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

## 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.

## 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

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

## References

1. Kim DS, Kim T: Daehee numbers and polynomials. Appl. Math. Sci. 2013,7(120):5969–5976.

2. Dere R, Simsek Y: Applications of umbral algebra to some special polynomials. Adv. Stud. Contemp. Math. 2012,22(3):433–438.

3. Kim DS, Kim T: Higher-order Cauchy of the first kind and poly-Cauchy of the first kind mixed type polynomials. Adv. Stud. Contemp. Math. 2013,33(4):621–636.

4. Roman S: The Umbral Calculus. Dover, New York; 2005.

5. Araci S, Acikgoz M: A note on the Frobenius-Euler numbers and polynomials associated with Bernstein polynomials. Adv. Stud. Contemp. Math. 2012,22(3):399–406.

6. Comtet L: Advanced Combinatorics. Reidel, Dordrecht; 1974.

7. Dolgy DV, Kim T, Lee B, Lee S-H: Some new identities on the twisted Bernoulli and Euler polynomials. J. Comput. Anal. Appl. 2013,15(3):441–451.

8. Hwang K-W, Dolgy DV, Kim DS, Kim T, Kee SH: Some theorems on Bernoulli and Euler numbers. Ars Comb. 2013, 109: 285–297.

9. Jeong J-H, Park J-W, Rim S-H: New approach to the analogue of Lebesgue-Radon-Nikodym theorem with respect to weighted p -adic q -measure on ${\mathbb{Z}}_{p}$ . J. Comput. Anal. Appl. 2013,15(7):1310–1316.

10. Kim T, Adiga C: Sums of products of generalized Bernoulli numbers. Int. Math. J. 2004,5(1):1–7.

11. Kim T, Kim DS, Dolgy DV, Rim SH: Some identities on the Euler numbers arising from Euler basis polynomials. Ars Comb. 2013, 109: 433–446.

12. Kim T: An identity of the symmetry for the Frobenius-Euler polynomials associated with the fermionic p -adic invariant q -integrals on ${\mathbb{Z}}_{p}$ . Rocky Mt. J. Math. 2011,41(1):239–247.

13. Kim T: q -Volkenborn integration. Russ. J. Math. Phys. 2002,9(3):288–299.

14. Komatsu T: Poly-Cauchy numbers. Kyushu J. Math. 2013, 67: 143–153.

15. Ozden H, Cangul IN, Simsek Y: Remarks on q -Bernoulli numbers associated with Daehee numbers. Adv. Stud. Contemp. Math. 2009,18(1):41–48.

16. Rim S-H, Jeong J: Identities on the modified q -Euler and q -Bernstein polynomials and numbers with weight. J. Comput. Anal. Appl. 2013,15(1):39–44.

17. Simsek Y: Generating functions of the twisted Bernoulli numbers and polynomials associated with their interpolation functions. Adv. Stud. Contemp. Math. 2008,16(2):251–278.

18. Zhang Z, Yang H: Some closed formulas for generalized Bernoulli-Euler numbers and polynomials. Proc. Jangjeon Math. Soc. 2008,11(2):191–198.

## 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).

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