For the integer sequence {1,13,314,9368,312411,11163022, ...}, each term is given by the function
$f(n)=\sum\_{k=0}^n{1\over2n+3k-1}{2n+3k-1\choose k}{6n-6k-3\choose2n-2k-2}$. Is there a method to determine the exact value of $\lim\_{n\rightarrow\infty}f(n+1)/f(n)$? The approximate value of $f(10001)/f(10000)$ is 47.7251; for $f(20001)/f(20000)$ it is 47.7287; for $f(30001)/f(30000)$ it is 47.7299. I unsuccessfully tested the sequence for a linear recurrence with Mathematica.



Growth constant limit for sum of products of two binomial coefficients
Let $\alpha\approx 0.0493932733732$ be the positive zero of $621\alpha^3+1242\alpha^2+585\alpha-32$. Then the limit you want is probably
$$ \rho = \frac{729(2+3\alpha)^2}{64(1+\alpha)^2}
 \left(\frac{4(8+9\alpha)}{621\alpha(1+\alpha)^2}\right)^{\textstyle\alpha}
 \approx 47.7322896460174547.
$$ 


The reason why I say "probably" is that I didn't prove it rigorously, but I'm sure that rigour is routine to add. First identify the largest terms in the sum by looking at the ratio $f(n,k+1)/f(n,k-1)$, where $f(n,k)$ is the term. This ratio is approximately 1 when $k\approx\alpha n$. In this range, the ratio of $f(n+1,\alpha(n+1))/f(n,\alpha n)$ tends to the quantity I have identified as $n\to\infty$.


To add rigour, find which range of terms are required for the asymptotic value of the sum by expanding $f(n,\alpha n+t)$ as a series in $t$. Probably you will find that the shape is Gaussian and $|t|\le n^{1/2+\epsilon}$ will suffice. With $k$ in that range, the ratio $f(n+1,k)/f(n,k)$ might always converge to the limit above as $n\to\infty$. If not, use the Euler-Maclaurin theorem to sum the terms in this range, make a crude bound on the terms outside this range, and you will have the asymptotic value of the sum.


ADDED:
The stuff inside the large parens simplifies to 1, and the stuff outside them simplifies to
$$ \rho = \frac{6561}{16} - \frac{452709}{64}\alpha - \frac{1358127}{256}\alpha^2,$$
which implies that $\rho$ is the smallest zero of
$$ 1048576\rho^3 - 4353564672\rho^2 + 4518872583696\rho - 205891132094649,$$
which also happens to be $3^{10}/2^8$ times the smallest zero of $y^3-18y^2+81y-16$, and I guess that's about as much simplification as possible.


