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\begin{center}
\vskip 1cm{\LARGE\bf Matrix Decomposition of the Unified\\
\vskip .05in
Generalized Stirling Numbers and Inversion \\
\vskip .12in
of the Generalized Factorial Matrices}
\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 give a matrix decomposition method used to
calculate unified generalized Stirling numbers in an explicit,
non-recursive mode, and some of its applications. Then, we define
generalized factorial matrices which may be regarded as a
generalization in the form of the Vandermonde matrices, and presents
some of their properties --- in particular, triangular matrix factors
of the inverse matrices of the generalized factorial matrices.
\end{abstract}


\section{Introduction}

The unified generalized Stirling numbers, defined by Hsu and Shuie
\cite{ref1}, 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), \quad
n=1,2,\ldots.
\end{equation}
Thus, the unified generalized Stirling numbers with real
parameters $\alpha,\beta,\gamma$, denoted by
$S(n,k;\alpha,\beta,\gamma)$, $n,k=0,1,2,\ldots$, may be defined
as (see \cite{ref1})
\begin{equation}\label{e:unidefinition0}
(x|\alpha)_n=\sum_{k=0}^{\infty}S(n,k;\alpha,\beta,\gamma)(x-\gamma|\beta)_k.
\end{equation}
We see from (\ref{e:unidefinition0}) that for any
$\alpha,\beta,\gamma$, $S(n,n;\alpha,\beta,\gamma)=1$; and
$S(n,k;\alpha,\beta,\gamma)=0$ if $k&gt;n$. Therefore, the upper
limit $\infty$ of the summation in the right side of equality
(\ref{e:unidefinition0}) may be replaced by $n$.

As Hsu and Shuie pointed, the definition (\ref{e:unidefinition0})
gives a more transparent unification of various Stirling-type
numbers studied previously by other authors (see \cite{ref1}). In
particularly, we see from (\ref{e:unidefinition0}) that
$S(n,k;0,0,0)=\delta_{n,k}$ (the Kronecker symbols),
$S(n,k;1,0,0)=s(n,k)$, $S(n,k;0,1,0)=S(n,k)$, and
$S(n,k;0,0,1)=\binom{n}{k}$, where $s(n,k)$ and $S(n,k)$ are the
Stirling numbers of the first and second kind respectively.

Hsu, Shuie and other authors then investigated basic properties of
the unified generalized Stirling numbers, such as recurrence
relations, generating functions, convolution formulas, congruence
properties, asymptotic expansion, $q$-analoque formulas and
corresponding combinatorial interpretation (see
\cite{ref1,ref2,ref3}).

\section{Matrix Decomposition of the Unified Generalized Stirling Numbers}



Now let's have the following definition.

\begin{definition}\label{d:UnifiedGSM}
\emph{the unified generalized Stirling (transform) matrix with
real parameters} $\alpha,\beta,\gamma$, denoted by
$\mathbf{S}_{\alpha,\beta,\gamma}$, is defined to be an
infinite-dimensional lower triangular matrix, the $(n,k)$th
entries of which are $S(n,k;\alpha,\beta,\gamma)$
$(n,k=0,1,2,\ldots)$, such that
\begin{equation}\label{e:unidefinition1}
\mathbf{v}_\alpha(x)=\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{v}_\beta(x-\gamma)
\end{equation}
where $\mathbf{v}_\alpha(x)$ is \emph{the vector of the
generalized factorials with real increment} $\alpha$, as follows
\begin{equation}\label{e:GFactorialV}
\mathbf{v}_\alpha(x)=(1,x,(x|\alpha)_2,\ldots,(x|\alpha)_n,\ldots)^T.
\end{equation}
\end{definition}

\begin{remark}\label{r:BasicStirling}
It can easily be shown that equations (\ref{e:unidefinition0}) and
(\ref{e:unidefinition1}) are equivalent. We may find immediately
that matrices $\mathbf{S}_{0,0,0}=\mathbf{E}$,
$\mathbf{S}_{0,0,1}=\mathbf{B}$,
$\mathbf{S}_{1,0,0}=\mathbf{S}_1$,
$\mathbf{S}_{0,1,0}=\mathbf{S}_2$ are, respectively, the
infinite-dimensional unit matrix and the matrices of the binomial
transform, the Stirling transforms of the first and second kind
for integer sequences.
\end{remark}

\begin{remark}\label{r:Inverse}
We see from the matrix definition of the unified generalized
Stirling numbers, (\ref{e:unidefinition1}), that $$
\mathbf{v}_\beta(x-\gamma)=\mathbf{S}_{\beta,\alpha,-\gamma}\mathbf{v}_\alpha(x).$$
This leads to that $$
\mathbf{v}_\alpha(x)=\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{S}_{\beta,\alpha,-\gamma}\mathbf{v}_\alpha(x),$$
namely
$(\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{S}_{\beta,\alpha,-\gamma}-\mathbf{E})\mathbf{v}_\alpha(x)=\mathbf{0}$
(the infinite-dimensional zero matrix). Because $x$ is an
arbitrary real or complex number, we have
$\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{S}_{\beta,\alpha,-\gamma}=\mathbf{E}$,
namely matrix $\mathbf{S}_{\beta,\alpha,-\gamma}$ is the inverse
of $\mathbf{S}_{\alpha,\beta,\gamma}$.
\end{remark}



We first investigate three basic types of the generalized Stirling
matrices: $\mathbf{S}_{0,0,\gamma}$, $\mathbf{S}_{\alpha,0,0}$,
and $\mathbf{S}_{0,\beta,0}$.

According to the binomial theorem, we have that
$$(x|0)_n=x^n=\sum_{k=0}^n
\binom{n}{k}\gamma^{n-k}(x-\gamma)^k=\sum_{k=0}^n\binom{n}{k}\gamma^{n-k}(x-\gamma|0)_k.$$
Hence, the $(n,k)$th entry of $\mathbf{S}_{0,0,\gamma}$ is that
\begin{equation}\label{e:Gamma}
S(n,k;0,0,\gamma)=\gamma^{n-k}\binom{n}{k}, \qquad
(n,k=0,1,2,\ldots).
\end{equation}

\begin{remark}\label{r:Gamma}
In case $\gamma$ is an integer, $\mathbf{S}_{0,0,\gamma}$ is just
the $\gamma$-fold binomial transform matrix for an integer
sequence, namely $\gamma$ successive binomial transform for an
integer sequence (see \cite{ref4}).
\end{remark}

Next, we look at the generalized Stirling numbers
$S(n,k;\alpha,0,0)$ ($\alpha\neq 0$) defined in expression
$(x|\alpha)_n=\sum_{k=0}^n S(n,k;\alpha,0,0)x^k$,
$(n,k=0,1,2,\ldots)$.

Because $(x|\alpha)_n=\alpha^n(\frac{x}{\alpha}|1)_n$ and
$x^k=\alpha^k(\frac{x}{\alpha})^k$, $$
\alpha^n(\frac{x}{\alpha}|1)_n=\sum_{k=0}^n
S(n,k;\alpha,0,0)\alpha^k(\frac{x}{\alpha})^k.$$ On the other
hand, $(\frac{x}{\alpha}|1)_n=\sum_{k=0}^n
S(n,k;1,0,0)(\frac{x}{\alpha})^k$. Hence, we have
$S(n,k;\alpha,0,0)\alpha^k=\alpha^n S(n,k;1,0,0)$, which implies
that
\begin{equation}\label{e:Alpha}
S(n,k;\alpha,0,0)=\alpha^{n-k}S(n,k;1,0,0)=\alpha^{n-k}s(n,k),
\qquad (n,k=0,1,2,\ldots).
\end{equation}
Namely, the $(n,k)$th entry of matrix $\mathbf{S}_{\alpha,0,0}$ is
the Stirling number $s(n,k)$ of the first kind, multiplied by
$\alpha^{n-k}$.

Similarly, we may obtain that when $\beta\neq 0$,
\begin{equation}\label{e:Beta}
S(n,k;0,\beta,0)=\beta^{n-k}S(n,k;0,1,0)=\beta^{n-k}S(n,k), \qquad
(n,k=0,1,2,\ldots).
\end{equation}
Namely, the $(n,k)$th entry of matrix $\mathbf{S}_{0,\beta,0}$ is
the Stirling number $S(n,k)$ of the second kind, multiplied by
$\beta^{n-k}$.

\begin{remark}\label{r:IntegerMatrices}
We see from (\ref{e:Gamma}), (\ref{e:Alpha}), and (\ref{e:Beta})
that, if $\theta$ is an integer, then each one of the generalized
Stirling numbers $S(n,k;\theta,0,0)$, $S(n,k;0,\theta,0)$ and
$S(n,k;0,0,\theta)$) is an integer number.
\end{remark}

\begin{remark}\label{r:GStirling}
In case $\alpha$ (or $\beta$) is an integer, matrix
$\mathbf{S}_{\alpha,0,0}$ (or $\mathbf{S}_{0,\beta,0}$) defines a
\emph{generalized Stirling transform matrix of the first (or
second) kind for an integer sequence}. They both are a pair of
transform and inverse transform of integer sequences, if and only
if $\alpha=\beta$.
\end{remark}



Now we may give a general decomposition formula of the unified
generalized Stirling matrices, $\mathbf{S}_{\alpha,\beta,\gamma}$.

\begin{theorem}\label{t:Decomposition}
Let $\mathbf{S}_{\alpha,\beta,\gamma}$ be a generalized Stirling
matrix with real parameters $\alpha,\beta,\gamma$. Then
\begin{equation}\label{e:Decomposition}
\mathbf{S}_{\alpha,\beta,\gamma}=\mathbf{S}_{\alpha,0,0}\mathbf{S}_{0,0,\gamma}\mathbf{S}_{0,\beta,0}.
\end{equation}
\end{theorem}

\begin{proof}
We see from (\ref{e:unidefinition1}) that $$
\mathbf{v}_\alpha(x)=\mathbf{S}_{\alpha,0,0}\mathbf{v}_0(x),\quad
\mathbf{v}_0(x)=\mathbf{S}_{0,0,\gamma}\mathbf{v}_0(x-\gamma),\quad
\mathbf{v}_0(x-\gamma)=\mathbf{S}_{0,\beta,0}\mathbf{v}_\beta(x-\gamma).$$
Hence we have $$
\mathbf{v}_\alpha(x)=\mathbf{S}_{\alpha,0,0}\mathbf{S}_{0,0,\gamma}\mathbf{S}_{0,\beta,0}
\mathbf{v}_\beta(x-\gamma).$$ On the other hand, we also have
$\mathbf{v}_\alpha(x)=\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{v}_\beta(x-\gamma)$.
Thus,
$$(\mathbf{S}_{\alpha,\beta,\gamma}-\mathbf{S}_{\alpha,0,0}\mathbf{S}_{0,0,\gamma}\mathbf{S}_{0,\beta,0})
\mathbf{v}_\beta(x-\gamma)=\mathbf{0}.$$ Because $x$ is an
arbitrary real or complex number, the equality
(\ref{e:Decomposition}) holds.
\end{proof}

\begin{remark}\label{r:Transform}
In case all of $\alpha,\beta,\gamma$ are integer numbers, we may
regard $\mathbf{S}_{\alpha,\beta,\gamma}$ as a transform matrix of
integer sequences. However, we see from (\ref{e:Decomposition})
that, it is better to regard $\mathbf{S}_{\alpha,\beta,\gamma}$ as
a composition of three successive transforms:
$\mathbf{S}_{0,\beta,0}$, $\mathbf{S}_{0,0,\gamma}$ and
$\mathbf{S}_{\alpha,0,0}$.
\end{remark}

\begin{remark}\label{r:BTransform}
As we know that, a linear homogeneous recurrent integer sequence
$a(n)$ of order $q$ has the general-term with the following form:
$a(n)=\sum^{q}_{i=1}c_i\lambda_i^n$, ($n=0,1,2,\ldots$), where
$\lambda_i$, ($i=1,\ldots,q$) are its $q$ real or complex
characteristic values. We see from \cite{ref5} that if $\gamma$ is
an integer, the $q$ characteristic values of the $\gamma$-fold
binomial transform $b(n)$ of integer sequence $a(n)$ are
$\lambda_i+\gamma$, and the general-term of integer sequence
$b(n)$ is $b(n)=\sum^{q}_{i=1}c_i(\lambda_i+\gamma)^n$. Now, we
may also obtain this conclusion by using the basic relation
(\ref{e:unidefinition1}). Denoting the vectors corresponding to
integer sequences $a(n)$ and $b(n)$ by
$\mathbf{a}=(a(0),a(1),\cdots,a(n),\cdots)^T$ and
$\mathbf{b}=(b(0),b(1),\cdots,b(n),\cdots)^T$, we have
$\mathbf{a}=\sum^{q}_{i=1}c_i\mathbf{v}_0(\lambda_i)$, and $$
\mathbf{b}=\mathbf{S}_{0,0,\gamma}\mathbf{a}=\mathbf{S}_{0,0,\gamma}
\sum^{q}_{i=1}c_i\mathbf{v}_0(\lambda_i)=\sum^{q}_{i=1}c_i\mathbf{S}_{0,0,\gamma}
\mathbf{v}_0(\lambda_i)=\sum^{q}_{i=1}c_i\mathbf{v}_0(\lambda_i+\gamma).$$
\end{remark}

Before we mention the next theorem, let us consider first the
following definition.

\begin{definition}\label{d:AlphaGST}
Let $\alpha$ be an integer, and $a(n)$ ($n=0,1,2,\ldots$) be a
linear homogeneous recurrent integer sequence of order $q$.
\emph{The $\alpha$-generalized Stirling transform of the first
kind} of $a(n)$ is defined by $$
b(n)=\sum_{k=0}^nS(n,k;\alpha,0,0)a(k). $$
\end{definition}

Now, let us give another property related to the recurrent integer
sequences, as follows.

\begin{theorem}\label{t:GSTransform}
The $\alpha$-generalized Stirling transform of the first kind of a
linear homogeneous recurrent integer sequence $$
a(n)=\sum^{q}_{i=1}c_i\lambda_i^n,\quad  n=0,1,2,\ldots, $$ is an
integer sequence with the general-term
$$b(n)=\sum^{q}_{i=1}c_i(\lambda_i|\alpha)_n,\quad
n=0,1,2,\ldots,$$ where $\lambda_i$, ($i=1,\ldots,q$) are $q$ real
or complex characteristic values of $a(n)$.
\end{theorem}

\begin{proof}
Denoting the vectors corresponding to integer sequences $a(n)$ and
$b(n)$ by $\mathbf{a}=(a(0),a(1),\cdots,a(n),\cdots)^T$ and
$\mathbf{b}=(b(0),b(1),\cdots,b(n),\cdots)^T$, we have $
\mathbf{a}=\sum^{q}_{i=1}c_i\mathbf{v}_0(\lambda_i)$, and $$
\mathbf{b}=\mathbf{S}_{\alpha,0,0}\mathbf{a}=\mathbf{S}_{\alpha,0,0}
\sum^{q}_{i=1}c_i\mathbf{v}_0(\lambda_i)=\sum^{q}_{i=1}c_i\mathbf{S}_{\alpha,0,0}
\mathbf{v}_0(\lambda_i)=\sum^{q}_{i=1}c_i\mathbf{v}_\alpha(\lambda_i),$$
that is, $b(n)=\sum^{q}_{i=1}c_i(\lambda_i|\alpha)_n$,
($n=0,1,2,\ldots$). At the same time, we see from (\ref{e:Alpha})
that if $\alpha$ is integer, each of entries of
$\mathbf{S}_{\alpha,0,0}$ is also an integer, which implies that
$b(n)$ is an integer sequences.
\end{proof}

\begin{example}\label{ex:GStirlingT}
For example, the general-term of the recurrent integer sequence of
order $2$ of Lucas numbers $L(n)=2$, $1$, $3$, $4$, $7$, $11$,
$18$, $\ldots$ (A000032\cite{ref6}) is
$L(n)=c_1\lambda_1^n+c_2\lambda_2^n=(\frac{1}{2}+\frac{\sqrt{5}}{2})^n+
(\frac{1}{2}-\frac{\sqrt{5}}{2})^n$. Hence, we find from Theorem
\ref{t:GSTransform} that the general-term of its Stirling
transform of the first kind is $$
 L_1(n)=(\frac{1}{2}+\frac{\sqrt{5}}{2}|1)_n+
(\frac{1}{2}-\frac{\sqrt{5}}{2}|1)_n, \qquad n=0,1,2,\ldots, $$
that is, $L_1(n)=2$, $1$, $2$, $-3$, $10$, $-45$, $250$, $-1645$,
$\ldots$ (A213593\cite{ref6}).
\end{example}


\section{Generalized Factorial Matrices}

We may see that if taking $\mathbf{S}_{\alpha,\beta,\gamma}$ to be
a $p$-dimensional ($p=1,2,3,\ldots$) matrix, which in fact is the
$p\times p$ upper-left sub-matrix of the (\emph{original})
infinite dimensional generalized Stirling matrix, then all of the
conclusions presented in the preceding section still hold. We need
to remember always this point of view while reading this section.



\begin{definition}\label{d:GFactorialMatrix}
Let $x_1,x_2,\ldots,x_p$ be $p$ distinct real or complex numbers,
and $\alpha$ be a given real parameter. The \emph{Generalized
Factorial Matrices} of order $p$ is a $p\times p$ matrix, denoted
by $\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)$, whose $i$th column
entries are the entries of the $p$-dimensional column vector
$\mathbf{v}_\alpha(x_i)$ ($i=1,\ldots,p$). That is
\begin{equation}\label{e:Vandermonde}
\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)=[\mathbf{v}_\alpha(x_1),\mathbf{v}_\alpha(x_2),\cdots,\mathbf{v}_\alpha(x_p)].
\end{equation}
In particular, $\mathbf{V}_0(x_1,x_2,\ldots,x_p)$ is just the
Vandermonde matrix of $p$ distinct parameters
$x_1,x_2,\ldots,x_p$. Hence we may regard the generalized
factorial matrices as a generalization in form of the Vandermonde
matrices.
\end{definition}

\begin{example}\label{ex:GVandermonde10}
For example, the $6\times 6$ matrices
$\mathbf{V}_\alpha(0,1,2,3,4,5)$, ($\alpha=0,1,2,3$) are,
respectively, $$
\mathbf{V}_0(0,1,2,3,4,5)=\left(\begin{array}{rrrrrr} 1 &amp; 1 &amp; 1 &amp;
1 &amp; 1 &amp; 1
\\ 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ 0 &amp; 1 &amp; 4 &amp; 9 &amp;
16 &amp; 25 \\ 0 &amp; 1 &amp; 8 &amp; 27 &amp; 64  &amp; 125 \\ 0 &amp; 1 &amp; 16 &amp; 81 &amp; 256 &amp;
625
\\ 0 &amp; 1 &amp; 32 &amp; 243 &amp; 1024 &amp; 3125
   \end{array} \right),
$$
 $$ \mathbf{V}_1(0,1,2,3,4,5)=\left(\begin{array}{rrrrrr} 1 &amp; 1 &amp;
1 &amp; 1 &amp; 1 &amp; 1
\\ 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ 0 &amp; 0 &amp; 2 &amp; 6 &amp;
12 &amp; 20 \\ 0 &amp; 0 &amp; 0 &amp; 6 &amp; 24  &amp; 60 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 24 &amp; 120
\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 120
   \end{array} \right),
$$
 $$ \mathbf{V}_2(0,1,2,3,4,5)=\left(\begin{array}{rrrrrr} 1 &amp; 1 &amp;
1 &amp; 1 &amp; 1 &amp; 1
\\ 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ 0 &amp; -1 &amp; 0 &amp; 3 &amp;
8 &amp; 5 \\ 0 &amp; 3 &amp; 0 &amp; -3 &amp; 0  &amp; 15 \\ 0 &amp; -15 &amp; 0 &amp; 9 &amp; 0 &amp; -15
\\ 0 &amp; 105 &amp; 0 &amp; -45 &amp; 0 &amp; 45
\end{array} \right),
$$
 and $$ \mathbf{V}_3(0,1,2,3,4,5)=\left(\begin{array}{rrrrrr} 1 &amp;
1 &amp; 1 &amp; 1 &amp; 1 &amp; 1
\\ 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ 0 &amp; -2 &amp; -2 &amp; 0 &amp;
4 &amp; 10 \\ 0 &amp; 10 &amp; 8 &amp; 0 &amp; -8  &amp; -10 \\ 0 &amp; -80 &amp; -56 &amp; 0 &amp; 40 &amp;
40
\\ 0 &amp; 880 &amp; 560 &amp; 0 &amp; -320 &amp; -280
   \end{array} \right).
$$
\end{example}



The next theorem gives basic properties of the generalized
factorial matrices.

\begin{theorem}\label{t:Vandermonde}
Let $x_1,x_2,\ldots,x_p$ be $p$ distinct real or complex numbers,
and $\alpha$,$\beta$,$\gamma$ be three real parameters. Then the
generalized factorial matrices have the following basic
properties.

(i) Two generalized factorial matrices can be connected with a
suitable generalized Stirling matrice, generally like
\begin{equation}\label{e:VSrelation}
\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)=\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{V}_\beta(x_1-\gamma,x_2-\gamma,\ldots,x_p-\gamma).
\end{equation}

(ii) Determinant of the generalized factorial matrix
$\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)$ is
\begin{equation}\label{e:Determinant}
\det\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)=\det\mathbf{V}_0(x_1,x_2,\ldots,x_p)=\prod_{1\leq
i&lt;j\leq p}(x_j-x_i).
\end{equation}
This implies that matrix $\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)$
is invertible.

(iii) The inverse of the generalized factorial matrix
$\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)$ is
\begin{equation}\label{e:InverseV}
\mathbf{V}^{-1}_\alpha(x_1,x_2,\ldots,x_p)=\mathbf{V}^{-1}_0(x_1,x_2,\ldots,x_p)\mathbf{S}_{0,\alpha,0},
\end{equation}
namely, the inverse of the Vandermonde matrix with the same
parameters, right-multiplied by the $\alpha$-generalized Stirling
matrix of the second kind.
\end{theorem}

\begin{proof}
The matrix equality (\ref{e:VSrelation}) is the matrix form of $p$
equalities
$\mathbf{v}_\alpha(x_i)=\mathbf{S}_{\alpha,\beta,\gamma}\mathbf{v}_\beta(x_i-\gamma)$,
($i=1,2,\ldots,p$). From (\ref{e:VSrelation}), we have
$\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)=\mathbf{S}_{\alpha,0,0}\mathbf{V}_0(x_1,x_2,\ldots,x_p)$.
Then, we obtain $$
\det\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)=\det\mathbf{S}_{\alpha,0,0}\det\mathbf{V}_0(x_1,x_2,\ldots,x_p).$$
Because $\det\mathbf{S}_{\alpha,0,0}=1$ and
$\det\mathbf{V}_0(x_1,x_2,\ldots,x_p)=\prod_{1\leq i&lt;j\leq
p}(x_j-x_i)$, the property (\ref{e:Determinant}) holds. Besides,
we also see that $$
\mathbf{V}^{-1}_\alpha(x_1,\ldots,x_p)=\mathbf{V}^{-1}_0(x_1,\ldots,x_p)\mathbf{S}^{-1}_{\alpha,0,0}
=\mathbf{V}^{-1}_0(x_1,\ldots,x_p)\mathbf{S}_{0,\alpha,0},$$ that
is, equality (\ref{e:InverseV}) holds.
\end{proof}

\begin{corollary}\label{c:InverseV}
Let $\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)$ be a generalized
factorial matrices of $p$ distinct real or complex numbers
$x_1,x_2,\ldots,x_p$, with real parameter $\alpha$. Then, we may
express the inverse of matrix
$\mathbf{V}_\alpha(x_1,x_2,\ldots,x_p)$ as
\begin{equation}\label{e:InverseV1}
\mathbf{V}^{-1}_\alpha(x_1,x_2,\ldots,x_p)=\mathbf{H}_p\mathbf{L}_p\mathbf{S}_{0,\alpha,0}.
\end{equation}
where factor matrix $\mathbf{L}_p$ is a lower triangular matrix,
the arbitrary entry $L_p(n,k)$ ($n,k=0,1,\ldots,p-1$) of which is
determined by the following linear relations
\begin{equation}\label{e:Lp}
L_p(0,0)=1, \quad \mbox{and} \quad
\prod_{i=1}^n(x-x_i)=\sum_{k=0}^n L_p(n,k)x^k,
\end{equation}
and the matrix factor $\mathbf{H}_p$ is a upper triangular matrix,
the arbitrary entry $h_p(n,k)$ ($n,k=0,1,\ldots,p-1$) of which is
given by
\begin{equation}\label{e:Hp}
h_p(0,0)=1, \quad h_p(n,k)=\left\{\begin{array}{ll} 0, &amp; \mbox{if}
\quad k&lt;n,
\\ \Big(\prod_{i=1,i\neq n+1}^{k+1}(x_{n+1}-x_i)\Big)^{-1}, &amp; \mbox{if} \quad
k\geq n.
\end{array} \right.
\end{equation}
\end{corollary}

\begin{proof}
By using a triangular-matrix decomposition method given by Hou and
Hou \cite{ref7}, we may obtain $$
\mathbf{V}_0^{-1}(x_1,x_2,\ldots,x_p)=\mathbf{H}_p\mathbf{L}_p,$$
where the lower triangular matrix $\mathbf{L}_p$ and the upper
triangular matrix $\mathbf{H}_p$ are calculated by using
(\ref{e:Lp}) and (\ref{e:Hp}). Hence, we see from
(\ref{e:InverseV}) that the Corollary holds.
\end{proof}

\begin{example}\label{ex:GVandermonde}
In case $x_1=0$, $x_2=1$, $x_3=2$,$\ldots$, $x_p=p-1$, we have
$\mathbf{L}_p=\mathbf{S}_{1,0,0}$, and
$\mathbf{H}_p=\mathbf{S}^T_{0,0,-1}\mathbf{D_p}$, where matrix
$\mathbf{D_p}$ is a diagonal matrix, the $k$th diagonal entry of
which is $\frac{1}{k!}$, ($k=0,1,\ldots,p-1$). Hence,
\begin{equation}\label{e:InverseVV}
\mathbf{V}_\alpha^{-1}(0,1,\ldots,p-1)=\mathbf{H}_p\mathbf{S}_{1,0,0}\mathbf{S}_{0,\alpha,0}
=\mathbf{S}^T_{0,0,-1}\mathbf{D_p}\mathbf{S}_{1,0,0}\mathbf{S}_{0,\alpha,0}.
\end{equation}
namely, $\mathbf{V}_\alpha^{-1}(0,1,\ldots,p-1)$ is a product of
several triangular matrices. For example, we may obtain the
inverse matrices of the $6\times 6$ generalized factorial matrix
$\mathbf{V}_\alpha(0,1,2,3,4,5)$ ($\alpha=0,1,2,3$) (see Example
\ref{ex:GVandermonde10}), as follows $$
\mathbf{V}_0^{-1}(0,1,2,3,4,5)=\mathbf{H}_6\mathbf{S}_{1,0,0}\mathbf{S}_{0,0,0}=\mathbf{H}_6\mathbf{S}_{1,0,0}
=\left(\begin{array}{rrrrrr} 5 &amp; -\frac{77}{12} &amp; \frac{71}{24} &amp;
-\frac{7}{12} &amp; \frac{1}{24} &amp;
\\ -10 &amp; \frac{107}{6} &amp; -\frac{59}{6} &amp; \frac{13}{6} &amp; -\frac{1}{6} &amp;  \\ 10 &amp; -\frac{39}{2} &amp; \frac{49}{4} &amp; -3 &amp;
\frac{1}{4} &amp;  \\ -5 &amp; \frac{61}{6} &amp; -\frac{41}{6} &amp; \frac{11}{6}
&amp; -\frac{1}{6}  &amp;
\\ 1 &amp; -\frac{25}{12} &amp; \frac{35}{24} &amp; -\frac{5}{12} &amp; \frac{1}{24} &amp;
\\  &amp;  &amp;  &amp;  &amp;  &amp;
   \end{array} \right),
$$ $$
\mathbf{V}^{-1}_1(0,1,2,3,4,5)=\mathbf{H_6}\mathbf{S}_{1,0,0}\mathbf{S}_{0,1,0}=\mathbf{H_6}
=\left(\begin{array}{rrrrrr} 1 &amp; -1 &amp; \frac{1}{2} &amp; -\frac{1}{6} &amp;
\frac{1}{24} &amp; -\frac{1}{120}
\\ 0 &amp; 1 &amp; -1 &amp; \frac{1}{2} &amp; -\frac{1}{6} &amp; \frac{1}{24} \\ 0 &amp; 0 &amp; \frac{1}{2} &amp; -\frac{1}{2} &amp;
\frac{1}{4} &amp; -\frac{1}{12} \\ 0 &amp; 0 &amp; 0 &amp; \frac{1}{6} &amp;
-\frac{1}{6}  &amp; \frac{1}{12} \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; \frac{1}{24} &amp;
-\frac{1}{24}
\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \frac{1}{120}
   \end{array} \right),
$$ $$
\mathbf{V}^{-1}_2(0,1,2,3,4,5)=\mathbf{H_6}\mathbf{S}_{1,0,0}\mathbf{S}_{0,2,0}
=\left(\begin{array}{rrrrrr} 1 &amp; -\frac{1}{2} &amp; \frac{1}{8} &amp;
-\frac{1}{24} &amp; -\frac{1}{24} &amp; -\frac{1}{120}
\\ 0 &amp; 0 &amp; 0 &amp; \frac{1}{8} &amp; \frac{1}{4} &amp; \frac{1}{24} \\ 0 &amp; \frac{1}{2} &amp; -\frac{1}{4} &amp; -\frac{1}{4} &amp;
-\frac{7}{12} &amp; -\frac{1}{12} \\ 0 &amp; 0 &amp; 0 &amp; \frac{5}{12} &amp;
\frac{2}{3}  &amp; \frac{1}{12} \\ 0 &amp; 0 &amp; \frac{1}{8} &amp; -\frac{3}{8}
&amp; -\frac{3}{8} &amp; -\frac{1}{24}
\\ 0 &amp; 0 &amp; 0 &amp; \frac{1}{8} &amp; \frac{1}{12} &amp; \frac{1}{120}
   \end{array} \right),
$$
and $$
 \mathbf{V}_3^{-1}(0,1,2,3,4,5)=\mathbf{H_6}\mathbf{S}_{1,0,0}\mathbf{S}_{0,3,0}=\left(\begin{array}{rrrrrr} 1 &amp;
-\frac{1}{3} &amp; 0 &amp; -\frac{1}{3} &amp; -\frac{1}{8} &amp; -\frac{1}{120}
\\ 0 &amp; 0 &amp; \frac{1}{3} &amp; \frac{11}{6} &amp; \frac{2}{3} &amp; \frac{1}{24} \\ 0 &amp; 0 &amp; -\frac{5}{6} &amp; -\frac{25}{6} &amp;
-\frac{17}{12} &amp; -\frac{1}{12} \\ 0 &amp; \frac{1}{3} &amp; 1 &amp;
\frac{29}{6} &amp; \frac{3}{2}  &amp; \frac{1}{12} \\ 0 &amp; 0 &amp; -\frac{5}{6}
&amp; -\frac{17}{6} &amp; -\frac{19}{24} &amp; -\frac{1}{24}
\\ 0 &amp; 0 &amp; \frac{1}{3} &amp; \frac{2}{3} &amp; \frac{1}{6} &amp; \frac{1}{120}
   \end{array} \right),
$$ where matrices $\mathbf{S}_{0,2,0}$ and $\mathbf{S}_{0,3,0}$
are the $2$- and $3$-generalized Stirling matrices of the second
kind, their arbitrary entries are $2^{n-k}S(n,k)$ and
$3^{n-k}S(n,k)$ respectively.
\end{example}


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



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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B73; Secondary 11C20.

\noindent {\it Keywords}: unified generalized Stirling numbers,
generalized Stirling matrices, generalized factorial matrices,
recurrence, integer sequence.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences \seqnum{A000032} and
\seqnum{A213593}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 15 2012;
revised version received  June 15 2012.
Published in {\it Journal of Integer Sequences}, June 26 2012.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                

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