What is the generating function of $f\_{m,n}$?


$ f\_{m,n} = \begin{cases} 0 , & \text{if $m<0 $ or $ n<0$ }; \\
 f\_{n,m} , & \text{ if $n<m$}; \\
 1, & \text{ if $0=m$ and $ n\in\{0,1\} $}; \\
 f\_{0 ,n-1}+ f\_{1,n-1}, & \text{ if $0=m$ and $ n>1 $}; \\ f\_{m-1 ,n}+ f\_{m ,n-1}+ f\_{m-1,n-1}, & \text{ if $0<m\in \{n,n-1\} $}; \\
 f\_{m-1 ,n}+ f\_{ m ,n-1}+ f\_{m-1,n-1} + f\_{m+1,n-1},& \text{ if $0<m<n-1$}. 
\end{cases}
$



A generating function related to the Delannoy numbers
Consider the following generating function:
$$F(x,y) := \sum\_{n=0}^\infty \sum\_{m=0}^n f\_{m,n} x^n y^m$$
and its diagonal
$$D(z) := \sum\_{n=0}^\infty f\_{n,n} z^n.$$
Then the recurrence implies the following functional equation:
$$2(1-x-y-xy)F(x,y) - 2x\frac{F(x,y)-F(x,0)}y = 1 + (1-xy-2y)D(xy).$$
I'm not yet sure if it can be solved explicitly.




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It's also worth to mention a couple of boundary properties for $F(x,y)$:
$$[y^1]\ F(x,y) = \left.\frac{F(x,y)-F(x,0)}y\right|\_{y=0} = \frac{(1-x)F(x,0)-1}x$$
and
$$\sum\_{n=1}^\infty f\_{n-1,n} z^n = \frac{(1-z)D(z)-1}2.$$




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**ADDED.** Introducing another generating function:
$$G(x,y) := F(x,\frac{y}{x}) = \sum\_{n=0}^\infty \sum\_{m=0}^n f\_{m,n} x^{n-m} y^m$$
we get a self-contained functional equation:
$$2(y-xy-y^2-xy^2-x) G(x,xy) = y + (y-2y^2-xy^2) G(0,xy) – 2x G(x,0).$$


