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\begin{center}
\vskip 1cm{\LARGE\bf
Bivariate Extension of Bell Polynomials}
\vskip 1cm
\large
Yuanping Zheng and Nadia N. Li\\
School of Mathematics and Statistics\\
Zhoukou Normal University\\
Zhoukou, Henan \\
P.~R.~China\\
\href{mailto:yuanpingzheng0@sohu.com}{\tt yuanpingzheng0@sohu.com}\\
\href{mailto:lina20190606@outlook.com}{\tt lina20190606@outlook.com}
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\begin{abstract}
In this paper, we study bivariate extensions of Bell polynomials and $r$-Bell
polynomials. Some identities related to the $r$-Stirling numbers and
Bell polynomials are presented as special cases.
\end{abstract}

\section{Introduction and motivation}

The Stirling number of the second kind is the number of ways to partition
a set of $n$ objects into $k$ non-empty subsets (cf.\ Comtet \cite[\S5.1]{comtet})
and denoted by $\{\mathop{}_k^n\}$ (see Graham et al.~\cite[\S6.1]{graham}).
The $n$th Bell number is the sum of the Stirling numbers of the second kind and denoted by
\[B_n=\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix},\ n=0,1,\ldots.\]
The corresponding Bell polynomials are
\begin{equation}\label{B(nx)}
B_n(x)=\sum_{k=0}^n\begin{Bmatrix}n\\
k\end{Bmatrix}x^k,\ n=0,1,\ldots.
\end{equation}
Recently, Spivey \cite[Eq.\ 3]{Spivey} discovered the following remarkable formula
\begin{equation}\label{bmn}
B_{m+n}=\sum_{k=0}^m\sum_{j=0}^nj^{n-k}
{n\choose k}\begin{Bmatrix}m\\j\end{Bmatrix}B_k
\end{equation}
and gave a short combinatorial proof.
Later, Gould and Quaintance \cite{Gould} extended this
Bell number to the following Bell polynomial
\begin{equation}\label{bm+nx}B_{m+n}(x)
=\sum_{k=0}^m\sum_{j=0}^nj^{n-k}
{n\choose k}\begin{Bmatrix}m\\j\end{Bmatrix}B_k(x)x^j.
\end{equation}

For any positive integer $r$, Carlitz \cite{carlitz} introduced the $r$-Stirling
number $\{\mathop{}_k^n\}_r$, which is the number of ways to partition a set of $n+r$
objects into $k+r$ non-empty subsets such that the first $r$ elements are in distinct
subsets. He  also found that the $r$-Stirling numbers and the Stirling numbers have
the following relationship
\begin{equation}\label{stirl-co}
   \begin{Bmatrix}n\\m\end{Bmatrix}_r=\sum_{k=0}^n
      {n\choose k}\begin{Bmatrix}k\\m\end{Bmatrix}r^{n-k}.
\end{equation}
Analogous to the definition of the $r$-Stirling number, Mez\H{o} \cite{Mezo2012}
extended the Bell number $B_n$ to the $r$-Bell number
\[B_{n,r}=\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}_r,\ n=0,1,\ldots\]
and derived the following identity
\begin{align*}B_{m+n,r}=\sum_{k=0}^m\sum_{j=0}^n
{(j+r)^{n-k}}{n \choose k}\begin{Bmatrix}m\\j\end{Bmatrix}_rB_k.
\end{align*}

The aim of this paper is to generalize further these identities
to the bivariate case. In the next section, we give the bivariate
extension of Bell polynomials. The bivariate extension of $r$-Bell
polynomials and some examples will be shown in the third section.

\section{Bivariate extension of Bell polynomials}

Firstly, for any positive integer $n$, we define the bivariate
Bell polynomials by
\[B_n(x,y)=\sum_{k=0}^n\begin{Bmatrix}n\\k\end{Bmatrix}(x)_ky^k,\ n=0,1,\ldots,\]
where the falling factorial is defined by $(x)_0=1$ and
\[(x)_k=x(x-1)\cdots(x-k+1)
\quad\text{for}\quad k=1,2,\ldots.\]
Letting ${y\rightarrow\frac{y}{x}}$ and then ${x\rightarrow\infty}$,
we can see that the bivariate Bell polynomial $B_n(x,y)$ reduces
to the univariate Bell polynomial in Eq.~\eqref{B(nx)}.

The exponential generating function of $B_n(x,y)$ is given by the following theorem.
\begin{theorem}\label{funcB(nxy)}
\begin{align*}\sum_{n=0}^{\infty}{B_n(x,y)}\frac{t^n}{n!}
=\{1+y(e^t-1)\}^x.
\end{align*}
\end{theorem}

\begin{proof}
Recalling the generating function \cite[\S5.2]{comtet}
(also see \cite[\S6.3]{graham})
\begin{equation}\label{stirl}
\sum_{n=0}^\infty
\begin{Bmatrix}n\\k\end{Bmatrix}
\frac{x^n}{n!}
=\sum_{n=0}^\infty
S(n,k)\frac{x^n}{n!}
=\frac{(e^x-1)^k}{k!},
\end{equation}
we get
\begin{align*}
\sum_{n=0}^{\infty}{B_n(x,y)}\frac{t^n}{n!}
&=\sum_{n=0}^{\infty}\sum_{k=0}^n
\begin{Bmatrix}n\\k\end{Bmatrix}
(x)_ky^k\frac{t^n}{n!}\\
&=\sum_{k=0}^{\infty}(x)_ky^k
\sum_{n=k}^{\infty}
\begin{Bmatrix}n\\k\end{Bmatrix}
\frac{t^n}{n!}\\
&=\sum_{k=0}^{\infty}
(x)_ky^k\frac{(e^t-1)^k}{k!}\\
&=\sum_{k=0}^{\infty}{x \choose k}\big\{y(e^t-1)\big\}^k.
\end{align*}
By means of the binomial formula \cite[\S 1.6]{comtet}
\begin{equation}\label{binm}
(x+y)^\alpha
=\sum_{k=0}^{\infty}{\alpha\choose k}x^{\alpha-k}y^k,
\end{equation}
we have the desired result.
\end{proof}

If the exponential generating function $f(x)$ of the sequence $\{A_n\}$ is
\begin{equation}\label{ang}
\sum_{n=0}^{\infty}A_n\frac{x^n}{n!}=f(x),
\end{equation}
then it is routine to get the generating function of the double
sequence $\{A_{m+n}\}$
\begin{equation}\label{amng}
\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}
A_{i+j}\frac{x^i}{i!}\frac{y^j}{j!}=f(x+y).
\end{equation}
From this, we can prove the following theorem.
\begin{theorem}\label{SB(mnxy)} Let $m, n\in\mathbb{N}$. We have
\begin{align*}
B_{m+n}(x,y)
=\sum_{k=0}^m\sum_{i=0}^n
k^{n-i}{n\choose i}\begin{Bmatrix}m\\k\end{Bmatrix}
{B_i(x-k,y)}(x)_ky^k.
\end{align*}
\end{theorem}

\begin{proof}
In view of Eqs.~\eqref{ang} and \eqref{amng}, we directly obtain
the following bivariate exponential generating function
of Bell polynomials $\{B_{m+n}(x,y)\}$
\begin{equation}\label{bmnxy}
\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
{B_{m+n}(x,y)}\frac{u^m}{m!}\frac{v^n}{n!}
={\{1+y(e^{u+v}-1)\}^x}.
\end{equation}
The right-hand side of Eq.~\eqref{bmnxy} can be reformulated to
\begin{align*}
{\{1+y(e^{u+v}-1)\}^x}
&={\{1+y(e^v-1)+ye^v(e^u-1)\}}^x\\
&=\sum_{k=0}^{\infty}{x\choose k}
{\{1+y(e^v-1)\}^{x-k}}{\{ye^v(e^u-1)\}^k},
\end{align*}
where we have employed the binomial theorem in Eq.~\eqref{binm}. According
to ${x\choose k}=\frac{(x)_k}{k!}$ and Theorem \ref{funcB(nxy)},
the last expression can be transformed into
\begin{align*}{\{1+y({e^{u+v}}-1)\}}^x
&=\sum_{k=0}^{\infty}\frac{(x)_k}{k!}
\sum_{i=0}^{\infty}{B_i(x-k,y)}
\frac{v^i}{i!}y^ke^{kv}(e^u-1)^k\\
&=\sum_{k=0}^{\infty}(x)_ky^k
\frac{(e^u-1)^k}{k!}
\sum_{i=0}^{\infty}{B_i(x-k,y)}\frac{v^i}{i!}
\sum_{j=0}^{\infty}\frac{k^jv^j}{j!}.
\end{align*}
By means of Eq.~\eqref{stirl}, we have
\begin{align*}
{\{1+y({e^{u+v}}-1)\}}^x
=\sum_{k=0}^{\infty}(x)_ky^k
\sum_{m=k}^{\infty}\begin{Bmatrix}m\\k\end{Bmatrix}
\frac{u^m}{m!}
\sum_{i=0}^{\infty}{B_i(x-k,y)}\frac{v^i}{i!}
\sum_{j=0}^{\infty}\frac{k^jv^j}{j!}.
\end{align*}
Letting $i+j=n$ and then changing the summation order with respect
to $k$ and $m$, the last expression can be reformulated to
\begin{align*}
{\{1+y(e^{u+v}-1)\}}^x
=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}
\frac{u^m}{m!}\frac{v^n}{n!}
\sum_{k=0}^m\sum_{i=0}^n
k^{n-i}(x)_ky^k
{n\choose i}\begin{Bmatrix}m\\k\end{Bmatrix}
{B_i(x-k,y)}.
\end{align*}
Now extracting the coefficient of $\frac{u^m}{m!}\frac{v^n}{n!}$
across Eq.~\eqref{bmnxy}, we get the desired result.
\end{proof}

Replacing $y$ by ${\frac{y}{x}}$ and then letting $x\rightarrow{\infty}$
in Theorem \ref{SB(mnxy)}, we recover Eq.~\eqref{bm+nx} for the Bell
polynomials of single variable.

\section{Bivariate extension of $r$-Bell polynomials}

In this section, we firstly define the following bivariate $r$-Bell polynomials
\begin{align*}
B_{n,r}(x,y)
=\sum_{k=0}^n{\begin{Bmatrix}n\\k\end{Bmatrix}_r}
(x)_ky^k,\ n=0,1,\ldots.
\end{align*}

\begin{theorem}\label{bnrxy} Let $r$ be any positive integer. We have
\begin{align*}
\sum_{n=0}^{\infty}{B_{n,r}(x,y)}\frac{t^n}{n!}
=e^{rt}\{1+y(e^t-1)\}^x.
\end{align*}
\end{theorem}

\begin{proof}
The univariate generating function of $B_{n,r}(x,y)$ can be written as
\begin{align*}
\sum_{n=0}^{\infty}B_{n,r}(x,y)\frac{t^n}{n!}
&=\sum_{n=0}^{\infty}\sum_{k=0}^n
\begin{Bmatrix}n\\k\end{Bmatrix}_r
(x)_ky^k\frac{t^n}{n!}\\
&=\sum_{k=0}^{\infty}(x)_k y^k
\sum_{n=k}^{\infty}\begin{Bmatrix}n\\k\end{Bmatrix}_r
\frac{t^n}{n!}.
\end{align*}
In view of Eq.~\eqref{stirl}, we have
\begin{align*}
\sum_{n=0}^{\infty}B_{n,r}(x,y)\frac{t^n}{n!}
&=\sum_{k=0}^{\infty}(x)_ky^k\frac{e^{rt}{(e^t-1)^k}}{k!}\\
&=e^{rt}\sum_{k=0}^{\infty}{x\choose k}\{y(e^t-1)\}^k.
\end{align*}
Applying the binomial theorem (Eq.~\eqref{binm}) in the right-hand side of
the above equation yields the desired result.
\end{proof}

\begin{remark}
The anonymous referee observed that
the last identity is equivalent to
\[B_{n,r}(x,y)=\sum_{j=0}^n{n\choose j}B_{n-j}(x,y),\]
which is meaningful for all $r$, not just the nonnegative integers.
It would be interesting to find combinatorial interpretations
of $B_{n,r}(x,y)$ (or $B_{n,r}(x,1)$) when $r$ is a negative
integer.
\end{remark}

\begin{theorem}
\label{bmnrxy-1} For $\{r, m, n\}\in\mathbb{N}$, we have
\[B_{m+n,r}(x,y)=\sum_{k=0}^m\sum_{i=0}^n
k^{n-i}{n\choose i}{\begin{Bmatrix}m\\k\end{Bmatrix}_r}
{B_{i,r}(x-k,y)}(x)_ky^k.\]
\end{theorem}

\begin{proof}
According to Eqs.~\eqref{ang} and \eqref{amng},
we can write the following bivariate generating function directly:
\begin{equation}\label{bmnrxy}
\sum_{m,n\geq0}B_{m+n,r}(x,y)
    \frac{u^m}{m!}\frac{v^n}{n!}
    =e^{r(u+v)}\{1+y(e^{(u+v)}-1)\}^x.
\end{equation}
Reformulating the right-hand side of Eq.~\eqref{bmnrxy}, we have
\begin{align*}
e^{r(u+v)}\{1+y({e^{u+v}}-1)\}^x
      &=e^{ru}e^{rv}\{1+y(e^v-1)+ye^v(e^u-1)\}^x\\
     &=e^{ru}e^{rv}\sum_{k=0}^{\infty}
     {x\choose k}\{1+y({e^v}-1)\}^{x-k}\{ye^v(e^u-1)\}^k\\
     &=\sum_{k=0}^{\infty}
    e^{rv}\{1+y(e^v-1)\}^{x-k}e^{kv}e^{ru}
    \frac{({e^u}-1)^k}{k!}{(x)_k}{y^k}.
\end{align*}
With the help of Theorem \ref{bnrxy} and Eq.~\eqref{stirl},
we get from the last expression
\begin{align*}
e^{r(u+v)}\{1+y(e^{(u+v)}-1)\}^x
   &=\sum_{j=0}^{\infty}\sum_{s=0}^\infty
 \sum_{i=0}^{\infty}\sum_{l=0}^\infty
\frac{u^j}{j!}\frac{u^s}{s!}
\frac{v^l}{l!}\frac{v^i}{i!}\\
   \times&\sum_{k=0}^jk^lr^s
   B_{i,r}(x-k,y)\begin{Bmatrix}j\\k\end{Bmatrix}(x)_ky^k.
\end{align*}

Letting $i+l=n$ and $s+j=m$, then extracting the coefficients of
$\frac{u^m}{m!}\frac{v^n}{n!}$ across Eq.~\eqref{bmnrxy},
we arrive, after having simplified the result by means of Eq.~\eqref{stirl-co},
at the desired identity.
\end{proof}

Using the same method as that for Theorem \ref{bmnrxy-1},
we get another expression for $B_{m+n,r}(x,y)$.
\begin{theorem}\label{bmnrxy-2}
For $\{r, m, n\}\in\mathbb{N}$, we have
\begin{align*}
B_{m+n,r}(x,y)
=\sum_{i=0}^n\sum_{k=0}^m(k+r)^{n-i}
    {n\choose i}\begin{Bmatrix}m\\k\end{Bmatrix}_r
    B_i(x-k,y)(x)_ky^k.
\end{align*}
\end{theorem}

When $r=0$,
the last two theorems reduce to Theorem \ref{SB(mnxy)}.
\begin{proposition}
\label{prop1}
\begin{align*}
B_{m+n,r}(y)
=\sum_{i=0}^n\sum_{k=0}^mk^{n-i}
    {n\choose i}\begin{Bmatrix}m\\k\end{Bmatrix}_r
    B_{i,r}(y)y^k.
\end{align*}
\end{proposition}
\begin{proof}
Let $y\to\frac{y}{x}$, $x\to\infty$ in Theorem \ref{bmnrxy-1}.
\end{proof}

\begin{proposition}
\label{prop2}
\begin{align*}
B_{m+n,r}(y)
=\sum_{i=0}^n\sum_{k=0}^m(k+r)^{n-i}{n\choose i}{\begin{Bmatrix}m\\k\end{Bmatrix}_r}
B_i(y)y^k.
\end{align*}
\end{proposition}
\begin{proof}
Let $y\to\frac{y}{x}$, $x\to\infty$ in Theorem \ref{bmnrxy-2}.
\end{proof}

\begin{corollary}
\label{corl1}
 \begin{align*}
 B_{m+n,r}
 =\sum_{i=0}^n\sum_{k=0}^mk^{n-i}
      {n\choose i}\begin{Bmatrix}m\\k\end{Bmatrix}_rB_{i,r}.
 \end{align*}
\end{corollary}
\begin{proof}
Let
$y\to\frac{y}{x}$, $x\to\infty$ and then set $y=1$ in Theorem \ref{bmnrxy-1}.
\end{proof}

\begin{corollary} \mbox{[Mez\H{o} \cite[Theorem\ 2]{Mezo2012}}]
\label{corl2}
\[B_{m+n,r}=\sum_{i=0}^n\sum_{k=0}^m
    (k+r)^{n-i}{n\choose i}\begin{Bmatrix}m\\k\end{Bmatrix}_rB_i.\]
\end{corollary}
\begin{proof}
Let $y\to\frac{y}{x}$, $x\to\infty$ and then set $y=1$
in Theorem \ref{bmnrxy-2}.
\end{proof}

It should be pointed out that Corollary \ref{corl1} and Corollary \ref{corl2} can be obtained
by setting $y=1$ in Propositions \ref{prop1} and \ref{prop2}, respectively.
When $r=0$, the last corollary reduces to Eq.~\eqref{bmn}.

\section{Acknowledgments}
The authors are sincerely grateful to Professor
Wenchang Chu for his valuable comments and suggestions.
The corresponding author is partially supported, during this work, by the
National Science foundation of China [Youth Grant No.~11601543] and
the Research Foundation of Zhoukou Normal University [No.~ZKNUB3201701].


\begin{thebibliography}{9}

\bibitem{carlitz} L. Carlitz, Weighted Stirling numbers of the first and
second kind---I, \textit{Fibonacci Quart.} \textbf{18} (1980), 147--162.

\bibitem{comtet} L. Comtet, \textit{Advanced Combinatorics},
Dordrecht-Holland, 1974.

\bibitem{Gould} H. W. Gould and J. Quaintance,
Implications of Spivey's Bell number formula,
\textit{J. Integer Sequences} \textbf{11} (2008),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL11/Gould/gould35.pdf}
{Article 08.3.7.}

\bibitem{graham} R. L. Graham, D. E. Knuth, and O. Patashnik,
\textit{Concrete Mathematics}, Addison-Wesley, 1989.

\bibitem{Mezo2012} I. Mez\H{o}, The dual of Spivey's Bell number formula,
\textit{J. Integer Sequences} \textbf{15} (2012),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Mezo/mezo14.pdf}
{Article 12.2.4.}

\bibitem{Spivey} M. Z. Spivey,
A generalized recurrence for Bell numbers,
\textit{J. Integer Sequences} \textbf{11} (2008),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL11/Spivey/spivey25.pdf}
{Article 08.2.5.}

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary 05A19; Secondary 05A10.

\bigskip
\noindent \emph{Keywords:}
Stirling number of the second kind, $r$-Stirling number, Bell polynomial.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  June 28 2019; 
revised versions received  December 3 2019; December 10 2019; 
December 17 2019.
Published in {\it Journal of Integer Sequences}, December 27 2019.

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