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\begin{center}
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{\LARGE\bf On the Expansion of $(XV)^n$ and Bessel Numbers \\
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} 
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\large
Abdelhay Benmoussa\\
Secteur Scolaire Ansis, Unit\'e Alanfoukt\\
Tiqqi 80324, Agadir Ida Outanane\\
Morocco\\
\href{mailto:abdelhay.benmoussa@taalim.ma}{\tt abdelhay.benmoussa@taalim.ma}
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\begin{abstract}
We expand the operator \((XV)^n\), where \(X\) is multiplication by \(x\) and \(V\) is the Volterra operator. The resulting coefficients are shown to be the Bessel numbers (OEIS \seqnum{A001498}). We also present two applications and a conjectural generalization.
\end{abstract}

\section{Introduction}

Let
\[
(Xf)(x)=xf(x), \qquad D=\frac{d}{dx}.
\]
A well-known result in combinatorics states that \cite{MSS}
\[
(XD)^n=
\sum_{k=0}^n
S(n,k)
X^kD^k,
\]
where
\(
S(n,k)
\)
denote the Stirling numbers of the second kind (OEIS \seqnum{A008277}).

In this note, we investigate an alternative setting obtained by replacing the derivative operator with the Volterra operator
\[
V (f)(x) = \int_0^x f(t)\,dt.
\]
We show that the expansion of the operator $(XV)^n$, defined by
\[
(XV)^n(f)(x) :=
\begin{cases}
f(x), & \mbox{if}\ n = 0; \\[4pt]
x \displaystyle\int_0^x (XV)^{\,n-1}(f)(t)\, dt, & \mbox{if }\ n \ge 1,
\end{cases}
\]
leads naturally to an explicit formula whose coefficients are the Bessel numbers (OEIS \seqnum{A001498}), given by
\[
a(n,k)=[x^k]\,y_n(x)=\frac{(n+k)!}{2^k\,k!\,(n-k)!},
\]
where \(y_n(x)\) is the $n$-th Bessel polynomial \cite{Grosswald}. Then we present two applications, and conclude with a conjecture that generalizes the obtained formula.

\section{Main result}
\begin{theorem}\label{th}
Let \(n\ge1\). Then
\begin{equation}\label{main}
(XV)^n = \sum_{k=0}^{n-1} (-1)^k\, a(n-1,k)\, X^{\,n-k} V^{\,n+k}.
\end{equation}
\end{theorem}
\begin{proof}
Applying \((XV)^n\) to a function \(f\) gives 
\begin{align*}
(XV)^n (f)(x)
&=x \int_0^x x_{n-1} \int_0^{x_{n-1}} \cdots x_1 \int_0^{x_1} f(t)\, dt \, dx_1 \cdots dx_{n-1} \\
&=x \int_0^x f(t) 
\left( \iiint_{t \le x_1 \le \cdots \le x_{n-1} \le x} x_1 x_2 \cdots x_{n-1} \, dx_1 \cdots dx_{n-1} \right) dt.
\end{align*}
Now, by induction we have
\[\iiint_{t \le x_1 \le \cdots \le x_{n-1} \le x} x_1 x_2 \cdots x_{n-1} \, dx_1 \cdots dx_{n-1} =
\frac{(x^2 - t^2)^{n-1}}{2^{\,n-1}(n-1)!}.
\]
Thus
\begin{align*}
(XV)^n (f)(x)&=\frac{x}{2^{\,n-1}(n-1)!}\int_{0}^{x}
f(t)
(x^2 - t^2)^{n-1}\, dt\\
&=\frac{x}{2^{\,n-1}(n-1)!}
\int_{0}^{x}
f(t)(x-t)^{n-1}(x+t)^{n-1}
\, dt \\
&=
\frac{x}{2^{\,n-1}(n-1)!}
\int_{0}^{x}
f(t)(x-t)^{n-1}
\sum_{k=0}^{n-1}
\binom{n-1}{k}
(t-x)^k (2x)^{\,n-1-k}
\, dt 
\end{align*}
\begin{align*}
&=
\frac{x}{2^{\,n-1}(n-1)!}
\sum_{k=0}^{n-1}
(-1)^k
\binom{n-1}{k}
(2x)^{\,n-1-k}
\int_{0}^{x}
f(t)(x-t)^{n+k-1}
\, dt \\
&=\sum_{k=0}^{n-1}
(-1)^k\frac{1}{2^k(n-1-k)!k!}
x^{\,n-k}
\int_{0}^{x}
f(t)(x-t)^{n+k-1}
\, dt .
\end{align*}
Applying Cauchy's formula for repeated integration, we obtain
\begin{align*}
(XV)^n (f)(x)&=
\sum_{k=0}^{n-1}
(-1)^k\frac{(n+k-1)!}{2^k(n-1-k)!k!}
x^{\,n-k}
\, V^{n+k}(f)(x)\\
&=
\sum_{k=0}^{n-1}
(-1)^k
a(n-1,k)
\,x^{\,n-k}
V^{n+k}(f)(x)\\
&=
\left(\sum_{k=0}^{n-1}
(-1)^k
a(n-1,k)
\,X^{\,n-k}
V^{n+k}\right)(f)(x).
\end{align*}
\end{proof}

\section{Two applications}

\subsection{Power functions}

Let \(\alpha>0\). We first compute
\begin{align*}
(XV)^{n+1}(t^{\alpha-1})(x)
&=x\int_0^x  \frac{(x^2 - t^2)^n}{2^n n!}t^{\alpha-1} \, dt \\
&= \frac{x}{2^n n!} \int_0^1 \big(x^2(1-u)\big)^n
   \big(x u^{1/2}\big)^{\alpha-1}\frac{x}{2}u^{-1/2}\,du
   \qquad (t=x\sqrt{u})\\
&= \frac{x}{2^n n!}\cdot \frac{x}{2}\, x^{2n} x^{\alpha-1}
   \int_0^1 (1-u)^n u^{(\alpha/2)-1}\,du \\
&= \frac{x^{\alpha+2n+1} }{2^{n+1}n!}B(\alpha/2,n+1)\\   
&= \frac{x^{\alpha+2n+1} \Gamma(\alpha/2)}{2^{n+1} \, \Gamma(\alpha/2+n+1)}\\
&=\frac{x^{\alpha+2n+1}}{\alpha(\alpha\,+2)\cdots(\alpha+2n)},
\end{align*}
where \(B(x,y)\) is the Beta function \cite{betagamma}.

On the other hand,
\begin{align*}
(XV)^{n+1}(t^{\alpha-1})(x)
&=\sum_{k=0}^{n} (-1)^k a(n,k)\,
   x^{n+1-k}\,V^{n+1+k}(t^{\alpha-1})(x)\\
&= \sum_{k=0}^{n} (-1)^k a(n,k)\,
   x^{n+1-k} \int_0^x \frac{(x-t)^{n+k}}{(n+k)!} t^{\alpha-1}\,dt \\
&= \sum_{k=0}^{n} \frac{(-1)^k a(n,k)}{(n+k)!}\,
   x^{n+1-k} \int_0^1 x^{n+k}(1-u)^{n+k}
   x^{\alpha-1}u^{\alpha-1}\,x\,du
   \qquad (t=xu)\\
&= \sum_{k=0}^{n} \frac{(-1)^k a(n,k)}{(n+k)!}\,
   x^{2n+\alpha+1}
   \int_0^1 (1-u)^{n+k}u^{\alpha-1}\,du \\
&= x^{2n+\alpha+1}\sum_{k=0}^{n}\,
   \frac{(-1)^k a(n,k)}{(n+k)!}B(\alpha,n+k+1)\\   
&= x^{2n+\alpha+1}\sum_{k=0}^{n}\,
   \frac{(-1)^k a(n,k)\Gamma(\alpha)}{\Gamma(\alpha+n+k+1)}\\
&= x^{\alpha+2n+1}\,
\sum_{k=0}^{n}\frac{(-1)^k\,a(n,k)}{\alpha(\alpha+1)\cdots(\alpha+n+k)}.
\end{align*}

By comparing the two expressions, we obtain the generating function
\begin{equation}
\frac{1}{\alpha(\alpha\,+2)\cdots(\alpha+2n)}=\sum_{k=0}^{n}\frac{(-1)^k\,a(n,k)}{\alpha(\alpha+1)\cdots(\alpha+n+k)}.
\end{equation}


\subsection{Exponential function}

Next, we consider the exponential function. Using Formula \label{main}, we obtain
\begin{align*}
(XV)^{n+1}(e^t)(x)
&=\sum_{k=0}^{n} (-1)^k a(n,k)\,
   x^{n+1-k}\,V^{n+1+k}(e^t)(x)\\
&= \sum_{k=0}^{n} (-1)^k a(n,k)\,
   x^{n+1-k} \int_0^x\frac{(x-t)^{n+k}}{(n+k)!}e^t\,dt\\
&= e^x \sum_{k=0}^{n} (-1)^k a(n,k)\, x^{n+1-k} \,\frac{\gamma(n+k+1,x)}{(n+k)!}\\
&= e^x \sum_{k=0}^{n} (-1)^k a(n,k)\, x^{n+1-k} 
   \left(1 - e^{-x}\sum_{j=0}^{n+k} \frac{x^j}{j!}\right) \\
&= e^x \sum_{k=0}^{n} (-1)^k a(n,k)\, x^{\,n+1-k}
   - \sum_{k=0}^{n}\sum_{j=0}^{n+k} (-1)^k a(n,k)\,\frac{x^{\,n+j+1-k}}{j!},
\end{align*}
Here, \(\gamma(x,y)\) is the lower incomplete gamma function \cite{betagamma}. Rosengren proved \cite{RosengrenMO} that
\begin{equation*}
\sum_{k=0}^{n}\sum_{j=0}^{n+k} (-1)^k a(n,k)\,\frac{x^{\,n+j-k}}{j!}=\sum_{k=0}^{n}
(-1)^{\,n-k}\,
\frac{(2(n-k)-1)!!}{(2k)!!}\,
x^{\,2k}.    
\end{equation*} 
Hence
\begin{equation}
(XV)^{n+1}(e^t)(x)= x^{\,n+1} e^{x}\,
y_n\!\left(-\frac1x\right)
-
\sum_{k=0}^{n}
(-1)^{\,n-k}\,
\frac{(2(n-k)-1)!!}{(2k)!!}\,
x^{\,2k+1}.    
\end{equation}

Evaluating at $x=1$ yields the following Dobinski-like identity for the sequence
$a(n)=
y_n(-1)$ (OEIS \seqnum{A000806}):
\begin{equation}
a(n)
=
\frac1e\left(
 (XV)^{n+1}(e^t)(1)
+
\sum_{k=0}^{n}
(-1)^{\,n-k}\frac{(2(n-k)-1)!!}{(2k)!!}
\right).
\end{equation}

\section{Generalization}
Using Cauchy's formula for repeated integration, we obtain
\[
XV-VX = V^2.
\]
Motivated by this identity, consider the general commutation relation \cite{MSS}
\begin{equation}\label{commutrel}
uv - vu = hv^s,
\qquad h \in \mathbb{C} \setminus \{0\},\ s \in \mathbb{R}.    
\end{equation}

We conclude with the following conjecture.

\begin{conjecture}
Let $u$ and $v$ be two variables satisfying \eqref{commutrel}. Then, for any integer $n \ge 1$,
\begin{equation}
(uv)^n
  = \sum_{k=1}^n \mathfrak{S}_{s;h}(n,k)\,
      u^{\,k}\, v^{\,s(n-k)+k},
\end{equation}
where $\mathfrak{S}_{s;h}(n,k)$ are the generalized Stirling numbers introduced by Mansour and Schork \cite{MSS}.    
\end{conjecture}

\section{Acknowledgments}
The author thanks the referee and the editor for their valuable comments, which improved the manuscript.

\begin{thebibliography}{9}

\bibitem{betagamma}
M. Abramowitz and I. A. Stegun, editors. 
\newblock \textit{Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing}. New York: Dover.

\bibitem{Grosswald} E. Grosswald. 
\newblock \textit{Bessel Polynomials}, Springer, 1978.

\bibitem{MSS}
T. Mansour, M. Schork and M. Shattuck.
\newblock The generalized Stirling and Bell numbers revisited.
\newblock \textit{J. Integer
Sequences} \textbf{15} (2012), \href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Schork/schork2.html}{Article 12.8.3}.

\bibitem{RosengrenMO}
H. Rosengren. 
\newblock
Proving that a certain factorial double sum collapses to a double--factorial series. 
\newblock 
MathOverflow posting (2025). Available at 
\url{https://mathoverflow.net/questions/499957}.

\end{thebibliography}

\bigskip
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\noindent 2020 \textit{Mathematics Subject Classification}: Primary 11B83; Secondary 11C08, 26A36.

\noindent \emph{Keywords:} Bessel number, Volterra operator.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A000806},
\seqnum{A001497},
\seqnum{A001498},
\seqnum{A008277},
\seqnum{A122848}, and
\seqnum{A122850}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received July 26 2025;
revised versions received July 30 2025; July 31 2025; December 6 2025;
December 7 2025; May 19 2026.  Published in
{\it Journal of Integer Sequences}, May 20 2026.

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

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