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# Some properties of higher-order Daehee polynomials of the second kind arising from umbral calculus

Journal of Inequalities and Applications20142014:195

https://doi.org/10.1186/1029-242X-2014-195

• Received: 17 November 2013
• Accepted: 30 April 2014
• Published:

## Abstract

In this paper, we study the higher-order Daehee polynomials of the second kind from the umbral calculus viewpoint and give various identities of the higher-order Daehee polynomials of the second kind arising from umbral calculus.

## Keywords

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

## 1 Introduction

Let $k\in {\mathbb{Z}}_{\ge 0}$. The Daehee polynomials of the second kind of order k are defined by the generating function to be
${\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{x}=\sum _{n=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}$
(1)

(see ).

When $x=0$, ${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}={\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(0\right)$ are called the Daehee numbers of the second kind of order k.

The Stirling number of the first kind is defined by the falling factorial to be
${\left(x\right)}_{n}=x\left(x-1\right)\cdots \left(x-n+1\right)=\sum _{l=0}^{n}{S}_{1}\left(n,l\right){x}^{l}.$
(2)
Thus, by (2), we get
${\left(log\left(1+t\right)\right)}^{m}=m!\sum _{l=m}^{\mathrm{\infty }}{S}_{1}\left(l,m\right)\frac{{t}^{l}}{l!}$
(3)

(see ), where $m\in {\mathbb{Z}}_{\ge 0}$.

For $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, the Frobenius-Euler polynomials of order s ($\in \mathbb{N}$) are given by
${\left(\frac{1-\lambda }{{e}^{t}-\lambda }\right)}^{s}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{H}_{n}^{\left(s\right)}\left(x|\lambda \right)\frac{{t}^{n}}{n!}$
(4)

(see ).

When $x=0$, ${H}_{n}^{\left(s\right)}\left(\lambda \right)={H}_{n}^{\left(s\right)}\left(\lambda |0\right)$ are called the Frobenius-Euler numbers of order s.

As is well known, the Bernoulli polynomials of order k ($\in \mathbb{N}$) are defined by the generating function to be
${\left(\frac{t}{{e}^{t}-1}\right)}^{k}{e}^{xt}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}^{\left(k\right)}\left(x\right)\frac{{t}^{n}}{n!}$
(5)

(see ).

When $x=0$, ${B}_{n}^{\left(k\right)}={B}_{n}^{\left(k\right)}\left(0\right)$ are called the Bernoulli numbers of order k.

In this paper, we study the higher-order Daehee polynomials of the second kind with umbral calculus viewpoint and give various identities of the higher-order Daehee polynomials of the second kind arising from umbral calculus.

## 2 Umbral calculus

Let be the complex number field and let be the set of all formal power series
$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}\frac{{t}^{k}}{k!}|{a}_{k}\in \mathbb{C}\right\}.$
Let $\mathbb{P}=\mathbb{C}\left[x\right]$, and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on . $〈L|p\left(x\right)〉$ indicates the action of the linear functional L on the polynomial $p\left(x\right)$. Then the vector space operations on ${\mathbb{P}}^{\ast }$ are given by $〈L+M|p\left(x\right)〉=〈L|p\left(x\right)〉+〈M|p\left(x\right)〉$, and $〈cL|p\left(x\right)〉=c〈L|p\left(x\right)〉$, where c is a complex constant in . For $f\left(t\right)\in \mathcal{F}$, the linear functional on is defined by $〈f\left(t\right)|{x}^{n}〉={a}_{n}$. Then, in particular, we have
$〈{t}^{k}|{x}^{n}〉=n!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(n,k\ge 0\right)$
(6)

(see [3, 18]), where ${\delta }_{n,k}$ is the Kronecker symbol.

Let ${f}_{L}\left(t\right)={\sum }_{k=0}^{\mathrm{\infty }}\frac{〈L|{x}^{k}〉}{k!}{t}^{k}$. By (6), we get $〈{f}_{L}\left(t\right)|{x}^{n}〉=〈L|{x}^{n}〉$. That is, $L={f}_{L}\left(t\right)$. The map $L↦{f}_{L}\left(t\right)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto . Henceforth, denotes both the algebra of the 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 the umbral algebra. The order $o\left(f\left(t\right)\right)$ of the power series $f\left(t\right)$ (≠0) is the smallest integer for which the coefficient of ${t}^{k}$ does not vanish. If $o\left(f\left(t\right)\right)=0$, then $f\left(t\right)$ is called an invertible series; if $o\left(f\left(t\right)\right)=1$, then $f\left(t\right)$ is called a delta series.

Let $f\left(t\right),g\left(t\right)\in \mathcal{F}$ with $o\left(f\left(t\right)\right)=1$ and $o\left(g\left(t\right)\right)=0$. Then there exists a unique sequence ${s}_{n}\left(x\right)$ ($deg{s}_{n}\left(x\right)=n$) such that $〈g\left(t\right)f{\left(t\right)}^{k}|{s}_{n}\left(x\right)〉=n!{\delta }_{n,k}$, for $n,k\ge 0$. 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 $f\left(t\right),g\left(t\right)\in \mathcal{F}$, we have
$〈f\left(t\right)g\left(t\right)|p\left(x\right)〉=〈f\left(t\right)|g\left(t\right)p\left(x\right)〉=〈g\left(t\right)|f\left(t\right)p\left(x\right)〉.$
(7)
From (6), we note that
$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)|{x}^{k}〉\frac{{t}^{k}}{k!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}|p\left(x\right)〉\frac{{x}^{k}}{k!}$
(8)
and, by (8), we get
${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)=\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}^{yt}p\left(x\right)=p\left(x+y\right)$
(9)

(see [3, 18]).

For ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$, we have
$\frac{d{s}_{n}\left(x\right)}{dx}=\sum _{l=0}^{n-1}\left(\genfrac{}{}{0}{}{n}{l}\right)〈\overline{f}\left(t\right)|{x}^{n-l}〉{s}_{l}\left(x\right),$
(10)
where $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$ with $\overline{f}\left(f\left(t\right)\right)=t$. We have
(11)
$f\left(t\right){s}_{n}\left(x\right)=n{s}_{n-1}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 1\right),\phantom{\rule{2em}{0ex}}{s}_{n}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈g{\left(\overline{f}\left(t\right)\right)}^{-1}\overline{f}{\left(t\right)}^{j}|{x}^{n}〉{x}^{j},$
(12)
${s}_{n}\left(x+y\right)=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){s}_{j}\left(x\right){p}_{n-j}\left(y\right),$
(13)
where ${p}_{n}\left(x\right)=g\left(t\right){s}_{n}\left(x\right)$.
$〈f\left(t\right)|xp\left(x\right)〉=〈{\partial }_{t}f\left(t\right)|p\left(x\right)〉,$
(14)
with ${\partial }_{t}f\left(t\right)=\frac{df\left(t\right)}{dt}$, and
${s}_{n+1}\left(x\right)=\left(x-\frac{{g}^{\prime }\left(t\right)}{g\left(t\right)}\right)\frac{1}{{f}^{\prime }\left(t\right)}{s}_{n}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right)$
(15)

(see [3, 18]).

Let us assume that ${s}_{n}\left(x\right)\sim \left(g\left(t\right),f\left(t\right)\right)$ and ${r}_{n}\left(x\right)\sim \left(h\left(t\right),l\left(t\right)\right)$. Then we see that
${s}_{n}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{r}_{m}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right),$
(16)
where
${C}_{n,m}=\frac{1}{m!}〈\frac{h\left(\overline{f}\left(t\right)\right)}{g\left(\overline{f}\left(t\right)\right)}l{\left(\overline{f}\left(t\right)\right)}^{m}|{x}^{n}〉$
(17)

(see [3, 18]).

## 3 Higher-order Daehee polynomials of the second kind

By (1), we see that
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t{e}^{t}}\right)}^{k},{e}^{t}-1\right).$
(18)
From (18), we have
${\left(\frac{{e}^{t}-1}{t{e}^{t}}\right)}^{k}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\sim \left(1,{e}^{t}-1\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right).$
(19)
By (19), we get
$\begin{array}{rl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& ={\left(\frac{t{e}^{t}}{{e}^{t}-1}\right)}^{k}{\left(x\right)}_{n}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right){\left(\frac{t{e}^{t}}{{e}^{t}-1}\right)}^{k}{x}^{m}\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right){e}^{kt}{B}_{n}^{\left(k\right)}\left(x\right)\\ =\sum _{m=0}^{n}{S}_{1}\left(n,m\right){B}_{m}^{\left(k\right)}\left(x+k\right).\end{array}$
(20)
From (12) and (18), we have
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=\sum _{j=0}^{n}\frac{1}{j!}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{j}|{x}^{n}〉{x}^{j},$
(21)
where
$\begin{array}{c}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{j}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\left(\frac{log\left(1+t\right)}{t}\right)}^{k+j}{\left(1+t\right)}^{k}|{t}^{j}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}={\left(n\right)}_{j}〈{\left(\frac{log\left(1+t\right)}{t}\right)}^{k+j}|\sum _{m=0}^{min\left\{k,n-j\right\}}\left(\genfrac{}{}{0}{}{k}{m}\right){t}^{m}{x}^{n-j}〉\hfill \\ \phantom{\rule{1em}{0ex}}={\left(n\right)}_{j}\sum _{m=0}^{n-j}\left(\genfrac{}{}{0}{}{k}{m}\right){\left(n-j\right)}_{m}\sum _{l=0}^{\mathrm{\infty }}\frac{\left(k+j\right)!}{\left(l+k+j\right)!}{S}_{1}\left(l+k+j,k+j\right)〈{t}^{l}|{x}^{n-j-m}〉\hfill \\ \phantom{\rule{1em}{0ex}}={\left(n\right)}_{j}\sum _{m=0}^{n-j}\left(\genfrac{}{}{0}{}{k}{m}\right){\left(n-j\right)}_{m}\frac{\left(k+j\right)!}{\left(n+k-m\right)!}{S}_{1}\left(n+k-m,k+j\right)\left(n-j-m\right)!\hfill \\ \phantom{\rule{1em}{0ex}}={\left(n\right)}_{j}\sum _{m=0}^{n-j}\left(\genfrac{}{}{0}{}{k}{m}\right){\left(n-j\right)}_{m}\frac{{S}_{1}\left(n+k-m,k+j\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}.\hfill \end{array}$
(22)

Therefore, by (21) and (22), we obtain the following theorem.

Theorem 1 For $n\in {\mathbb{Z}}_{\ge 0}$ and $k\ge 1$, we have
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=\sum _{j=0}^{n}\left\{\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{m=0}^{n-j}\left(\genfrac{}{}{0}{}{k}{m}\right){\left(n-j\right)}_{m}\frac{{S}_{1}\left(n+k-m,k+j\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\right\}{x}^{j}.$
By (1) and (6), we get
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(y\right)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\left(y\right)\frac{{t}^{l}}{l!}|{x}^{n}〉\\ =& 〈{\left(\frac{log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{y}|{\left(1+t\right)}^{k}{x}^{n}〉\\ =& \sum _{0\le r\le min\left\{k,n\right\}}\left(\genfrac{}{}{0}{}{k}{r}\right){\left(n\right)}_{r}〈{\left(\frac{log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{y}|{x}^{n-r}〉\\ =& \sum _{0\le r\le min\left\{k,n\right\}}\left(\genfrac{}{}{0}{}{k}{r}\right){\left(n\right)}_{r}\sum _{0\le m\le n-r}\left(\genfrac{}{}{0}{}{y}{m}\right){\left(n-r\right)}_{m}\\ ×\sum _{0\le l\le n-r-m}\frac{k!{S}_{1}\left(l+k,k\right)}{\left(l+k\right)!}〈{t}^{l}|{x}^{n-r-m}〉\\ =& \sum _{0\le r\le n}\sum _{0\le m\le n-r}\frac{{\left(n\right)}_{r}\left(\genfrac{}{}{0}{}{k}{r}\right)\left(\genfrac{}{}{0}{}{n-r}{m}\right)}{\left(\genfrac{}{}{0}{}{n-r-m+k}{k}\right)}{S}_{1}\left(n-r-m+k,k\right){\left(y\right)}_{m}.\end{array}$
(23)

Therefore, by (23), we obtain the following theorem.

Theorem 2 For $n\ge 0$, we have
$\begin{array}{c}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le m\le n}\left\{\sum _{0\le r\le n-m}\frac{{\left(n\right)}_{r}\left(\genfrac{}{}{0}{}{k}{r}\right)\left(\genfrac{}{}{0}{}{n-r}{m}\right)}{\left(\genfrac{}{}{0}{}{n-r-m+k}{k}\right)}{S}_{1}\left(n-r-m+k,k\right)\right\}{\left(x\right)}_{m}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le m\le n}\left\{\sum _{0\le r\le n-m}\frac{{\left(n\right)}_{r}\left(\genfrac{}{}{0}{}{k}{r}\right)\left(\genfrac{}{}{0}{}{n-r}{n-m}\right)}{\left(\genfrac{}{}{0}{}{m-r+k}{k}\right)}{S}_{1}\left(m-r+k,k\right)\right\}{\left(x\right)}_{n-m}.\hfill \end{array}$
From (12) and (18), we have
$\left({e}^{t}-1\right){\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=n{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(x\right)$
(24)
and
$\left({e}^{t}-1\right){\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)={\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x+1\right)-{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right).$
Thus, by (24), we get
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x+1\right)-{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=n{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 1\right).$
(25)
From (15) and (18), we derive the following equation:
$\begin{array}{rl}{\stackrel{ˆ}{D}}_{n+1}^{\left(k\right)}\left(x\right)& =\left(x+k\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}\right){e}^{-t}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\\ =x{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x-1\right)+k{e}^{-t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right),\end{array}$
(26)
where
$\begin{array}{c}{e}^{-t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}={e}^{-t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}\sum _{0\le j\le n}\left\{\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0}{}{k}{m}\right)\left(\genfrac{}{}{0}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}×{S}_{1}\left(n+k-m,k+j\right)\right\}{x}^{j}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le j\le n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0}{}{k}{m}\right)\left(\genfrac{}{}{0}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}×{S}_{1}\left(n+k-m,k+j\right){e}^{-t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}{x}^{j}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le j\le n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0}{}{m+k}{m}\right)\left(\genfrac{}{}{0}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}×{S}_{1}\left(n+k-m,k+j\right){e}^{-t}\left(\frac{{e}^{t}-1-t}{{e}^{t}-1}\right)\frac{{x}^{j+1}}{j+1}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le j\le n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0}{}{m+k}{m}\right)\left(\genfrac{}{}{0}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}×\frac{{S}_{1}\left(n+k-m,k+j\right)}{j+1}{e}^{-t}\left({x}^{j+1}-{B}_{j+1}\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le j\le n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0}{}{m+k}{m}\right)\left(\genfrac{}{}{0}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}×\frac{{S}_{1}\left(n+k-m,k+j\right)}{j+1}{e}^{-t}\left({\left(x-1\right)}^{j+1}-{B}_{j+1}\left(x-1\right)\right).\hfill \end{array}$
(27)

Therefore, from (26) and (27), we obtain the following theorem.

Theorem 3 For $n\ge 0$, $k\ge 1$, we have
$\begin{array}{c}{\stackrel{ˆ}{D}}_{n+1}^{\left(k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=x{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x-1\right)+k\sum _{0\le j\le n}\left(\genfrac{}{}{0}{}{n}{j}\right)\sum _{0\le m\le n-j}\frac{m!\left(\genfrac{}{}{0}{}{m+k}{m}\right)\left(\genfrac{}{}{0}{}{n-j}{m}\right)}{\left(\genfrac{}{}{0}{}{n+k-m}{k+j}\right)}\hfill \\ \phantom{\rule{2em}{0ex}}×\frac{{S}_{1}\left(n+k-m,k+j\right)}{j+1}\left\{{\left(x-1\right)}^{j+1}-{B}_{j+1}\left(x-1\right)\right\}.\hfill \end{array}$
Now, we observe that
$\begin{array}{c}{e}^{-t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{j}\right)}{\left(\genfrac{}{}{0}{}{n+k}{j+k}\right)}{S}_{1}\left(n+k,j+k\right){e}^{-t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}{\left(x+k\right)}^{j}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{j}\right)}{\left(\genfrac{}{}{0}{}{n+k}{j+k}\right)}{S}_{1}\left(n+k,j+k\right){e}^{\left(k-1\right)t}\frac{{e}^{t}-1-t}{t\left({e}^{t}-1\right)}{x}^{j}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{j}\right)}{\left(\genfrac{}{}{0}{}{n+k}{j+k}\right)}\frac{{S}_{1}\left(n+k,j+k\right)}{j+1}{e}^{\left(k-1\right)t}\left({x}^{j+1}-{B}_{j+1}\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{j}\right)}{\left(\genfrac{}{}{0}{}{n+k}{j+k}\right)}\frac{{S}_{1}\left(n+k,j+k\right)}{j+1}\left({\left(x+k-1\right)}^{j+1}-{B}_{j+1}\left(x+k-1\right)\right).\hfill \end{array}$
(28)
Thus, by (28), we get
${\stackrel{ˆ}{D}}_{n+1}^{\left(k\right)}\left(x\right)=x{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x-1\right)+k\sum _{j=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{j}\right)}{\left(\genfrac{}{}{0}{}{n+k}{j+k}\right)}\frac{{S}_{1}\left(n+k,j+k\right)}{j+1}\left({\left(x+k-1\right)}^{j+1}-{B}_{j+1}\left(x+k-1\right)\right).$
From (10) and (18), we note that
$\frac{d}{dx}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=n!\sum _{l=0}^{n-1}\frac{{\left(-1\right)}^{n-l-1}}{l!\left(n-l\right)}{\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\left(x\right).$
(29)
By (6) and (18), we see that
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(y\right)& =& 〈\sum _{l=0}^{\mathrm{\infty }}{\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\left(y\right)\frac{{t}^{l}}{l!}|{x}^{n}〉\phantom{\rule{1em}{0ex}}\left(n\ge 1\right)\\ =& 〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{y}|{x}^{n}〉\\ =& 〈{\partial }_{t}\left({\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{y}\right)|{x}^{n-1}〉\\ =& 〈\left({\partial }_{t}{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}\right){\left(1+t\right)}^{y}|{x}^{n-1}〉\\ +y〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{y-1}|{x}^{n-1}〉\\ =& y{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(y-1\right)\\ +k〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(1+t\right)}^{y}|\left(log\left(1+t\right)+1-\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)\frac{{x}^{n}}{n}〉\\ =& y{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(y-1\right)+\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(1+t\right)}^{y}|log\left(1+t\right){x}^{n}〉\\ +\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(1+t\right)}^{y}|{x}^{n}〉\\ -\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{y}|{x}^{n}〉\\ =& y{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(y-1\right)+\frac{k}{n}{\stackrel{ˆ}{D}}_{n}^{\left(k-1\right)}\left(y\right)-\frac{k}{n}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(y\right)\\ +\frac{k}{n}\sum _{1\le l\le n}\frac{{\left(-1\right)}^{l-1}{\left(n\right)}_{l}}{l}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(1+t\right)}^{y}|{x}^{n-l}〉.\end{array}$
(30)
Thus, by (30), we get
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& =& \frac{n}{n+k}x{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(x-1\right)+\frac{k}{n+k}{\stackrel{ˆ}{D}}_{n}^{\left(k-1\right)}\left(x\right)\\ +\frac{k}{n+k}\sum _{1\le l\le n}{\left(-1\right)}^{l-1}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(l-1\right)!{\stackrel{ˆ}{D}}_{n-l}^{\left(k-1\right)}\left(x\right).\end{array}$
(31)

Therefore, by (31), we obtain the following theorem.

Theorem 4 For $n\ge 0$, $k\ge 1$, we have
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& =& \frac{n}{n+k}x{\stackrel{ˆ}{D}}_{n-1}^{\left(k\right)}\left(x-1\right)+\frac{k}{n+k}{\stackrel{ˆ}{D}}_{n}^{\left(k-1\right)}\left(x\right)\\ +\frac{k}{n+k}\sum _{1\le l\le n}{\left(-1\right)}^{l-1}\left(\genfrac{}{}{0}{}{n}{l}\right)\left(l-1\right)!{\stackrel{ˆ}{D}}_{n-l}^{\left(k-1\right)}\left(x\right).\end{array}$
Now, we compute $〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉$ in two different ways:
$\begin{array}{c}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le l\le n-m}\frac{m!}{\left(l+m\right)!}{S}_{1}\left(l+m,m\right){\left(n\right)}_{l+m}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}|{x}^{n-l-m}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0}{}{n}{l+m}\right){S}_{1}\left(l+m,m\right){\stackrel{ˆ}{D}}_{n-l-m}^{\left(k\right)}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}.\hfill \end{array}$
(32)
On the other hand,
$\begin{array}{c}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈{\partial }_{t}\left({\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}\right)|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=k〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}\left(\frac{log\left(1+t\right)+1-\frac{\left(1+t\right)log\left(1+t\right)}{t}}{t}\right){\left(log\left(1+t\right)\right)}^{m}|{x}^{n-1}〉\hfill \\ \phantom{\rule{2em}{0ex}}+m〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{-1}{\left(log\left(1+t\right)\right)}^{m-1}|{x}^{n-1}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(log\left(1+t\right)\right)}^{m+1}|{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}+m〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{-1}{\left(log\left(1+t\right)\right)}^{m-1}|{x}^{n-1}〉.\hfill \end{array}$
(33)
Thus, by (33), we get
$\begin{array}{c}\frac{n+k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(log\left(1+t\right)\right)}^{m+1}|{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{k}{n}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-1}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\hfill \\ \phantom{\rule{2em}{0ex}}+m〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(1+t\right)}^{-1}{\left(log\left(1+t\right)\right)}^{m-1}|{x}^{n-1}〉.\hfill \end{array}$
(34)
From (34), we derive the following equation:
$\begin{array}{c}\frac{n+k}{k}\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{k}{n}\sum _{0\le l\le n-m-1}\left(m+1\right)!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m+1\right){\stackrel{ˆ}{D}}_{l}^{\left(k-1\right)}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{k}{n}\sum _{0\le l\le n-m}m!\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k-1\right)}\hfill \\ \phantom{\rule{2em}{0ex}}+m\sum _{0\le l\le n-m}\left(m-1\right)!\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\left(-1\right).\hfill \end{array}$
(35)

Therefore, by (35), we obtain the following theorem.

Theorem 5 For $n-1\ge m\ge 1$, we have
$\begin{array}{c}\sum _{l=0}^{n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{k\left(m+1\right)}{n+k}\sum _{0\le l\le n-m-1}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m+1\right){\stackrel{ˆ}{D}}_{l}^{\left(k-1\right)}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{k}{n+k}\sum _{0\le l\le n-m}\left(\genfrac{}{}{0}{}{n}{l}\right){S}_{1}\left(n-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k-1\right)}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{n}{n+k}\sum _{0\le l\le n-m}\left(\genfrac{}{}{0}{}{n-1}{l}\right){S}_{1}\left(n-l-1,m-1\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\left(-1\right).\hfill \end{array}$
For ${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t{e}^{t}}\right)}^{k},{e}^{t}-1\right)$, and ${\left(x\right)}_{n}\sim \left(1,{e}^{t}-1\right)$, let us assume that
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{\left(x\right)}_{m}.$
(36)
Then, by (16) and (17), we get
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}|{t}^{m}{x}^{n}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right)〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}|{x}^{n-m}〉\\ =\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{D}}_{n-m}^{\left(k\right)}.\end{array}$
(37)

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

Theorem 6 For $n\ge 0$, we have
$\begin{array}{rl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& =\sum _{0\le m\le n}\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{D}}_{n-m}^{\left(k\right)}{\left(x\right)}_{m}\\ =\sum _{0\le m\le n}m!\left(\genfrac{}{}{0}{}{n}{m}\right){\stackrel{ˆ}{D}}_{n-m}^{\left(k\right)}\left(\genfrac{}{}{0}{}{x}{m}\right).\end{array}$
Now, we consider the following two Sheffer sequences:
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t{e}^{t}}\right)}^{k},{e}^{t}-1\right)$
(38)
and
Let
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{H}_{m}^{\left(s\right)}\left(x|\lambda \right).$
(39)
Here
$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!{\left(1-\lambda \right)}^{s}}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}{\left(1-\lambda +t\right)}^{s}|{x}^{n}〉\\ =& \frac{1}{m!{\left(1-\lambda \right)}^{s}}\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(1-\lambda \right)}^{s-j}{\left(n\right)}_{j}\\ ×〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n-j}〉\\ =& \sum _{j=0}^{n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(1-\lambda \right)}^{-j}{\left(n\right)}_{j}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0}{}{n-j}{l+m}\right){S}_{1}\left(l+m,m\right){\stackrel{ˆ}{D}}_{n-j-l-m}^{\left(k\right)}\\ =& \sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{l}\right){\left(n\right)}_{j}{\left(1-\lambda \right)}^{-j}{S}_{1}\left(n-j-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}.\end{array}$
(40)

Therefore, by (39) and (40), we obtain the following theorem.

Theorem 7 For $n\ge 0$, $k\ge 1$ and $\lambda \in \mathbb{C}$ with $\lambda \ne 1$, we have
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& =& \sum _{m=0}^{n}\left\{\sum _{j=0}^{n-m}\sum _{l=0}^{n-m-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{l}\right){\left(n\right)}_{j}\\ ×{\left(1-\lambda \right)}^{-j}{S}_{1}\left(n-j-l,m\right){\stackrel{ˆ}{D}}_{l}^{\left(k\right)}\right\}{H}_{m}^{\left(s\right)}\left(x|\lambda \right).\end{array}$
We consider the following two Sheffer sequences:
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t{e}^{t}}\right)}^{k},{e}^{t}-1\right),\phantom{\rule{2em}{0ex}}{B}_{n}^{\left(s\right)}\left(x\right)\sim \left({\left(\frac{{e}^{t}-1}{t}\right)}^{s},t\right).$
Let
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=\sum _{m=0}^{n}{C}_{n,m}{B}_{m}^{\left(s\right)}\left(x\right).$
(41)
Here
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈\frac{{\left(\frac{t}{log\left(1+t\right)}\right)}^{s}}{{\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{k}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉\\ =\frac{1}{m!}〈{\left(1+t\right)}^{s}\frac{{\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{s}}{{\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{k}}{\left(log\left(1+t\right)\right)}^{m}|{x}^{n}〉.\end{array}$
(42)
Case 1. For $s>k$, we have
$\begin{array}{rcl}{C}_{n,m}& =& \frac{1}{m!}〈{\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{s-k}{\left(log\left(1+t\right)\right)}^{m}|{\left(1+t\right)}^{s}{x}^{n}〉\\ =& \frac{1}{m!}\sum _{0\le j\le n}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(n\right)}_{j}〈{\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{s-k}|{\left(log\left(1+t\right)\right)}^{m}{x}^{n-j}〉\\ =& \sum _{0\le j\le n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(n\right)}_{j}\sum _{m\le l\le n-j}{S}_{1}\left(l,m\right)\\ ×\left(\genfrac{}{}{0}{}{n-j}{l}\right)〈{\left(\frac{t}{\left(1+t\right)log\left(1+t\right)}\right)}^{s-k}|{x}^{n-j-l}〉\\ =& \sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{l}\right){\left(n\right)}_{j}{S}_{1}\left(l,m\right){\stackrel{ˆ}{C}}_{n-j-l}^{\left(s-k\right)},\end{array}$
(43)

where ${\stackrel{ˆ}{C}}_{i}^{\left(s-k\right)}$ is the i th Cauchy number of the second kind of order $s-k$ (see ).

Case 2. For $s=k$, we have
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(log\left(1+t\right)\right)}^{m}|{\left(1+t\right)}^{s}{x}^{n}〉\\ =\frac{1}{m!}〈{\left(log\left(1+t\right)\right)}^{m}|\sum _{j=0}^{s}\left(\genfrac{}{}{0}{}{s}{j}\right){t}^{j}{x}^{n}〉\\ =\sum _{0\le j\le n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(n\right)}_{j}\sum _{m\le l<\mathrm{\infty }}\frac{{S}_{1}\left(l,m\right)}{l!}〈{t}^{l}|{x}^{n-j}〉\\ =\sum _{0\le j\le n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(n\right)}_{j}{S}_{1}\left(n-j,m\right).\end{array}$
(44)
Case 3. For $s, we have
$\begin{array}{rl}{C}_{n,m}& =\frac{1}{m!}〈{\left(\frac{\left(1+t\right)log\left(1+t\right)}{t}\right)}^{k-s}{\left(log\left(1+t\right)\right)}^{m}|{\left(1+t\right)}^{s}{x}^{n}〉\\ =\sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{l}\right){\left(n\right)}_{j}{S}_{1}\left(l,m\right){\stackrel{ˆ}{D}}_{n-j-l}^{\left(k-s\right)}.\end{array}$
(45)

Therefore, by (41), (42), (43), (44), and (45), we obtain the following theorem.

Theorem 8 Let $n\ge 0$, we have:
1. (I)
For $s>k$, we have
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& =& \sum _{0\le m\le n}\left\{\sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{l}\right)\\ ×{\left(n\right)}_{j}{S}_{1}\left(l,m\right){\stackrel{ˆ}{C}}_{n-j-l}^{\left(s-k\right)}\right\}{B}_{m}^{\left(s\right)}\left(x\right).\end{array}$

2. (II)
For $s=k$, we have
${\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)=\sum _{0\le m\le n}\left\{\sum _{0\le j\le n-m}\left(\genfrac{}{}{0}{}{s}{j}\right){\left(n\right)}_{j}{S}_{1}\left(n-j,m\right)\right\}{B}_{m}^{\left(s\right)}\left(x\right).$

3. (III)
For $s, we have
$\begin{array}{rcl}{\stackrel{ˆ}{D}}_{n}^{\left(k\right)}\left(x\right)& =& \sum _{0\le m\le n}\left\{\sum _{0\le j\le n-m}\sum _{m\le l\le n-j}\left(\genfrac{}{}{0}{}{s}{j}\right)\left(\genfrac{}{}{0}{}{n-j}{l}\right)\\ ×{\left(n\right)}_{j}{S}_{1}\left(l,m\right){\stackrel{ˆ}{D}}_{n-j-l}^{\left(k-s\right)}\right\}{B}_{m}^{\left(s\right)}\left(x\right).\end{array}$

## Declarations

### Acknowledgements

The authors would like to thank the referees for their valuable comments. This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korean government (MOE) (No. 2012R1A1A2003786).

## Authors’ Affiliations

(1)
Department of Mathematics, Sogang University, Seoul, 121-742, Republic of Korea
(2)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, Republic of Korea

## References 