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\pagestyle{myheadings}
\markboth{CHU Wenchang: Double Binomial Summations}
         {CHU Wenchang: Double Binomial Summations}{}
\title{Inverse Series Relations, Formal Power Series\\
       and Blodgett-Gessel's Type Binomial Identities}
\author{{\bf CHU Wenchang}\thanks{Partially spported 
		 by Alexander von Humboldt Foundation.}\\
 \rm             Istituto di Matematica\, 
               \, ``Guido Castelnuovo''\\
               Universit\a`{a} degli Studi di Roma 
               \,``La Sapienza''\\
               00185 Roma,\hspace{2.5mm}Italia
}
\date{}   

\begin{document}

\maketitle

\thispagestyle{empty}

\begin{abstract}
A pair of simple bivariate inverse series relations are used 
by embedding machinery to produce several double summation 
formulae on shifted factorials (or binomial coefficients), 
including the evaluation due to Blodgett-Gessel \cite{kn:bl}. 
Their $q$-analogues are established in the second section. 
Some generalized convolutions are presented through formal 
power series manipulation.
\end{abstract}

\section{Bivariate Inverse Series Relations}
%\setcounter{equation}{-1}

For any given bivariate function $f(x,y)$,  define its difference by 
\bn{a}
f_{\Delta}(x,y)=f(x,y)-f(x-1,y)-f(x,y-1),
\en{-1}\\[-3mm]
where it is supposed that $f(x,y)=0$ if $x$ or/and $y$ is less than zero. 
Then it is trival to verify 

{\bf Lemma}\hspace{2mm} Bivariate inverse series relations
\bns{1}
f_{\Delta}(m,n)&=&\sum_{i=0}^{m}\sum_{j=0}^{n}
          \:\left(\ba{c}m+n-i-j-2\\m-i,\:n-j\ea\right)
          \:f(i,j), \label{eq:Ia}\\[3mm]
f(m,n)&=&\sum_{i=0}^{m}\sum_{j=0}^{n}
          \:\left(\ba{c}m+n-i-j\\m-i,\:n-j\ea\right)
          \:f_{\Delta}(i,j); \label{eq:Ib}
\ens

where the multinomial coefficient is given by 
\[\left(\ba{c}x\\m,\,n\ea\right)\:=
  \:\left(\ba{c}x\\m+n\ea\right)
  \,\left(\ba{c}m+n\\m,\,n\ea\right).\]

Actually, Eq.(1.1b) is a restatement of difference (1.1a). 
But Eq.(1.1c) would produce innumerable summation formulae 
for various specified bivariate function $f(x,y)$.

Denote the rising-factorial by
\[(x)_n=x(x+1)\cdots(x+n-1)\hspace{3mm}(n>0),
 \hspace{5mm} (x)_0=1.\]
Some examples may be displayed here.

\begin{exam} For
\bn{a}
A(m,n)=\frac{(1+2x)_{\bp{\frac{m+n}{2}}}\,
               (\frac{1}{2}+x)_{\tp{\frac{m+n}{2}}}}
              {{\bp{\frac{m}{2}}}!\,
               {\bp{\frac{n}{2}}}!\,
               (\frac{1}{2}+x)_{\tp{\frac{m}{2}}}\,
               (\frac{1}{2}+x)_{\tp{\frac{n}{2}}}},
\en{-1}\\[-3mm]
its difference is given by
\bn{b}
A_{\Delta}(m,n)=\left\{\ba{lr}
     \frac{4x}{4x+m+n}\:A(m,n),&m,n\hzl{3.5mm}even;\\
      0,& otherwise.\ea\right.
\en{-1}\\[-3mm]
The corresponding summation formula
\bn{c}
\sum_{i\geq 0}\sum_{j\geq 0}
          \:\left(\ba{c}m+n-2i-2j\\m-2i,\:n-2j\ea\right)
          \:\frac{(2x)_{i+j}\,(\frac{1}{2}+x)_{i+j}}
            {i!\,j!\,(\frac{1}{2}+x)_i\,(\frac{1}{2}+x)_j}\:=
\en{0}
\[\hspace{-15mm}=\;\frac{(1+2x)_{\bp{\frac{m+n}{2}}}\,
               (\frac{1}{2}+x)_{\tp{\frac{m+n}{2}}}}
              {{\bp{\frac{m}{2}}}!\,
               {\bp{\frac{n}{2}}}!\,
               (\frac{1}{2}+x)_{\tp{\frac{m}{2}}}\,
               (\frac{1}{2}+x)_{\tp{\frac{n}{2}}}},\]
is Gessel's extension of Blodgett's identity {\em\cite{kn:bl}}.
\end{exam}
\begin{exam} For 
\bn{a}
B(m,n)=\frac{(1+x)_{m+n}}{(1+y)_m\,(1+z)_n},
\en{-1}\\[-3mm]
its difference is given by
\bn{b}
B_{\Delta}(m,n)=\frac{x-y\cdot \tau(m)-z\cdot \tau(n)}{x+m+n}\:B(m,n);
\en{-1}\\[-3mm]
where the sign function is defined by 
\[\tau(x)\:=\:\left\{\ba{rl}+1,&x>0\\             
                            -1,&x<0\\                 
                             0,&x=0.\ea\right.\]
Then the corresponding summation formula is
\bn{c}
\sum_{i=0}^m\sum_{j=0}^{n}
          \,\left(\ba{c}m+n-i-j\\m-i,\:n-j\ea\right)
          \,\frac{(1+x)_{i+j}}
                 {(1+y)_i\,(1+z)_j}
          \,\frac{x-y\cdot \tau(i)-z\cdot \tau(j)}{x+i+j}
          \;=\;\frac{(1+x)_{m+n}}
                          {(1+y)_{m}\,
                           (1+z)_n}.
\en{0}
\end{exam}


\begin{exam} For
\bn{a}
C(m,n)=\frac{(1+x)_{m+n}\,(1+y)_{m+n}}
            {(1+x)_m\,(1+y)_m\,(1+x)_n\,(1+y)_n},
\en{-1}\\[-3mm]
its difference is given by
\bn{b}
C_{\Delta}(m,n)=\left\{\ba{lc}
               \frac{2mn-xy}{(x+m+n)\,(y+m+n)}\:C(m,n),&m>0,\:n>0\\
               0,&m \; or \; n=0\\
               1,& m=n=0.\ea\right.
\en{-1}\\[-3mm]
Then the corresponding summation formula is
\bn{c}
\sum_{i=1}^m\sum_{j=1}^n
          \:\left(\ba{c}m+n-i-j\\m-i,\:n-j\ea\right)
          \:\frac{(x)_{i+j}\,(y)_{i+j}}
                 {(1+x)_i\,(1+x)_j\,(1+y)_i\,(1+y)_j}
          \:\frac{2ij-xy}{xy}\:=
\en{0}
\[\hspace*{-13.5mm}=\:\frac{(1+x)_{m+n}\,(1+y)_{m+n}}
            {(1+x)_m\,(1+y)_m\,(1+x)_n\,(1+y)_n}
           \:-\:\left(\ba{c}m+n\\m,\,n\ea\right).\]
\end{exam}

\section{$q$-Analogues}
\setcounter{equation}{0}
\setcounter{exam}{0}

Recall the notation:
\begin{itemize}
\item $q$-factorial
\[(x;q)_n=\prod_{k=1}^n(1-xq^{k-1})\hspace{3mm}(n>0),
\hspace{5mm}(x;q)_0=1.\]
\item Gaussian binomial coefficient
\[\left[\ba{c}x\\n\ea\right]=\frac{(q^{x-n+1};q)_n}{(q;q)_n}.\]
\item Gaussian multinomial coefficients  
\[\left[\ba{c}x\\m,\,n\ea\right]=
\left[\ba{c}x\\m+n\ea\right]
\:\left[\ba{c}m+n\\m,\,n\ea\right].\]
\end{itemize}

Then one possible $q$-analogue for the lemma may be formulated as follows:
For any given bivariate function $f(x,y)$, define its $q$-difference by 
\bn{a}
f_{\delta}(x,y)=f(x,y)-f(x-1,y)-q^xf(x,y-1),
\en{-1}\\[-3mm]
where it is supposed either that $f(x,y)=0$ 
if $x$ or/and $y$ is less than zero. It is 
routine to check the following reciprocity 

{\bf Proposition}\hspace{2mm} Bivariate inverse series relations
\bns{1}
f_{\delta}(m,n)&=&\sum_{i=0}^{m}\sum_{j=0}^{n}
          \:\left[\ba{c}m+n-i-j-2\\m-i,\:n-j\ea\right]
          \:q^{m+n-i-j+m(n-j)}
          \:f(i,j), \label{eq:Iqa}\\[3mm]
f(m,n)&=&\sum_{i=0}^{m}\sum_{j=0}^{n}
          \:\left[\ba{c}m+n-i-j\\m-i,\:n-j\ea\right]
          \:q^{i(n-j)}
          \:f_{\delta}(i,j). \label{eq:Iqb}
\ens
 

Similarly, this proposition admits the following specifications.

\begin{exam} For
\bn{a}
A(m,n)=x^{-n}\:\frac{(q^2x^2;q^2)_{\bp{\frac{m+n}{2}}}\,
               (qx;q^2)_{\tp{\frac{m+n}{2}}}}
              {(q^2;q^2)_{\bp{\frac{m}{2}}}\,
               (q^2;q^2)_{\bp{\frac{n}{2}}}\,
               (qx;q^2)_{\tp{\frac{m}{2}}}\,
               (qx;q^2)_{\tp{\frac{n}{2}}}},
\en{-1}\\[-3mm]
its $q$-difference is given by\\[3mm]
\begin{minipage}{17.5cm}
\bn{b}
A_{\delta}(m,n)=\left\{\ba{lr}
          q^m\,x^{-n}\,\frac{1+x\,(-q)^n}{1+x}
       \,\frac{(x^2;q^2)_{\tp{\frac{m+n}{2}}}\,
               (qx;q^2)_{\bp{\frac{m+n}{2}}}}
              {(q^2;q^2)_{m/2}\,
               (q^2;q^2)_{\bp{\frac{n}{2}}}\,
               (qx;q^2)_{m/2}\,
               (qx;q^2)_{\tp{\frac{n}{2}}}},
               &m\hzl{3.5mm}even;\\
      0,&m\hzl{3.5mm}odd.\ea\right.
\en{-1}
\end{minipage}

\begin{minipage}{17cm}
The corresponding summation formula
\bn{c}
\mbox{}\hspace{-1mm}\sum_{i=0}^{\bp{m/2}}\sum_{j=0}^n
          \,x^{n-j}
          \,\left[\ba{c}m+n-2i-j\\m-2i,\:n-j\ea\right]
           \:\frac{(x^2;q^2)_{i+\tp{j/2}}\,
               (qx;q^2)_{i+\bp{j/2}}}
              {(q^2;q^2)_i\,
               (q^2;q^2)_{\bp{j/2}}\,
               (qx;q^2)_i\,
               (qx;q^2)_{\tp{j/2}}}\:\times
\en{0}
\[\hspace{5.5mm}\times\,q^{2i(n-j-1)}\:
  \frac{1+x\,(-q)^j}{1+x}\;=\;
  \frac{(q^2x^2;q^2)_{\bp{\frac{m+n}{2}}}\,
               (qx;q^2)_{\tp{\frac{m+n}{2}}}}
              {(q^2;q^2)_{\bp{\frac{m}{2}}}\,
               (q^2;q^2)_{\bp{\frac{n}{2}}}\,
               (qx;q^2)_{\tp{\frac{m}{2}}}\,
               (qx;q^2)_{\tp{\frac{n}{2}}}},\]
is the $q$-analogue of the extended Blodgett's identity {\em\cite{kn:bl}}.
\end{minipage}
\end{exam}

\begin{exam} For 
\bn{a}
B(m,n)=y^{-n}\,\frac{(qx;q)_{m+n}}{(qy;q)_m\,(qz;q)_n},
\en{-1}\\[-3mm]
its difference is given by
\bn{b}
B_{\delta}(m,n)=\left\{\ba{ll}
  \left\{q^{m+n}\frac{y^{\tau(m)}\,z^{\tau(n)}-x}
                     {1-xq^{m+n}}
   +(1-\tau(m))\frac{(1-y)\,(1-z\,q^n)}
    {1-x\,q^{m+n}}\right\}\:B(m,n),& m+n\not=0\\
    1,&m=n=0.\ea\right.
\en{-1}\\[-3mm]
Then the corresponding summation formula is
\bns{3}
&&\sum_{i=0}^m\sum_{j=0}^{n}
          y^{n-j}\,q^{i+j+i(n-j)}
          \,\left[\ba{c}m+n-i-j\\m-i,\,n-j\ea\right]
          \,\frac{(x;q)_{i+j}}{(qy;q)_i\,(qz;q)_j}
          \,\frac{y^{\tau(i)}\,z^{\tau(j)}-x}{1-x}\:=\\
&&\hspace{5mm}=\:\frac{(qx;q)_{m+n}}{(qy;q)_m\,(qz;q)_n}\:
-\:\frac{(1-y)\,(1-z)}{1-x}\sum_{k=1}^n\, y^{n-k}
\left[\ba{c}m+n-k\\m\ea\right]\:\frac{(x;q)_k}{(z;q)_k}.\nonumber
\ens
\end{exam}

\begin{exam} For
\bn{a}
C(m,n)=\frac{(qx;q)_{m+n}\,(qy;q)_{m+n}}
            {(qx;q)_m\,(qy;q)_m\,(qx;q)_n\,(qy;q)_n},
\en{-1}\\[-3mm]
its difference is given by
\bn{b}
C_{\delta}(m,n)=\left\{\ba{lc}q^m\:\frac{xy(1-q^m)\,(1-q^{2n})-(1-x)\,(1-y)}
                 {(1-xq^{m+n})\,(1-yq^{m+n})}\:C(m,n),&m>0,\; n>0\\
                 0,&m \;or \; n=0\\
                  1,& m=n=0.
   \ea\right.
\en{-1}\\[-3mm]
Then the corresponding summation formula is
\bns{3}
&&\sum_{i=1}^m\sum_{j=1}^n
          \:q^{i(n-j+1)}
          \:\left[\ba{c}m+n-i-j\\m-i,\:n-j\ea\right]
          \:\frac{(x;q)_{i+j}\,(y;q)_{i+j}}
            {(qx;q)_i\,(qy;q)_i\,(qx;q)_j\,(qy;q)_j}\:\times\\
&&\hspace{5.5mm}\times\:{\sss \frac{xy(1-q^i)\,(1-q^{2j})-(1-x)\,(1-y)}
     {(1-x)\,(1-y)}}\:=
\:\frac{(qx;q)_{m+n}\,(qy;q)_{m+n}}
            {(qx;q)_m\,(qy;q)_m\,(qx;q)_n\,(qy;q)_n}
           -\left[\ba{c}m+n\\m\ea\right].\nonumber
\ens
\end{exam}
This is a q-analogue of Eq.(1.4c), although it is not as compact as desired.

In the same way as demonstrated in {\bf Exs.(1.1-1.3)} and {\bf Exs.(2.1-2.3)}, 
the reader is expected to try to invent some more summation formulae.

\section{Formal Power Series}
\setcounter{equation}{0}
\setcounter{exam}{0}

Consider the formal power series defined by 
\[f_z(x,y)\:=\:\sum_{i,j\,=0}^{\infty}
\:\left(\ba{c}-z\\i,\,j\ea\right)
\:\left(\ba{c}i+j\\i,\,j\ea\right)
\:x^i\,y^j.\]
Introducing the redundant variable $t$ and denoting by $[t^n]\,p(t)$ 
the coefficient of $t^n$ in the power series expansion of $p(t)$, 
we can perform the following manipulation.
\bean
f_z(x,y)
&=&\sum_{i,j\,=0}^{\infty}
\:\left(\ba{c}-z\\i+j\ea\right)
\:\left(\ba{c}i+j\\i,\,j\ea\right)^2
\:x^i\,y^j\\[2mm]
&=&\sum_{k=0}^{\infty}\:\left(\ba{c}-z\\k\ea\right)
\:y^k\,[t^0]\,(1+tx)^k\,(1+\frac{1}{ty})^k\;
=\;[t^0]\{1+y(1+tx)(1+\frac{1}{ty})\}^{-z}\\
&=&(1+x+y)^{-z}\:[t^0]\left(1+\frac{t^{-1}+txy}{1+x+y}\right)^{-z}\;
=\;(1+x+y)^{-z}\:\sum_{k=0}^{\infty}
\:\frac{\left(\ba{c}-z\\2n\ea\right)\,\left(\ba{c}2n\\n\ea\right)\:x^ny^n}
       {(1+x+y)^{2n}}\\[2mm]
&=&(1+x+y)^{-z}\:_2F_1\left[\ba{cc}\frac{z}{2},&\frac{1+z}{2}\\&1\ea
                    \, ;\,\frac{4xy}{(1+x+y)^2}\right].
\eean
Some examples may be displayed as follows.
\bns{0}
f_1(x,y)&=&\frac{1}{\sqrt{(1+x+y)^2-4xy}}.\\[3mm]
f_2(x,y)&=&\frac{1+x+y}{\sqrt{\{(1+x+y)^2-4xy\}^3}}.\\[3mm]
f_3(x,y)&=&(1+x+y)^2+\frac{2xy}{\sqrt{\{(1+x+y)^2-4xy\}^5}}.
\ens\\[-3mm]
Putting
\[f(x,y)=\{(1-x-y)^2-4xy\}^{-1},\]
we can make the expansion further:
\bean
(1-x-y)^{2\eps}\,f^{\alpha}(x,y)
&=&\:(1-x-y)^{2\eps}\{(1-x-y)^2-4xy\}^{-\alpha}\\[2mm]
&=&(1-x-y)^{2\eps-2\alpha}\:\sum_{k=0}^{\infty}
\:\left(\ba{c}-\alpha\\k\ea\right)
\:\frac{(-4xy)^k}{(1-x-y)^{2k}}\\[2mm]
&=&\:\sum_{k=0}^{\infty}
\:\left(\ba{c}-\alpha\\k\ea\right)
\:(-4xy)^k
\:\sum_{i,\,j=0}^{\infty}
\:\left(\ba{c}2\eps-2\alpha-2k\\i,\,j\ea\right)
\:(-x)^i\,(-y)^j\\[2mm]
&=&\sum_{m,n=0}^{\infty}\:x^my^n
\:\sum_{k}\:(-1)^{m+n+k}\,\left(\ba{c}-\alpha\\k\ea\right)\,
\left(\ba{c}2\eps-2\alpha-2k\\m-k,n-k\ea\right)\:4^k\\[2mm]
&=&\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha-2\eps)_{m+n}}{m!\,n!}\:x^my^n
\:_3F_2\left[\ba{ccc}\alpha,&-m,&-n\\
             &\alpha-\eps,&\frac{1}{2}+\alpha-\eps\ea\right].
\eean

For $\eps=0,\,1/2,\,1,\,3/2$, the corresponding formulae 
may be displayed, in view of the Chu-Van\-der\-monde summation theorem 
\cite[p.3]{kn:bai}, as follows:
\bns{0}
\{(1-x-y)^2-4xy\}^{-\alpha}&=
&\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha)_{m+n}}{m!\,n!}
\,\frac{(\frac{1}{2}+\alpha)_{m+n}}
       {(\frac{1}{2}+\alpha)_{m}(\frac{1}{2}+\alpha)_{n}}
\:x^my^n, \label{eq:fa}\\[3mm]
\frac{1-x-y}{\{(1-x-y)^2-4xy\}^{\alpha+\frac{1}{2}}}&=
&\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha)_{m+n}}{m!\,n!}
\,\frac{(\alpha)_{m+n}}
       {(\alpha)_{m}(\alpha)_{n}}
\:x^my^n,\\[3mm]
\frac{(1-x-y)^2}{\{(1-x-y)^2-4xy\}^{\alpha+1}}&=
&\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha)_{m+n}}{m!\,n!}
\,\frac{(\frac{1}{2}+\alpha)_{m+n}}
       {(\frac{1}{2}+\alpha)_{m}(\frac{1}{2}+\alpha)_{n}}\:\times\\
&\times&\left\{1-\frac{m\,n}
         {\alpha\,(\frac{1}{2}-\alpha-m-n)}\right\}
\:x^my^n,\nonumber \\[3mm]
\frac{(1-x-y)^3}{\{(1-x-y)^2-4xy\}^{\alpha+\frac{3}{2}}}&=
&\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha)_{m+n}}{m!\,n!}
\,\frac{(\alpha)_{m+n}}
       {(\alpha)_{m}\,(\alpha)_{n}}\:\times\\
&\times&\left\{1-\frac{m\,n}
           {(\frac{1}{2}+\alpha)\,(1-\alpha-m-n)}\right\}
\:x^my^n,\nonumber \\[3mm]
\frac{1-y}{\{(1-x-y)^2-4xy\}^{\alpha+\frac{1}{2}}}&=
&\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha)_{m+n}}{m!\,n!}
\,\frac{(\alpha)_{m+n}}
       {(\alpha)_{m}\,(1+\alpha)_{n}}\:\times\\
&\times&\left\{\frac{2\alpha+m+2n}{2\alpha}\right\}
\:x^my^n;\nonumber
\ens\\[-3mm]
where the last one follows from the simple series manipulation.

 
From the generating functions exhibited above, 
now we can demonstrate some hypergeometric evaluations.
\begin{exam}
Note that $f(x,y)=(1-v/u)^{-2}\:f(u,v)$, where $u=x/(1-y)^2$ and 
$v=xy/(1-y)^2$. From 
\bean
f^{\alpha}(x,y)&=&(1-y)^{-2\alpha}\:f^{\alpha}(u,v),\\[3mm]
(1-y)\:f^{\alpha}(x,y)&=&(1-y)^{1-2\alpha}\:f^{\alpha}(u,v);
\eean
we have a pair of Saalsch\"{u}tzian summation formulae {\em\cite[page 9]{kn:bai}}
\bns{0}
&&_3F_2\left[\ba{ccc}-m,&-n,&\frac{1}{2}-\alpha-m\\
                      &\frac{1}{2}+\alpha,&1-2\alpha-2m-n\ea\right]
=\frac{(2\alpha+m)_{n}\,(\frac{1}{2}+\alpha+m)_{n}}
        {(\frac{1}{2}+\alpha)_n\,(2\alpha+2m)_n},\\[3mm]
&&_3F_2\left[\ba{ccc}-m,&-n,&\frac{1}{2}-\alpha-m\\
                      &\frac{1}{2}+\alpha,&2-2\alpha-2m-n\ea\right]
=\frac{(2\alpha+m)_{n}\,(\frac{1}{2}+\alpha+m)_{n}}
        {(\frac{1}{2}+\alpha)_n\,(2\alpha+2m)_n}\:\times\\[1mm]
&&\hspace{4.3cm}\times\;\frac{(-1+2\alpha+m+2n)\,(-1+2\alpha+2m+n)}
                             {(-1+2\alpha+m+n)\,(-1+2\alpha+2m+2n)}.\nonumber
\ens
\end{exam}

When $\alpha=1/2$, the first one reduces 
to the well-known Le Jen-shoo's identity
\[\left(\ba{c}m+n\\m\ea\right)^2\:
=\:\sum_{k\geq 0}\,\left(\ba{c}n\\k\ea\right)^2
 \:\left(\ba{c}m+2n-k\\2n\ea\right).\]

\begin{exam}
From the functional equations
\bean
f^{\alpha+\gamma}(x,y)&=&f^{\alpha}(x,y)\cdot f^{\gamma}(x,y),\\[5mm]
(1-y)\:f^{\alpha+\gamma+\frac{1}{2}}(x,y)&=&f^{\alpha}(x,y)\cdot 
(1-y)\:f^{\gamma+\frac{1}{2}}(x,y),\\[5mm]
(1-x-y)\:f^{\alpha+\gamma+\frac{1}{2}}(x,y)&=&f^{\alpha}(x,y)\cdot 
(1-x-y)\:f^{\gamma+\frac{1}{2}}(x,y);
\eean
we obtain the following double-convolution formulae
\bns{0}
\sum_{i=0}^{m}\sum_{j=0}^n\:
     \left(\ba{c}m\\i\ea\right)\,
     \left(\ba{c}n\\j\ea\right)
     \:\frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}&\times&\\
\times\:\frac{(2\gamma)_{m+n-i-j}\,
                            (\frac{1}{2}+\gamma)_{m+n-i-j}}
       {(\frac{1}{2}+\gamma)_{m-i}(\frac{1}{2}+\gamma)_{n-j}}
&=&\frac{(2\alpha+2\gamma)_{m+n}\,
       (\frac{1}{2}+\alpha+\gamma)_{m+n}}
       {(\frac{1}{2}+\alpha+\gamma)_{m}\,
        (\frac{1}{2}+\alpha+\gamma)_{n}},\nonumber\\[5mm]
\sum_{i=0}^{m}\sum_{j=0}^n\:
     \left(\ba{c}m\\i\ea\right)\,
     \left(\ba{c}n\\j\ea\right)
     \:\frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}\!&\cdot&\!       
     \frac{(2\gamma+m+2n-i-2j)}
            {(2\alpha+2\gamma+m+2n)}\:\times\\
     \times\:\frac{(2\alpha+2\gamma)}{2\gamma}
           \:\cdot\:\frac{(2\gamma)_{m+n-i-j}\,
                            (\gamma)_{m+n-i-j}}
       {(\gamma)_{m-i}(1+\gamma)_{n-j}}
&=&\frac{(2\alpha+2\gamma)_{m+n}\,(\alpha+\gamma)_{m+n}}
       {(\alpha+\gamma)_{m}(1+\alpha+\gamma)_{n}},\nonumber\\[5mm]
\sum_{i=0}^{m}\sum_{j=0}^n\:
     \left(\ba{c}m\\i\ea\right)\,
     \left(\ba{c}n\\j\ea\right)
     \:\frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}&\times&\\
\times\:\frac{(2\gamma)_{m+n-i-j}\,
                            (\gamma)_{m+n-i-j}}
       {(\gamma)_{m-i}(\gamma)_{n-j}}
&=&\frac{(2\alpha+2\gamma)_{m+n}\,(\alpha+\gamma)_{m+n}}
      {(\alpha+\gamma)_{m}(\alpha+\gamma)_{n}}.\nonumber
\ens
\end{exam}

%special examples.


Taking $\gamma=1/2$, we deduce, from the above, the double summations:
\bns{0}
\sum_{i=0}^{m}\sum_{j=0}^n\hspace{3.5mm}
     \left(\ba{c}m+n-i-j\\m-i,\,n-j\ea\right)^2\hspace{3.5mm}
     \frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {i!\,j!\,(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}&=
&\frac{(1+2\alpha)_{m+n}\,
       (1+\alpha)_{m+n}}
       {m!\,n!\,(1+\alpha)_{m}\,
        (1+\alpha)_{n}},\label{eq:dualG}\\[5mm]
\sum_{i=0}^{m}\sum_{j=0}^n\,
     \left(\ba{c}2m+2n-2i-2j\\2m-2i,\,2n-2j\ea\right)\,
     \frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {i!\,j!\,(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}&\times&\\
\times\;\frac{1+m+2n-i-2j}{1+2n-2j}\:
     \frac{1+2\alpha}{1+2\alpha+m+2n}&=
&\frac{(1+2\alpha)_{m+n}\,(\frac{1}{2}+\alpha)_{m+n}}
       {m!\,n!\,(\frac{1}{2}+\alpha)_{m}\,
                (\frac{3}{2}+\alpha)_{n}},\nonumber\\[5mm]
\sum_{i=0}^{m}\sum_{j=0}^n\,
     \left(\ba{c}2m+2n-2i-2j\\2m-2i,\,2n-2j\ea\right)\,
     \frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {i!\,j!\,(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}&=
&\frac{(1+2\alpha)_{m+n}\,(\frac{1}{2}+\alpha)_{m+n}}
       {m!\,n!\,(\frac{1}{2}+\alpha)_{m}\,
                (\frac{1}{2}+\alpha)_{n}}. \label{eq:gessel}
\ens

Among them, the last one has been indicated by Gessel \cite{kn:bl}. 
When $\alpha=1/2$\, further, Eqs.\,(\ref{eq:dualG}-\ref{eq:gessel})
will reduce to the binomial identities due to Blodgett and Jagers 
\cite{kn:bl}, respectively. 

\begin{exam}
Consider the expansion
\bean
\frac{f^{\alpha}(x^2,y^2)}{\{1-(x+y)^2\}^{\eps}}
&=&{\{1-(x+y)^2\}^{-\eps}}\;
    \left\{[1-(x+y)^2]^2+4xy[1-(x+y)^2]\right\}^{-\alpha}\\[2mm]
&=&\sum_{k}\:\left(\ba{c}-\alpha\\k\ea\right)\:
   (4xy)^k\:[1-(x+y)^2]^{-\eps-2\alpha-k}\\[2mm]
&=&\sum_{k}\:\left(\ba{c}-\alpha\\k\ea\right)\:4^k\:
   \sum_                                                                                                                                                            {i,j\;(i+j\hzl{2.5mm}evan)}\:(-1)^{(i+j)/2}
   \left(\ba{c}-\eps-2\alpha-k\\(i+j)/2\ea\right)\:
   \left(\ba{c}i+j\\i,\,j\ea\right)\:x^{k+i}\,y^{k+j}\\[2mm]
&=&\sum_{m,n\;(m+n\hzl{2.5mm}even)}\:x^m\,y^n
   \sum_{k}\:(-1)^{k+(m+n)/2}\:
   \left(\ba{c}-\alpha\\k\ea\right)\:4^k\:\times\\
&&\hspace{37.5mm}\times\:\left(\ba{c}-\eps-2\alpha-k\\
                       \frac{m+n}{2}-k\ea\right)\:
   \left(\ba{c}m+n-2k\\m-k,\,n-k\ea\right)\\[2mm]
&=&\sum_{m,n\;(m+n\hzl{2.5mm}even)}\:x^m\,y^n\:
   \left(\ba{c}m+n\\m,\,n\ea\right)\:
   \frac{(\eps+2\alpha)_{\frac{m+n}{2}}}{(\frac{m+n}{2})!}\:
   _3F_2\left[\ba{cc}-m,\;-n,&\alpha\\
              \frac{1-m-n}{2},&\eps+2\alpha\ea\right].
\eean
Then a special version of Watson's formulae {\em\cite[page 16]{kn:bai}}
\bn{a}
_3F_2\left[\ba{cc}-m,\;-n,&\alpha\\
              \frac{1-m-n}{2},&2\alpha\ea\right]\:
=\:\left\{\ba{lr}\frac{(\frac{1}{2}+\alpha)_{\frac{m+n}{2}}\,
                       (\frac{1}{2})_{\frac{m}{2}}\,
                       (\frac{1}{2})_{\frac{n}{2}}}
                      {(\frac{1}{2})_{\frac{m+n}{2}}\,
                       (\frac{1}{2}+\alpha)_{\frac{m}{2}}\,
                       (\frac{1}{2}+\alpha)_{\frac{n}{2}}},
                 &m,n\hzl{3.5mm}even;\\
                 0,&m,n\hzl{3.5mm}odd.\ea\right.\label{eq:wat}
\en{-1}\\[-3mm]
is recovered by recalling the expansion from {\em (\ref{eq:fa})},
\bn{b}
f^{\alpha}(x^2,y^2)=
\sum_{m,n=0}^{\infty}
\:\frac{(2\alpha)_{m+n}}{m!\,n!}
\,\frac{(\frac{1}{2}+\alpha)_{m+n}}
       {(\frac{1}{2}+\alpha)_{m}(\frac{1}{2}+\alpha)_{n}}
\:x^{2m}y^{2n}.
\en{-0}


From this formula and the decomposition  
$1=\frac{k+2\alpha}{2\alpha}-\frac{k}{2\alpha}$
on the summation-index, 
we get another hypergeometric identity
\bn{a}
_3F_2\left[\ba{ccc}-m,&-n,&\alpha\\
         &\frac{1-m-n}{2},&1+2\alpha\ea\right]\:
=\:\frac{(\frac{1}{2}+\alpha)_{\frac{m+n}{2}}\,
         (\frac{1}{2})_{\tp{\frac{m}{2}}}\,
         (\frac{1}{2})_{\tp{\frac{n}{2}}}}
        {(\frac{1}{2})_{\frac{m+n}{2}}\,
         (\frac{1}{2}+\alpha)_{\tp{\frac{m}{2}}}\,
         (\frac{1}{2}+\alpha)_{\tp{\frac{n}{2}}}},
\en{-1}
\vspace{-3mm}
\[\hspace*{8.5cm}(m+n\hzl{3.5mm}even),\]
which yields the expansion corresponding to $\eps=1$
\bn{b}
\frac{f^{\alpha}(x^2,y^2)}{1-(x-y)^2}\:=\:
\sum_{m,n\;(m+n\hzl{2.5mm}even)}
\:\frac{(1+2\alpha)_{\frac{m+n}{2}}\:
        (\frac{1}{2}+\alpha)_{\frac{m+n}{2}}}
       {\bp{\frac{m}{2}}!\,
        \bp{\frac{n}{2}}!\,
        (\frac{1}{2}+\alpha)_{\tp{\frac{m}{2}}}\,
        (\frac{1}{2}+\alpha)_{\tp{\frac{n}{2}}}}
        \:x^{m}y^{n}.
\en{0}

\pagebreak

Next perform expansion
\bean
\frac{f^{\alpha}(x^2,y^2)}{1-x-y}
&=&\frac{\left\{[1-(x+y)^2]^2+4xy[1-(x+y)^2]\right\}^{-\alpha}}
        {1-x-y}\\
&=&\sum_{k}\:\left(\ba{c}-\alpha\\k\ea\right)\:
   \frac{(4xy)^k}{[1-(x+y)^2]^{1+2\alpha+k}}\:(1+x+y)\\[2mm]
&=&\sum_{k}\:\left(\ba{c}-\alpha\\k\ea\right)\:4^k\:
   \sum_{i,j\;(i+j\hzl{2.5mm}evan)}\:(-1)^{(i+j)/2}
   \left(\ba{c}-1-2\alpha-k\\
               (i+j)/2\ea\right)\:\times\nonumber\\
&&\hspace{22.5mm}\times\:\left(\ba{c}i+j\\i,\,j\ea\right)
                    \:x^{k+i}\,y^{k+j}\:(1+x+y)\\[2mm]
&=&\sum_{m,n=0}^{\infty}\:x^m\,y^n
   \sum_{k}\:(-1)^{k+\bp{\frac{m+n}{2}}}\:
   \left(\ba{c}-\alpha\\k\ea\right)\:
   \left(\ba{c}-1-2\alpha-k\\\bp{\frac{m+n}{2}}-k\ea\right)\:
   \left(\ba{c}m+n-2k\\m-k,\,n-k\ea\right)\:4^k\\[2mm]
\hspace*{10mm}&=&\sum_{m,n=0}^{\infty}\:x^m\,y^n\:
   \left(\ba{c}m+n\\m,\,n\ea\right)\:
   \frac{(1+2\alpha)_{\bp{\frac{m+n}{2}}}}{\bp{\frac{m+n}{2}}!}\:
   _4F_3\left[\ba{cccc}-\bp{\frac{m+n}{2}},\red -m,\red -n,\red \alpha\\
       \red -\frac{m+n}{2},\red -\frac{m+n-1}{2},\red 1+2\alpha\ea\right],
\eean
where the hypergeometric summation can be evaluated as 
\bn{a}
_4F_3\left[\ba{cccc}-\bp{\frac{m+n}{2}},&-m,&-n,&\alpha\\
      &-\frac{m+n}{2},&-\frac{m+n-1}{2},&1+2\alpha\ea\right]\:
=\:\frac{(\frac{1}{2}+\alpha)_{\tp{\frac{m+n}{2}}}\,
         (\frac{1}{2})_{\tp{\frac{m}{2}}}\,
         (\frac{1}{2})_{\tp{\frac{n}{2}}}}
        {(\frac{1}{2})_{\tp{\frac{m+n}{2}}}\,
         (\frac{1}{2}+\alpha)_{\tp{\frac{m}{2}}}\,
         (\frac{1}{2}+\alpha)_{\tp{\frac{n}{2}}}}
\en{-1}\\[-3mm]
by {\em (\ref{eq:wat})} via decomposing 
the summation-index\\[2.5mm]
\hspace*{50mm}$1=\frac{k+2\alpha}{2\alpha}-\frac{k}{2\alpha}$ 
when $m+n$ is even\\ and \\ 
\hspace*{35mm}$1=\frac{-1-m}{-1-m+k}\,
\frac{k+2\alpha}{2\alpha}+\frac{1+2\alpha+m}{2\alpha}\:
\frac{k}{-1-m+k}$ when $m+n$ is odd.

Thus, we establish the expansion 
\bn{b}
\frac{f^{\alpha}(x^2,y^2)}{1-x-y}\:=\:
\sum_{m,n=0}^{\infty}
\:\frac{(1+2\alpha)_{\bp{\frac{m+n}{2}}}\:
        (\frac{1}{2}+\alpha)_{\tp{\frac{m+n}{2}}}}
       {\bp{\frac{m}{2}}!\,
        \bp{\frac{n}{2}}!\,
        (\frac{1}{2}+\alpha)_{\tp{\frac{m}{2}}}\,
        (\frac{1}{2}+\alpha)_{\tp{\frac{n}{2}}}}
        \:x^{m}y^{n}.
\en{0}\\[-1mm]
The identity {\bf (1.2c)} may be revisited through the combination 
of the expansion and the equation displayed as follows:
\vspace{-2mm}
\bean  
(1-x-y)^{-1}&=&\sum_{m,n=0}^{\infty}\:
\left(\ba{c}m+n\\m,\,n\ea\right)\:x^m\,y^n,\\
\frac{f^{\alpha}(x^2,y^2)}{1-x-y}&=
&f^{\alpha}(x^2,y^2)\cdot (1-x-y)^{-1}.
\eean

Two other relations 
\bean
\frac{f^{\alpha+\gamma}(x^2,y^2)}{1-x-y}&=&  
f^{\alpha}(x^2,y^2)\;\cdot\;\frac{f^{\gamma}(x^2,y^2)}{1-x-y},\\[3mm]
\frac{f^{\alpha+\gamma}(x^2,y^2)}{1-(x+y)^2}&=&
\frac{f^{\alpha}(x^2,y^2)}{1+x+y}\cdot 
\frac{f^{\gamma}(x^2,y^2)}{1-x-y};
\eean
are equivalent to the convolution formulae
\bns{0}
%This identity would not be repeated here.
%\sum_{i,\,j\geq 0}\;
%     \left(\ba{c}{\sss m+n-2i-2j}\\{\sss m-2i,\,n-2j}\ea\right)\;
%     \frac{(2\alpha)_{i+j}\,
%             (\frac{1}{2}+\alpha)_{i+j}}
%       {i!\,j!\,(\frac{1}{2}+\alpha)_{i}\,
%        (\frac{1}{2}+\alpha)_{j}}&=
%&\frac{(1+2\alpha)_{\bp{\frac{m+n}{%2}}}\,
%       (\frac{1}{2}+\alpha)_{\tp{\frac{m+n}{2}}}}
%       {\bp{\frac{m}{2}}!\,\bp{\frac{n}{2}}!\,
%        (\frac{1}{2}+\alpha)_{\tp{\frac{m}{2}}}\,
%        (\frac{1}{2}+\alpha)_{\tp{\frac{n}{2}}}}\\[3mm]
\sum_{i,\,j\geq 0}
       \left(\ba{c}\!{\sss \bp{m/2}}\!\\i\ea\right)
       \left(\ba{c}\!{\sss \bp{n/2}}\!\\j\ea\right)
\frac{(2\alpha)_{i+j}\,
             (\frac{1}{2}+\alpha)_{i+j}}
       {(\frac{1}{2}+\alpha)_{i}\,
        (\frac{1}{2}+\alpha)_{j}}&\times&\\
\times\;\frac{(1+2\gamma)_{\bp{\frac{m+n}{2}}-i-j}\,
       (\frac{1}{2}+\gamma)_{\tp{\frac{m+n}{2}}-i-j}}
       {(\frac{1}{2}+\gamma)_{\tp{\frac{m}{2}}-i}\,
        (\frac{1}{2}+\gamma)_{\tp{\frac{n}{2}}-j}}&=
&\frac{(1+2\alpha+2\gamma)_{\bp{\frac{m+n}{2}}}\,
       (\frac{1}{2}+\alpha+\gamma)_{\tp{\frac{m+n}{2}}}}
       {(\frac{1}{2}+\alpha+\gamma)_{\tp{\frac{m}{2}}}\,
        (\frac{1}{2}+\alpha+\gamma)_{\tp{\frac{n}{2}}}},\nonumber\\[3mm]
\sum_{i,\,j\geq 0}\:       
       \left(\ba{c}\bp{m/2}\\
           {\sss \bp{i/2},\,\bp{(m-i)/2}}\ea\right)\,
       \left(\ba{c}\bp{n/2}\\
           {\sss \bp{j/2},\,\bp{(n-j)/2}}\ea\right)&\cdot
&(-1)^{i+j}\,
\frac{(1+2\alpha)_{\bp{\frac{i+j}{2}}}\,
       (\frac{1}{2}+\alpha)_{\tp{\frac{i+j}{2}}}}
       {(\frac{1}{2}+\alpha)_{\tp{\frac{i}{2}}}\,
        (\frac{1}{2}+\alpha)_{\tp{\frac{j}{2}}}}\;\times\\
\times\;\frac{(1+2\gamma)_{\bp{\frac{m+n-i-j}{2}}}\,
       (\frac{1}{2}+\gamma)_{\tp{\frac{m+n-i-j}{2}}}}
        {(\frac{1}{2}+\gamma)_{\tp{\frac{m-}{2}}}\,
        (\frac{1}{2}+\gamma)_{\tp{\frac{n-j}{2}}}}&=
&\left\{\ba{r}\frac{(1+2\alpha+2\gamma)_{\frac{m+n}{2}}\,
       (\frac{1}{2}+\alpha+\gamma)_{\frac{m+n}{2}}}
        {(\frac{1}{2}+\alpha+\gamma)_{\tp{\frac{m}{2}}}\,
        (\frac{1}{2}+\alpha+\gamma)_{\tp{\frac{n}{2}}}},\\
          (m+n\hzl{3.5mm}even);\\[3mm]
        0,\hspace{1.5cm}(m+n\hzl{3.5mm}odd).\ea\right.\nonumber
\ens
\end{exam}

\appendix

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\vspace{-5.5mm}
{\bf Remark \,} In an article by Carlitz {\bf \cite{kn:car}}, 
the bivariate generating functions 
\bns{0}
&&\sum_{m,n\,=0}^{\infty}
\:A_{m,n}(\alpha,\gamma)
\:x^m\,y^n\;=
\;(1+u)^{\alpha}\,(1+v)^{\gamma},\\
&&\sum_{m,n\,=0}^{\infty}
\:\left(\ba{c}\alpha+am+cn\\m\ea\right)
\:\left(\ba{c}\gamma+bm+dn\\n\ea\right)
\:x^m\,y^n\;=
\;\frac{(1+u)^{1+\alpha}\,(1+v)^{1+\gamma}}{\Delta};
\ens\\[-3mm]
where\vspace{-2mm}
\bns{0}
A_{m,n}(\alpha,\gamma)&
  =&\frac{\alpha\gamma+\alpha bm+\gamma cn}
          {(\alpha+am+cn)\,(\gamma+bm+dn)}
\:\left(\ba{c}\alpha+am+cn\\m\ea\right)
\:\left(\ba{c}\gamma+bm+dn\\n\ea\right),\nonumber\\
x&=&u\,(1+u)^{-a}\,(1+v)^{-b},\\[3mm]
y&=&v\,(1+u)^{-c}\,(1+v)^{-d},\\[3mm]
\Delta&=&(1+u-au)\,(1+v-dv)-bcuv;
\ens\\[-3mm]
are established via MacMahon's master theorem, 
which lead to a pair of convolution identities:
\bns{0}
&&\sum_{i=0}^m\sum_{j=0}^n
\:A_{i,j}(\alpha,\gamma)
\:A_{m-i,n-j}(\alpha',\gamma')\;=
\;A_{m,n}(\alpha+\alpha',\gamma+\gamma'),\\[3mm]
&&\sum_{i=0}^m\sum_{j=0}^n
\:\left(\ba{c}\alpha+ai+cj\\i\ea\right)
\:\left(\ba{c}\gamma+bi+dj\\j\ea\right)
\:A_{m-i,n-j}(\alpha',\gamma')\;=\\
&&\hspace*{4cm}=\;\left(\ba{c}\alpha+\alpha'+am+cn\\m\ea\right)
\:\left(\ba{c}\gamma+\gamma'+bm+dn\\n\ea\right)\nonumber.
\ens\\[-3mm]
However, the convolution formulae demonstrated 
in the last section are different essentially from Carlitz' summation, 
though {\bf Eqs.(3.1)} fit in the general relations stated in {\bf Eqs.(A.1)}.

\pagebreak

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{\bf Addendum \,}After having written the first version of this paper, 
the author was encouraged by V. Strehl to check whether the approach 
presented here works for Brock numbers. 

The original problem proposed by Brock (1960, cf.\cite{kn:str}) 
is to show that the summations defined by 
\be
H(m,n)\,=\,\sum_{i=0}^m\sum_{j=0}^n
         \:\left(\ba{c}m+n-i-j\\m-i,\,n-j\ea\right)
         \:\left(\ba{c}i+j\\i,\,j\ea\right)
         \:\left(\ba{c}m-i+j\\m-i,\,j\ea\right)
         \:\left(\ba{c}i+n-j\\i,\,n-j\ea\right)
\ee
satisfy the recurrence relation
\be
\left(\ba{c}m+n\\m,\,n\ea\right)^2\:=\:H(m,n)-H(m-1,n)-H(m,n-1).
\ee
In fact, an equivalent expression  of Brock summation defined 
by the recurrence can be recovered by {\bf Eqs.(\ref{eq:Ia},\ref{eq:Ib})}
\be
H(m,n)\,=\,\sum_{i=0}^m\sum_{j=0}^n
         \:\left(\ba{c}m+n-i-j\\m-i,\,n-j\ea\right)
         \:\left(\ba{c}i+j\\i,\,j\ea\right)^2.
\ee
In view of the expansion of $(1-x-y)^{2\eps}\,f^{\alpha}(x,y)$ 
displayed in the last section, we find that the expression 
for $H(m,n)$ can be simplified further
\bean
H(m,n)&=&[x^my^n]\{\frac{(1-x-y)^{-1}}{\sqrt{(1-x-y)^2-4xy}}\}\\
&=&\frac{(1+m+n)!}{m!\,n!}\:_3F_2\left[\ba{ccc}-m,&-n,&1/2\\
                       &1,&3/2 \ea\right],
\eean
which may be reformulated as follows (cf.\cite{kn:str})
\be
H(m,n)=\frac{(1+m+n)!}{m!\,n!}\sum_k
    \:\frac{\left(\ba{c}m\\k\ea\right)
    \,\left(\ba{c}n\\k\ea\right)}{1+2k}
   =\sum_k\:\left(\ba{c}1+m+n\\m-k,\,n-k\ea\right)
          \:\left(\ba{c}2k\\k\ea\right).
\ee

{\bf Acknowledgement} The author is grateful to V. Strehl 
and J. Zeng for their suggstions in the $29^{th}$ S\'{e}minaire 
Lotharingien de Combinatoire (September, 1992; Thurnau), 
which have made the present version essentially improved.  



\begin{thebibliography}{99}
\bibitem{kn:bai} W. N. Bailey, {\em Generalized Hypergeometric Series},  Cambridge University Press, London, 1935.

\bibitem{kn:bl} J. Blodgett, {\em A higher-degree binomial coefficient identity},  Amer. Math. Month.99:1 (1992), E3376.

\bibitem{kn:car} L. Carlitz, {\em Some expansion and convolution formulas related to MacMahon's master theorem},  SIAM J. Math. Anal.8 (1977), 320-336.

\bibitem{kn:mac} J. Riordan, {\em Combinatorial Identities}, John Wiley \& Sons, Inc., New York, 1968.

\bibitem{kn:str} V. Strehl, {\em Zykel-Enumeration bei Lokal-strukturierten Functionen} (Preprint). Univ. Erlangen-N\"urnberg, 1990.

\end{thebibliography}


\end{document}
