Closed form of the following Recurrence Relation Let $L\colon\mathbb{N}^3 \to \mathbb{N}$ satisfy the following recurrence relationship,
$$
L(a,b,c) = 1 + \sum_{i=0}^{a-1} \sum_{j=0}^{b-1} \sum_{k=0}^{c-1} L(i,j,k),
$$
With "initial conditions" $L(0,a,b) = L(c,0,d) = L(0,e,f) = 0$.  I am interested in knowing a closed form of $L$. 
Work I have done:
I have investigated a simpler case
$
G\colon \mathbb{N}^2\to \mathbb{N}
$, satisfying 
$$G(a,b) = 1 + \sum_{i=0}^{a-1} \sum_{j=0}^{b-1} G(i,j)$$
with similar "initial conditions" and can obtain $$G(a,b) = \binom{a+b-2}{a-1}= \binom{a+b-2}{b-1}= \frac{(a-b-2)!}{(a-1)!(b-1)!},$$ so I would have guessed that $$L(a,b,c) = \frac{(a+b+c-3)!}{(a-1)!(b-1)!(c-1)!}, $$ but this isn't correct.  I would appreciate any hints on how to find a closed form solution for this.
 A: Continuing according to Fred's very clever deduction of the multiple z-tranform
 ( the credit should go to him, so I invite him to post it, and votes to be casted on it as well)
$$
F(x,y,z) = \frac{{\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}}}
{{1 - \frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}}} = \sum\limits_{1\, \leqslant \,n} {\left( {\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}} \right)^{\,n} } 
$$
and considering that
$$
\frac{{z^n }}
{{\left( {1 - z} \right)^n }} = \sum\limits_{0\, \leqslant \,k} {\left( \begin{gathered}
  n - 1 + k \\ 
  k \\ 
\end{gathered}  \right)\,\,z^{\,n + k} }  = \sum\limits_{0\, \leqslant \,j} {\left( \begin{gathered}
  j - 1 \\ 
  j - n \\ 
\end{gathered}  \right)\,\,z^{\,j} } 
$$
then we have
$$
\begin{gathered}
  \sum\limits_{1\, \leqslant \,n} {\left( {\frac{{x\,y\,z}}
{{\left( {1 - x} \right)\left( {1 - y} \right)\left( {1 - z} \right)}}} \right)^n }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,a,\;b,\;c\;} {\left( {\sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  a - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  b - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  c - 1 \\ 
  c - n \\ 
\end{gathered}  \right)} } \right)x^{\,a} \,y^{\,b} \,z^{\,c} }  \hfill \\ 
\end{gathered} 
$$
i.e.
$$
\begin{gathered}
  L(a,b,c) = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  a - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  b - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  c - 1 \\ 
  c - n \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  n - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  n - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  c - 1 \\ 
  n - 1 \\ 
\end{gathered}  \right)}  \hfill \\ 
\end{gathered} 
$$
Note that: 


*

*in 1D it becomes $L(a,b,c) = 2^{\,a - 1} $

*in 2D
$$
\begin{gathered}
  L(a,b) = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b} \right)} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  a - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  b - n \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b} \right)} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  n - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  b - n \\ 
\end{gathered}  \right)}  = \left( \begin{gathered}
  a + b - 2 \\ 
  b - 1 \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$

*but in 3D I do not know if there is a closed form.


Addendum
Again thanks to Fred's hint, actually $L(a,b,c)$ can also be expressed in terms of Hypergeometric Function, as
$$
\begin{gathered}
  L(a,b,c) = \sum\limits_{1\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right)} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  a - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  b - n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  c - 1 \\ 
  c - n \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right) - 1} \right)} {\left( \begin{gathered}
  a - 1 \\ 
  n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  b - 1 \\ 
  n \\ 
\end{gathered}  \right)\left( \begin{gathered}
  c - 1 \\ 
  n \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right) - 1} \right)} {\left( { - 1} \right)^{\,n} \left( \begin{gathered}
  n - a \\ 
  n \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,n} \left( \begin{gathered}
  n - b \\ 
  n \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,n} \left( \begin{gathered}
  n - c \\ 
  n \\ 
\end{gathered}  \right)}  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,n\,\left( { \leqslant \,\min \left( {a,b,c} \right) - 1} \right)} {\frac{{\left( {1 - a} \right)^{\,\overline {\,n\,} } \left( {1 - b} \right)^{\,\overline {\,n\,} } \left( {1 - c} \right)^{\,\overline {\,n\,} } }}
{{1^{\,\overline {\,n\,} } \;1^{\,\overline {\,n\,} } }}\frac{{\left( { - 1} \right)^{\,n} }}
{{n!}}}  =  \hfill \\
   = {}_3F_{\,2} \left( {\left( {1 - a} \right),\left( {1 - b} \right),\left( {1 - c} \right);\;\;1,1;\;\; - 1} \right) \hfill \\ 
\end{gathered} 
$$
although, having the variable $z$ fixed at $-1$, we are not much exploiting that function.
