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

Please a proof of:

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

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

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  • $\begingroup$ @robinpemantle Here Robin Pemantle's standard theory does not work $\endgroup$ Apr 22, 2021 at 20:04
  • $\begingroup$ This should lead to a straightforward polynomial equation for the generating function that would let you determine coefficients explicitly, unless there's something that I'm missing... $\endgroup$ Apr 22, 2021 at 20:27
  • $\begingroup$ $$ J(x,y)=-\frac{x y}{4}-\frac{x}{2}-\frac{y}{3}+1$$ $$J(x,y)=0\land x J^{(1,0)}(x,y)=y J^{(0,1)}(x,y)$$ $$ \left\{x=\frac{2}{3} \left(\sqrt{10}-2\right),y=\sqrt{10}-2\right\}$$ $$\frac{1}{x y}=\frac{1}{12} \left(7+2 \sqrt{10}\right)=1.11038 ...$$ $\endgroup$ Apr 22, 2021 at 20:42
  • $\begingroup$ If I'm doing my generatingfunctionology right then the GF satisfies $f(w,z) = \frac1{1-w}+\frac1{1-z}-1+\frac z2f(w,z) + \frac w3 f(w,z) + \frac{zw}4 f(w,z)$; I suspect that gives a very different result than what you claim there. $\endgroup$ Apr 22, 2021 at 21:42
  • $\begingroup$ @StevenStadnicki I was just computing the generating function, and I got something slightly different for the last three summands. $\endgroup$ Apr 22, 2021 at 22:00

1 Answer 1

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We will compute the generating function, and use the method described in section 2 of this paper.

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 $F(m-1,m)$.

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