A definite integral of a hypergeometric series related to the enumeration of fusenes

Question

If my calculations are correct, the number of (not necessarily planar) fusenes of perimeter $2n$ grows asymptotically like $\mu^n$ where \begin{equation} \mu = \frac{6}{\int_0^{1/6} H(t)\mathrm{d}t} = 3.5460\ldots, \end{equation} and $H(t)$ is the hypergeometric series with radius of convergence $1/6$ given by \begin{equation} H(t) = \sum_{k=1}^\infty \frac{(3k)!}{(k!)^3} (1+2t)^{k}t^{2k-2} = \frac{_2F_1\left(\frac13,\frac23;1;27t^2(1+2t)\right)-1}{t^2}. \end{equation} Alternatively $H(t)$ can be expressed in terms of a complete elliptic integral. I am curious to know: does $\mu$ admit a more explicit expression in terms of known constants?

Self-avoiding polygons of perimeter $2n$ on the honeycomb lattice form a strict subclass of fusenes. The number of these polygons is famously known, since the prediction by Nienhuis and the proof by Duminil-Copin and Smirnov, to grow with slightly lower rate $\mu_{SAP} = 2+\sqrt{2} = 3.4142\ldots$. So it is natural to ask: is there an easy way (beyond numeric evaluation) to confirm from the formula above that $\mu > \mu_{SAP}$?

Answer 1

The integral you wrote down evaluates to $$ I \, = \, \int_0^{1/6} H(t) \, \mathrm{d}t \, = \, 5 - \frac{6\sqrt{3}}{\pi} \, .$$

You can derive this result by expressing the integral as an integral of a modular form and below I give a brief summary of the computation.

First we substitute $t = \frac{h}{1+6 \, h}$ with the Hauptmodul $h$ of $\Gamma_0(6)$ given by $$ h(\tau) \, = \, \frac{\eta(2\tau) \, \eta(6\tau)^5}{\eta(\tau)^5 \, \eta(3\tau)} \, , $$ where $\eta$ denotes the Dedekind eta function. The map $\tau \mapsto t(\tau)$ gives a bijection between $i \, \mathbb{R}_{\geq 0}$ and $(0,1/6]$. Pulling back one obtains $$ I \, = \, \lim_{\epsilon \rightarrow 0} \left( -2\pi i \int_0^{t^{-1}(\epsilon)} f(\tau) \, \mathrm{d}\tau + \left(6-\frac{1}{\epsilon}\right)\right) \, ,$$ with $$ f(\tau) \, = \, \frac{1}{2\pi i}\frac{_2F_1\left(\frac13,\frac23;1;27\, t(\tau)^2(1+2\, t(\tau))\right)}{t(\tau)^2} t'(\tau) \, . $$ Now we can use that $f$ is a modular form of weight $3$ under $\Gamma_0(6)$ (and with the non-trivial Dirichlet character of modulus 6). Even stronger, $f$ is the second derivative of a modular form of weight $-1$. More concretely, we have $$ f \, = \, \frac{1}{(2\pi i)^2} g'' $$ with $$ g(\tau) \, = \, \frac{1}{t(\tau) \, {_2F_1}\left(\frac13,\frac23;1;27\, t(\tau)^2(1+2\, t(\tau))\right)} \, . $$ This allows us to simplify the integral to $$ I \, = \, \lim_{\epsilon \rightarrow 0} \left( -\frac{1}{2\pi i}g'(t^{-1}(\epsilon))+\left( 6-\frac{1}{\epsilon}\right) \right) + \frac{1}{2\pi i}g'(0)$$ and it only remains to expand around the cusps $0$ and $\infty$. In terms of $q = e^{2\pi i \, \tau}$ we have \begin{align*} t(\tau) \, &= \, q - q^2 + O(q^3) \\ g(\tau) \, &= \, q^{-1} + O(1) \\ g(-1/6\tau) \, &= \, \frac{1}{6\tau} \, \left(12\sqrt{-3}+O(q)\right) \end{align*} for $\tau \rightarrow \infty$ and the result follows.