<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}}
\newcommand{\sstirling}[2]{\genfrac\{\}{0pt}{}{#1}{#2}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf Generalization of an Identity of Apostol}
\vskip 1cm
\large 
Abdelmoum\`ene Zekiri and Farid Bencherif \\
LA3C Laboratory \\
Faculty of Mathematics \\
USTHB \\
Algiers \\
Algeria \\
\href{mailto:azekiri@usthb.dz}{\tt azekiri@usthb.dz} \\
\href{mailto:fbencherif@usthb.dz}{\tt fbencherif@usthb.dz} \\
\ \\
Rachid Boumahdi\\
LA3C-USTHB  \\
\'Ecole Nationale Sup\'erieure d'Informatique \\
Algiers \\
Algeria\\
\href{mailto:r_boumehdi@esi.dz}{\tt r\_boumehdi@esi.dz} \\

\end{center}

\vskip .2 in

\begin{abstract}
We extend and generalize an identity of Apostol,
involving Bernoulli numbers, to every sequence of complex numbers. Moreover,
our result allows us to obtain other relations involving Appell
polynomial sequences and second-order linear recurrence sequences.
 \end{abstract}
 
\begin{center} \emph{This paper is dedicated in memory
 of professor  \\Tom~M.~Apostol $($1923 -- 2016$)$} 
 \end{center}

\section{Introduction and statement of results}

With each complex sequence $u$ we associate the sequence $T(u)=u^*$ defined 
as follows:
\begin{equation}
u_m^*=\sum_{k=0}^m (-1)^k{m \choose k}u_{k}, \quad  (m\geq 0).  \label{t2}
\end{equation}
The mapping $T$ is called a {\it binomial transformation} \cite{Sun:2001}, and the sequence $u^*$ is called the {\it dual sequence\/} of $u$.
It is easy to prove that $(u^*)^*=u$, and if $u$ is linearly
recurrent over $\mathbb{C}$,
then $u^*$ is also linearly recurrent over $\mathbb{C}$.
Moreover, 
if $C(x)$ is a characteristic polynomial of $u$,
then $C(1-x)$ is a characteristic polynomial of $u^*$. The sequence $u$ is called {\it invariant\/} (resp., {\it inverse invariant}) under the binomial transformation if $u^*=u$ (resp., if $u^*=-u$).

Let $(r)_j$ be the real sequence defined by $(r)_j=j! {r\choose j}$, where $r$ and $j$ are positive integers, and $B_m$ is the $m$'th Bernoulli number
defined by:
\begin{equation}
\frac{z}{e^{z}-1}=\sum_{m=0}^{+\infty }B_{m}\frac{z^{m}}{m!}. \label{Bernoulli}
\end{equation} 
In 2008, Apostol \cite{Apostol:2008} proved the following identity:
\begin{equation}
\sum_{k=0}^{n}{n \choose k}\frac{B_{k}}{n+2-k}=\frac{B_{n+1}}{n+1},\quad n\geq 1.  \label{b3}
\end{equation}
The main result of this paper is the following theorem, which gives a simplified expression for the sum 
\begin{equation*}
\sum_{k=0}^{n}(-1)^k {n \choose k}\frac{u_k}{n+r+1-k},
\end{equation*}
and allows us to generalize relation~\eqref{b3}.
\begin{theorem}
\label{th1} For every sequence $u$ of complex numbers,
and all non-negative integers $r$ and $n$, the following holds:
\begin{equation}
\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{u_{k}}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{j}(r)_j\frac{u_{n+j+1}^*}{(n+j+1)_{j+1}} 
+(-1)^{n}r!\frac{u_{n+r+1}}{(n+r+1)_{r+1}}.  \label{n10}
\end{equation}
\end{theorem}
We can easily prove, from relation~\eqref{Bernoulli} and the equality $\frac{-z}{e^{-z}-1}=\frac{z}{e^z-1}e^z$, 
that $\left((-1)^{n}B_{n}\right)_{n\geq 0}$ is  invariant under the binomial transformation. Thus, we deduce, from Theorem~\ref{th1}, the following relation for integers $r\geq 0$ and $n\geq 1$.
\begin{equation}
\sum_{k=0}^{n}{n \choose k}\frac{B_{k}}{n+r+1-k}=\sum_{j=0}^{r-1}(-1)^{n+1}(r)_{j}\frac{B_{n+j+1}}{(n+j+1)_{j+1}}.\label{b1}
\end{equation}
It is clear that the relation~\eqref{b1} is a generalization of identity~\eqref{b3}.

Theorem~\ref{th1} can also be applied to the sequence of Bell numbers.
Recall that for $n\geq 1$, the $n$th Bell number $b_{n}$ is the number
of distinct partitions of a set of $n$ elements  (sequence
\seqnum{A000110} in the OEIS \cite{oeis}). By convention, we set $b_{0}=1$. If
$u_{n}=(-1)^{n}b_{n}$ for $n\geq 0$, the well-known relation
$\sum_{k=0}^{n}{n \choose k}b_{k}=b_{n+1}$ for $n\geq 0$ yields $u_{n}^*=b_{n+1}$ for $n\geq 0$.
By applying Theorem~\ref{th1} for $r=1$ to the sequence $u$, we obtain an identity similar to relation~\eqref{b3} for the Bell numbers. 
\begin{equation*}
\sum_{k=0}^{n} {n \choose k}\frac{b_{k}}{n+2-k}=\frac{(n+3)b_{n+2}-b_{n+3}}{(n+2)(n+1)}.
\end{equation*}
Theorem~\ref{th1} enables us to obtain some identities such as the following: 
\begin{equation}
\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2^{k}(n+k+1)}=\frac{2^{n}}{2n+1}{2n \choose n}^{-1},  \label{e0}
\end{equation}
due to Sun  \cite[Relation (1.13)]{Sun:2003}. To do this, it is enough to apply Theorem~\ref{th1} for $r=n$ to the sequence 
$u=\left( 2^{m}\right)_{m\geq 0}$ for which we have $u^*=\left( (-1)^{m}\right)_{m\geq 0}$. One gets the following:
\[
2^{n}\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{1}{2^{k}(n+k+1)}
=-\frac{n!n!}{(2n+1)!}\sum_{j=0}^{n}{2n+1 \choose j}+\frac{n!n!}{(2n+1)!}2^{2n+1}.
\]
From this we can deduce relation~\eqref{e0} by using the well-known identity $\sum_{j=0}^{n}{2n+1 \choose j}=2^{2n}$.

The following corollary gives us some identities when $u$ is a
second-order linear recurrent complex sequence such as the Fibonacci
numbers \seqnum{A000045}, the Lucas numbers \seqnum{A000032}, the Pell numbers
\seqnum{A000129}, the companion Pell numbers \seqnum{A002203}, the Jacobsthal
numbers \seqnum{A001045}, and the Jacobsthal-Lucas numbers
\seqnum{A014551}.

\begin{corollary}
\label{co2}If $u$ is a second-order linear recurrent sequence of
complex numbers with $x^{2}-ax-b$ as characteristic polynomial, then the following relations are satisfied for all non-negative integers $r$ and $n$:
\begin{equation}
\sum_{k=0}^{n}{n \choose k}\frac{a^{k}b^{n+r+1-k}u_{k}}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{j}(r)_{j}\frac{b^{r-j}u_{2n+2j+2}}{(n+j+1)_{j+1}}+(-1)^{r+1}\frac{r!a^{n+r+1}u_{n+r+1}}{( n+r+1)_{r+1}}.
\label{p1}
\end{equation}
\begin{equation}
\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{b^{n+r+1-k}u_{2k}}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{n+1}(r)_{j}\frac{a^{n+j+1}b^{r-j}u_{n+j+1}}{(n+j+1)_{j+1}}+(-1)^{n}\frac{r!u_{2n+2r+2}}{(n+r+1) _{r+1}}.
\label{p2}
\end{equation}
If $u_{0}=0$, then
\begin{equation}
\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{a^{n+r-k+1}u_{k}}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{j+1}(r)_j\frac{a^{r-j}u_{n+j+1}}{(n+j+1)_{j+1}}+(-1)^n \frac{r!u_{n+r+1}}{(n+r+1)_{r+1}}.  \label{p3}
\end{equation}
If $( u_{0},u_{1}) =(2,a)$, then
\begin{equation}
\sum_{k=0}^{n}(-1)^{k} {n \choose k}\frac{a^{n+r+1-k}u_{k}}{n+r+1-k}=\sum_{j=0}^{r}(-1)^{j}(r)_{j}\frac{a^{r-j}u_{n+j+1}}{(n+j+1)_{j+1}}\\
+(-1)^{n}\frac{r!u_{n+r+1}}{(n+r+1)_{r+1}}.  \label{p4}
\end{equation}
\end{corollary}
Theorem~\ref{th1} enables us also to get an identity for Appell polynomial sequences. 
Recall that an Appell polynomial sequence \cite{Appell:1880} associated with a formal series 
$S(z)\in \mathbb{C}[[z]] $ is the polynomial sequence $\left(A_{n}(x)\right)_{n\geq 0}$ of 
$\mathbb{C}[x]$ given by the generating relation $\sum_{n=0}^\infty A_{n}(x)\frac{z^{n}}{n!}=S(z)e^{xz}$.  
\begin{corollary}
\label{co3}If $\left( A_{m}(x)\right)_{m\geq 0}$ is an Appell polynomial sequence, then for all complex numbers $\lambda$ and all non-negative integers $r,n$, the following holds: 
\begin{equation}
\sum_{k=0}^{n}{n \choose k}\frac{\lambda ^{n+r+1-k}A_{k}(x)}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^j (r)_j\lambda ^{r-j}\frac{A_{n+j+1}(x+\lambda)}{(n+j+1)_{j+1}} +(-1)^{r+1}\frac{r!\ A_{n+r+1}(x)}{(n+r+1)_{r+1}}.  \label{p9}
\end{equation}
\end{corollary}
We now consider some examples. Let $\alpha$ be a complex number. We let $B^{(\alpha)}_{n}(x)$,  $E^{(\alpha)}_{n}(x)$, 
$H_{n}(x)$, and $L^{(\alpha)}_{n}(x)$ denote, respectively, the generalized Bernoulli polynomial, the generalized Euler polynomial, the Hermite polynomial, and the generalized Laguerre polynomial of degree $n$ defined as follows:
\begin{align*}
\left(\frac{z}{e^z-1}\right)^{\alpha}e^{zx}&amp;=\sum_{n=0}^{+\infty}B_n^{(\alpha)}(x)\frac{z^n}{n!},\\
\left(\frac{2}{e^z+1}\right)^{\alpha}e^{zx}&amp;=\sum_{n=0}^{+\infty}E_n^{(\alpha)}(x)\frac{z^n}{n!},\\
e^{2xz-z^2}&amp;=\sum_{n=0}^{+\infty}H_n(x)\frac{z^n}{n!},\\
L_{n}^{(\alpha)}(x) &amp;=\sum_{j=0}^n {\alpha+n \choose n-j}(-1)^j\frac{x^j}{j!}.
\end{align*}
Define, for non-negative integers $n$,  $K_{n}^{(\alpha)}(x) =\frac{n!}{(\alpha+n)_n}{x^{n}L_{n}^{(\alpha)}\left(\frac{-1}{x}\right)}$.
It is not difficult to see that the polynomial sequences $\left( B_{n}^{(\alpha )}(x)\right)_{n\geq 0}$, 
$\left(E_{n}^{(\alpha )}(x)\right)_{n\geq 0}$, 
$\left( \frac{1}{2^n}H_{n}(x)\right)_{n\geq 0}$, and $\left( K_{n}^{(\alpha)}(x)\right) _{n\geq 0}$ are the Appell
polynomial sequences associated, respectively, with the formal series $\left(\frac{z}{e^z-1}\right)^\alpha$, $\left(\frac{2}{e^z+1}\right)^\alpha$, $e^{\frac{-z^2}{4}}$, and $\sum_{n=0}^{+\infty }\frac{1}{(\alpha + n)_n} \frac{z^{n}}{n!}$. 
Applying Corollary \ref{co3}, we get,  for all complex numbers $\alpha$ and $\lambda$ and all non-negative integers $r$ and $n$ the following identities: 
\begin{align*}
\sum_{k=0}^{n}{n \choose k}\frac{\lambda ^{n+r+1-k}B^{(\alpha)}_{k}(x)}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{j}(r)_{j}\lambda^{r-j}\frac{B^{(\alpha)}_{n+j+1}(x+\lambda )}{(n+j+1)_{j+1}}
 +(-1)^{r+1}\frac{r!\ B^{(\alpha)}_{n+r+1}(x)}{(n+r+1)_{r+1}}, \\ 
\sum_{k=0}^{n}{n \choose k}\frac{\lambda^{n+r+1-k}E^{(\alpha)}_{k}(x)}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{j}(r)_j \lambda^{r-j}\frac{E^{(\alpha)}_{n+j+1}(x+\lambda )}{(n+j+1)_{j+1}}
+(-1)^{r+1}\frac{r!E^{(\alpha)}_{n+r+1}(x)}{(n+r+1)_{r+1}},  \\
\sum_{k=0}^{n}{n \choose k}\frac{(2\lambda) ^{n+r+1-k}H_{k}(x)}{(n+r+1-k)}
=\sum_{j=0}^{r}(-1)^j (r)_j(2\lambda)^{r-j}\frac{H_{n+j+1}(x+\lambda)}{(n+j+1)_{j+1}} +(-1)^{r+1}\frac{r!\ H_{n+r+1}(x)}{(n+r+1)_{r+1}}.
\end{align*}
\begin{multline*}
\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{\alpha +k}{k}^{-1}\frac{(\lambda x)^{n+r+1-k}L_{k}^{(\alpha )}(x)}{n!(n+r+1-k)}\\
=\sum_{j=0}^{r}(-1)^{j}\frac{(r)_{j}(\lambda x)^{r-j}(1+x\lambda )^{n+j+1}L_{n+j+1}^{(\alpha )}\left(\frac{x}{1+x\lambda}\right) }
{(\alpha +n+j+1)_{n+j+1}}+(-1)^{r+1}r!\frac{L_{n+r+1}^{(\alpha )}(x) }{(\alpha+n+r+1)_{n+r+1}}.
\end{multline*}
\section{Proofs}
The proof of Theorem~\ref{th1} is mainly based  on the following lemma.
\begin{lemma}
For all non-negative integers $n$, $r$, we have
\begin{equation}
\sum_{k=0}^{n}(-1)^{k}{n \choose k}\frac{x^{k}}{n+r+1-k}
=\sum_{j=0}^{r}(-1)^{j}(r)_{j}\frac{(1-x)^{n+j+1}}{(n+j+1)_{j+1}}+(-1)^{n}r!\frac{x^{n+r+1}}{(n+r+1)_{r+1}}.  \label{s9}
\end{equation}
\end{lemma}

\begin{proof}
Let $r$, $n$ be non-negative integers, $P(x)=x^{r}$, $Q(x)=\frac{(1+x)^{n+r}}{(n+r)_{r}}$, and 
\[
I(x)=\int_{0}^{x}P(t)Q^{(r)}(t)dt.
\] 
By a direct computation, we get 
\begin{equation}
I(x)=\int_{0}^{x}t^{r}(1+t) ^{n}dt=\sum_{k=0}^{n}{n \choose k}\frac{x^{n+r+1-k}}{n+r+1-k}.  \label{n1}
\end{equation}
Using generalized integration by parts, we have
\begin{align*}
I(x) &amp;=\left[ \sum_{j=0}^{r-1}(-1)^{j}P^{(j)}(t)Q^{(r-j-1)}(t)\right]
_{0}^{x}+(-1)^{r}\int_{0}^{x}P^{(r)}(t)Q(t)dt  \\
&amp;=\sum_{j=0}^{r-1}(-1)^{j}(r)_{j}x^{r-j}\frac{(1+x)^{n+j+1}}{(n+j+1) _{j+1}}+(-1)^{r}r!\frac{(1+x)^{n+r+1}}{(n+r+1)_{r+1}} +\frac{(-1)^{r+1}r!}{(n+r+1) _{r+1}}.  
\end{align*}
Hence
\begin{equation}
I(x) =\sum_{j=0}^{r}(-1)^{j}(r)_{j}\frac{x^{r-j}(1+x)^{n+j+1}}{(n+j+1)_{j+1}}+\frac{(-1)^{r+1}r!}{(n+r+1)_{r+1}}.  \label{N2}
\end{equation}
According to relation~\eqref{n1}, it is obvious that $I(x)$ is a polynomial of degree 
$n+r+1$. We  let $J(x)$ denote  the reciprocal polynomial of $I(x)$. We have $J(x)=x^{n+r+1}I\left( \frac{1}{x}\right)$. Using the two expressions for $I(x)$ obtained in relations~\eqref{n1} and \eqref{N2}, we get two expressions for $J(-x)$.
By equating both expressions of $J(-x)$, we obtain relation~\eqref{s9}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{th1}]
Let $u$ be a complex sequence.
Consider the linear function $L_{u}$ from $\mathbb{C}[x] $ to 
$\mathbb{C}^{\mathbb{N}}$ defined for all $n \geq 0$ by $L_{u}(x^{n})=u_{n}$. 
Note that for all $n\geq 0$, we have $L_{u}\left((1-x)^{n}\right) =u_{n}^*$.
Applying $L_{u}$ to both sides of relation~\eqref{s9}, we get relation~\eqref{n10}. The proof of Theorem~\ref{th1} is then complete.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{co2}]
It can be clearly seen that relations~\eqref{p1} and \eqref{p2} are 
satisfied in the case where $b=0$. We can assume that $b\neq 0$. We note that for $n\geq 0$, $x^{2}-(ax+b)$ divides $x^{2n}-(ax+b)^{n}$. Then $x^{2n}-(ax+b) ^{n}$ is a characteristic polynomial of $u$. Hence $L_{u}\left( x^{2n}-(ax+b)^{n}\right) =0$. We conclude that 
\[
u_{2n}=\sum_{k=0}^{n}{n \choose k}b^{n-k}a^{k}u_{k}.
\]
Thus $\left( \left( -\frac{a}{b}\right) ^{n}u_{n}\right) ^*=\frac{u_{2n}}{b^{n}}$ 
and $\left( \frac{u_{2n}}{b^{n}}\right)^*=\left(-\frac{a}{b}\right)^{n}u_{n}$. By applying Theorem~\ref{th1} to the sequences 
$\left(\left( -\frac{a}{b}\right) ^{n}u_{n}\right) _{n\geq 0}$ and $\left( \frac{u_{2n}}{b^{n}}\right) _{n\geq 0}$ 
we obtain relations~\eqref{p1} and \eqref{p2}. To prove relations~\eqref{p3} and \eqref{p4}, consider the two sequences $v$ and $w$ defined by $v_{n}=\frac{u_{n}}{a^{n-1}}$ and $w_{n}=\frac{u_{n}}{a^{n}}$ $(n\geq 0)$. 
We have $( v_{0}^*,v_{1}^*)=(-v_{0},-v_{1}) $ and $(w_{0}^*,w_{1}^*)=(w_{0},w_{1})$. Moreover, the sequences $v$, $v^*$, $w$, and $w^*$ have the
same characteristic polynomial, i.e., $x^{2}-x-\frac{b}{a^{2}}$. Then $v$ is inverse invariant under the binomial transformation and $w$ is invariant under the binomial transformation. Finally, by applying Theorem~\ref{th1} to the sequences $v$ and $w$ we get relations~\eqref{p3} and \eqref{p4}.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{co3}]
Let $\left( A_{n}(x)\right) _{n}$ be an Appell polynomial sequence defined by 
\[
\sum_{n=0}^{\infty }A_{n}(x)\frac{z^{n}}{n!}=S(z)e^{xz}
\]
and $\lambda $ a complex number. Obviously, for $\lambda =0$, the relation~\eqref{p9} is satisfied. Suppose now that $\lambda \neq 0$, we consider the sequence $u$ defined for a fixed complex number $x$ 
by $u_{n}=(-1)^{n}\frac{A_{n}(x)}{\lambda ^{n}}$. We have 
\begin{align*}
\sum_{n=0}^{\infty }\lambda ^{n}u_{n}^*\frac{z^{n}}{n!} =&amp;\left(\sum_{n=0}^{\infty }\lambda ^{n}\frac{z^{n}}{n!}\right) \left(\sum_{n=0}^{\infty }A_{n}(x)\frac{z^{n}}{n!}\right)  \\
=&amp;S(z)e^{( \lambda +x) z}=\sum_{n=0}^{\infty }A_{n}(x+\lambda )
\frac{z^{n}}{n!}.
\end{align*}
It follows that $u_{n}^*=\frac{A_{n}(x+\lambda )}{\lambda ^{n}}$. 
By application of Theorem~\ref{th1} to the sequence $u$,
we get relation~\eqref{p9}.
\end{proof}

\section{Acknowledgment}
The authors express their gratitude to the anonymous referee for his careful reading, kind words of
praise, and  constructive suggestions,
which  improved the quality of the paper. 

\begin{thebibliography}{4}

\bibitem{Apostol:2008}
T.~M.~Apostol, A primer on {B}ernoulli numbers and polynomials, {\em Math.
  Mag.} {\bf 81} (2008), 178--190.

\bibitem{Appell:1880}
P.~Appell, Sur une classe de polyn\^omes, {\em Ann. Sci. \'Ecole Norm. Sup.
  (2)} {\bf 9} (1880), 119--144.

\bibitem{oeis}
N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences,
2018, available at \url{https://oeis.org}.

\bibitem{Sun:2001}
Z.-H.~Sun, Invariant sequences under binomial transformation, {\em Fibonacci
  Quart.} {\bf 39} (2001), 324--333.

\bibitem{Sun:2003}
Z.-W.~Sun, Combinatorial identities in dual sequences, {\em European J.
  Combin.} {\bf 24} (2003), 709--718.

\end{thebibliography}

\bigskip
\hrule
\bigskip


\noindent 2010 {\it Mathematics Subject Classification}: 
Primary 11B39; Secondary 11B68, 11B50.

\noindent \emph{Keywords:} Binomial transformation, Bernoulli number, Appell polynomial sequence.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000110},
\seqnum{A000129},
\seqnum{A001045},
\seqnum{A002203}, and
\seqnum{A014551}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 4 2017;
revised versions received April 1 2018; April 2 2018.
Published in {\it Journal of Integer Sequences}, May 9 2018.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}
</pre></body></html>