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\begin{center}
\vskip 1cm{\LARGE\bf 
Convolution Properties of the Generalized \\
\vskip .02in
Stirling Numbers and the Jacobi-Stirling \\
\vskip .11in
Numbers of the First Kind
}
\vskip 1cm
\large
Jiaqiang Pan\\ 
School of Biomedical Engineering and Instrumental Science\\ 
Zhejiang University\\
Hangzhou 210027\\ 
China\\ 
\href{mailto:jqpan@zju.edu.cn} {\tt jqpan@zju.edu.cn}\\
\end{center}

\vskip .2 in


\begin{abstract}
In this paper, we establish several properties of the unified
generalized Stirling numbers of the first kind, and the
Jacobi-Stirling numbers of the first kind, by means of the
convolution principle of sequences. Obtained results include
generalized Vandermonde convolution for the unified generalized
Stirling numbers of the first kind, triangular recurrence relation
for general Stirling-type numbers of the first kind, and linear
recurrence formula for the Jacobi-Stirling numbers of the first
kind, and so forth, thereby extending and supplementing known
knowledge to the existent literature about these Stirling-type
numbers.
\end{abstract}


\section{Introduction}

We know that \cite{ref1}, if generating functions of two sequences
$\langle a(k)\rangle\triangleq (a(0),a(1),a(2),a(3),\ldots)$ and
$\langle b(k)\rangle\triangleq (b(0),b(1),b(2),a(3),\ldots)$ are
$a(x)$ and $b(x)$ respectively, namely,
 $$ a(x)=\sum_{k=0}^\infty
a(k)x^k, \quad b(x)=\sum_{k=0}^\infty b(k)x^k,$$
 then product
function $a(x)b(x)$ is generating function of convolution
(sequence) $\langle c(k)\rangle$ of the two sequences $\langle
a(k)\rangle$ and $\langle b(k)\rangle$, where each term of the
sequence $\langle c(k)\rangle$ is calculated by the following
formula:
\begin{equation}\label{e:ConvPrinciple}
c(k)=\sum_{i=0}^k a(i)b(k-i)=\sum_{i=0}^k a(k-i)b(i), \quad
k=0,1,2,3,\ldots.
\end{equation}
For convenience, we occasionally denote the convolution operation
by symbol "$\ast$". For example, equality (\ref{e:ConvPrinciple})
is also expressed as
 $$\langle c(k)\rangle=\langle a(k)\rangle\ast\langle
b(k)\rangle=\langle b(k)\rangle\ast\langle a(k)\rangle.$$ In this
paper, we will call this property of sequences \emph{the
convolution principle of sequences}.

The generating functions of several well-known sequences in
combinatorics have product form. \emph{Unified generalized
Stirling numbers of the first kind} and \emph{Jacobi-Stirling
numbers of the first kind} are two examples of such sequences.

The unified generalized Stirling numbers, defined first by Hsu and
Shuie \cite{ref2}, are the connection coefficients of linear
relations between generalized factorial functions. The generalized
factorial functions of a real or complex number $x$ with real
increment $\alpha$, denoted by $(x|\alpha)_n$, are special
polynomials in $x$ of degree $n$, as
\begin{equation}\label{e:GFactorial}
(x|\alpha)_0=1, \quad \mbox{and} \quad
(x|\alpha)_n=x(x-\alpha)\cdots(x-n\alpha+\alpha)=\prod_{i=0}^{n-1}(x-i\alpha),
\quad n=1,2,\ldots.
\end{equation}
Thus, the unified generalized Stirling numbers with real
parameters $\alpha,\beta,\gamma$, denoted by
$S(n,k;\alpha,\beta,\gamma)$, $n,k=0,1,2,\ldots$, are defined as
(see \cite{ref2})
\begin{equation}\label{e:unidefinition0}
S(0,k;\alpha,\beta,\gamma)=\delta_{0,k}, \, \mbox{and} \,
(x|\alpha)_n=\sum_{k=0}^{\infty}S(n,k;\alpha,\beta,\gamma)(x-\gamma|\beta)_k,
\quad n=1,2,\ldots,
\end{equation}
or
\begin{equation}\label{e:unidefinition1}
(x+\gamma|\alpha)_n=\sum_{k=0}^{\infty}S(n,k;\alpha,\beta,\gamma)(x|\beta)_k,
\quad n=1,2,3,\ldots.
\end{equation}
We see from (\ref{e:unidefinition1}) that for any
$\alpha,\beta,\gamma$, when $k&gt;n$, $S(n,k;\alpha,\beta,\gamma)=0$;
and when $k=n$ $S(n,n;\alpha,\beta,\gamma)=1$. Therefore, the
upper limit $\infty$ of the summation in the right side of
equalities (\ref{e:unidefinition0}) and (\ref{e:unidefinition1})
may be replaced by $n$.

The most popular special cases of $S(n,k;\alpha,\beta,\gamma)$ are
the Kronecker delta $\delta_{n,k}$ ($S(n,k;0,0,0)$), the binomial
coefficients $\binom{n}{k}$ ($S(n,k;0,0,1)$), and two kinds of the
classical Stirling numbers $s(n,k)$ and $S(n,k)$ ($S(n,k;1,0,0)$
and $S(n,k;0,1,0)$).

Taking $S(n,k;\alpha,\beta,\gamma)$ $(n,k=0,1,2,\ldots)$ as
entries, we may obtain a $\infty$-dimensional, lower triangular
matrix
$\mathbf{S}_{\alpha,\beta,\gamma}=\big(S(n,k;\alpha,\beta,\gamma)\big)_{n,k=0.1.2.\ldots}$,
named \emph{the Generalized Stirling matrix} with parameters
$\alpha,\beta,\gamma$ \cite{ref7}. We also name the sequence
$$\langle S(n,k;\alpha,\beta,\gamma)\rangle\triangleq
(S(n,0;\alpha,\beta,\gamma), S(n,1;\alpha,\beta,\gamma),
S(n,2;\alpha,\beta,\gamma),S(n,3;\alpha,\beta,\gamma), \ldots)$$
\emph{the $n$-th row sequence of the unified generalized Stirling
numbers $S(n,k;\alpha,\beta,\gamma)$}.

We call $S(n,k;\alpha,0,\gamma)$ \emph{the unified generalized
Stirling numbers of the first kind}. For $S(n,k;\alpha,0,\gamma)$,
\begin{equation}\label{e:unidefinition2}
S(0,k;\alpha,0,\gamma)=\delta_{0,k}, \, \mbox{and}
\,\prod_{i=0}^{n-1}(x+\gamma-i\alpha)
=\sum_{k=0}^{\infty}S(n,k;\alpha,0,\gamma)x^k, \quad
n=1,2,3,\ldots,
\end{equation}
which shows that the (horizontal) generating function,
$\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$, of the $n$-th row sequence
$\langle S(n,k;\alpha,0,\gamma)\rangle$ has product form.

For an excellent account of the unified generalized Stirling
numbers, see \cite{ref2}.

The Jacobi-Stirling numbers of the first kind, $J(n,k;\zeta)$
($n,k=0,1,2,\ldots$, and $\zeta&gt;-1$ is a fixed constant
parameter), are another special case. In this case, the $n$-th row
sequence $\langle J(n,k;\zeta)\rangle$ also has a (horizontal)
generating function of product form, such as
$\prod_{i=0}^{n-1}(x-i(i+\zeta))$. Thus,
\begin{equation}\label{e:JacobiStirling1}
\prod_{i=0}^{n-1}(x-i(i+\zeta))=\sum_{k=0}^\infty J(n,k;\zeta)x^k,
\quad n=1,2,3,\ldots.
\end{equation}
(Note: $J(0,k;\zeta)=\delta_{0,k}, k=0,1,2,\ldots$). We see from
(\ref{e:JacobiStirling1}) that for any $\zeta$, when $k&gt;n$,
$J(n,k;\zeta)=0$; and when $k=n$, $J(n,n;\zeta)=1$. Therefore, the
upper limit $\infty$ of the summation in the right side of
equality (\ref{e:JacobiStirling1}) may be replaced by $n$.
Particularly, we name $J(n,k;1)$ \emph{the Legendre-Stirling
numbers of the first kind}.

For the initial definition, elementary properties (explicit
expressions, triangular recurrence relations, similarity between
the Jacobi-Stirling and classical Stirling numbers, etc.) and
different combinatorial interpretations of special cases of the
Jacobi-Stirling numbers, see 
\cite{ref3,ref4,ref5,ref6,ref8}, respectively.

Because the unified generalized Stirling numbers of the first
kind, and the Jacobi-Stirling numbers of the first kind, both have
generating functions of product form, thus it is reasonable to
investigate their several properties by means of the convolution
principle of sequences. In the following sections, we will present
the obtained results, including generalized Vandermonde
convolution for the unified generalized Stirling numbers of the
first kind, triangular recurrence relation for general
Stirling-type numbers of the first kind, and linear recurrence
formulae for the Jacobi-Stirling numbers of the first kind, and so
forth.


\section{Generalized Vandermonde convolution}

For the unified generalized Stirling numbers of the first kind, we
may obtain the following theorem by means of the convolution
principle of sequences.

\begin{theorem}\label{t:CovUGSN1}
Let $r$, $t$ and $n$ be three positive integers, and $n=r+t$. Then
\begin{equation}\label{e:CovUGSN1}
\langle S(n,k;\alpha,0,\gamma)\rangle = \langle
S(r,k;\alpha,0,\gamma)\rangle \ast \langle
S(t,k;\alpha,0,\gamma-r\alpha)\rangle.
\end{equation}
namely for $k=0,1,2,3,\ldots$,
\begin{equation}\label{e:CovUGSN2}
S(n,k;\alpha,0,\gamma)= \sum_{i=0}^k S(r,i;\alpha,0,\gamma)
S(t,k-i;\alpha,0,\gamma-r\alpha).
\end{equation}
We name this convolution formula \emph{the Generalized Vandermonde
convolution}.
\end{theorem}

\begin{proof}
We see from (\ref{e:unidefinition2}) that the generalized
factorial $\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$ is the generating
function of sequence $\langle S(n,k;\alpha,0,\gamma)\rangle$. On
the other hand, $\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$ is a product
of two factorial functions, $\prod_{i=0}^{r-1}(x+\gamma-i\alpha)$
and $\prod_{i=0}^{t-1}(x+\gamma-r\alpha-i\alpha)$, which are
generating functions of sequences $\langle
S(r,k;\alpha,0,\gamma)\rangle$ and $\langle
S(t,k;\alpha,0,\gamma-r\alpha)\rangle$ respectively. Hence,
$\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$ is also the generating
function of convolution $\langle S(r,k;\alpha,0,\gamma)\rangle\ast
\langle S(t,k;\alpha,0,\gamma-r\alpha)\rangle$. Thus, $\langle
S(n,k;\alpha,0,\gamma)\rangle = \langle
S(r,k;\alpha,0,\gamma)\rangle \ast \langle
S(t,k;\alpha,0,\gamma-r\alpha)\rangle$.
\end{proof}

\begin{remark}\label{r:CovUGSN1}
We know that $S(n,k;0,0,1)=\binom{n}{k}$,
$S(r,k;0,0,1)=\binom{r}{k}$ and $S(t,k;0,0,1)=\binom{t}{k}$, (or
$S(n,k;0,0,-1)=(-1)^{n-k}\binom{n}{k}$,
$S(r,k;0,0,-1)=(-1)^{r-k}\binom{r}{k}$ and
$S(t,k;0,0,-1)=(-1)^{t-k}\binom{t}{k}$). In this special case, we
may find that (whether $\gamma=1$ or $\gamma=-1$) formula
(\ref{e:CovUGSN2}) lead to the classical Vandermonde
convolution \cite{ref1} (also named Vandermonde's identity or
Vandermonde formula) as
\begin{equation}\label{e:Binom}
\binom{n}{k}
=\sum_{i=0}^k\binom{r}{i}\binom{t}{k-i}=\sum_{i=0}^k\binom{r}{k-i}\binom{t}{i},
\end{equation}
where $n=r+t$.
\end{remark}

\begin{remark}\label{r:CovUGSN2}
The most simple case of formula (\ref{e:CovUGSN1}) or
(\ref{e:CovUGSN2}) is $\alpha=\gamma=0$. In this case,
$S(n,k;0,0,0)=\delta_{n,k}$, $S(r,k;0,0,0)=\delta_{r,k}$, and
$S(t,k;0,0,0)=\delta_{t,k}$. Thus, we obtain self-convolution
property of the kronecker delta, as
\begin{equation}\label{e:Kronecker1}
\langle\delta_{n,k}\rangle =
\langle\delta_{r,k}\rangle\ast\langle\delta_{t,k}\rangle
\end{equation}
or
\begin{equation}\label{e:Kronecker2}
\delta_{n,k} = \sum_{i=0}^k\delta_{r,i}\delta_{t,k-i}=
\sum_{i=0}^k\delta_{r,k-i}\delta_{t,i}
\end{equation}
where $n=r+s$.
\end{remark}

\begin{remark}\label{r:CovUGSN3}
The unified generalized Stirling number $S(r,k;\alpha,0,\gamma)$
of the first kind is the $(r,k)$-th entry of the generalized
Stirling matrix $\mathbf{S}_{\alpha,0,\gamma}$, and
$S(t,k;\alpha,0,\gamma-r\alpha)$ is the $(t,k)$-th entry of the
generalized Stirling matrix
$\mathbf{S}_{\alpha,0,\gamma-r\alpha}$. We know from 
\cite[Theorem 7]{ref7} that, $S(r,k;\alpha,0,\gamma)$ is the scalar product
of the $r$-th row of the matrix $\mathbf{S}_{\alpha,0,0}$ and the
$k$-th column of the matrix $\mathbf{S}_{0,0,\gamma}$; and
$S(t,k;\alpha,0,\gamma-r\alpha)$ is the scalar product of the
$t$-th row of the matrix $\mathbf{S}_{\alpha,0,0}$ and the $k$-th
column of the matrix $\mathbf{S}_{0,0,\gamma-r\alpha}$. Hence,
$S(r,k;\alpha,0,\gamma)$ and $S(t,k;\alpha,0,\gamma-r\alpha)$ in
(\ref{e:CovUGSN2}) may be calculated by using the classical
Stirling numbers $s(n,k)$ of the first kind, and the binomial
coefficients $\binom{n}{k}$ ($n,k=0,1,2,\ldots$), as that
 $$
 S(r,k;\alpha,0,\gamma)=
 \sum_{i=k}^r \gamma^{i-k}\alpha^{r-i}s(r,i)\binom{i}{k},
$$ and
 $$
 S(t,k;\alpha,0,\gamma-r\alpha)=
 \sum_{i=k}^s(\gamma-r\alpha)^{i-k}\alpha^{t-i}s(t,i)\binom{i}{k}.
$$
\end{remark}

\begin{remark}\label{r:CSNF}
For the classical Stirling numbers $s(n,k)$ of the first kind,
namely $S(n,k;1,0,0)$, we have that $$\langle
S(n,k;1,0,0)\rangle=\langle S(r,k;1,0,0)\rangle \ast \langle
S(t,k;1,0,-r)\rangle,$$ or $$ S(n,k;1,0,0)= \sum_{i=0}^k
S(r,k-i;1,0,0)S(t,i;1,0,-r).$$ According to Remark
\ref{r:CovUGSN3}, $$
S(t,i;1,0,-r)=\sum_{j=i}^t(-r)^{j-i}s(t,j)\binom{j}{i}.$$ Finally,
we may write the convolution as
$$s(n,k)=\sum_{i=0}^k\sum_{j=i}^t(-r)^{j-i}\binom{j}{i}s(r,k-i)s(t,j),
\quad n=r+t. $$ This is just the Vandermonde convolution for the
classical Stirling numbers of the first kind.
\end{remark}


\section{Triangular recurrence relations of the Stirling-type numbers of the first kind}

\begin{remark}\label{r:Reccurence1}
We see from Equation (\ref{e:CovUGSN1}) that when $r=n-1$ and
$t=1$,
\begin{equation*}
\langle S(n,k;\alpha,0,\gamma)\rangle = \langle
S(n-1,k;\alpha,0,\gamma)\rangle \ast \langle
S(1,k;\alpha,0,\gamma-(n-1)\alpha)\rangle,
\end{equation*}
namely,
\begin{eqnarray}\label{e:Reccurence1}
S(n,k;\alpha,0,\gamma)=\sum_{i=0}^k
S(n-1,k-i;\alpha,0,\gamma)S(1,i;\alpha,0,\gamma-(n-1)\alpha)
\nonumber\\
=S(n-1,k;\alpha,0,\gamma)S(1,0;\alpha,0,\gamma-(n-1)\alpha)
\nonumber\\
+S(n-1,k-1;\alpha,0,\gamma)S(1,1;\alpha,0,\gamma-(n-1)\alpha)
\nonumber\\
=(\gamma-(n-1)\alpha)S(n-1,k;\alpha,0,\gamma)+S(n-1,k-1;\alpha,0,\gamma)
\end{eqnarray}
This is the triangular recurrence relation of
$S(n,k;\alpha,0,\gamma)$ shown in \cite[Theorem 1]{ref2}.
Therefore, the triangular recurrence relation of the unified
generalized Stirling numbers of the first kind just is a
convolution in essence.
\end{remark}

We may generalize this conclusion to a more general case of
\emph{the Stirling-type numbers of the first kind}, denoted by
$S(n,k;\alpha_i,i=0,1,2,\ldots)$
($\alpha_0,\alpha_1,\alpha_2,\alpha_3,\ldots$ is a given monotonic
non-decreasing or non-increasing sequence). The (horizontal)
generating function of row-sequence $\langle
S(n,k;\alpha_i,i=0,1,2,\ldots)\rangle$ of the Stirling-type
numbers of the first kind is $\prod_{i=0}^{n-1}(x-\alpha_i)$,
namely, $S(0,k;\alpha_i,i=0,1,2,3,\ldots)=\delta_{0,k}$, and
\begin{equation}\label{e:Stirling-type}
\prod_{i=0}^{n-1}(x-\alpha_i)=\sum_{k=0}^n
S(n,k;\alpha_i,i=0,1,2,\ldots)x^k, \quad n=1,2,3,\ldots.
\end{equation}

For the Stirling-type numbers of the first kind, we may obtain
corresponding triangular recurrence relations by means of the
convolution principle of sequences.

\begin{theorem}\label{t:Recurrence}
Let $S(n,k;\alpha_i,i=0,1,2,\ldots)$ be the Stirling-type numbers
of the first kind defined in (\ref{e:Stirling-type}). Then
$S(n,k;\alpha_i,i=0,1,2,\ldots)$ satisfies the following
triangular recurrence relation, namely, for $n,k=1,2,3,\ldots,$
\begin{equation}\label{e:Recurrence2}
S(n,k;\alpha_i,i=0,1,2,\ldots)=-\alpha_{n-1}S(n-1,k;\alpha_i,i=0,1,2,\ldots)\\
+S(n-1,k-1;\alpha_i,i=0,1,2,\ldots)
\end{equation}
\end{theorem}

\begin{proof}
We see from (\ref{e:Stirling-type}) that the generating function
of sequence $\langle S(n,k;\alpha_i,i=0,1,2,\ldots)\rangle$ is
$\prod_{i=0}^{n-1}(x-\alpha_i)$. On the other hand,
$\prod_{i=0}^{n-2}(x-\alpha_i)$ is the generating function of
sequence $\langle S(n-1,k;\alpha_i,i=0,1,2,\ldots)\rangle$, and
$(x-\alpha_{n-1})$ is the generating function of sequence
$(-\alpha_{n-1},1,0,0,0,\ldots)$. Hence, according to the
convolution principle of sequences we have that
\begin{eqnarray*}\label{e:ReccurenceM}
 &amp; &amp;S(n,k;\alpha_i,i=0,1,2,\ldots) \\
 &amp;=&amp;S(n-1,k;\alpha_i,i=0,1,2,\ldots)\cdot(-\alpha_{n-1})+
S(n-1,k-1;\alpha_i,i=0,1,2,\ldots)\cdot1 \\
 &amp;=&amp;-\alpha_{n-1}S(n-1,k;\alpha_i,i=0,1,2,\ldots)
+S(n-1,k-1;\alpha_i,i=0,1,2,\ldots).
\end{eqnarray*}
\end{proof}

This theorem proves that for the most general Stirling-type
numbers of the first kind, exists a triangular recurrence
relation, and the triangular recurrence relation is a convolution
in essence.


\section{Convolution of the Jacobi-Stirling numbers of the first kind}

The Jacobi-Stirling numbers of the first kind, $J(n,k;\zeta)$ are
a special case of the Stirling-type numbers of the first kind, in
which $\alpha_i$ corresponds to $i(i+\zeta)$ ($i=0,1,2,\ldots$).

Because for the Jacobi-Stirling numbers of the first kind,
$\alpha_{n-1}=(n-1)(n+\zeta-1)$, according to
(\ref{e:Recurrence2}), $J(n,k;\zeta)$ satisfy the following
triangular recurrence relation (also see \cite{ref3, ref4, ref6}):
\begin{equation}\label{e:Recurrance3}
J(n,k;\zeta)=-(n-1)(n+\zeta-1)J(n-1,k;\zeta)+J(n-1,k-1;\zeta).
\end{equation}

Furthermore, we may establish several other properties of the
Jacobi-Stirling numbers of the first kind by means of the
convolution principle of sequences, as shown in the subsections
following.


\subsection{Convolution of the degenerate Jacobi-Stirling numbers of the
first kind}

We first investigate $J(n,k;0)$. In this paper, we name $J(n,k;0)$
\emph{the degenerate Jacobi-Stirling numbers of the first kind}.
In fact, they are just so-called \emph{central factorial numbers
of the first kind with even indices} \cite{ref8}.

In this case, the (horizontal) generating function of the $n$-th
row sequence $\langle J(n,k;0)\rangle$ is $\prod_{i=0}^{n-1}
(x-i^2)$, that is,
\begin{equation}\label{e:JS00}
J(0,k;0)=\delta_{0,k}, \quad\mbox{and}\quad \prod_{i=0}^{n-1}
(x-i^2) = \sum_{k=0}^n J(n,k;0)x^k, \quad n=1,2,3,\ldots.
\end{equation}

For the degenerate Jacobi-Stirling numbers of the first kind, we
may obtain the following property by means of the convolution
principle of sequences.

\begin{lemma}\label{l:JS0}
Let $n$ be a given positive integer, and $\langle J(n,k;0)\rangle$
be the $n$-th row sequence of the degenerate Jacobi-Stirling
numbers of the first kind, whose generating function is shown in
(\ref{e:JS00}). Then defining a sequence
$\langle\bar{J}(n,k)\rangle$ derived from $\langle
J(n,k;0)\rangle$ as
$$\langle\bar{J}(n,k)\rangle\triangleq(J(n,0;0),0,J(n,1;0),0,J(n,2;0),0,\ldots),$$
we have that
\begin{equation}\label{e:CJSG0}
\langle\bar{J}(n,k)\rangle = \langle s(n,k)\rangle\ast\langle
(-1)^{n-k}s(n,k)\rangle
\end{equation}
where $s(n,k)$ are the classical Stirling numbers of the first
kind.
\end{lemma}

\begin{proof}
Replacing $x$ in (\ref{e:JS00}) by $y^2$, we have that $$
\prod_{i=0}^{n-1} (y^2-i^2) = \prod_{i=0}^{n-1}
(y-i)\prod_{i=0}^{n-1} (y+i)\\ =\sum_{k=0}^n
 J(n,k;0)y^{2k}= \sum_{k=0}^{2n}
 \bar{J}(n,k)y^{k}
$$ We know that $\prod_{i=0}^{n-1} (y-i)$ and $\prod_{i=0}^{n-1}
(y+i)$ both are the (horizontal) generating functions of two
sequences $\langle s(n,k)\rangle$ and $\langle
(-1)^{n-k}s(n,k)\rangle$, respectively, Hence according to the
convolution principle of sequences, formula (\ref{e:CJSG0}) holds.
\end{proof}

\begin{theorem}\label{t:JS0}
The degenerate Jacobi-Stirling numbers of the first kind,
$J(n,k;0)$ may be calculated by the classical Stirling numbers
$s(n,k)$ of the first kind, as follows,
\begin{equation}\label{e:CJSG1}
J(n,k;0) = \sum_{i=0}^{2k}(-1)^{n-i}s(n,i)s(n,2k-i), \quad
n,k=0,1,2,\ldots.
\end{equation}
\end{theorem}

\begin{proof}
According to (\ref{e:CJSG0}), we have that $$
 \bar{J}(n,k)=\sum_{i=0}^k (-1)^{n-i}s(n,i)s(n,k-i).
$$ Because $J(n,k;0)=\bar{J}(n,2k)$, thus formula (\ref{e:CJSG1})
holds.
\end{proof}

\begin{example}
For example, $$
  J(4,2;0)=\sum_{i=0}^4(-1)^{4-i}s(4,i)s(4,4-i)=49,$$
and $$  J(5,2;0)=\sum_{i=0}^4(-1)^{5-i}s(5,i)s(5,4-i)=-820, $$ and
so forth.
\end{example}

\begin{remark}\label{r:Stirling1}
Because $\bar{J}(n,2k+1)\equiv 0$ ($k=0,1,2,\ldots$), from
(\ref{e:CJSG0}) we have the following identity:
\begin{equation}\label{e:Stirling1}
\sum_{i=0}^{2k+1}(-1)^{n-i}s(n,i)s(n,2k+1-i) = 0, \quad
(k=0,1,2,\ldots).
\end{equation}
In fact, this is a trivial identity, for its first $k+1$ terms
corresponding to $i=0$, $i=1$, $\ldots$, $i=k$ are the contrary
numbers of the rest $k+1$ terms corresponding to $i=2k+1$, $i=2k$,
$\ldots$, $i=k+1$, respectively.
\end{remark}


\subsection{Linear recurrence formula of the Legendre-Stirling numbers of the
first kind}

The Jacobi-Stirling numbers of the first kind with $\zeta=1$,
$J(n,k;1)$ also are named \emph{the Legendre-Stirling numbers of
the first kind} \cite{ref3}. For $J(n,k;1)$, we may obtain a
non-homogeneous linear recurrence relation by means of the
convolution principle of sequences.

\begin{theorem}\label{t:LStirling}
Let $n$ be a given non-negative integer. Then the $n$-th row
sequence, $\langle J(n,k;1)\rangle$, of the Legendre-Stirling
numbers of the first kind satisfies the following non-homogeneous
linear recurrence formulae:
\begin{equation}\label{e:LStirling0}
J(n,0;1)=\delta_{n,0}, \quad  J(n,1;1)=
\sum_{i=0}^n(-1)^{n-i}s(n,i)s(n,1),
\end{equation}
and for $k=2,3,\ldots,n$,
\begin{equation}\label{e:LStirling1}
J(n,k;1)=-\sum_{i=[\frac{k+1}{2}]}^{k-1}\binom{i}{k-i}J(n,i;1)+
\sum_{i=0}^{k}\sum_{j=i}^n(-1)^{n-j}\binom{j}{i}s(n,j)s(n,k-i),
\end{equation}
where $[\cdot]$ is the floor function, and $s(n,k)$s are the
classical Stirling numbers of the first king.
\end{theorem}

\begin{proof}
We see from (\ref{e:JacobiStirling1}) that for the
Legendre-Stirling numbers $J(n,k;1)$ of the first kind,
\begin{equation}\label{e:LeStirling}
\prod_{i=0}^{n-1}(x-i(i+1))=\sum_{k=0}^nJ(n,k;1)x^k.
\end{equation}
Thus, $J(n,0;1)=0$ for $n=1,2,\ldots$. Besides, we know
$J(0,0;1)=1$. Hence, $J(n,0;1)=\delta_{n,0}$. We note that
$y(y+1)-i(i+1)=(y-i)(y+(i+1))$. Hence, replacing $x$ in
(\ref{e:LeStirling}) with $y(y+1)$ we may express the left side on
(\ref{e:LeStirling}) as product of two factorial functions in $y$,
$\prod_{i=0}^{n-1}(y-i)$ and $\prod_{i=0}^{n-1}(y+1+i)$, which are
the (horizontal) generating functions in $y$ of sequences $\langle
s(n,k)\rangle$ and $\langle S(n,k;-1,0,1)\rangle$ respectively.
Because according to Remark \ref{r:CovUGSN3},
$S(n,k;-1,0,1)=\sum_{i=k}^n(-1)^{n-i}\binom{i}{k}s(n,i)$, by means
of the convolution principle of sequences, we may express the left
side on (\ref{e:LeStirling}) as
$$\sum_{k=0}^n\{\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\binom{i}{j}s(n,k-j)s(n,i)\}y^k$$
On the other hand, now we may express the right side on
(\ref{e:LeStirling}) as $\sum_{j=0}^nJ(n,j;1)y^j(y+1)^j$. In the
latter, coefficients of the terms with monomial $y^k$ are
respectively $J(n,k;1)\binom{k}{0}$, $J(n,k-1;1)\binom{k-1}{1}$,
$J(n,k-2;1)\binom{k-2}{2}$, $\ldots$,
$J(n,k-[\frac{k}{2}];1)\binom{k-[\frac{k}{2}]}{[\frac{k}{2}]}$.
Hence, noting $k=[\frac{k}{2}]+[\frac{k+1}{2}]$ we also may
express the right side as
$$\sum_{k=0}^n\{\sum_{i=[\frac{k+1}{2}]}^k\binom{i}{k-i}J(n,i;1)\}y^k.$$
Because $y$ is arbitrary, by comparison of coefficients on both
sides we obtain that $$
\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\binom{i}{j}s(n,k-j)s(n,i)
=\sum_{i=[\frac{k+1}{2}]}^k\binom{i}{k-i}J(n,i;1).$$ Hence,
(\ref{e:LStirling0}) and (\ref{e:LStirling1}) both hold.
\end{proof}

\begin{example}\label{ex:LStirling}
Substituting $0,720,-1764,1624,-735,175,-21,1$ for the Stirling
numbers of the first kind, $s(7,0)$, $s(7,1)$, $s(7,2)$, $\ldots$,
$s(7,7)$ respectively , we find the Legendre-Stirling numbers of
the first kind, $J(7,0;1)=0$, $J(7,k;1)$ ($k=1,2,\ldots,6$), and
$J(7,7;1)=1$ by using formulae (\ref{e:LStirling0}) and
(\ref{e:LStirling1}), where $J(7,k;1)$ ($k=1,2,\ldots,6$) are
listed as follows,
$$J(7,1;1)=\sum_{j=1}^7(-1)^{7-j}s(7,j)s(7,1)=3628800$$
$$J(7,2;1)=-J(7,1;1)+\sum_{i=0}^{2}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,2-i)=-3110400,$$
$$J(7,3;1)=-2J(7,2;1)+\sum_{i=0}^{3}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,3-i)=808848,$$
$$J(7,4;1)=-3J(7,3;1)-J(7,2;1)+\sum_{i=0}^{4}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,4-i)=-89280,$$
$$J(7,5;1)=-4J(7,4;1)-3J(7,3;1)+\sum_{i=0}^{5}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,5-i)=4648,$$
$$J(7,6;1)=-5J(7,5;1)-6J(7,4;1)-J(7,3;1)+\sum_{i=0}^{6}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,6-i)=-112.$$
(see sequence \seqnum{A191936} in \cite{ref9}, 
and also \cite[Table 2]{ref3}).
\end{example}


\subsection{Linear recurrence formula of the Jacobi-Stirling numbers of the
first kind}

For general cases of the Jacobi-Stirling numbers $J(n,k;\zeta)$ of
the first kind, we may obtain a similar linear recurrence relation
for its $n$-th row sequence $\langle J(n,k;\zeta)\rangle$, by
means of the convolution principle of sequences.

\begin{theorem}\label{t:GJStirling}
Let $n$ be a given non-negative integer, and $\zeta$ $(&gt;-1)$ be a
real number. Then the $n$-th row sequence $\langle
J(n,k;\zeta)\rangle$ of the Jacobi-Stirling numbers of the first
kind satisfies the following non-homogeneous linear recurrence
formulae:
\begin{equation}\label{e:JStirling0}
J(n,0;\zeta)=\delta_{n,0}, \quad J(n,1;\zeta)=
\sum_{i=0}^n(-1)^{n-i}\zeta^{i-1}s(n,i)s(n,1),
\end{equation}
and for $k=2,3,\ldots,n$,
\begin{equation}\label{e:JStirling1}
\zeta^kJ(n,k;\zeta)=-\sum_{i=[\frac{k+1}{2}]}^{k-1}\zeta^{2i-k}\binom{i}{k-i}J(n,i;\zeta)\\
+\sum_{i=0}^{k}\sum_{j=i}^n(-1)^{n-j}\zeta^{j-i}\binom{j}{i}s(n,j)s(n,k-i),
\end{equation}
where $[\cdot]$ is the floor function, and $s(n,k)$s are the
classical Stirling numbers of the first kind.
\end{theorem}

\begin{proof}
We see from (\ref{e:JacobiStirling1}) that $J(n,0;\zeta)=0$ for
$n=1,2,$. Besides, we know $J(0,0;\zeta)=1$. Hence,
$J(n,0;\zeta)=\delta_{n,0}$. We note that
$y(y+\zeta)-i(i+\zeta)=(y-i)(y+(i+\zeta))$. Hence, replacing $x$
in (\ref{e:JacobiStirling1}) by $y(y+\zeta)$ we may express the
left side on (\ref{e:JacobiStirling1}) as a product of two
factorial functions in $y$, $\prod_{i=0}^{n-1}(y-i)$ and
$\prod_{i=0}^{n-1}(y+\zeta+i)$, which are the (horizontal)
generating functions in $y$ of sequences, $\langle s(n,k)\rangle$
and $\langle S(n,k;-1,0,\zeta)\rangle$, respectively. Noting that
$S(n,k;-1,0,\zeta)=\sum_{i=k}^n(-1)^{n-i}\zeta^{i-k}\binom{i}{k}s(n,i)$,
by means of the convolution principle of sequences, we may express
the left side on (\ref{e:JacobiStirling1}) as
$$\sum_{k=0}^n\{\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\zeta^{i-j}\binom{i}{j}s(n,k-j)s(n,i)\}y^k.$$
On the other hand, we may express the right side of
(\ref{e:JacobiStirling1}) as
$$\sum_{j=0}^nJ(n,j;\zeta)x^j=\sum_{j=0}^nJ(n,j;\zeta)y^j(y+\zeta)^j.$$
In the latter, coefficients of the terms with monomial $y^k$ are
respectively $\zeta^k\binom{k}{0}J(n,k;\zeta)$,
$\zeta^{k-2}\binom{k-1}{1}J(n,k-1;\zeta)$,
$\zeta^{k-4}\binom{k-2}{2}J(n,k-2;\zeta)$, $\ldots$,
$\zeta^{k-2[\frac{k}{2}]}\binom{k-[\frac{k}{2}]}{[\frac{k}{2}]}J(n,k-[\frac{k}{2}];\zeta)$.
Therefore, noting $k=[\frac{k}{2}]+[\frac{k+1}{2}]$, we may
rewrite the sum on the right side as
$$\sum_{k=0}^n\{\sum_{i=[\frac{k+1}{2}]}^k\zeta^{2i-k}\binom{i}{k-i}J(n,i;\zeta)\}y^k.$$
Because $y$ is arbitrary, by comparison of coefficients on both
sides we obtain that $$
\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\zeta^{i-j}\binom{i}{j}s(n,k-j)s(n,i)\\
=\sum_{i=[\frac{k+1}{2}]}^k\zeta^{2i-k}\binom{i}{k-i}J(n,i;\zeta).
$$ Hence, (\ref{e:JStirling0}) and (\ref{e:JStirling1}) both hold.
\end{proof}

\begin{example}\label{ex:JStirling}
Substituting $0,24,-50,35,-10,1$ for the row sequence of the
classical Stirling numbers of the first kind,
($s(5,0),s(5,1),\ldots,s(5,5)$), we may obtain the row sequence of
the Jacobi-Stirling numbers of the first kind, ($J(5,0;\zeta)=0,
J(5,1;\zeta),\ldots,J(5,4;\zeta),J(5,5;\zeta)=1$) according to
(\ref{e:JStirling0}) and (\ref{e:JStirling1}). $J(5,1;\zeta)$,
$\ldots$, $J(5,4;\zeta)$ are listed as follows. $$
J(5,1;\zeta)=\sum_{i=0}^n(-1)^{5-i}\zeta^{i-1}s(5,i)s(5,1),$$
$$\zeta^2J(5,2;\zeta)=-J(5,1;\zeta)+\sum_{i=0}^2\sum_{j=i}^5(-1)^{5-j}\zeta^{j-i}\binom{j}{i}s(5,j)s(5,2-i).$$
$$\zeta^3J(5,3;\zeta)=-2\zeta
J(5,2;\zeta)+\sum_{i=0}^3\sum_{j=i}^5(-1)^{5-j}\zeta^{j-i}\binom{j}{i}s(5,j)s(5,3-i).$$
$$\zeta^4J(5,4;\zeta)=-3\zeta^2J(5,3;\zeta)-J(5,2;\zeta)+\sum_{i=0}^4\sum_{j=i}^5(-1)^{5-j}\zeta^{j-i}\binom{j}{i}s(5,j)s(5,4-i),$$
which then lead to that
$$J(5,1;\zeta)=576+1200\zeta+840\zeta^2+240\zeta^3+24\zeta^4,$$
$$J(5,2;\zeta)=-(820+1030\zeta+404\zeta^2+50\zeta^3),$$
$$J(5,3;\zeta)=273+200\zeta+35\zeta^2,$$ $$
J(5,4;\zeta)=-(30+10\zeta).$$ (see \cite[Table 2]{ref8}).
\end{example}

\begin{remark}\label{r:JSNC}
We may find that the linear recurrence formulae
(\ref{e:JStirling0}) and (\ref{e:JStirling1}) also verify
\cite[Theorem 1]{ref8}, that is, $J(n,k;\zeta)$ is a polynomial in
$\zeta$ of degree $n-k$, the coefficient of the first term with
$\zeta^{n-k}$ is $s(n,k)$, and the last terms (constant term) is
the central factorial numbers $u(n,k)$ of the first kind with even
indices, which is identical to $J(n,k;0)$. By the way, we may see
that the sum of coefficients of the polynomial is $J(n,k;1)$.
\end{remark}


\section{Acknowledgement}
The author would like to thank the referee for his/her very useful
suggestions.



\begin{thebibliography}{9}

\bibitem{ref1}
Richard~A. Brualdi,
\newblock {\em Introductory Combinatorics}, 5th ed.,
\newblock Pearson Prentice Hall, 2009.

\bibitem{ref2}
Leetsch~C.~Hsu, and Peter~Jau-Shyong~Shiue,
\newblock A unified approach to generalized Stirling numbers,
\newblock {\em Advances in Appl. Math.} {\bf 20} (1998), 366--384.

\bibitem{ref3}
George~E.~Andrews, Wolfgang~Gawronski, and Lance~L.~Littlejohn,
\newblock The Legendre-Stirling numbers,
\newblock {\em Discrete Math.} {\bf 311} (2011),
1255--1272.

\bibitem{ref4}
Eric~S.~Egge,
\newblock Legendre-Stirling permutations,
\newblock {\em European J. Combin.} {\bf 31} (2010),
1735--1750.

\bibitem{ref5}
Pietro~Mongelli,
\newblock Combinatorial interpretations of particular evaluations of
 complete and elementary symmetric functions,
\newblock {\em Electron. J. Combin.} {\bf 19} (2012), 
\href{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v19i1p60}{\#R60}.


\bibitem{ref6}
George~E.~Andrews, Eric~S.~Egge, Wolfgang~Gawronski, and
Lance~L.~Littlejohn,
\newblock The Jacobi-Stirling numbers,
\newblock {\em J. Combin. Theory Ser. A} {\bf 120} (2013),
288--303.

\bibitem{ref7}
Jiaqiang~Pan,
\newblock Matrix decomposition of the unified
generalized Stirling numbers and inversion of the generalized
factorial matrices,
\newblock {\em J. Integer Seq.} {\bf 15} (2012),
Article 12.6.6.

\bibitem{ref8}
Yoann~Gelineau, and Jiang~Zeng,
\newblock Combinatorial interpretations of the Jacobi-Stirling numbers,
\newblock {\em Electron. J. Combin.} {\bf 17}(2) (2010),
\#R70.

\bibitem{ref9}
Neil J.~A. Sloane,
\newblock {\em The On-Line Encyclopedia of Integer Sequences},
\newblock published electronically at
\url{http://oeis.org/}.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B73; Secondary 05A15.

\noindent {\it Keywords}: convolution, unified generalized
Stirling numbers of the first kind, Jacobi-Stirling numbers of the
first kind, generalized Vandermonde convolution, triangular
recurrence relation, non-homogeneous linear recurrence relation.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A191936}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received April ** 2013; revised version
received September ** 2013.


\bigskip
\hrule
\bigskip

\noindent
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%\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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\begin{document}

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\begin{center}
\vskip 1cm{\LARGE\bf Convolution Properties of the Generalized
Stirling Numbers and the Jacobi-Stirling Numbers of the first
kind} \vskip 1cm \large Jiaqiang Pan\\ School of Biomedical
Engineering and Instrumental Science\\ Zhejiang University\\
Hangzhou 210027\\ China\\ \href{mailto:jqpan@zju.edu.cn} {\tt
jqpan@zju.edu.cn}\\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, we establish several properties of the unified
generalized Stirling numbers of the first kind, and the
Jacobi-Stirling numbers of the first kind, by means of the
convolution principle of sequences. Obtained results include
generalized Vandermonde convolution for the unified generalized
Stirling numbers of the first kind, triangular recurrence relation
for general Stirling-type numbers of the first kind, and linear
recurrence formula for the Jacobi-Stirling numbers of the first
kind, and so forth, thereby extending and supplementing known
knowledge to the existent literature about these Stirling-type
numbers.
\end{abstract}


\section{Introduction}

We know that \cite{ref1}, if generating functions of two sequences
$\langle a(k)\rangle\triangleq (a(0),a(1),a(2),a(3),\ldots)$ and
$\langle b(k)\rangle\triangleq (b(0),b(1),b(2),a(3),\ldots)$ are
$a(x)$ and $b(x)$ respectively, namely,
 $$ a(x)=\sum_{k=0}^\infty
a(k)x^k, \quad b(x)=\sum_{k=0}^\infty b(k)x^k,$$
 then product
function $a(x)b(x)$ is generating function of convolution
(sequence) $\langle c(k)\rangle$ of the two sequences $\langle
a(k)\rangle$ and $\langle b(k)\rangle$, where each term of the
sequence $\langle c(k)\rangle$ is calculated by the following
formula:
\begin{equation}\label{e:ConvPrinciple}
c(k)=\sum_{i=0}^k a(i)b(k-i)=\sum_{i=0}^k a(k-i)b(i), \quad
k=0,1,2,3,\ldots.
\end{equation}
For convenience, we occasionally denote the convolution operation
by symbol "$\ast$". For example, equality (\ref{e:ConvPrinciple})
is also expressed as
 $$\langle c(k)\rangle=\langle a(k)\rangle\ast\langle
b(k)\rangle=\langle b(k)\rangle\ast\langle a(k)\rangle.$$ In this
paper, we will call this property of sequences \emph{the
convolution principle of sequences}.

The generating functions of several well-known sequences in
combinatorics have product form. \emph{Unified generalized
Stirling numbers of the first kind} and \emph{Jacobi-Stirling
numbers of the first kind} are two examples of such sequences.

The unified generalized Stirling numbers, defined first by Hsu and
Shuie \cite{ref2}, are the connection coefficients of linear
relations between generalized factorial functions. The generalized
factorial functions of a real or complex number $x$ with real
increment $\alpha$, denoted by $(x|\alpha)_n$, are special
polynomials in $x$ of degree $n$, as
\begin{equation}\label{e:GFactorial}
(x|\alpha)_0=1, \quad \mbox{and} \quad
(x|\alpha)_n=x(x-\alpha)\cdots(x-n\alpha+\alpha)=\prod_{i=0}^{n-1}(x-i\alpha),
\quad n=1,2,\ldots.
\end{equation}
Thus, the unified generalized Stirling numbers with real
parameters $\alpha,\beta,\gamma$, denoted by
$S(n,k;\alpha,\beta,\gamma)$, $n,k=0,1,2,\ldots$, are defined as
(see \cite{ref2})
\begin{equation}\label{e:unidefinition0}
S(0,k;\alpha,\beta,\gamma)=\delta_{0,k}, \, \mbox{and} \,
(x|\alpha)_n=\sum_{k=0}^{\infty}S(n,k;\alpha,\beta,\gamma)(x-\gamma|\beta)_k,
\quad n=1,2,\ldots,
\end{equation}
or
\begin{equation}\label{e:unidefinition1}
(x+\gamma|\alpha)_n=\sum_{k=0}^{\infty}S(n,k;\alpha,\beta,\gamma)(x|\beta)_k,
\quad n=1,2,3,\ldots.
\end{equation}
We see from (\ref{e:unidefinition1}) that for any
$\alpha,\beta,\gamma$, when $k&gt;n$, $S(n,k;\alpha,\beta,\gamma)=0$;
and when $k=n$ $S(n,n;\alpha,\beta,\gamma)=1$. Therefore, the
upper limit $\infty$ of the summation in the right side of
equalities (\ref{e:unidefinition0}) and (\ref{e:unidefinition1})
may be replaced by $n$.

The most popular special cases of $S(n,k;\alpha,\beta,\gamma)$ are
the Kronecker delta $\delta_{n,k}$ ($S(n,k;0,0,0)$), the binomial
coefficients $\binom{n}{k}$ ($S(n,k;0,0,1)$), and two kinds of the
classical Stirling numbers $s(n,k)$ and $S(n,k)$ ($S(n,k;1,0,0)$
and $S(n,k;0,1,0)$).

Taking $S(n,k;\alpha,\beta,\gamma)$ $(n,k=0,1,2,\ldots)$ as
entries, we may obtain a $\infty$-dimensional, lower triangular
matrix
$\mathbf{S}_{\alpha,\beta,\gamma}=\big(S(n,k;\alpha,\beta,\gamma)\big)_{n,k=0.1.2.\ldots}$,
named \emph{the Generalized Stirling matrix} with parameters
$\alpha,\beta,\gamma$ \cite{ref7}. We also name the sequence
$$\langle S(n,k;\alpha,\beta,\gamma)\rangle\triangleq
(S(n,0;\alpha,\beta,\gamma), S(n,1;\alpha,\beta,\gamma),
S(n,2;\alpha,\beta,\gamma),S(n,3;\alpha,\beta,\gamma), \ldots)$$
\emph{the $n$-th row sequence of the unified generalized Stirling
numbers $S(n,k;\alpha,\beta,\gamma)$}.

We call $S(n,k;\alpha,0,\gamma)$ \emph{the unified generalized
Stirling numbers of the first kind}. For $S(n,k;\alpha,0,\gamma)$,
\begin{equation}\label{e:unidefinition2}
S(0,k;\alpha,0,\gamma)=\delta_{0,k}, \, \mbox{and}
\,\prod_{i=0}^{n-1}(x+\gamma-i\alpha)
=\sum_{k=0}^{\infty}S(n,k;\alpha,0,\gamma)x^k, \quad
n=1,2,3,\ldots,
\end{equation}
which shows that the (horizontal) generating function,
$\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$, of the $n$-th row sequence
$\langle S(n,k;\alpha,0,\gamma)\rangle$ has product form.

For an excellent account of the unified generalized Stirling
numbers, see \cite{ref2}.

The Jacobi-Stirling numbers of the first kind, $J(n,k;\zeta)$
($n,k=0,1,2,\ldots$, and $\zeta&gt;-1$ is a fixed constant
parameter), are another special case. In this case, the $n$-th row
sequence $\langle J(n,k;\zeta)\rangle$ also has a (horizontal)
generating function of product form, such as
$\prod_{i=0}^{n-1}(x-i(i+\zeta))$. Thus,
\begin{equation}\label{e:JacobiStirling1}
\prod_{i=0}^{n-1}(x-i(i+\zeta))=\sum_{k=0}^\infty J(n,k;\zeta)x^k,
\quad n=1,2,3,\ldots.
\end{equation}
(Note: $J(0,k;\zeta)=\delta_{0,k}, k=0,1,2,\ldots$). We see from
(\ref{e:JacobiStirling1}) that for any $\zeta$, when $k&gt;n$,
$J(n,k;\zeta)=0$; and when $k=n$, $J(n,n;\zeta)=1$. Therefore, the
upper limit $\infty$ of the summation in the right side of
equality (\ref{e:JacobiStirling1}) may be replaced by $n$.
Particularly, we name $J(n,k;1)$ \emph{the Legendre-Stirling
numbers of the first kind}.

For the initial definition, elementary properties (explicit
expressions, triangular recurrence relations, similarity between
the Jacobi-Stirling and classical Stirling numbers, etc.) and
different combinatorial interpretations of special cases of the
Jacobi-Stirling numbers, see 
\cite{ref3,ref4,ref5,ref6,ref8}, respectively.

Because the unified generalized Stirling numbers of the first
kind, and the Jacobi-Stirling numbers of the first kind, both have
generating functions of product form, thus it is reasonable to
investigate their several properties by means of the convolution
principle of sequences. In the following sections, we will present
the obtained results, including generalized Vandermonde
convolution for the unified generalized Stirling numbers of the
first kind, triangular recurrence relation for general
Stirling-type numbers of the first kind, and linear recurrence
formulae for the Jacobi-Stirling numbers of the first kind, and so
forth.


\section{Generalized Vandermonde convolution}

For the unified generalized Stirling numbers of the first kind, we
may obtain the following theorem by means of the convolution
principle of sequences.

\begin{theorem}\label{t:CovUGSN1}
Let $r$, $t$ and $n$ be three positive integers, and $n=r+t$. Then
\begin{equation}\label{e:CovUGSN1}
\langle S(n,k;\alpha,0,\gamma)\rangle = \langle
S(r,k;\alpha,0,\gamma)\rangle \ast \langle
S(t,k;\alpha,0,\gamma-r\alpha)\rangle.
\end{equation}
namely for $k=0,1,2,3,\ldots$,
\begin{equation}\label{e:CovUGSN2}
S(n,k;\alpha,0,\gamma)= \sum_{i=0}^k S(r,i;\alpha,0,\gamma)
S(t,k-i;\alpha,0,\gamma-r\alpha).
\end{equation}
We name this convolution formula \emph{the Generalized Vandermonde
convolution}.
\end{theorem}

\begin{proof}
We see from (\ref{e:unidefinition2}) that the generalized
factorial $\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$ is the generating
function of sequence $\langle S(n,k;\alpha,0,\gamma)\rangle$. On
the other hand, $\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$ is a product
of two factorial functions, $\prod_{i=0}^{r-1}(x+\gamma-i\alpha)$
and $\prod_{i=0}^{t-1}(x+\gamma-r\alpha-i\alpha)$, which are
generating functions of sequences $\langle
S(r,k;\alpha,0,\gamma)\rangle$ and $\langle
S(t,k;\alpha,0,\gamma-r\alpha)\rangle$ respectively. Hence,
$\prod_{i=0}^{n-1}(x+\gamma-i\alpha)$ is also the generating
function of convolution $\langle S(r,k;\alpha,0,\gamma)\rangle\ast
\langle S(t,k;\alpha,0,\gamma-r\alpha)\rangle$. Thus, $\langle
S(n,k;\alpha,0,\gamma)\rangle = \langle
S(r,k;\alpha,0,\gamma)\rangle \ast \langle
S(t,k;\alpha,0,\gamma-r\alpha)\rangle$.
\end{proof}

\begin{rem}\label{r:CovUGSN1}
We know that $S(n,k;0,0,1)=\binom{n}{k}$,
$S(r,k;0,0,1)=\binom{r}{k}$ and $S(t,k;0,0,1)=\binom{t}{k}$, (or
$S(n,k;0,0,-1)=(-1)^{n-k}\binom{n}{k}$,
$S(r,k;0,0,-1)=(-1)^{r-k}\binom{r}{k}$ and
$S(t,k;0,0,-1)=(-1)^{t-k}\binom{t}{k}$). In this special case, we
may find that (whether $\gamma=1$ or $\gamma=-1$) formula
(\ref{e:CovUGSN2}) lead to the classical Vandermonde
convolution \cite{ref1} (also named Vandermonde's identity or
Vandermonde formula) as
\begin{equation}\label{e:Binom}
\binom{n}{k}
=\sum_{i=0}^k\binom{r}{i}\binom{t}{k-i}=\sum_{i=0}^k\binom{r}{k-i}\binom{t}{i},
\end{equation}
where $n=r+t$.
\end{rem}

\begin{rem}\label{r:CovUGSN2}
The most simple case of formula (\ref{e:CovUGSN1}) or
(\ref{e:CovUGSN2}) is $\alpha=\gamma=0$. In this case,
$S(n,k;0,0,0)=\delta_{n,k}$, $S(r,k;0,0,0)=\delta_{r,k}$, and
$S(t,k;0,0,0)=\delta_{t,k}$. Thus, we obtain self-convolution
property of the kronecker delta, as
\begin{equation}\label{e:Kronecker1}
\langle\delta_{n,k}\rangle =
\langle\delta_{r,k}\rangle\ast\langle\delta_{t,k}\rangle
\end{equation}
or
\begin{equation}\label{e:Kronecker2}
\delta_{n,k} = \sum_{i=0}^k\delta_{r,i}\delta_{t,k-i}=
\sum_{i=0}^k\delta_{r,k-i}\delta_{t,i}
\end{equation}
where $n=r+s$.
\end{rem}

\begin{rem}\label{r:CovUGSN3}
The unified generalized Stirling number $S(r,k;\alpha,0,\gamma)$
of the first kind is the $(r,k)$-th entry of the generalized
Stirling matrix $\mathbf{S}_{\alpha,0,\gamma}$, and
$S(t,k;\alpha,0,\gamma-r\alpha)$ is the $(t,k)$-th entry of the
generalized Stirling matrix
$\mathbf{S}_{\alpha,0,\gamma-r\alpha}$. We know from 
\cite[Theorem 7]{ref7} that, $S(r,k;\alpha,0,\gamma)$ is the scalar product
of the $r$-th row of the matrix $\mathbf{S}_{\alpha,0,0}$ and the
$k$-th column of the matrix $\mathbf{S}_{0,0,\gamma}$; and
$S(t,k;\alpha,0,\gamma-r\alpha)$ is the scalar product of the
$t$-th row of the matrix $\mathbf{S}_{\alpha,0,0}$ and the $k$-th
column of the matrix $\mathbf{S}_{0,0,\gamma-r\alpha}$. Hence,
$S(r,k;\alpha,0,\gamma)$ and $S(t,k;\alpha,0,\gamma-r\alpha)$ in
(\ref{e:CovUGSN2}) may be calculated by using the classical
Stirling numbers $s(n,k)$ of the first kind, and the binomial
coefficients $\binom{n}{k}$ ($n,k=0,1,2,\ldots$), as that
 $$
 S(r,k;\alpha,0,\gamma)=
 \sum_{i=k}^r \gamma^{i-k}\alpha^{r-i}s(r,i)\binom{i}{k},
$$ and
 $$
 S(t,k;\alpha,0,\gamma-r\alpha)=
 \sum_{i=k}^s(\gamma-r\alpha)^{i-k}\alpha^{t-i}s(t,i)\binom{i}{k}.
$$
\end{rem}

\begin{rem}\label{r:CSNF}
For the classical Stirling numbers $s(n,k)$ of the first kind,
namely $S(n,k;1,0,0)$, we have that $$\langle
S(n,k;1,0,0)\rangle=\langle S(r,k;1,0,0)\rangle \ast \langle
S(t,k;1,0,-r)\rangle,$$ or $$ S(n,k;1,0,0)= \sum_{i=0}^k
S(r,k-i;1,0,0)S(t,i;1,0,-r).$$ According to Remark
\ref{r:CovUGSN3}, $$
S(t,i;1,0,-r)=\sum_{j=i}^t(-r)^{j-i}s(t,j)\binom{j}{i}.$$ Finally,
we may write the convolution as
$$s(n,k)=\sum_{i=0}^k\sum_{j=i}^t(-r)^{j-i}\binom{j}{i}s(r,k-i)s(t,j),
\quad n=r+t. $$ This is just the Vandermonde convolution for the
classical Stirling numbers of the first kind.
\end{rem}


\section{Triangular recurrence relations of the Stirling-type numbers of the first kind}

\begin{rem}\label{r:Reccurence1}
We see from Equation (\ref{e:CovUGSN1}) that when $r=n-1$ and
$t=1$,
\begin{equation*}
\langle S(n,k;\alpha,0,\gamma)\rangle = \langle
S(n-1,k;\alpha,0,\gamma)\rangle \ast \langle
S(1,k;\alpha,0,\gamma-(n-1)\alpha)\rangle,
\end{equation*}
namely,
\begin{eqnarray}\label{e:Reccurence1}
S(n,k;\alpha,0,\gamma)=\sum_{i=0}^k
S(n-1,k-i;\alpha,0,\gamma)S(1,i;\alpha,0,\gamma-(n-1)\alpha)
\nonumber\\
=S(n-1,k;\alpha,0,\gamma)S(1,0;\alpha,0,\gamma-(n-1)\alpha)
\nonumber\\
+S(n-1,k-1;\alpha,0,\gamma)S(1,1;\alpha,0,\gamma-(n-1)\alpha)
\nonumber\\
=(\gamma-(n-1)\alpha)S(n-1,k;\alpha,0,\gamma)+S(n-1,k-1;\alpha,0,\gamma)
\end{eqnarray}
This is the triangular recurrence relation of
$S(n,k;\alpha,0,\gamma)$ shown in \cite[Theorem 1]{ref2}.
Therefore, the triangular recurrence relation of the unified
generalized Stirling numbers of the first kind just is a
convolution in essence.
\end{rem}

We may generalize this conclusion to a more general case of
\emph{the Stirling-type numbers of the first kind}, denoted by
$S(n,k;\alpha_i,i=0,1,2,\ldots)$
($\alpha_0,\alpha_1,\alpha_2,\alpha_3,\ldots$ is a given monotonic
non-decreasing or non-increasing sequence). The (horizontal)
generating function of row-sequence $\langle
S(n,k;\alpha_i,i=0,1,2,\ldots)\rangle$ of the Stirling-type
numbers of the first kind is $\prod_{i=0}^{n-1}(x-\alpha_i)$,
namely, $S(0,k;\alpha_i,i=0,1,2,3,\ldots)=\delta_{0,k}$, and
\begin{equation}\label{e:Stirling-type}
\prod_{i=0}^{n-1}(x-\alpha_i)=\sum_{k=0}^n
S(n,k;\alpha_i,i=0,1,2,\ldots)x^k, \quad n=1,2,3,\ldots.
\end{equation}

For the Stirling-type numbers of the first kind, we may obtain
corresponding triangular recurrence relations by means of the
convolution principle of sequences.

\begin{theorem}\label{t:Recurrence}
Let $S(n,k;\alpha_i,i=0,1,2,\ldots)$ be the Stirling-type numbers
of the first kind defined in (\ref{e:Stirling-type}). Then
$S(n,k;\alpha_i,i=0,1,2,\ldots)$ satisfies the following
triangular recurrence relation, namely, for $n,k=1,2,3,\ldots,$
\begin{equation}\label{e:Recurrence2}
S(n,k;\alpha_i,i=0,1,2,\ldots)=-\alpha_{n-1}S(n-1,k;\alpha_i,i=0,1,2,\ldots)\\
+S(n-1,k-1;\alpha_i,i=0,1,2,\ldots)
\end{equation}
\end{theorem}

\begin{proof}
We see from (\ref{e:Stirling-type}) that the generating function
of sequence $\langle S(n,k;\alpha_i,i=0,1,2,\ldots)\rangle$ is
$\prod_{i=0}^{n-1}(x-\alpha_i)$. On the other hand,
$\prod_{i=0}^{n-2}(x-\alpha_i)$ is the generating function of
sequence $\langle S(n-1,k;\alpha_i,i=0,1,2,\ldots)\rangle$, and
$(x-\alpha_{n-1})$ is the generating function of sequence
$(-\alpha_{n-1},1,0,0,0,\ldots)$. Hence, according to the
convolution principle of sequences we have that
\begin{eqnarray*}\label{e:ReccurenceM}
 &amp; &amp;S(n,k;\alpha_i,i=0,1,2,\ldots) \\
 &amp;=&amp;S(n-1,k;\alpha_i,i=0,1,2,\ldots)\cdot(-\alpha_{n-1})+
S(n-1,k-1;\alpha_i,i=0,1,2,\ldots)\cdot1 \\
 &amp;=&amp;-\alpha_{n-1}S(n-1,k;\alpha_i,i=0,1,2,\ldots)
+S(n-1,k-1;\alpha_i,i=0,1,2,\ldots).
\end{eqnarray*}
\end{proof}

This theorem proves that for the most general Stirling-type
numbers of the first kind, exists a triangular recurrence
relation, and the triangular recurrence relation is a convolution
in essence.


\section{Convolution of the Jacobi-Stirling numbers of the first kind}

The Jacobi-Stirling numbers of the first kind, $J(n,k;\zeta)$ are
a special case of the Stirling-type numbers of the first kind, in
which $\alpha_i$ corresponds to $i(i+\zeta)$ ($i=0,1,2,\ldots$).

Because for the Jacobi-Stirling numbers of the first kind,
$\alpha_{n-1}=(n-1)(n+\zeta-1)$, according to
(\ref{e:Recurrence2}), $J(n,k;\zeta)$ satisfy the following
triangular recurrence relation (also see \cite{ref3, ref4, ref6}):
\begin{equation}\label{e:Recurrance3}
J(n,k;\zeta)=-(n-1)(n+\zeta-1)J(n-1,k;\zeta)+J(n-1,k-1;\zeta).
\end{equation}

Furthermore, we may establish several other properties of the
Jacobi-Stirling numbers of the first kind by means of the
convolution principle of sequences, as shown in the subsections
following.


\subsection{Convolution of the degenerate Jacobi-Stirling numbers of the
first kind}

We first investigate $J(n,k;0)$. In this paper, we name $J(n,k;0)$
\emph{the Degenerate Jacobi-Stirling numbers of the first kind}.
In fact, they are just so-called \emph{central factorial numbers
of the first kind with even indices} \cite{ref8}.

In this case, the (horizontal) generating function of the $n$-th
row sequence $\langle J(n,k;0)\rangle$ is $\prod_{i=0}^{n-1}
(x-i^2)$, that is,
\begin{equation}\label{e:JS00}
J(0,k;0)=\delta_{0,k}, \quad\mbox{and}\quad \prod_{i=0}^{n-1}
(x-i^2) = \sum_{k=0}^n J(n,k;0)x^k, \quad n=1,2,3,\ldots.
\end{equation}

For the degenerate Jacobi-Stirling numbers of the first kind, we
may obtain the following property by means of the convolution
principle of sequences.

\begin{lemma}\label{l:JS0}
Let $n$ be a given positive integer, and $\langle J(n,k;0)\rangle$
be the $n$-th row sequence of the degenerate Jacobi-Stirling
numbers of the first kind, whose generating function is shown in
(\ref{e:JS00}). Then defining a sequence
$\langle\bar{J}(n,k)\rangle$ derived from $\langle
J(n,k;0)\rangle$ as
$$\langle\bar{J}(n,k)\rangle\triangleq(J(n,0;0),0,J(n,1;0),0,J(n,2;0),0,\ldots),$$
we have that
\begin{equation}\label{e:CJSG0}
\langle\bar{J}(n,k)\rangle = \langle s(n,k)\rangle\ast\langle
(-1)^{n-k}s(n,k)\rangle
\end{equation}
where $s(n,k)$ are the classical Stirling numbers of the first
kind.
\end{lemma}

\begin{proof}
Replacing $x$ in (\ref{e:JS00}) by $y^2$, we have that $$
\prod_{i=0}^{n-1} (y^2-i^2) = \prod_{i=0}^{n-1}
(y-i)\prod_{i=0}^{n-1} (y+i)\\ =\sum_{k=0}^n
 J(n,k;0)y^{2k}= \sum_{k=0}^{2n}
 \bar{J}(n,k)y^{k}
$$ We know that $\prod_{i=0}^{n-1} (y-i)$ and $\prod_{i=0}^{n-1}
(y+i)$ both are the (horizontal) generating functions of two
sequences $\langle s(n,k)\rangle$ and $\langle
(-1)^{n-k}s(n,k)\rangle$, respectively, Hence according to the
convolution principle of sequences, formula (\ref{e:CJSG0}) holds.
\end{proof}

\begin{theorem}\label{t:JS0}
The degenerate Jacobi-Stirling numbers of the first kind,
$J(n,k;0)$ may be calculated by the classical Stirling numbers
$s(n,k)$ of the first kind, as follows,
\begin{equation}\label{e:CJSG1}
J(n,k;0) = \sum_{i=0}^{2k}(-1)^{n-i}s(n,i)s(n,2k-i), \quad
n,k=0,1,2,\ldots.
\end{equation}
\end{theorem}

\begin{proof}
According to (\ref{e:CJSG0}), we have that $$
 \bar{J}(n,k)=\sum_{i=0}^k (-1)^{n-i}s(n,i)s(n,k-i).
$$ Because $J(n,k;0)=\bar{J}(n,2k)$, thus formula (\ref{e:CJSG1})
holds.
\end{proof}

\begin{example}
For example, $$
  J(4,2;0)=\sum_{i=0}^4(-1)^{4-i}s(4,i)s(4,4-i)=49,$$
and $$  J(5,2;0)=\sum_{i=0}^4(-1)^{5-i}s(5,i)s(5,4-i)=-820, $$ and
so forth.
\end{example}

\begin{rem}\label{r:Stirling1}
Because $\bar{J}(n,2k+1)\equiv 0$ ($k=0,1,2,\ldots$), from
(\ref{e:CJSG0}) we have the following identity:
\begin{equation}\label{e:Stirling1}
\sum_{i=0}^{2k+1}(-1)^{n-i}s(n,i)s(n,2k+1-i) = 0, \quad
(k=0,1,2,\ldots).
\end{equation}
In fact, this is a trivial identity, for its first $k+1$ terms
corresponding to $i=0$, $i=1$, $\ldots$, $i=k$ are the contrary
numbers of the rest $k+1$ terms corresponding to $i=2k+1$, $i=2k$,
$\ldots$, $i=k+1$, respectively.
\end{rem}


\subsection{Linear recurrence formula of the Legendre-Stirling numbers of the
first kind}

The Jacobi-Stirling numbers of the first kind with $\zeta=1$,
$J(n,k;1)$ also are named \emph{the Legendre-Stirling numbers of
the first kind} \cite{ref3}. For $J(n,k;1)$, we may obtain a
non-homogeneous linear recurrence relation by means of the
convolution principle of sequences.

\begin{theorem}\label{t:LStirling}
Let $n$ be a given non-negative integer. Then the $n$-th row
sequence, $\langle J(n,k;1)\rangle$, of the Legendre-Stirling
numbers of the first kind satisfies the following non-homogeneous
linear recurrence formulae:
\begin{equation}\label{e:LStirling0}
J(n,0;1)=\delta_{n,0}, \quad  J(n,1;1)=
\sum_{i=0}^n(-1)^{n-i}s(n,i)s(n,1),
\end{equation}
and for $k=2,3,\ldots,n$,
\begin{equation}\label{e:LStirling1}
J(n,k;1)=-\sum_{i=[\frac{k+1}{2}]}^{k-1}\binom{i}{k-i}J(n,i;1)+
\sum_{i=0}^{k}\sum_{j=i}^n(-1)^{n-j}\binom{j}{i}s(n,j)s(n,k-i),
\end{equation}
where $[\cdot]$ is the floor function, and $s(n,k)$s are the
classical Stirling numbers of the first king.
\end{theorem}

\begin{proof}
We see from (\ref{e:JacobiStirling1}) that for the
Legendre-Stirling numbers $J(n,k;1)$ of the first kind,
\begin{equation}\label{e:LeStirling}
\prod_{i=0}^{n-1}(x-i(i+1))=\sum_{k=0}^nJ(n,k;1)x^k.
\end{equation}
Thus, $J(n,0;1)=0$ for $n=1,2,\ldots$. Besides, we know
$J(0,0;1)=1$. Hence, $J(n,0;1)=\delta_{n,0}$. We note that
$y(y+1)-i(i+1)=(y-i)(y+(i+1))$. Hence, replacing $x$ in
(\ref{e:LeStirling}) with $y(y+1)$ we may express the left side on
(\ref{e:LeStirling}) as product of two factorial functions in $y$,
$\prod_{i=0}^{n-1}(y-i)$ and $\prod_{i=0}^{n-1}(y+1+i)$, which are
the (horizontal) generating functions in $y$ of sequences $\langle
s(n,k)\rangle$ and $\langle S(n,k;-1,0,1)\rangle$ respectively.
Because according to Remark \ref{r:CovUGSN3},
$S(n,k;-1,0,1)=\sum_{i=k}^n(-1)^{n-i}\binom{i}{k}s(n,i)$, by means
of the convolution principle of sequences, we may express the left
side on (\ref{e:LeStirling}) as
$$\sum_{k=0}^n\{\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\binom{i}{j}s(n,k-j)s(n,i)\}y^k$$
On the other hand, now we may express the right side on
(\ref{e:LeStirling}) as $\sum_{j=0}^nJ(n,j;1)y^j(y+1)^j$. In the
latter, coefficients of the terms with monomial $y^k$ are
respectively $J(n,k;1)\binom{k}{0}$, $J(n,k-1;1)\binom{k-1}{1}$,
$J(n,k-2;1)\binom{k-2}{2}$, $\ldots$,
$J(n,k-[\frac{k}{2}];1)\binom{k-[\frac{k}{2}]}{[\frac{k}{2}]}$.
Hence, noting $k=[\frac{k}{2}]+[\frac{k+1}{2}]$ we also may
express the right side as
$$\sum_{k=0}^n\{\sum_{i=[\frac{k+1}{2}]}^k\binom{i}{k-i}J(n,i;1)\}y^k.$$
Because $y$ is arbitrary, by comparison of coefficients on both
sides we obtain that $$
\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\binom{i}{j}s(n,k-j)s(n,i)
=\sum_{i=[\frac{k+1}{2}]}^k\binom{i}{k-i}J(n,i;1).$$ Hence,
(\ref{e:LStirling0}) and (\ref{e:LStirling1}) both hold.
\end{proof}

\begin{example}\label{ex:LStirling}
Substituting $0,720,-1764,1624,-735,175,-21,1$ for the Stirling
numbers of the first kind, $s(7,0)$, $s(7,1)$, $s(7,2)$, $\ldots$,
$s(7,7)$ respectively , we find the Legendre-Stirling numbers of
the first kind, $J(7,0;1)=0$, $J(7,k;1)$ ($k=1,2,\ldots,6$), and
$J(7,7;1)=1$ by using formulae (\ref{e:LStirling0}) and
(\ref{e:LStirling1}), where $J(7,k;1)$ ($k=1,2,\ldots,6$) are
listed as follows,
$$J(7,1;1)=\sum_{j=1}^7(-1)^{7-j}s(7,j)s(7,1)=3628800$$
$$J(7,2;1)=-J(7,1;1)+\sum_{i=0}^{2}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,2-i)=-3110400,$$
$$J(7,3;1)=-2J(7,2;1)+\sum_{i=0}^{3}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,3-i)=808848,$$
$$J(7,4;1)=-3J(7,3;1)-J(7,2;1)+\sum_{i=0}^{4}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,4-i)=-89280,$$
$$J(7,5;1)=-4J(7,4;1)-3J(7,3;1)+\sum_{i=0}^{5}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,5-i)=4648,$$
$$J(7,6;1)=-5J(7,5;1)-6J(7,4;1)-J(7,3;1)+\sum_{i=0}^{6}\sum_{j=i}^7(-1)^{7-j}\binom{j}{i}s(7,j)s(7,6-i)=-112.$$
(see sequence \seqnum{A191936} in \cite{ref9}, 
and also \cite[Table 2]{ref3}).
\end{example}


\subsection{Linear recurrence formula of the Jacobi-Stirling numbers of the
first kind}

For general cases of the Jacobi-Stirling numbers $J(n,k;\zeta)$ of
the first kind, we may obtain a similar linear recurrence relation
for its $n$-th row sequence $\langle J(n,k;\zeta)\rangle$, by
means of the convolution principle of sequences.

\begin{theorem}\label{t:GJStirling}
Let $n$ be a given non-negative integer, and $\zeta$ $(&gt;-1)$ be a
real number. Then the $n$-th row sequence $\langle
J(n,k;\zeta)\rangle$ of the Jacobi-Stirling numbers of the first
kind satisfies the following non-homogeneous linear recurrence
formulae:
\begin{equation}\label{e:JStirling0}
J(n,0;\zeta)=\delta_{n,0}, \quad J(n,1;\zeta)=
\sum_{i=0}^n(-1)^{n-i}\zeta^{i-1}s(n,i)s(n,1),
\end{equation}
and for $k=2,3,\ldots,n$,
\begin{equation}\label{e:JStirling1}
\zeta^kJ(n,k;\zeta)=-\sum_{i=[\frac{k+1}{2}]}^{k-1}\zeta^{2i-k}\binom{i}{k-i}J(n,i;\zeta)\\
+\sum_{i=0}^{k}\sum_{j=i}^n(-1)^{n-j}\zeta^{j-i}\binom{j}{i}s(n,j)s(n,k-i),
\end{equation}
where $[\cdot]$ is the floor function, and $s(n,k)$s are the
classical Stirling numbers of the first kind.
\end{theorem}

\begin{proof}
We see from (\ref{e:JacobiStirling1}) that $J(n,0;\zeta)=0$ for
$n=1,2,$. Besides, we know $J(0,0;\zeta)=1$. Hence,
$J(n,0;\zeta)=\delta_{n,0}$. We note that
$y(y+\zeta)-i(i+\zeta)=(y-i)(y+(i+\zeta))$. Hence, replacing $x$
in (\ref{e:JacobiStirling1}) by $y(y+\zeta)$ we may express the
left side on (\ref{e:JacobiStirling1}) as a product of two
factorial functions in $y$, $\prod_{i=0}^{n-1}(y-i)$ and
$\prod_{i=0}^{n-1}(y+\zeta+i)$, which are the (horizontal)
generating functions in $y$ of sequences, $\langle s(n,k)\rangle$
and $\langle S(n,k;-1,0,\zeta)\rangle$, respectively. Noting that
$S(n,k;-1,0,\zeta)=\sum_{i=k}^n(-1)^{n-i}\zeta^{i-k}\binom{i}{k}s(n,i)$,
by means of the convolution principle of sequences, we may express
the left side on (\ref{e:JacobiStirling1}) as
$$\sum_{k=0}^n\{\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\zeta^{i-j}\binom{i}{j}s(n,k-j)s(n,i)\}y^k.$$
On the other hand, we may express the right side of
(\ref{e:JacobiStirling1}) as
$$\sum_{j=0}^nJ(n,j;\zeta)x^j=\sum_{j=0}^nJ(n,j;\zeta)y^j(y+\zeta)^j.$$
In the latter, coefficients of the terms with monomial $y^k$ are
respectively $\zeta^k\binom{k}{0}J(n,k;\zeta)$,
$\zeta^{k-2}\binom{k-1}{1}J(n,k-1;\zeta)$,
$\zeta^{k-4}\binom{k-2}{2}J(n,k-2;\zeta)$, $\ldots$,
$\zeta^{k-2[\frac{k}{2}]}\binom{k-[\frac{k}{2}]}{[\frac{k}{2}]}J(n,k-[\frac{k}{2}];\zeta)$.
Therefore, noting $k=[\frac{k}{2}]+[\frac{k+1}{2}]$, we may
rewrite the sum on the right side as
$$\sum_{k=0}^n\{\sum_{i=[\frac{k+1}{2}]}^k\zeta^{2i-k}\binom{i}{k-i}J(n,i;\zeta)\}y^k.$$
Because $y$ is arbitrary, by comparison of coefficients on both
sides we obtain that $$
\sum_{j=0}^k\sum_{i=j}^n(-1)^{n-i}\zeta^{i-j}\binom{i}{j}s(n,k-j)s(n,i)\\
=\sum_{i=[\frac{k+1}{2}]}^k\zeta^{2i-k}\binom{i}{k-i}J(n,i;\zeta).
$$ Hence, (\ref{e:JStirling0}) and (\ref{e:JStirling1}) both hold.
\end{proof}

\begin{example}\label{ex:JStirling}
Substituting $0,24,-50,35,-10,1$ for the row sequence of the
classical Stirling numbers of the first kind,
($s(5,0),s(5,1),\ldots,s(5,5)$), we may obtain the row sequence of
the Jacobi-Stirling numbers of the first kind, ($J(5,0;\zeta)=0,
J(5,1;\zeta),\ldots,J(5,4;\zeta),J(5,5;\zeta)=1$) according to
(\ref{e:JStirling0}) and (\ref{e:JStirling1}). $J(5,1;\zeta)$,
$\ldots$, $J(5,4;\zeta)$ are listed as follows. $$
J(5,1;\zeta)=\sum_{i=0}^n(-1)^{5-i}\zeta^{i-1}s(5,i)s(5,1),$$
$$\zeta^2J(5,2;\zeta)=-J(5,1;\zeta)+\sum_{i=0}^2\sum_{j=i}^5(-1)^{5-j}\zeta^{j-i}\binom{j}{i}s(5,j)s(5,2-i).$$
$$\zeta^3J(5,3;\zeta)=-2\zeta
J(5,2;\zeta)+\sum_{i=0}^3\sum_{j=i}^5(-1)^{5-j}\zeta^{j-i}\binom{j}{i}s(5,j)s(5,3-i).$$
$$\zeta^4J(5,4;\zeta)=-3\zeta^2J(5,3;\zeta)-J(5,2;\zeta)+\sum_{i=0}^4\sum_{j=i}^5(-1)^{5-j}\zeta^{j-i}\binom{j}{i}s(5,j)s(5,4-i),$$
which then lead to that
$$J(5,1;\zeta)=576+1200\zeta+840\zeta^2+240\zeta^3+24\zeta^4,$$
$$J(5,2;\zeta)=-(820+1030\zeta+404\zeta^2+50\zeta^3),$$
$$J(5,3;\zeta)=273+200\zeta+35\zeta^2,$$ $$
J(5,4;\zeta)=-(30+10\zeta).$$ (see \cite[Table 2]{ref8}).
\end{example}

\begin{rem}\label{r:JSNC}
We may find that the linear recurrence formulae
(\ref{e:JStirling0}) and (\ref{e:JStirling1}) also verify
\cite[Theorem 1]{ref8}, that is, $J(n,k;\zeta)$ is a polynomial in
$\zeta$ of degree $n-k$, the coefficient of the first term with
$\zeta^{n-k}$ is $s(n,k)$, and the last terms (constant term) is
the central factorial numbers $u(n,k)$ of the first kind with even
indices, which is identical to $J(n,k;0)$. By the way, we may see
that the sum of coefficients of the polynomial is $J(n,k;1)$.
\end{rem}


\section{Acknowledgement}
The author would like to thank the referee for his/her very useful
suggestions.



\begin{thebibliography}{9}

\bibitem{ref1}
Richard~A. Brualdi,
\newblock {\em Introductory Combinatorics}, 5th ed.,
\newblock Pearson Prentice Hall, 2009.

\bibitem{ref2}
Leetsch~C.~Hsu, and Peter~Jau-Shyong~Shiue,
\newblock A unified approach to generalized Stirling numbers,
\newblock {\em Advances in Appl. Math.} {\bf 20} (1998), 366--384.

\bibitem{ref3}
George~E.~Andrews, Wolfgang~Gawronski, and Lance~L.~Littlejohn,
\newblock The Legendre-Stirling numbers,
\newblock {\em Discrete Math.} {\bf 311} (2011),
1255--1272.

\bibitem{ref4}
Eric~S.~Egge,
\newblock Legendre-Stirling permutations,
\newblock {\em European J. Combin.} {\bf 31} (2010),
1735--1750.

\bibitem{ref5}
Pietro~Mongelli,
\newblock Combinatorial interpretations of particular evaluations of
 complete and elementary symmetric functions,
\newblock {\em Electron. J. Combin.} {\bf 19} (2012), \#R60.


\bibitem{ref6}
George~E.~Andrews, Eric~S.~Egge, Wolfgang~Gawronski, and
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\newblock The Jacobi-Stirling numbers,
\newblock {\em J. Combin. Theory Ser. A} {\bf 120} (2013),
288--303.

\bibitem{ref7}
Jiaqiang~Pan,
\newblock Matrix decomposition of the unified
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\newblock {\em J. Integer Seq.} {\bf 15} (2012),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Pan/pan19.html}{Article 12.6.6}.

\bibitem{ref8}
Yoann~Gelineau, and Jiang~Zeng,
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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B73; Secondary 05A15.

\noindent {\it Keywords}: convolution, unified generalized
Stirling number of the first kind,
Jacobi-Stirling number of the
first kind, generalized Vandermonde convolution, triangular
recurrence relation, non-homogeneous linear recurrence relation.

\bigskip
\hrule
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\noindent (Concerned with sequence \seqnum{A191936}.)


\bigskip
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\vspace*{+.1in} \noindent 
Received April 1 2013;
revised version received September 24 2013.
Published in {\it Journal of Integer Sequences}, 
October 13 2013.

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