Asymptotics of q-Catalan numbers - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T07:54:18Zhttp://mathoverflow.net/feeds/question/7258http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/7258/asymptotics-of-q-catalan-numbersAsymptotics of q-Catalan numbersLeonid Petrov2009-11-30T11:11:17Z2010-10-30T17:59:50Z
<p>q-Catalan numbers are defined recurrently as C<sub>0</sub>=1, $C_{N+1}=\sum_{k=0}^N q^k C_k C_{N-k}$.</p>
<p>What can be said about the asymptotics of C<sub>n</sub> when <code>0<q<1</code>?</p>
<p>P.S. In the case q>1 it is known that as n goes to infinity, $q^{-{n\choose 2}}C_n(q)$ tends to the partition function $\prod_{i=1}^\infty\frac1{1-q^{-i}}$. However, this doesn't help in the case <code>0<q<1</code>.</p>
http://mathoverflow.net/questions/7258/asymptotics-of-q-catalan-numbers/7280#7280Answer by Scott Carter for Asymptotics of q-Catalan numbersScott Carter2009-11-30T14:54:01Z2009-11-30T14:54:01Z<p>Frohman and Bartoszynska did a lot of work on the asymptotics of the quantum $6j$-symbols over the last 5 to 7 years. I think their papers on these matters are found on the arxiv. This is where one should look first.</p>
http://mathoverflow.net/questions/7258/asymptotics-of-q-catalan-numbers/7295#7295Answer by Michael Lugo for Asymptotics of q-Catalan numbersMichael Lugo2009-11-30T17:29:59Z2009-11-30T17:29:59Z<p>It's not hard to compute numerical values. If you do this, in the regime $0 < q < 1$ it looks like $C_n$ grows exponentially, i. e. $C_n \sim \alpha_q \beta_q^n$ for some constants $\alpha_q$ and $\beta_q$ which depend on q. </p>
<p>Unfortunately, I don't know what $\alpha_q$ and $\beta_q$ are. For example, when q = 1/2 the ratio $C_n/C_{n-1}$ approaches a constant which is approximately 1.6022827223; I claim this is $\beta_{1/2}$. Then $C_{50}/\beta_{1/2}^{50} = 0.5757566503$, which I claim is $\alpha_{1/2}$. Neither of these constants appears in the <a href="http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html" rel="nofollow">inverse symbolic calculator</a>.</p>
<p>The generating function $C(q,z) = C_0 + C_1 z + C_2 z^2 + \ldots$, where the $C_n$ are $q$-Catalan numbers, ought to satisfy some functional equation, and then one could use techniques from singularity analysis (see, for example, <i>Analytic Combinatorics</i> by Flajolet and Sedgewick). But I am having trouble finding that functional equation.</p>
http://mathoverflow.net/questions/7258/asymptotics-of-q-catalan-numbers/43981#43981Answer by Didier Piau for Asymptotics of q-Catalan numbersDidier Piau2010-10-28T15:01:57Z2010-10-28T15:01:57Z<p>Indeed, $C_n^{1/n}$ converges. Call the limit $\beta_q$ like Michael Lugo did. One can show that $\beta_q\ge 1+q$ for every positive $q$, that $\beta_q\le 2(1+q)$ and $\beta_q\le 1/(1-q)$ for every $q$ in $(0,1)$, that $\beta_q$ is related to the smallest positive zero of a given $q$-hypergeometric function, and various other estimates. The $q$-Catalan numbers are related to some properties of products of correlated Wigner matrices just like the ordinary Catalan numbers describe the (statistical properties of the) spectrum of (large random) Wigner matrices. This is explained in <a href="http://www.springerlink.com/content/awmrn12yprgrrujr/" rel="nofollow">this paper</a> (caveat: I am one of the authors).</p>
http://mathoverflow.net/questions/7258/asymptotics-of-q-catalan-numbers/44260#44260Answer by Didier Piau for Asymptotics of q-Catalan numbersDidier Piau2010-10-30T17:59:50Z2010-10-30T17:59:50Z<p>Re Leonid's comment on a previous answer.</p>
<p>If the ratios $C_{n+1}/C_n$ converge, their limit $c(q)$ is such that $C(q,q/c(q))=c(q)$.
Equivalently, $1/c(q)$ is the radius of convergence of the series $z\mapsto C(q,z)$.
Or, writing $C(q,\cdot)$ as the ratio of two $q$-hypergeometric functions, one can show that $F(q,1/c(q))=0$, where
$$
F(q,z)=\sum_{n\ge0}(-1)^nq^{n^2-n}z^n/(q)_n.
$$
This implies that $c(q)$ is the sum of a series in $q$ with integer coefficients, whose signs seem to be alternating starting with the coefficient of $q$. The first terms are
$$
c(q)=1+q+q^3-q^4+2q^5-3q^6+6q^7-12q^8+25q^9-52q^{10}+111q^{11}+\ldots
$$
The function $q\mapsto c(q)$ is nondecreasing on $q\ge0$, obvious values are $c(0)=1$ and $c(1)=4$, and as a holomorphic function, $c(\cdot)$ might have a pole inside the unit disk at about $q\approx-.4$.
But apart from that...</p>