What is the limit of $a (n + 1) / a (n)$? Let $a(n) = f(n,n)$ where $f(m,n) = 1$ if $m < 2  $ or  $ n < 2$ and $f(m,n) = f(m-1,n-1) + f(m-1,n-2) + 2 f(m-2,n-1)$ otherwise.
What is the limit of $a(n + 1) / a (n)$?  $(2.71...)$
 A: Here is a derivation for an explicit formula for $a(n)$.
The generating function for $f(m,n)$ is
$$F(x,y) := \sum_{m,n\geq 0} f(m,n)x^m y^n = \big(1 + \frac{3x^2y^2}{1-xy(1+2x+y)}\big)\frac{1}{1-x}\frac{1}{1-y}.$$
It follows that
\begin{split}
a(n) &= 1 + \sum_{i,j=0}^n [x^iy^j]\ \frac{3x^2y^2}{1-xy(1+2x+y)} \\
&= 1 + 3\sum_{i,j=0}^{n-2} [x^iy^j]\ \frac{1}{1-xy(1+2x+y)} \\
&= 1 + 3\sum_{i,j=0}^{n-2} [x^iy^j]\ \sum_{k=0}^{n-2} x^ky^k(1+2x+y)^k \\
&= 1 + 3\sum_{k=0}^{n-2} \sum_{i,j=0}^{n-2-k} [x^iy^j]\ (1+2x+y)^k \\
&= 1 + 3\sum_{k=0}^{n-2} \sum_{i,j=0}^{n-2-k} \binom{k}{i,j,k-i-j} 2^i \\
&= 1 + 3\sum_{i,j=0}^{n-2} \binom{i+j}{i} 2^i \sum_{k=i+j}^{n-2-\max(i,j)} \binom{k}{i+j} \\
&= 1 + 3\sum_{i,j=0}^{n-2} \binom{i+j}{i}\binom{n-1-\max(i,j)}{i+j+1}2^i.
\end{split}
A: The value is close to $e$ but not.  It's actually the positive real root of $p(t) := t^3 - 2t^2 + t - 8$.  This is solvable via ACSV (see book by Pemantle and Wilson 2013).  To summarize, the bivariate generating function is $1/Q := 1 / (1-xy-xy^2-2x^2y)$. The intersection of this in the positive quadrant with $xQ_x = yQ_y$ is a point $(x_0,y_0)$, where $\frac{1}{(x_0 y_0)}$ has minimal polynomial $p(t)$. Diagonal growth rate of $(x_0 y_0)^{-n}$ follows from some standard stuff.  Happy to explain more if anyone is curious.
