$$F(m,n)= \begin{cases}
1, & \text{if $m n=0$ }; \\
 \frac{1}{2} F(m ,n-1) + \frac{1}{3} F(m-1,n )+ \frac{1}{4} F(m-1,n-1), & \text{ if $m n>0$. }%
\end{cases}$$


Please a proof of:


$$\lim\_{n\rightarrow \infty}\frac{F(n,n)}{F(n-1,n-1)} =\lim\_{n\rightarrow \infty}\frac{F(n,n-1)}{F(n-1,n-1)}=\frac{9}{8}$$


$$\lim\_{n\rightarrow \infty}\frac{F(n-1,n)}{F(n-1,n-1)}=1$$



Why $\lim_{n\rightarrow \infty}\frac{F(n,n)}{F(n-1,n-1)} =\frac{9}{8}$?
We will compute the generating function, and use the method described in section 2 of [this paper](https://arxiv.org/abs/math/0702595).


Let $F\_{m,n}=F(m,n)$. Consider the generating function
$$G(x,y)=\sum\_{m=0}^\infty\sum\_{n=0}^\infty F\_{m,n}x^my^n.$$
Then the recurrence gives
\begin{align\*}
&G(x,y)=\sum\_{m=0}^\infty F\_{m,0}x^m+\sum\_{n=1}^\infty F\_{0,n}y^n+\sum\_{m=1}^\infty\sum\_{n=1}^\infty F\_{m,n}x^my^n\\
&=\frac{1}{1-x}+\frac{y}{1-y}+\sum\_{m=1}^\infty\sum\_{n=1}^\infty\left(\frac{1}{2}F\_{m,n-1}+\frac{1}{3}F\_{m-1,n}+\frac{1}{4}F\_{m-1,n-1}\right)x^my^n\\
&=\frac{1-xy}{(1-x)(1-y)}+\frac{y}{2}\sum\_{m=1}^\infty\sum\_{n=0}^\infty F\_{m,n}x^my^n+\frac{x}{3}\sum\_{m=0}^\infty\sum\_{n=1}^\infty F\_{m,n}x^my^n+\frac{xy}{4}G(x,y)\\
&=\frac{1-xy}{(1-x)(1-y)}+\frac{y}{2}\left(G(x,y)-\frac{1}{1-y}\right)+\frac{x}{3}\left(G(x,y)-\frac{1}{1-x}\right)+\frac{xy}{4}G(x,y)\\
&=\frac{1-xy-\frac{y}{2}(1-x)-\frac{x}{3}(1-y)}{(1-x)(1-y)}+\left(\frac{x}{3}+\frac{y}{2}+\frac{xy}{4}\right)G(x,y)\\
&=\frac{1-\frac{x}{3}-\frac{y}{2}-\frac{xy}{6}}{(1-x)(1-y)}+\left(\frac{x}{3}+\frac{y}{2}+\frac{xy}{4}\right)G(x,y).
\end{align\*}
Solving for $G(x,y)$ gives
$$G(x,y)=\frac{1-\frac{x}{3}-\frac{y}{2}-\frac{xy}{6}}{(1-x)(1-y)\left(1-\frac{x}{3}-\frac{y}{2}-\frac{xy}{4}\right)}.$$
Let $R(x,y)$ denote this rational function.
We have shown that $G(x,y)$ converges to $R(x,y)$ in some neighborhood of the origin.
Then for fixed small $x$, the Laurent series $G(x/y,y)$ will converge to $R(x/y,y)$ in some annulus around $y=0$.
Furthermore, $H(x)=\sum\_{m=0}^\infty F\_{m,m}x^m$ is the constant term of $G(x/y,y)$, and can be found via residue calculus as
\begin{align\*}
H(x)&=\frac{1}{2\pi i}\int\_\gamma\frac{1}{y}G(x/y,y)\,dy\\
&=\frac{1}{2\pi i}\int\_\gamma\frac{1}{y}R(x/y,y)\,dy\\
&=\sum\_k\mathrm{Res}\left[\frac{1}{y}R(x/y,y),y=z\_k\right]
\end{align\*}
where $\gamma$ is a counterclockwise contour in the annulus, and where $z\_k$ are the singularities of $R(x/y,y)$ lying inside of $\gamma$.
We can compute
\begin{align\*}
\frac{1}{y}R(x/y,y)&=\frac{y-\frac{x}{3}-\frac{y^2}{2}-\frac{xy}{6}}{(y-x)(1-y)\left(y-\frac{x}{3}-\frac{y^2}{2}-\frac{xy}{4}\right)}.
\end{align\*}
This rational function has poles at the following points:


* $y=1$. This pole does not lie inside of $\gamma$.
* $y=x$. This pole lies inside of $\gamma$, and has residue $\frac{8}{8-9x}$.
* $y=(1-\frac{x}{4})+\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}$. This pole does not lie inside of $\gamma$.
* $y=(1-\frac{x}{4})-\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}$. This pole lies inside of $\gamma$, and has residue
$$\frac{\frac{x}{12}\left((1-\frac{x}{4})-\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)}{\left((1-\frac{5x}{4})-\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)\left(\frac{x}{4}+\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)\left(\sqrt{(\frac{x}{4}-1)^2-\frac{2}{3}x}\right)}$$
which Wolfram Alpha can simplify to
$$\frac{x\left(13\sqrt3\,x-12\sqrt3+\sqrt{3x^2-56x+48}\right)}{(7x-6)(9x-8)\sqrt{3x^2-56x+48}}.$$


Putting this all together gives
$$H(x)=\frac{8}{8-9x}+\frac{x\left(13\sqrt3\,x-12\sqrt3+\sqrt{3x^2-56x+48}\right)}{(7x-6)(9x-8)\sqrt{3x^2-56x+48}}.$$
The second summand is actually holomorphic at $x=6/7$ and $x=8/9$.
Then the singularity of $H(x)$ closest to the origin is $x=8/9$, and we obtain the asymptotic
$$F\_{m,m}\sim\left(\frac{9}{8}\right)^m$$
which proves the first limit.


The remaining limits can be solved in a similar way, by first determining the asymptotics of $F(m,m-1)$ and $F(m-1,m)$.


