12
$\begingroup$

q-Catalan numbers are defined recurrently as C0=1, $C_{N+1}=\sum_{k=0}^N q^k C_k C_{N-k}$.

What can be said about the asymptotics of Cn when 0<q<1?

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 0<q<1.

$\endgroup$
2
  • $\begingroup$ Could you provide a reference for the asymptotic in the $q>1$ case? $\endgroup$
    – HMPanzo
    Commented Nov 12, 2019 at 14:21
  • $\begingroup$ @HMPanzo I cannot immediately provide a reference, but it should follow from considering coefficient by coefficient expansion and by interpreting q-Catalan numbers as generating functions of broken lines. When n grows, the diagonal boundary bounding the broken lines disappears, and we obtain generating function of all partitions. $\endgroup$ Commented Nov 13, 2019 at 15:03

4 Answers 4

3
$\begingroup$

Re Leonid's comment on a previous answer.

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...

$\endgroup$
1
  • $\begingroup$ Well, this really counts as a nice answer. Numerically I perfectly see the above series for $c(q)$, and also the pole. The coefficients of the Taylor expansion of $c(q)$ at zero stabilize. $\endgroup$ Commented Oct 31, 2010 at 5:41
6
$\begingroup$

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.

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 inverse symbolic calculator.

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, Analytic Combinatorics by Flajolet and Sedgewick). But I am having trouble finding that functional equation.

$\endgroup$
2
  • 3
    $\begingroup$ Looks to me like the functional equation is C(q,z)= 1 + zC(q,qz)C(q,z). I don't know anything about how to extract information about asymptotics from this, though. $\endgroup$ Commented Nov 30, 2009 at 23:08
  • 1
    $\begingroup$ Slightly off topic: may I advertise the guessing package included in FriCAS again? guessADE(q)([c n for n in 0..10], debug==true) finds the functional equation given by Hugh... $\endgroup$ Commented Oct 28, 2010 at 17:05
4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ But does that include $q$-Catalan numbers? $\endgroup$ Commented Nov 30, 2009 at 15:19
  • $\begingroup$ Good question, and I don't know for sure. As I recall they did an extensive bit of analysis in that work. $\endgroup$ Commented Nov 30, 2009 at 16:01
3
$\begingroup$

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 this paper (caveat: I am one of the authors).

$\endgroup$
5
  • $\begingroup$ Thank you for the answer, I will try to read the paper. However, the fact that $C_n^{1/n}$ converges also follows from random trees (the theory of Aldous' CRT) and I already knew it. It does not help, however. But nevertheless, thanks for the interest in this old question. $\endgroup$ Commented Oct 29, 2010 at 4:19
  • $\begingroup$ OK. Sooo... you might care to state more precisely the kind of property of the $C_n$s you are interested in. :-) $\endgroup$
    – Did
    Commented Oct 29, 2010 at 15:30
  • $\begingroup$ Actually, the time I asked the question I wanted to know the limit $C_{n+1}/C_n$ for a fixed $q$. Now you say that this lies in $q$-hypergeometric matters, and I know very little about these. So I think I need to investigate into that direction. $\endgroup$ Commented Oct 30, 2010 at 16:37
  • $\begingroup$ The link to the paper at springerlink.com is broken. I'm also unable to find any copy saved on the Wayback Machine. $\endgroup$ Commented Sep 11, 2022 at 17:49
  • 1
    $\begingroup$ @TheAmplitwist The title of that paper mentioned by Did is Products of correlated symmetric matrices and q-Catalan numbers by Mazza and Piau. You can find a pdf by searching for it on Google Scholar. $\endgroup$
    – HMPanzo
    Commented May 20, 2023 at 19:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .