MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.