$
(7)

where $\lambda \in \mathbb{C}$ with $\lambda \ne 1$ and $\alpha \in {\mathbb{Z}}_{\ge 0}$.

When $x=0$, $C{P}_{n}^{\left(k\right)}\left(0;\lambda ,\mu \right)$ (or $\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(0;\lambda ,\mu \right)$) are called the poly-Cauchy of the first kind and Peters (or the poly-Cauchy of the second kind and Peters) mixed-type numbers.

The higher-order Cauchy polynomials of the first kind are defined by the generating function to be

${\left(\frac{t}{log\left(1+t\right)}\right)}^{\alpha }{\left(1+t\right)}^{-x}=\sum _{n=0}^{\mathrm{\infty }}{\mathbb{C}}_{n}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\alpha \in {\mathbb{Z}}_{\ge 0}\right),$
(8)

and the higher-order Cauchy polynomials of the second kind are given by

${\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{\alpha }{\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{\mathbb{C}}}_{n}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\alpha \in {\mathbb{Z}}_{\ge 0}\right).$
(9)

The Stirling number of the first kind is given by

${\left(x\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l}.$
(10)

Thus, by (10), we get

(11)

It is easy to show that

${x}^{\left(n\right)}=x\left(x+1\right)\cdots \left(x+n-1\right)={\left(-1\right)}^{n}{\left(-x\right)}_{n}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{n-l}{x}^{l}.$
(12)

Let be the complex number field and let be the algebra of all formal power series in the variable t over as follows:

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}\frac{{t}^{k}}{k!}|{a}_{k}\in \mathbb{C}\right\}.$
(13)

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

For $f\left(t\right)\in \mathcal{F}$, let us define the linear functional on by setting

$〈f\left(t\right)\mid {x}^{n}〉={a}_{n}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(14)

Then, by (13) and (14), we get

(15)

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

Let ${f}_{L}\left(t\right)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L\mid {x}^{k}〉}{k!}{t}^{k}.$ Then, by (14), we see that $〈{f}_{L}\left(t\right)\mid {x}^{n}〉=〈L\mid {x}^{n}〉$. The map $L↦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto . Henceforth, denotes both the algebra of formal power series in t and the vector space of all linear functionals on , and so an element $f\left(t\right)$ of will be thought of as both a formal power series and a linear functional. We call the umbral algebra, and the umbral calculus is the study of umbral algebra. The order $O\left(f\right)$ of the power series $f\left(t\right)$ (≠0) is the smallest integer for which the coefficient of ${t}^{k}$ does not vanish. If $O\left(f\left(t\right)\right)=1$, then $f\left(t\right)$ is called a delta series; if $O\left(f\left(t\right)\right)=0$, then $f\left(t\right)$ is called an invertible series. For $f\left(t\right),g\left(t\right)\in \mathcal{F}$ with $O\left(f\left(t\right)\right)=1$ and $O\left(g\left(t\right)\right)=0$, there exists a unique sequence ${s}_{n}\left(x\right)$ of polynomials such that $〈g\left(t\right)f{\left(t\right)}^{k}\mid {s}_{n}\left(x\right)〉=n!{\delta }_{n,k}$ ($n,k\ge 0$).

The sequence ${s}_{n}\left(x\right)$ is called the Sheffer sequence for $\left(g\left(t\right),f\left(t\right)\right)$ which is denoted by ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$.

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

$〈f\left(t\right)g\left(t\right)\mid p\left(x\right)〉=〈f\left(t\right)\mid g\left(t\right)p\left(x\right)〉=〈g\left(t\right)\mid f\left(t\right)p\left(x\right)〉$

and

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)\mid {x}^{k}〉\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}\mid p\left(x\right)〉\frac{{x}^{k}}{k!}.$
(16)

By (16), we get

${p}^{\left(k\right)}\left(0\right)=〈{t}^{k}\mid p\left(x\right)〉=〈1\mid {p}^{\left(k\right)}\left(x\right)〉\phantom{\rule{1em}{0ex}}\left(k\ge 0\right),$
(17)

where ${p}^{\left(k\right)}\left(0\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}{|}_{x=0}$.

Thus, by (17), we have

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{(see [1–3])}.$
(18)

Let ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$. Then the following equations are known in [1]:

(19)

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

${s}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈\frac{{\left(\overline{f}\left(t\right)\right)}^{j}}{g\left(\overline{f}\left(t\right)\right)}|{x}^{n}〉{x}^{j},$
(20)
(21)

and

${s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right),\phantom{\rule{1em}{0ex}}f\left(t\right){s}_{n}\left(x\right)=n{s}_{n-1}\left(x\right)\phantom{\rule{0.25em}{0ex}}\left(n\ge 0\right),$
(22)

and

$\frac{d}{dx}{s}_{n}\left(x\right)=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\overline{f}\left(t\right)\mid {x}^{n-l}〉{s}_{l}\left(x\right).$
(23)

As is well known, the transfer formula for ${p}_{n}\left(x\right)\sim \left(1,f\left(t\right)\right)$, ${q}_{n}\left(x\right)\sim \left(1,g\left(t\right)\right)$ ($n\ge 1$) is given by

${q}_{n}\left(x\right)=x{\left(\frac{f\left(t\right)}{g\left(t\right)}\right)}^{n}{x}^{-1}{p}_{n}\left(x\right).$
(24)

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, ${r}_{n}\left(x\right)\sim \left(h\left(t\right),l\left(t\right)\right)$, let

${s}_{n}\left(x\right)=\sum _{m=0}^{\mathrm{\infty }}{C}_{n,m}{r}_{n}\left(x\right),$
(25)

where

${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}{\left(l\left(\overline{f}\left(t\right)\right)\right)}^{m}|{x}^{n}〉\phantom{\rule{1em}{0ex}}\text{(see [1])}.$
(26)

It is known that

$〈f\left(t\right)\mid xp\left(x\right)〉=〈{\partial }_{t}f\left(t\right)\mid p\left(x\right)〉,\phantom{\rule{2em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right),$
(27)

where $f\left(t\right)\in \mathcal{F}$ and $p\left(x\right)\in \mathbb{P}$ (see [13]).

In this paper, we consider Peters and poly-Cauchy mixed-type polynomials with umbral calculus viewpoint and investigate the properties of those polynomials which are derived from umbral calculus. Finally, we give some interesting identities of those polynomials associated with special polynomials.

## 2 Poly-Cauchy and Peters mixed-type polynomials

From (2), (3), and (19), we note that

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left({\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)},{e}^{-t}-1\right)$
(28)

and

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left({\left(1+{e}^{\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)},{e}^{t}-1\right).$
(29)

It is not difficult to show that

$\begin{array}{rl}{\left(1+{e}^{-\lambda t}\right)}^{\mu }& ={2}^{\mu }{\left(1+\frac{1}{2}\sum _{j=1}^{\mathrm{\infty }}\frac{{\left(-\lambda t\right)}^{j}}{j!}\right)}^{\mu }\\ =\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}\sum _{{j}_{1}+\cdots +{j}_{i}=j}{2}^{\mu -i}\left(\genfrac{}{}{0}{}{\mu }{i}\right)\left(\genfrac{}{}{0}{}{j+i}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\frac{{\left(-\lambda t\right)}^{j+i}}{\left(j+i\right)!}\end{array}$
(30)

and

$\begin{array}{rl}{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }& ={2}^{-\mu }{\left(1+\frac{1}{2}\sum _{j=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{\lambda }{j+1}\right){t}^{j+1}\right)}^{-\mu }\\ =\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}\sum _{{j}_{1}+\cdots +{j}_{i}=j}{2}^{-\left(\mu +i\right)}\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{\lambda }{{j}_{1}+1}\right)\cdots \left(\genfrac{}{}{0}{}{\lambda }{{j}_{i}+1}\right){t}^{j+i}.\end{array}$
(31)

From (14), we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(y;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=〈\sum _{l=0}^{\mathrm{\infty }}C{P}_{l}^{\left(k\right)}\left(y;\lambda ,\mu \right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-y}\mid {x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }|\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){C}_{l}^{\left(k\right)}\left(y\right){x}^{n-l}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){C}_{l}^{\left(k\right)}\left(y\right)〈\sum _{m=0}^{\mathrm{\infty }}{S}_{m}\left(0;\lambda ,\mu \right)\frac{{t}^{m}}{m!}|{x}^{n-l}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{n-l}\left(0;\lambda ,\mu \right){C}_{l}^{\left(k\right)}\left(y\right).\end{array}$
(32)

Therefore, by (32), we obtain the following theorem.

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

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{n-l}\left(0;\lambda ,\mu \right){C}_{l}^{\left(k\right)}\left(x\right).$

Alternatively,

$\begin{array}{rcl}C{P}_{n}^{\left(k\right)}\left(y;\lambda ,\mu \right)& =& 〈\sum _{l=0}^{\mathrm{\infty }}C{P}_{l}^{\left(k\right)}\left(y;\lambda ,\mu \right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ =& 〈{Lif}_{k}\left(log\left(1+t\right)\right)\mid {\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{\left(1+t\right)}^{-y}{x}^{n}〉\\ =& 〈{Lif}_{k}\left(log\left(1+t\right)\right)|\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{l}\left(-y;\lambda ,\mu \right){x}^{n-l}〉\\ =& \sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{l}\left(-y;\lambda ,\mu \right)〈{Lif}_{k}\left(log\left(1+t\right)\right)\mid {x}^{n-l}〉\\ =& \sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{l}\left(-y;\lambda ,\mu \right){C}_{n-l}^{\left(k\right)}\left(0\right).\end{array}$
(33)

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

Theorem 2 For $n\ge 0$, let ${C}_{n-l}^{\left(k\right)}\left(0\right)={C}_{n-l}^{\left(k\right)}$. Then we have

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){C}_{n-l}^{\left(k\right)}{S}_{l}\left(-x;\lambda ,\mu \right).$

Remark By the same method, we get

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{n-l}\left(0;\lambda ,\mu \right){\stackrel{ˆ}{C}}_{l}^{\left(k\right)}\left(x\right)$
(34)

and

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){\stackrel{ˆ}{C}}_{n-l}^{\left(k\right)}{S}_{l}\left(x;\lambda ,\mu \right).$
(35)

From (20) and (28), we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\frac{1}{j!}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(-log\left(1+t\right)\right)}^{j}\mid {x}^{n}〉{x}^{j}.\end{array}$
(36)

From (31), we note that

$\begin{array}{r}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(-log\left(1+t\right)\right)}^{j}\mid {x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{j}}{m!{\left(m+1\right)}^{k}}\sum _{l=0}^{n-j-m}\frac{\left(m+j\right)!}{\left(l+m+j\right)!}{S}_{1}\left(l+m+j,m+j\right)\\ \phantom{\rule{2em}{0ex}}×{\left(n\right)}_{l+m+j}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\mid {x}^{n-l-m-j}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n-j}\frac{{\left(-1\right)}^{j}}{m!{\left(m+1\right)}^{k}}\sum _{l=0}^{n-j-m}\frac{\left(m+j\right)!}{\left(l+m+j\right)!}\\ \phantom{\rule{2em}{0ex}}×{S}_{1}\left(l+m+j,m+j\right){\left(n\right)}_{l+m+j}\sum _{i=0}^{n-j-m-l}\sum _{r=0}^{\mathrm{\infty }}\sum _{{r}_{1}+\cdots +{r}_{i}=r}{2}^{-\left(\mu +i\right)}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{\lambda }{{r}_{1}+1}\right)\cdots \left(\genfrac{}{}{0}{}{\lambda }{{r}_{i}+1}\right)〈{t}^{r+i}\mid {x}^{n-l-m-j}〉\\ \phantom{\rule{1em}{0ex}}={2}^{-\mu }n!\sum _{m=0}^{n-j}\sum _{l=0}^{n-j-m}\sum _{i=0}^{n-j-m-l}\sum _{{r}_{1}+\cdots +{r}_{i}=n-j-m-l-i}\frac{{2}^{-i}{\left(-1\right)}^{j}\left(m+j\right)!}{m!{\left(m+1\right)}^{k}\left(l+m+j\right)!}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{\lambda }{{r}_{1}+1}\right)\cdots \left(\genfrac{}{}{0}{}{\lambda }{{r}_{i}+1}\right){S}_{1}\left(l+m+j,m+j\right).\end{array}$
(37)

Therefore, by (36) and (37), we obtain the following theorem.

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

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}={2}^{-\mu }n!\sum _{j=0}^{n}\frac{{\left(-1\right)}^{j}}{j!}\left\{\sum _{m=0}^{n-j}\sum _{l=0}^{n-j-m}\sum _{i=0}^{n-j-m-l}\sum _{{r}_{1}+\cdots +{r}_{i}=n-j-m-l-i}\frac{{2}^{-i}}{m!{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}×\frac{\left(m+j\right)!}{\left(l+m+j\right)!}\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{\lambda }{{r}_{1}+1}\right)\cdots \left(\genfrac{}{}{0}{}{\lambda }{{r}_{i}+1}\right){S}_{1}\left(l+m+j,m+j\right)\right\}{x}^{j}.\end{array}$

Remark By the same method as Theorem 3, we get

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}={2}^{-\mu }n!\sum _{j=0}^{n}\frac{1}{j!}\left\{\sum _{m=0}^{n-j}\sum _{l=0}^{n-j-m}\sum _{i=0}^{n-j-m-l}\sum _{{r}_{1}+\cdots +{r}_{i}=n-j-m-l-i}\frac{{2}^{-i}{\left(-1\right)}^{m}}{m!{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}×\frac{\left(m+j\right)!}{\left(l+m+j\right)!}\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{\lambda }{{r}_{1}+1}\right)\cdots \left(\genfrac{}{}{0}{}{\lambda }{{r}_{i}+1}\right){S}_{1}\left(l+m+j,m+j\right)\right\}{x}^{j}.\end{array}$
(38)

From (28), we note that

${\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left(1,{e}^{-t}-1\right)$
(39)

and

${x}^{n}\sim \left(1,t\right).$
(40)

By (24), (39), and (40), we get

$\begin{array}{r}{\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=x{\left(\frac{t}{{e}^{-t}-1}\right)}^{n}{x}^{n-1}\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}x{\left(\frac{-t}{{e}^{-t}-1}\right)}^{n}{x}^{n-1}\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}\sum _{l=0}^{n-1}{\left(-1\right)}^{l}{B}_{l}^{\left(n\right)}\left(\genfrac{}{}{0}{}{n-1}{l}\right){x}^{n-l}.\end{array}$
(41)

Thus, by (41), we see that

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}\sum _{l=0}^{n-1}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n-1}{l}\right){B}_{l}^{\left(n\right)}{\left(1+{e}^{-\lambda t}\right)}^{-\mu }{Lif}_{k}\left(-t\right){x}^{n-l}\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}\sum _{l=0}^{n-1}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n-1}{l}\right){B}_{l}^{\left(n\right)}\sum _{m=0}^{n-l}\frac{{\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{n-l}{m}\right)}{{\left(m+1\right)}^{k}}{\left(1+{e}^{-\lambda t}\right)}^{-\mu }{x}^{n-l-m}\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}\sum _{l=0}^{n-1}{\left(-1\right)}^{l}\left(\genfrac{}{}{0}{}{n-1}{l}\right){B}_{l}^{\left(n\right)}\sum _{m=0}^{n-l}\frac{{\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{n-l}{m}\right)}{{\left(m+1\right)}^{k}}\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}\sum _{{j}_{1}+\cdots +{j}_{n}=j}{2}^{-\mu -i}\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{j+i}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\frac{{\left(-\lambda t\right)}^{j+i}}{\left(j+i\right)!}{x}^{n-l-m}\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}\sum _{l=0}^{n}\sum _{m=0}^{n-l}\sum _{i=0}^{n-l-m}\sum _{j=0}^{n-l-m-i}\sum _{{j}_{1}+\cdots +{j}_{i}=n-l-m-i-r}{\left(-1\right)}^{n-r}\frac{{2}^{-\mu -i}{\lambda }^{n-l-m-r}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n-1}{l}\right)\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{n-l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{n-l-m-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{n-l-m}{r}\right){B}_{l}^{\left(n\right)}{x}^{r}.\end{array}$
(42)

Therefore, by (42), we obtain the following theorem.

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

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\frac{{\lambda }^{n}}{{2}^{\mu }}\sum _{r=0}^{n}{\left(-{\lambda }^{-1}\right)}^{r}\left\{\sum _{l=0}^{n-r}\sum _{m=0}^{n-r-l}\sum _{i=0}^{n-r-l-m}\sum _{{j}_{1}+\cdots +{j}_{i}=n-r-l-m-i}\frac{{2}^{-i}{\lambda }^{-l-m}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n-1}{l}\right)\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{n-l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{n-r-l-m}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{n-l-m}{r}\right){B}_{l}^{\left(n\right)}\right\}{x}^{r}.\end{array}$

Remark By the same method as Theorem 4, we get

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\frac{{\lambda }^{n}}{{2}^{\mu }}\sum _{r=0}^{n}{\lambda }^{-r}\left\{\sum _{l=0}^{n-r}\sum _{m=0}^{n-l-r}\sum _{i=0}^{n-l-m-r}\sum _{{j}_{1}+\cdots +{j}_{i}=n-r-l-m-i}\frac{{\left(-1\right)}^{m}{2}^{-i}{\lambda }^{-l-m}}{{\left(m+1\right)}^{k}}\left(\genfrac{}{}{0}{}{n-1}{l}\right)\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{n-l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{n-r-l-m}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{n-l-m}{r}\right){B}_{l}^{\left(m\right)}\right\}{x}^{r}.\end{array}$
(43)

From (12), we note that

${x}^{\left(n\right)}=x\left(x+1\right)\cdots \left(x+n-1\right)\sim \left(1,1-{e}^{-t}\right).$
(44)

Thus, by (44), we see that

${\left(-1\right)}^{n}{x}^{\left(n\right)}={\left(-x\right)}_{n}=\sum _{m=0}^{n}{S}_{1}\left(n,m\right){\left(-x\right)}^{m}\sim \left(1,{e}^{-t}-1\right)$
(45)

and

${\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left(1,{e}^{-t}-1\right).$
(46)

From (45) and (46), we have

$\begin{array}{r}{\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{n}{x}^{\left(n\right)}\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-x\right)}^{l}.\end{array}$
(47)

Thus, by (47), we get

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}{\left(1+{e}^{-\lambda t}\right)}^{-\mu }{Lif}_{k}\left(-t\right){x}^{l}\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}\sum _{m=0}^{l}\frac{{\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{l}{m}\right)}{{\left(m+1\right)}^{k}}{\left(1+{e}^{-\lambda t}\right)}^{-\mu }{x}^{l-m}\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){\left(-1\right)}^{l}\sum _{m=0}^{l}\frac{{\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{l}{m}\right)}{{\left(m+1\right)}^{k}}\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}\sum _{{j}_{1}+\cdots +{j}_{i}=j}{2}^{-\mu -i}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{j+i}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\frac{{\left(-\lambda \right)}^{j+i}}{\left(j+i\right)!}{t}^{j+i}{x}^{l-m}\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n}\sum _{m=0}^{l}\sum _{i=0}^{l-m}\sum _{r=0}^{l-m-i}\sum _{{j}_{1}+\cdots +{j}_{i}=l-m-i-r}{\left(-1\right)}^{r}\frac{{2}^{-\mu -i}{\lambda }^{l-m-r}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{l-m-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l-m}{r}\right){S}_{1}\left(n,l\right){x}^{r}\\ \phantom{\rule{1em}{0ex}}={2}^{-\mu }\sum _{r=0}^{n}{\left(-{\lambda }^{-1}\right)}^{r}\left\{\sum _{l=r}^{n}\sum _{m=0}^{l-r}\sum _{i=0}^{l-r-m}\sum _{{j}_{1}+\cdots +{j}_{i}=l-r-m-i}\frac{{2}^{-i}{\lambda }^{l-m}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{l-m-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l-m}{r}\right){S}_{1}\left(n,l\right)\right\}{x}^{r}.\end{array}$
(48)

Therefore, by (48), we obtain the following theorem.

Theorem 5 For $n\ge 0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}={2}^{-\mu }\sum _{r=0}^{n}{\left(-{\lambda }^{-1}\right)}^{r}\left\{\sum _{l=r}^{n}\sum _{m=0}^{l-r}\sum _{i=0}^{l-r-m}\sum _{{j}_{1}+\cdots +{j}_{i}=l-r-m-i}\frac{{2}^{-i}{\lambda }^{l-m}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{l-m-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l-m}{r}\right){S}_{1}\left(n,l\right)\right\}{x}^{r}.\end{array}$

It is easy to see that

${\left(1+{e}^{\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left(1,{e}^{t}-1\right)$
(49)

and

${\left(x\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l}\sim \left(1,{e}^{t}-1\right).$
(50)

By the same method as Theorem 5, we get

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}={2}^{-\mu }\sum _{r=0}^{n}{\lambda }^{-r}\left\{\sum _{l=r}^{n}\sum _{m=0}^{l-r}\sum _{i=0}^{l-r-m}\sum _{{j}_{1}+\cdots +{j}_{i}=l-r-m-i}\frac{{\left(-1\right)}^{m}{2}^{-i}{\lambda }^{l-m}}{{\left(m+1\right)}^{k}}\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{l}{m}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{l-m-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l-m}{r}\right){S}_{1}\left(n,l\right)\right\}{x}^{r}.\end{array}$
(51)

From (20) and (28), we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\frac{1}{j!}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(-log\left(1+t\right)\right)}^{j}\mid {x}^{n}〉{x}^{j}.\end{array}$
(52)

Now, we observe that

$\begin{array}{r}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(-log\left(1+t\right)\right)}^{j}\mid {x}^{n}〉\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{j}〈log{\left(1+t\right)}^{j}|\sum _{m=0}^{\mathrm{\infty }}C{P}_{m}^{\left(k\right)}\left(0;\lambda ,\mu \right)\frac{{t}^{m}}{m!}{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{j}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{m}^{\left(k\right)}\left(0;\lambda ,\mu \right)〈{\left(log\left(1+t\right)\right)}^{j}\mid {x}^{n-m}〉\\ \phantom{\rule{1em}{0ex}}={\left(-1\right)}^{j}\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{m}^{\left(k\right)}\left(0;\lambda ,\mu \right)j!{S}_{1}\left(n-m,j\right).\end{array}$
(53)

Therefore, by (52) and (53), we obtain the following theorem.

Theorem 6 For $n\ge 0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{j=0}^{n}{\left(-1\right)}^{j}\left\{\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){S}_{1}\left(n-m,j\right)C{P}_{m}^{\left(k\right)}\left(0;\lambda ,\mu \right)\right\}{x}^{j}.$

Remark By the same method as Theorem 6, we get

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{j=0}^{n}\left\{\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right){S}_{1}\left(n-m,j\right)\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(0;\lambda ,\mu \right)\right\}{x}^{j}.$
(54)

From (21), we have

$C{P}_{n}^{\left(k\right)}\left(x+y;\lambda ,\mu \right)=\sum _{j=0}^{n}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{n}{j}\right)C{P}_{n-j}^{\left(k\right)}\left(x;\lambda ,\mu \right){y}^{\left(j\right)}$
(55)

and

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x+y;\lambda ,\mu \right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)\stackrel{ˆ}{C}{P}_{n-j}^{\left(k\right)}\left(x;\lambda ,\mu \right){\left(y\right)}_{j}.$
(56)

By (22) and (28), we get

$\left({e}^{-t}-1\right)C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=nC{P}_{n-1}^{\left(k\right)}\left(x;\lambda ,\mu \right)$
(57)

and

$\left({e}^{-t}-1\right)C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=C{P}_{n}^{\left(k\right)}\left(x-1;\lambda ,\mu \right)-C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right).$
(58)

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

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

$C{P}_{n}^{\left(k\right)}\left(x-1;\lambda ,\mu \right)-C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=nC{P}_{n-1}^{\left(k\right)}\left(x;\lambda ,\mu \right).$

Remark By the same method as Theorem 7, we get

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x+1;\lambda ,\mu \right)-\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=n\stackrel{ˆ}{C}{P}_{n-1}^{\left(k\right)}\left(x;\lambda ,\mu \right).$
(59)

From (22), (28), and (29), we have

$\begin{array}{rl}C{P}_{n+1}^{\left(k\right)}\left(x;1,\mu \right)=& -xC{P}_{n}^{\left(k\right)}\left(x+1;1,\mu \right)+\mu \sum _{m=0}^{n}{\left(-\frac{1}{2}\right)}^{m+1}{\left(n\right)}_{m}C{P}_{n-m}^{\left(k\right)}\left(x;1,\mu \right)\\ +{2}^{-\mu }\sum _{r=0}^{n}{\left(-1\right)}^{r}\left\{\sum _{m=r}^{n}\sum _{l=r}^{m}\sum _{i=0}^{l-r}\sum _{{j}_{1}+\cdots +{j}_{i}=l-i-r}\frac{{2}^{-i}}{{\left(m-l+2\right)}^{k}}\left(\genfrac{}{}{0}{}{m}{l}\right)\\ ×\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\left(\genfrac{}{}{0}{}{l-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l}{r}\right){S}_{1}\left(n,m\right)\right\}{\left(x+1\right)}^{r}\end{array}$
(60)

and

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n+1}^{\left(k\right)}\left(x;1,\mu \right)\\ \phantom{\rule{1em}{0ex}}=x\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x-1;1,\mu \right)+\mu \sum _{m=0}^{n}{\left(-\frac{1}{2}\right)}^{m+1}{\left(n\right)}_{m}\stackrel{ˆ}{C}{P}_{n-m}^{\left(k\right)}\left(x;1,\mu \right)\\ \phantom{\rule{2em}{0ex}}-{2}^{-\mu }\sum _{r=0}^{n}\left\{\sum _{m=r}^{n}\sum _{l=r}^{m}\sum _{i=0}^{l-r}\sum _{{j}_{1}+\cdots +{j}_{i}=l-i-r}\frac{{\left(-1\right)}^{m-l}{2}^{-i}}{{\left(m-l+2\right)}^{k}}\left(\genfrac{}{}{0}{}{m}{l}\right)\left(\genfrac{}{}{0}{}{-\mu }{i}\right)\\ \phantom{\rule{2em}{0ex}}×\left(\genfrac{}{}{0}{}{l-r}{{j}_{1}+1,\dots ,{j}_{i}+1}\right)\left(\genfrac{}{}{0}{}{l}{r}\right){S}_{1}\left(n,m\right)\right\}{\left(x-1\right)}^{r}.\end{array}$
(61)

By (14) and (27), we get

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(y;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=〈\sum _{l=0}^{\mathrm{\infty }}C{P}_{l}^{\left(k\right)}\left(y;\lambda ,\mu \right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-y}\mid x\cdot {x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\left({\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-y}\right)\mid {x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈\left({\partial }_{t}{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\right){Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-y}\mid {x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\left({\partial }_{t}{Lif}_{k}\left(log\left(1+t\right)\right)\right){\left(1+t\right)}^{-y}\mid {x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right)\left({\partial }_{t}{\left(1+t\right)}^{-y}\right)\mid {x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\mu \lambda 〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu -1}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-\left(y-\lambda +1\right)}\mid {x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}-y〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\left({Lif}_{k}\left(log\left(1+t\right)\right)\right){\left(1+t\right)}^{-y-1}\mid {x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\left({\partial }_{t}{Lif}_{k}\left(log\left(1+t\right)\right)\right){\left(1+t\right)}^{-y}\mid {x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=-\mu \lambda C{P}_{n-1}^{\left(k\right)}\left(y-\lambda +1;\lambda ,\mu +1\right)-yC{P}_{n-1}^{\left(k\right)}\left(y+1;\lambda ,\mu \right)\\ \phantom{\rule{2em}{0ex}}+〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\frac{{Lif}_{k-1}\left(log\left(1+t\right)\right)-{Lif}_{k}\left(log\left(1+t\right)\right)}{\left(1+t\right)log\left(1+t\right)}{\left(1+t\right)}^{-y}\mid {x}^{n-1}〉.\end{array}$
(62)

Now, we observe that

$\begin{array}{r}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\frac{{Lif}_{k-1}\left(log\left(1+t\right)\right)-{Lif}_{k}\left(log\left(1+t\right)\right)}{\left(1+t\right)log\left(1+t\right)}{\left(1+t\right)}^{-y}|{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\frac{{Lif}_{k-1}\left(log\left(1+t\right)\right)-{Lif}_{k}\left(log\left(1+t\right)\right)}{t}{\left(1+t\right)}^{-y}|\\ \phantom{\rule{2em}{0ex}}\frac{t}{\left(1+t\right)log\left(1+t\right)}{x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{l}\right){\stackrel{ˆ}{\mathbb{C}}}_{n-1-l}^{\left(1\right)}\left(0\right)\\ \phantom{\rule{2em}{0ex}}×〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\frac{{Lif}_{k-1}\left(log\left(1+t\right)\right)-{Lif}_{k}\left(log\left(1+t\right)\right)}{t}{\left(1+t\right)}^{-y}|{x}^{l}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n-1}{l}\right){\stackrel{ˆ}{\mathbb{C}}}_{n-1-l}^{\left(1\right)}\left(0\right)\\ \phantom{\rule{2em}{0ex}}×〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\frac{{Lif}_{k-1}\left(log\left(1+t\right)\right)-{Lif}_{k}\left(log\left(1+t\right)\right)}{t}{\left(1+t\right)}^{-y}|t\frac{{x}^{l+1}}{l+1}〉\\ \phantom{\rule{1em}{0ex}}=\frac{1}{n}\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){\stackrel{ˆ}{\mathbb{C}}}_{n-1-l}^{\left(1\right)}\left(0\right)\left\{C{P}_{l+1}^{\left(k-1\right)}\left(y;\lambda ,\mu \right)-C{P}_{l+1}^{\left(k\right)}\left(y;\lambda ,\mu \right)\right\}.\end{array}$
(63)

Therefore, by (62) and (63), we obtain the following theorem.

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

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=-\mu \lambda C{P}_{n-1}^{\left(k\right)}\left(x-\lambda +1;\lambda ,\mu +1\right)-xC{P}_{n-1}^{\left(k\right)}\left(x+1;\lambda ,\mu \right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){\stackrel{ˆ}{C}}_{n-1-l}\left\{C{P}_{l+1}^{\left(k-1\right)}\left(x;\lambda ,\mu \right)-C{P}_{l+1}^{\left(k\right)}\left(x;\lambda ,\mu \right)\right\},\end{array}$

where ${\stackrel{ˆ}{C}}_{n-1-l}={\stackrel{ˆ}{\mathbb{C}}}_{n-1-l}^{\left(1\right)}\left(0\right)$.

Remark By the same method as Theorem 8, we get

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=-\mu \lambda \stackrel{ˆ}{C}{P}_{n-1}^{\left(k\right)}\left(x+\lambda -1;\lambda ,\mu +1\right)+x\stackrel{ˆ}{C}{P}_{n-1}^{\left(k\right)}\left(x-1;\lambda ,\mu \right)\\ \phantom{\rule{2em}{0ex}}+\frac{1}{n}\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l+1}\right){\stackrel{ˆ}{C}}_{n-1-l}\left(\stackrel{ˆ}{C}{P}_{l+1}^{\left(k-1\right)}\left(x;\lambda ,\mu \right)-\stackrel{ˆ}{C}{P}_{l+1}^{\left(k\right)}\left(x;\lambda ,\mu \right)\right).\end{array}$
(64)

By (23), we get

$\begin{array}{r}\frac{d}{dx}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right)〈-log\left(1+t\right)\mid {x}^{n-l}〉C{P}_{l}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{m}}{m}{t}^{m}|{x}^{n-l}〉C{P}_{l}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right)\sum _{m=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{m}}{m}〈{t}^{m}\mid {x}^{n-l}〉C{P}_{l}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right){\left(-1\right)}^{n-l}C{P}_{l}^{\left(k\right)}\left(x;\lambda ,\mu \right)\left(n-l-1\right)!\\ \phantom{\rule{1em}{0ex}}=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l}}{\left(n-l\right)l!}C{P}_{l}^{\left(k\right)}\left(x;\lambda ,\mu \right).\end{array}$
(65)

By the same method as (65), we get

$\begin{array}{r}\frac{d}{dx}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{\left(n-l\right)l!}\stackrel{ˆ}{C}{P}_{l}^{\left(k\right)}\left(x;\lambda ,\mu \right).\end{array}$
(66)

Now, we compute the following equation in two different ways:

$〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}\mid {x}^{n}〉.$

On the one hand,

$\begin{array}{r}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}\mid {x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right)\mid {\left(log\left(1+t\right)\right)}^{m}{x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right)〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right)\mid {x}^{n-l-m}〉\\ \phantom{\rule{1em}{0ex}}=\sum _{l=0}^{n-m}m!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right)\stackrel{ˆ}{C}{P}_{l}^{\left(k\right)}\left(0;\lambda ,\mu \right).\end{array}$
(67)

On the other hand,

$\begin{array}{r}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}\mid {x}^{n}〉\\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\left({\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}\right)\mid {x}^{n-1}〉\\ \phantom{\rule{1em}{0ex}}=〈\left({\partial }_{t}{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\right){Lif}_{k}\left(-log\left(1+t\right)\right){\left(log\left(1+t\right)\right)}^{m}\mid {x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }\left({\partial }_{t}{Lif}_{k}\left(-log\left(1+t\right)\right)\right){\left(log\left(1+t\right)\right)}^{m}\mid {x}^{n-1}〉\\ \phantom{\rule{2em}{0ex}}+〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(-log\left(1+t\right)\right)\left({\partial }_{t}{\left(log\left(1+t\right)\right)}^{m}\right)\mid {x}^{n-1}〉.\end{array}$
(68)

Therefore, by (67) and (68), we obtain the following theorem.

Theorem 9 For $n\in \mathbb{N}$ with $n\ge 2$, let $n-1\ge m\ge 1$. Then we have

$\begin{array}{r}m\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right)\stackrel{ˆ}{C}{P}_{l}^{\left(k\right)}\left(0;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=-\mu \lambda m\sum _{l=0}^{n-1-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-1-l,m\right)\stackrel{ˆ}{C}{P}_{l}^{\left(k\right)}\left(\lambda -1;\lambda ,\mu +1\right)\\ \phantom{\rule{2em}{0ex}}+\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-1-l,m-1\right)\stackrel{ˆ}{C}{P}_{l}^{\left(k-1\right)}\left(-1;\lambda ,\mu \right)\\ \phantom{\rule{2em}{0ex}}+\left(m-1\right)\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-1-l,m-1\right)\stackrel{ˆ}{C}{P}_{l}^{\left(k\right)}\left(-1;\lambda ,\mu \right).\end{array}$

Remark By the same method as Theorem 9, we get

$\begin{array}{r}m\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right)C{P}_{l}^{\left(k\right)}\left(0;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=-\mu \lambda m\sum _{l=0}^{n-1-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-1-l,m\right)C{P}_{l}^{\left(k\right)}\left(1-\lambda ;\lambda ,\mu +1\right)\\ \phantom{\rule{2em}{0ex}}+\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-1-l,m-1\right)C{P}_{l}^{\left(k-1\right)}\left(1;\lambda ,\mu \right)\\ \phantom{\rule{2em}{0ex}}+\left(m-1\right)\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-1-l,m-1\right)C{P}_{l}^{\left(k\right)}\left(-1;\lambda ,\mu \right),\end{array}$

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

Let us consider the following two Sheffer sequences:

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left({\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)},{e}^{-t}-1\right)$
(69)

and

${B}_{n}^{\left(s\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{s},t\right)\phantom{\rule{1em}{0ex}}\left(s\in {\mathbb{Z}}_{\ge 0}\right).$
(70)

Let

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{m=0}^{n}{C}_{n,m}{B}_{m}^{\left(s\right)}\left(x\right).$
(71)

Then, by (26), we get

$\begin{array}{rl}{C}_{n,m}=& \frac{1}{m!}〈\frac{{\left(\frac{{e}^{-log\left(1+t\right)}-1}{-log\left(1+t\right)}\right)}^{s}}{{\left(1+{e}^{\lambda log\left(1+t\right)}\right)}^{\mu }}{Lif}_{k}\left(log\left(1+t\right)\right){\left(-log\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ =& \frac{{\left(-1\right)}^{m}}{m!}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right)\\ ×{\left(1+t\right)}^{-s}{\left(\frac{t}{log\left(1+t\right)}\right)}^{s}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n}〉\\ =& \frac{{\left(-1\right)}^{m}}{m!}\sum _{l=0}^{n-m}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right)\sum _{i=0}^{n-l-m}\left(\genfrac{}{}{0}{}{n-l-m}{i}\right){\mathbb{C}}_{i}^{\left(s\right)}\\ ×〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-s}\mid {x}^{n-l-m-i}〉\\ =& {\left(-1\right)}^{m}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right)\sum _{i=0}^{l}\left(\genfrac{}{}{0}{}{l}{i}\right){\mathbb{C}}_{i}^{\left(s\right)}C{P}_{l-i}^{\left(k\right)}\left(s;\lambda ,\mu \right).\end{array}$
(72)

Therefore, by (71) and (72), we obtain the following theorem.

Theorem 10 For $n\ge 0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{m=0}^{n}{\left(-1\right)}^{m}\left\{\sum _{l=0}^{n-m}\sum _{i=0}^{l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(n-l,m\right){\mathbb{C}}_{i}^{\left(s\right)}C{P}_{l-i}^{\left(k\right)}\left(s;\lambda ,\mu \right)\right\}{B}_{n}^{\left(s\right)}\left(x\right).$

Remark By the same method as Theorem 10, we have

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left\{\sum _{l=0}^{n-m}\sum _{i=0}^{l}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{l}{i}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{\mathbb{C}}}_{i}^{\left(s\right)}\stackrel{ˆ}{C}{P}_{l-i}^{\left(k\right)}\left(s;\lambda ,\mu \right)\right\}{B}_{m}^{\left(s\right)}\left(x\right).\end{array}$
(73)

For $C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left({\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)},{e}^{-t}-1\right)$, ${H}_{n}^{\left(s\right)}\left(x\mid \lambda \right)\sim \left({\left(\frac{{e}^{t}-\lambda }{1-\lambda }\right)}^{s},t\right)$, $s\in {\mathbb{Z}}_{\ge 0}$, $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, let us assume that

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{m=0}^{n}{C}_{n,m}{H}_{m}^{\left(s\right)}\left(x;\lambda \right).$
(74)

From (26), we have

$\begin{array}{rl}{C}_{n,m}=& \frac{{\left(-1\right)}^{m}}{m!}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right)\\ ×{\left(1+t\right)}^{-s}{\left(1+\frac{\lambda }{\lambda -1}t\right)}^{s}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n}〉\\ =& \frac{{\left(-1\right)}^{m}}{m!}\sum _{l=0}^{n-m}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right)\sum _{i=0}^{min\left\{s,n-l-m\right\}}\left(\genfrac{}{}{0}{}{s}{i}\right){\left(\frac{\lambda }{\lambda -1}\right)}^{i}\\ ×〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(1+t\right)}^{-s}\mid {t}^{i}{x}^{n-l-m}〉\\ =& {\left(-1\right)}^{m}\sum _{l=0}^{n-m}\sum _{i=0}^{min\left\{s,n-l-m\right\}}\left(\genfrac{}{}{0}{}{n}{l+m}\right)\left(\genfrac{}{}{0}{}{s}{i}\right)\\ ×{\left(n-l-m\right)}_{i}{\left(\frac{\lambda }{\lambda -1}\right)}^{i}{S}_{1}\left(l+m,m\right)C{P}_{n-l-m-i}^{\left(k\right)}\left(s;\lambda ,\mu \right)\\ =& {\left(-1\right)}^{m}\sum _{l=0}^{n-m}\sum _{i=0}^{min\left\{s,l\right\}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{s}{i}\right){\left(l\right)}_{i}{\left(\frac{\lambda }{\lambda -1}\right)}^{i}{S}_{1}\left(n-l,m\right)C{P}_{l-i}^{\left(k\right)}\left(s;\lambda ,\mu \right).\end{array}$
(75)

Therefore, by (75) and (76), we obtain the following theorem.

Theorem 11 For $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, $n\ge 0$, we have

$\begin{array}{r}C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}{\left(-1\right)}^{m}\left\{\sum _{l=0}^{n-m}\sum _{i=0}^{min\left\{s,l\right\}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{s}{i}\right){\left(l\right)}_{i}\\ \phantom{\rule{2em}{0ex}}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{i}{S}_{1}\left(n-l,m\right)C{P}_{l-i}^{\left(k\right)}\left(s;\lambda ,\mu \right)\right\}{H}_{m}^{\left(s\right)}\left(x;\lambda \right).\end{array}$

Remark By the same method as Theorem 11, we get

$\begin{array}{r}\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{n}\left\{\sum _{l=0}^{n-m}\sum _{i=0}^{min\left\{s,l\right\}}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(\genfrac{}{}{0}{}{s}{i}\right){\left(l\right)}_{i}\\ \phantom{\rule{2em}{0ex}}\cdot {\left(\frac{1}{1-\lambda }\right)}^{i}{S}_{1}\left(n-l,m\right)\stackrel{ˆ}{C}{P}_{l-i}^{\left(k\right)}\left(0;\lambda ,\mu \right)\right\}{H}_{m}^{\left(s\right)}\left(x;\lambda \right).\end{array}$

For $C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)\sim \left({\left(1+{e}^{-\lambda t}\right)}^{\mu }\frac{1}{{Lif}_{k}\left(-t\right)},{e}^{-t}-1\right)$ and ${x}^{\left(n\right)}\sim \left(1,1-{e}^{-t}\right)$, let us assume that

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{m=0}^{\mathrm{\infty }}{C}_{n,m}{x}^{\left(m\right)}.$
(76)

By (26), we get

$\begin{array}{rl}{C}_{n,m}=& \frac{1}{m!}〈\frac{1}{{\left(1+{e}^{\lambda log\left(1+t\right)}\right)}^{\mu }}{Lif}_{k}\left(log\left(1+t\right)\right){\left(1-{e}^{log\left(1+t\right)}\right)}^{m}|{x}^{n}〉\\ =& \frac{1}{m!}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right){\left(-t\right)}^{m}\mid {x}^{n}〉\\ =& \frac{{\left(-1\right)}^{m}}{m!}〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right)\mid {t}^{m}{x}^{n}〉\\ =& {\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{n}{m}\right)〈{\left(1+{\left(1+t\right)}^{\lambda }\right)}^{-\mu }{Lif}_{k}\left(log\left(1+t\right)\right)\mid {x}^{n-m}〉\\ =& {\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{n-m}^{\left(k\right)}\left(0;\lambda ,\mu \right).\end{array}$
(77)

Therefore, by (77) and (78), we obtain the following theorem.

Theorem 12 For $n\ge 0$, we have

$C{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{m=0}^{n}{\left(-1\right)}^{m}\left(\genfrac{}{}{0}{}{n}{m}\right)C{P}_{n-m}^{\left(k\right)}\left(0;\lambda ,\mu \right){x}^{\left(m\right)}.$

Remark By the same method as Theorem 12, we get

$\stackrel{ˆ}{C}{P}_{n}^{\left(k\right)}\left(x;\lambda ,\mu \right)=\sum _{m=0}^{n}\left(\genfrac{}{}{0}{}{n}{m}\right)\stackrel{ˆ}{C}{P}_{n-m}^{\left(k\right)}\left(0;\lambda ,\mu \right){\left(x\right)}_{m}.$
(78)

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## Acknowledgements

This work was supported by Kwangwoon University 2013.

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, D.S., Kim, T. Poly-Cauchy and Peters mixed-type polynomials. Adv Differ Equ 2014, 4 (2014). https://doi.org/10.1186/1687-1847-2014-4

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• DOI: https://doi.org/10.1186/1687-1847-2014-4

### Keywords

• Formal Power Series
• Linear Functional
• Bernoulli Polynomial
• Stirling Number
• Special Polynomial