Simple recurrence relation in three dimensions I have the following recurrence relation:
$$f[i,j,k] = f[i-1,j,k] + f[i,j-1,k] + f[i,j,k-1],\quad \mbox{for } i \geq j+k,$$
starting with $f[0,0,0]=1$, for $i$, $j$, and $k$ non-negative.
Is there any way to find a closed form expression for $f[i,j,k]$?
Note that this basically is a three dimensional version of the Catalan triangle, for which $f[i,j] = f[i-1,j] + f[i,j-1]$, for $i \geq j$, starting with $f[0,0]=1$. For this, a closed form expression is known: $f[i,j] = \frac{(i+j)!(i-j+1)}{j!(i+1)!}$.
Appreciate your help!

With the constraint $i \geq j+k$ I got following formula (inspired by the Fuss-Catalan tetrahedra formula page 10 and with my thanks to Brian M. Scott for pointing out my errancies...)  :
$$f[i,j,k]=\binom{i+1+j}{j} \binom{i+j+k}{k} \frac{i+1-j-k}{i+1+j}\ \ \text{for}\ i \geq j+k\ \ \text{and}\ \ 0\ \ \text{else}$$
plane $k=0$
$ 
\begin{array} {lllll|lllll}
1\\
1 & 1\\
1 & 2 & 2\\  
1 & 3 & 5 & 5\\
1 & 4 & 9 & 14 & 14\\
\end{array}
$
plane $k=1$
$ 
\begin{array} {l}
0\\
1 \\
2 & 4\\
3 & 10 & 15\\
4 & 18 & 42 & 56\\
5 & 28 & 84 & 168 & 210\\
\end{array}
$
plane $k=2$
$ 
\begin{array} {l}
0\\
0\\
2 \\
5 & 15\\
9 & 42 & 84\\
14 & 84 & 252 & 420\\
20 & 144 & 540 & 1320 & 1980\\
\end{array}
$

Without the $i \geq j+k$ constrains we get the simple : $$f[i,j,k]=\frac{(i+j+k)!}{i!j!k!}$$
That is the Trinomial expansion (extension of Pascal triangle in 3D : Pascal tetrahedron).
At least with the rules :


*

*$f[0,0,0]=1$

*$f[i,j,k]=0$ if $i<0$ or $j<0$ or $k<0$ 

*$f[i,j,k] = f[i-1,j,k] + f[i,j-1,k] + f[i,j,k-1]$ in the remaining cases  

