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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.

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Why are you interested in this specific sequence? -- I think it would be helpful if you could say a little more about the background of the question. – Stefan Kohl Jan 29 '13 at 20:57
It is the number of rooted $n$-ominoes of the regular tiling $\{4,3,\infty\}$ with a plane of symmetry midway between two opposite facets for odd $n$. – Robert A. Russell Jan 31 '13 at 1:50
up vote 6 down vote accepted

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.

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