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\begin{center}
\vskip 1cm{\LARGE\bf A New Identity for Appell Polynomials
}
\vskip 1cm
\large
Farid Bencherif\\
LA3C, Faculty of Mathematics\\
USTHB\\
Algiers\\
Algeria\\
\href{mailto:fbencherif@usthb.dz}{\texttt{fbencherif@usthb.dz}}\\
\ \\  
Nouara Mokhtari\footnote{
Corresponding author.} and Schehrazade Zerroukhat \\
LATN, Faculty of Mathematics\\
USTHB\\
Algiers\\
Algeria\\
\href{mailto:nmokhtari1@usthb.dz}{\texttt{nmokhtari1@usthb.dz}}\\
\href{mailto:mokhtari.nouara@gmail.com}{\texttt{mokhtari.nouara@gmail.com}} 
\\
\href{mailto:szerroukhat@usthb.dz}{\texttt{szerroukhat@usthb.dz}}\\
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\vskip .2 in

\begin{abstract}
In the present paper, we generalize an identity for some Appell
polynomials, from which we deduce many explicit formulas for generalized
Bernoulli and Euler numbers and polynomials.
\end{abstract}

\section{Introduction and main result}

The classical and generalized Bernoulli numbers and polynomials as well as the classical and generalized Euler numbers and polynomials have been extensively studied \cite{gou, luo1}. In the present paper, we attempt to improve some results related to Bernoulli and Euler polynomials and extend a generalization of the 
Appell polynomials identity published in \cite{ben}.
Let $\mathbb{Z}$, $\mathbb{N}$ and $\mathbb{C}$ denote the set of integers, non-negative integers and complex numbers respectively. A polynomial sequence $(A_{n}(x))_{n\in \mathbb{N}}$ of $\mathbb{C}[x]$ is an \textit{Appell polynomial sequence\/} \cite{app} if $A_{0}(x)$ is a non-zero constant polynomial and $A_{n}^{\prime}(x)=nA_{n-1}(x)$ for $n\geq1$. The exponential generating series of the sequence $(A_{n}(x))_{n\in \mathbb{N}}$ can be formulated as follows: 
\begin{equation*}
\sum_{n=0}^{\infty}A_{n}(x)\frac{z^{n}}{n!}=S(z)e^{xz},
\end{equation*}
where $S(z)$ is a formal power series of $\mathbb{C}[[z]]$ with non-zero constant term. We focus now on the following theorem which generalizes \cite[Theorem 1.1]{ben} and adds other properties. 
\begin{theorem}\label{theo}
Let $S(z)=\displaystyle \sum_{n=0}^{\infty}a_{n}\frac{z^{n}}{n!}$ be a formal series of $\mathbb{C}[[z]]$ with $a_{0}=1$.
For every $\alpha \in \mathbb{C}$ and 
$(A_{n}^{(\alpha)}(x))_{n\in \mathbb{N}}$ 
consider the Appell polynomial sequence defined by
\begin{equation}
\sum_{n=0}^{\infty}A_{n}^{(\alpha)}(x)\frac{z^{n}}{n!}=S^{\alpha}(z)e^{xz}.\label{r0}
\end{equation}
Then for all $\lambda$, $\alpha \in \mathbb{C}$, $\ell \in \mathbb{Z}$,
and non-negative integers $n$, $m$, with $m\geq n$, we have
\begin{equation}
A_{n}^{(\alpha \ell )}(x)=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}A_{n}^{(-k\ell)}(x+\lambda \ell (\alpha+k))
\label{r1}
\end{equation}
and for $m\geq \lfloor \frac{n}{2} \rfloor $, we have
\begin{equation}
A_{n}^{(\alpha \ell )}(x)=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}A_{n}^{(-k\ell )}(x+a_{1}\ell (\alpha+k))
.\label{r2}
\end{equation}	
\end{theorem}
By using Relation (\ref{r1}) for $\lambda=0$ and $\ell=1$, we obtain 
\cite[Theorem 1.1]{ben}.

\section{Applications}

In this section, we give some applications of Theorem \ref{theo}. Let
us first recall that several sequences of remarkable polynomials in
$\mathbb{C}[x]$ are Appell polynomial sequences. Indeed, it is clear
that for $\alpha\in \mathbb{C}$, the sequences of generalized Bernoulli
polynomials $(B_{n}^{(\alpha)}(x))_{n\in \mathbb{N}}$ and generalized
Euler polynomials $(E_{n}^{(\alpha )}(x))_{n\in \mathbb{N}}$ defined by
(see \cite{nor, sri}) 
\begin{equation}
\sum_{n=0}^{\infty}B_{n}^{(\alpha)}(x)\frac{z^{n}}{n!}=\left(\frac{z}{e^{z}-1 }\right)^{\alpha}e^{xz}  \label{r3}
\end{equation}
and 
\begin{equation}
\sum_{n=0}^{\infty}E_{n}^{(\alpha )}(x)\frac{z^{n}}{n!}=\left(\frac{2}{e^{z}+1}\right)^{\alpha}e^{xz}  \label{r4}
\end{equation}
are Appell polynomial sequences, associated with the formal series $\left( \frac{z}{e^{z}-1}\right)^{\alpha}$, $\left(\frac {2}{e^{z}+1}
\right)^{\alpha} $ respectively. The classical Bernoulli polynomials $B_{n}(x)$ and classical Euler polynomials $E_{n}(x)$ are defined by $
B_{n}(x) =B_{n}^{(1)}(x)$ and $E_{n}(x)=E_{n}^{(1)}(x)$.

For $k \in \mathbb{N}$, from (\ref{r3}) and (\ref{r4}) we deduce the explicit expressions for $B_{n}^{(-k)}(x)$ and $E_{n}^{(-k)}(x)$: 
\begin{equation*}
B_{n}^{(-k)}(x)=\frac{1}{k!}\dbinom{n+k}{k}^{-1}\sum_{j=0}^{k}(-1)^{k-j} \dbinom{k}{j}(x+j)^{n+k}
\end{equation*}
and 
\begin{equation*}
E_{n}^{(-k)}(x)=\frac{1}{2^{k}}\sum_{j=0}^{k}\dbinom{k}{j}(x+j)^{n}.
\end{equation*}

The Stirling numbers of second kind $S(n,k)$ 
(see sequence \seqnum{A008277} in the \textit{On-Line Encyclopedia of Integer Sequences} (OEIS) \cite{Slo}) are defined by 
\begin{equation*}
\sum_{n=0}^{\infty}S(n,k)\frac{z^{n}}{n!}=\frac{1}{k!}(e^{z}-1)^{k}.
\end{equation*}

The generalized Stirling numbers of second kind $S(n,k,x)$ \cite[Eq.\ (3.9), p.\ 152]{car} are defined by 
\begin{equation*}
\sum_{n=0}^{\infty}S(n,k,x)\frac{z^{n}}{n!}=\frac{1}{k!}e^{xz}(e^{z}-1)^{k}.
\end{equation*}
We have 
\begin{equation*}
\sum_{n=0}^{\infty}S(n,k,x)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}
(x+j)^{n}.
\end{equation*}
and 
\begin{equation*}
S(n,k)=S(n,k,0)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^{n}.
\end{equation*}
by using the generalized Stirling numbers of second kind, we can express $B_{n}^{(-k)}(x)$ as follows: 
\begin{equation}
B_{n}^{(-k)}(x)=\binom{n+k}{k}^{-1}S(n+k,k,x).  \label{r10}
\end{equation}

\begin{corollary}\label{cor1} Let $\alpha,\lambda \in \mathbb{C}$, $\ell \in \mathbb{Z}$ and $n,m\in \mathbb{N}$. For $m\geq n$, we have
\begin{equation}
B_{n}^{(\alpha \ell)}(x)=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}B_{n}^{(-k\ell)}(x+\lambda \ell(\alpha+k)),\label{r11}
\end{equation}
and for $m\geq \left \lfloor \frac{n}{2}\right \rfloor$, we have
\begin{equation}
B_{n}^{(\alpha \ell )}(x)=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}B_{n}^{(-k\ell )}(x-\frac{\ell}{2}(\alpha+k)). \label{r12}
\end{equation}
\end{corollary}

\begin{corollary}\label{cor2} Let $\alpha,\lambda \in \mathbb{C}$, $\ell \in \mathbb{Z}$, $n,m\in \mathbb{N}$. For $m\geq n$, we have
\begin{equation*}
E_{n}^{(\alpha \ell )}(x)=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}E_{n}^{(-k\ell )}(x+\lambda \ell (\alpha+k)),
\end{equation*}
and for $m\geq \left \lfloor \frac{n}{2}\right \rfloor$, we have
\begin{equation}
E_{n}^{(\alpha \ell )}(x)=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}E_{n}^{(-k\ell )}(x-\frac{\ell}{2}(\alpha+k)).\label{r14}
\end{equation}	
\end{corollary}

Corollaries \ref{cor1} and \ref{cor2} are deduced from Theorem \ref{theo} by choosing successively $S(z)=\frac{z}{e^{z}-1}$ then $S(z)=\frac{2}{e^{z}+1}$ and using Definitions (\ref{r3}) and (\ref{r4}). The generalized Bernoulli numbers $B_{n}^{(\alpha)}$ and Euler numbers $E_{n}^{(\alpha)}$ are defined by $B_{n}^{(\alpha )}=B_{n}^{(\alpha)}(0)$ and $E_{n}^{(\alpha)}=2^{n}E_{n}^{(\alpha )}(\frac{\alpha}{2})$.
The classical Bernoulli numbers $B_{n}$ and classical Euler numbers $E_{n}$ \seqnum{A122045} are defined by $B_{n}=B_{n}^{(1)}=B_{n}^{(1)}(0)$ and $E_{n}=E_{n}^{(1)}=2^{n}E_{n}(\frac{1}{2})$. We also define the polynomials $\widehat{B}_{n}^{(\alpha )}$ by $\widehat{B}_{n}^{(\alpha)}=B_{n}^{(\alpha)}(\alpha/2)$ \cite[p.\ 259]{wan}.

\begin{corollary}
Let $\ell,n,m\in \mathbb{N} $. For $m\geq n$, we have
\begin{equation}
B_{n}^{(\ell )}(x)=\sum_{k=0}^{m}(-1)^{k}\frac{\dbinom{m+1}{k+1}}{\dbinom{n+k\ell}{n}}\, S(n+k\ell,k\ell,x)  .\label{r5}
\end{equation}	
\end{corollary}

This follows by using Relations (\ref{r10}) and (\ref{r11}) for $\alpha=1$.

Replacing $x=0$ and $m=n$ in Relation (\ref{r5}) we obtain the following relation which can be found in \cite[p.\ 60]{jeo}: 
\begin{equation*}
B_{n}^{(\ell )}=\sum_{k=0}^{n}(-1)^{k}\frac{\dbinom{n+1}{k+1}}{\dbinom{
n+k\ell}{n}}S(n+k\ell,k\ell).
\end{equation*}
Replacing $x=0$, $m=n$, and $\ell=1$ in Relation (\ref{r5}) it follows that
\begin{equation*}
B_{n}=\sum_{k=0}^{n}(-1)^{k}\frac{\dbinom{n+1}{k+1}}{\dbinom{n+k}{n}}
S(n+k,k).
\end{equation*}
Note that this formula has been proven by numerous authors:  see, for
example, \cite[Eq.\ (11), p.\ 48]{gou}, \cite[Eq.\ (6), p.\ 27]{guo}, \cite[p.\ 59]{jeo}, \cite[p.\ 219 ]{jor}, \cite[Eq.\ (1.3), p.\ 91]{qi}, \cite[p.\ 140]{shi}.

\begin{corollary}
Let $\alpha \in \mathbb{C}$, $\ell \in \mathbb{Z}$ and $n,m\in \mathbb{N}$. For $m\geq \left \lfloor \frac{n}{2}\right \rfloor $, the following hold:
\begin{equation}
\widehat{B}_{n}^{(\ell \alpha)}=\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}\widehat{B}_{n}^{(-k\ell)},\label{r6}
\end{equation}
and
\begin{equation*}
\widehat{B}_{2n}^{(\ell \alpha)}=\sum_{k=0}^{n}(-1)^{k}\dbinom{\alpha+n}{n-k}\dbinom{\alpha+k-1}{k}\widehat{B}_{2n}^{(-k\ell)}.
\end{equation*}
\end{corollary}

This follows immediately from Relation (\ref{r12}) for $x=\frac{1}{2}\alpha
\ell$. For $\ell=1$ and $m=n$, Relation (\ref{r6}) can be written as 
\begin{equation}
\widehat{B}_{n}^{(\alpha )}=\sum_{k=0}^{n}(-1)^{k}\dbinom{\alpha+n}{n-k} 
\dbinom{\alpha+k-1}{k}\widehat{B}_{n}^{(-k)}.  \label{r13}
\end{equation}
Note that (\ref{r13}) is exactly Relation (6.12)
of \cite[Theorem 6.3]{wan}. 
\begin{corollary}
For $\alpha \in \mathbb{C} $ , $\ell \in \mathbb{N} $, $n$, $m\in \mathbb{N}$, with $m\geq \left \lfloor \frac{n}{2}\right \rfloor$, we have
\begin{equation}
E_{n}^{(\alpha \ell)}=\sum_{k=0}^{m}\frac{(-1)^{k}}{2^{k\ell}}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}\sum_{j=0}^{k\ell
}\dbinom{k\ell}{j}(k\ell-2j)^{n},\label{r7}
\end{equation}
\begin{equation*}
E_{2n}^{(\alpha \ell )}=\sum_{k=0}^{n}\frac{(-1)^{k}}{2^{k\ell}}\dbinom{\alpha+n}{n-k}\dbinom{\alpha+k-1}{k}\sum_{j=0}^{k\ell}\dbinom{k\ell}{j}(k\ell-2j)^{2n},
\end{equation*}
\begin{equation*}
E_{2n}^{(\ell)}=\sum_{k=0}^{n}\frac{(-1)^{k}}{2^{k\ell}}\dbinom{n+1}{k+1}\sum_{j=0}^{k\ell}\binom{k\ell}{j}(k\ell-2j)^{2n}.
\end{equation*}	
\end{corollary}

This follows immediately from Relation (\ref{r14}) for $x=\frac{1}{2}\alpha\ell$, and by noticing that $E_{n}^{(\alpha \ell )}=0$ for $n$ odd. For $m=n$ and $\ell=1$ the relation (\ref{r7}) can be written as follows: 
\begin{equation}
E_{n}^{(\alpha )}=\sum_{k=0}^{n}\frac{(-1)^{k}}{2^{k}}\dbinom{\alpha+n}{n-k} 
\dbinom{\alpha+k-1}{k}\sum_{j=0}^{k}\dbinom {k}{j}(k-2j)^{n}.  \label{r8}
\end{equation}
We notice that (\ref{r8}) is nothing else than the relation obtained by Luo \cite{luo1}.

\section{Proof of Theorem \protect \ref{theo}}

The following lemma, proved in \cite{ben}, will be useful for the proof of the main result. 
\begin{lemma}\label{lem1}For $m$, $q\in \mathbb{N}$, we have
\begin{equation*}
1=\sum_{k=0}^{m}(-1)^{k}\dbinom{q+m}{m-k}\dbinom{q+k-1}{k}x^{q+k}-(-1)^{m}\dbinom{q+m}{q}\sum_{k=1}^{q}\frac{k}{m+k}\dbinom{q}{k}(x-1)^{m+k}.
\end{equation*}	
\end{lemma}

Let us consider the formal series $S(z)=1+\sum_{n=1}^{\infty}a_{n}\frac{z^{n} }{n!}$. For every $\alpha \in \mathbb{C}$, let $(A_{n}^{(
\alpha)}(x))_{n\in \mathbb{N}}$ be the Appell polynomial sequence defined by (\ref{r0}). For non-negative integers $n$, $m$, a given integer $\ell$ and for every $\lambda \in \mathbb{C}$, we consider the polynomial 
\begin{equation*}
P_{\lambda}(x,\alpha)=A_{n}^{(\alpha \ell)}(x)-\sum_{k=0}^{m}(-1)^{k}\dbinom{\alpha+m}{m-k}\dbinom{\alpha+k-1}{k}A_{n}^{(-k\ell)}( x+\lambda \ell
(\alpha+k)).
\end{equation*}
Let $n_{\lambda}=n$ if $\lambda \neq a_{1} $ and $n_{\lambda}= \lfloor \frac{n}{2} \rfloor $ if $\lambda=a_{1}$. To prove our theorem it is equivalent to show that $P_{\lambda}(x,\alpha)=0$ for $m\geq n_{\lambda}$. For this, since $P_{\lambda}(x,\alpha)$ is a polynomial in $x$ and $\alpha$, this amounts to proving that $P_{\lambda}(x,q)=0$ for every non-negative integer $q$, provided that $m\geq n_{\lambda}$.

Let $D$ denote the derivation operator of $\mathbb{C}[x]$. In the commutative $\mathbb{C}$-algebra $\mathbb{C}[[D]]$ of operators of
composition of $\mathbb{C}[x]$, consider the operator of translation $T_{\beta \text{ }}=\exp(\beta D)=\sum_{n=0}^{\infty}\beta^{n}\frac{D^{n}}{n!}$ for $\beta \in \mathbb{C}$ and the automorphism $\Omega=S(D)=1+\sum_{n=1}^{\infty}a_{n}\frac{D^{n}}{n!}$ \cite[p.\ 200]{rob}, we can write $T_{\beta}(x^{n})=(x+\beta)^{n}$ and $\Omega^{q}(x^{n})=A_{n}^{(q)}(x)$. We remark that for all non-negative integer $n$, $P_{\lambda}(x,q)=\Lambda_{\lambda}(x^{n})$ with 
\begin{equation*}
\Lambda_{\lambda}=\Omega^{q\ell}-\sum_{k=0}^{m}(-1)^{k}\dbinom{q+m}{m-k} 
\dbinom{q+k-1}{k}\Omega^{-k\ell}\circ T_{\lambda \ell(q+k)}.
\end{equation*}
Then, to get the desired result we must show that $\Lambda_{\lambda}(x^{n})=0
$ for $m\geq n_{\lambda}$. We can write 
\begin{equation}
\Lambda_{\lambda}=\Omega^{q\ell}\circ \Psi_{\lambda}  \label{r9}
\end{equation}
with 
\begin{equation*}
\Psi_{\lambda}=1-\sum_{k=0}^{m}(-1)^{k}\dbinom{q+m}{m-k}\dbinom{q+k-1}{k}
(\Omega^{-\ell}\circ T_{\lambda}^{\ell})^{q+k}.
\end{equation*}
Lemma \ref{lem1} allows us to express $\Psi_{\lambda}$ as follows: 
\begin{equation*}
\Psi_{\lambda}=(-1)^{m+1}\dbinom{q+m}{q}\sum_{k=1}^{q}\frac{k}{m+k}\dbinom{q 
}{k}(\Omega^{-\ell}\circ T_{\lambda}^{\ell}-1)^{m+k}.
\end{equation*}

Let $T$ be a nonzero composition operator. We can write $T = \sum_{j\geq i}a_{j}D^j$ with a first nonzero coefficient $a_i \neq 0$. In this case, we say that the composition operator $T$ has order $i$. Noting then that for every $\lambda,$ $\ord(\Omega^{-\ell}\circ T_{\lambda}^{\ell}-1)\geq1$, and for the particular case $\lambda=a_{1}$, we have $\ord(\Omega^{-\ell}\circ T_{\lambda}^{\ell}-1) \geq2$, we deduce (for $k\geq 1$) that  $\ord(\Omega^{-\ell}\circ T_{\lambda }^{\ell}-1)^{m+k}>m$ and $\ord(\Omega^{-\ell}\circ T_{\lambda}^{\ell}-1)^{m+k}>2m$ in the special case where $\lambda=a_{1}$. Thus for every integer $m\geq n_{\lambda}$, we have $
\Psi_{\lambda}(x^{n})=0$. Using (\ref{r9}), we obtain the equality $\Lambda _{\lambda}(x^{n})=0$ for $m\ge n_{\lambda}$, which completes the
proof.

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\bibitem{luo1} Q.-M. Luo, An explicit formula for the Euler numbers of
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\end{thebibliography}

\bigskip 
\hrule
\bigskip

\noindent 2010 \textit{Mathematics Subject Classification}: Primary 11B68;
Secondary 05A10.

\noindent \emph{Keywords:} Appell polynomial, Bernoulli number, Bernoulli
polynomial, Euler number, Euler polynomial.

\bigskip 
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A008277} and
\seqnum{A122045}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 21 2022; 
revised versions received  June 30 2022; September 16 2022; September 23 2022.
Published in {\it Journal of Integer Sequences}, October 8 2022.

\bigskip
\hrule
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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