3.1 The weight sequence from [Reference Bernardi, Curien and Miermont6] with the edge parameter

In [Reference Bernardi, Curien and Miermont6, Section 2.2], it was established that for a random site-percolated triangulation (conditioned on the event where both ends of the root edge are colored in black), the cluster of the root is a random Boltzmann map with weight sequence $\mathbf q (p,t) = \left (q_k(p,t) \right )_{k \geq 1}$ given by

(3.1) $$ \begin{align} q_k(p,t) = \frac{1}{p} \left((pt)^{3/2} \delta_{\{k=3\}} + (p t^3)^{k/2} \sum_{l \geq 0} \binom{k+l-1}{k-1} [y^l] T(1-p,t,ty) \right)\!, \end{align} $$

for $k \geq 1$ (see [Reference Bernardi, Curien and Miermont6, equation (9)]). We briefly recall here what this statement means and how to obtain it.

Figure 2:

Gasket decomposition: A face of the root cluster (black vertices in the figure) is filled with a triangulation with arbitrary white boundary (upper-right figure, gray vertices can be either black or white) and a necklace (lower-right figure) triangulating the region between the two.

For every percolated rooted planar triangulation $\mathfrak t$ , we denote, respectively, by $\mathrm v_{\circ } (\mathfrak t)$ and $\mathrm v_{\bullet } (\mathfrak t)$ the number of white (resp. black) vertices. Let $\mathcal {T}^{\mathrm {perc}}$ be the set of all percolated rooted triangulations with both end vertices of the root edge black. For every $\mathfrak t \in \mathcal {T}^{\mathrm {perc}}$ , we denote by ${\mathfrak C} (\mathfrak t)$ the cluster of its root edge viewed as a planar rooted map (whose root is the root edge of the triangulation). The fact that the root cluster is a random Boltzmann map comes from the following identity (see equation (7) of [Reference Bernardi, Curien and Miermont6]): for every nonatomic map $\mathfrak m$ , one has

(3.2) $$ \begin{align} \sum_{\mathfrak t \in \mathcal {T}^{\mathrm{perc}} \, : \, {\mathfrak C} (\mathfrak t) = \mathfrak m} t^{e(\mathfrak t)} p^{\mathrm v_{\bullet} (\mathfrak t)} (1-p)^{\mathrm v_{\circ} (\mathfrak t)} = p^2 \,\cdot \prod_{f \in F(\mathfrak m)} q_{\mathrm{deg}(f)} (p,t). \end{align} $$

This identity stems from the classical gasket decomposition [Reference Borot, Bouttier and Guitter9] sometimes called island decomposition [Reference Bernardi, Curien and Miermont6], and we briefly explain how it is obtained in the following lines (see Figure 2 for an illustration of these arguments). Fix $\mathfrak m$ a nonatomic map and $\mathfrak t$ a site-percolated triangulation with root cluster $\mathfrak m$ colored black. For such a triangulation $\mathfrak t$ , each face of its root cluster is filled with a triangulation with an arbitrary boundary of white vertices, and a necklace of triangles with no additional vertices between this triangulation with a boundary and the cluster. For each cluster face of degree $k \geq 1$ filled with a triangulation with perimeter $l\geq 1$ , there are $\binom {k+l-1}{k-1}$ different possible necklaces, and each of these necklaces requires an additional $k+l$ edges. Taking into account the case where the cluster face has degree $3$ and can be part of the triangulation, this writes

$$ \begin{align*} & \sum_{\mathfrak t \in \mathcal {T}^{\mathrm{perc}} \, : \, {\mathfrak C} (\mathfrak t) = \mathfrak m} t^{e(\mathfrak t)} p^{\mathrm v_{\bullet} (\mathfrak t)} (1-p)^{\mathrm v_{\circ} (\mathfrak t)}\\ & \quad = p^{|V(\mathfrak m)|} \, t^{|e(\mathfrak m)|} \\ & \hspace{2cm} \times \prod_{f \in F(\mathfrak m)} \left( \mathbf{1}_{\mathrm{deg}(f) = 3} + \sum_{l \geq 0} \binom{\mathrm{deg}(f)+l-1}{\mathrm{deg}(f)-1} t^{\mathrm{deg}(f)+l} [y^l] T(1-p,t,y) \right)\!,\\ &\quad = p^2 \cdot \prod_{f \in F(\mathfrak m)} \frac{1}{p} (pt)^{\mathrm{deg}(f)/2} \\ & \hspace{2cm} \times \left( \mathbf{1}_{\mathrm{deg}(f) = 3} + \sum_{l \geq 0} \binom{\mathrm{deg}(f)+l-1}{\mathrm{deg}(f)-1} t^{\mathrm{deg}(f)+l} [y^l] T(1-p,t,y) \right)\!, \end{align*} $$

where we used Euler’s relation $|V(\mathfrak m)| -2 = \sum _{f \in F(\mathfrak m)} \left ( \mathrm {deg}(f)/2 - 1\right )$ in the last display. The expression (3.2) follows easily, and we refer the reader to [Reference Bernardi, Curien and Miermont6, Section 2.2] for additional details. Note for future reference that for any $k \geq 1$ , the weight $q_k(p,t)$ is $1/p$ times the generating series of all triangulations of the k-gon with a weight t per inner edge, a weight $\sqrt {pt}$ per boundary edge, and a weight $1-p$ per inner vertex adjacent to the boundary.

We can define the partition function of our percolated triangulations by

(3.3) $$ \begin{align} \mathcal Z (p,t) = \frac{1}{p^2} \cdot \sum_{\mathfrak t \in \mathcal {T}^{\mathrm{perc}}} t^{e(\mathfrak t)} p^{\mathrm v_{\bullet} (\mathfrak t)} (1-p)^{\mathrm v_{\circ} (\mathfrak t)}. \end{align} $$

From identity (3.2), denoting by $\mathcal M$ the set of all nonatomic rooted planar maps, we see that

(3.4) $$ \begin{align} \mathcal Z (p,t) = \sum_{\mathfrak m \in \mathcal M} \prod_{f \in F(\mathfrak m)} q_{\mathrm{deg}(f)} (p,t). \end{align} $$

Notice that this sum is finite when $p \in (0,1)$ and $t \in (0,t_c]$ , meaning that the weight sequence $\mathbf q(p,t)$ is admissible in the sense of [Reference Miermont25]. We will need the asymptotic behavior of the coefficients in $t^3$ of the series $\mathcal Z(p,t)$ .

Proposition 3.1 Fix $p \in (0,1)$ , we have

$$\begin{align*}[t^{3n}] \mathcal Z(p,t) \underset{n\to\infty}{\sim} \frac{\sqrt{2}\, \left(3 p-3+2 \sqrt{3}\right)}{2 p \sqrt{\pi}} \, t_c^{-3n} \, n^{-5/2}. \end{align*}$$

Proof By opening the root edge of the triangulations appearing in the sum (3.3), we can see that, for $p \in (0,1)$ and $|t| \leq t_c$ , the partition function $\mathcal Z(p,t)$ is given by

(3.5) $$ \begin{align} \mathcal Z (p,t) = \frac{1}{p^2 \, t} T_2(p,t) = \frac{1}{p^2 \, t^3} \, t^2 T_2(p,t) = \frac{1}{t^3} tT_1(p,t) \left( 1 +\frac{1-p}{p^3} tT_1(p,t) \right)\!. \end{align} $$

From Lemma 2.1, the function $tT_1(t)$ seen as a series in $t^3$ has a unique dominant singularity at $t^3_c$ and we can obtain its asymptotic expansion at $t_c^3$ from the expansion of U. As a consequence, the function $\mathcal Z (p,t)$ also has a unique dominant singularity at $t_c^3$ and we can easily obtain the following asymptotic expansion:

$$ \begin{align*} \mathcal Z(p,t) &= \frac{3 \sqrt{3}\, \left(7-4 \sqrt{3}\right) + p}{4 p} -\frac{\sqrt{3}\, \left(3 p-27+16 \sqrt{3}\right)} {4 p} \, \left( 1- t^3/t_c^3 \right) \\ & \quad + \frac{2 \sqrt{2}\, \left(3 p-3+2 \sqrt{3}\right)}{3 p} \, \left( 1- (t/t_c)^3)^{3/2} \right) + \mathcal O \left( \left( 1- t^3/t_c^3)^2 \right) \right)\!. \end{align*} $$

The asymptotic behavior of $[t^{3n}] \mathcal Z (p,t)$ then follows from the classical transfer theorem [Reference Flajolet and Sedgewick19, Theorem VI.4].

3.2 Generating series of the weights and of the associated Boltzmann maps

The generating series of the weight sequence $\mathbf q (p,t)$ is straightforward to compute. As shown by the following lines, the expression we obtain is valid for every $p\in (0,1)$ , t in the greater domain of analyticity of $U(t^3)$ , and z close enough to $0$ . More precisely, we need $\sqrt {p t^3} z \in \mathbb C \setminus [1 , + \infty )$ and $\left ( 1 - \sqrt {p t^3} z\right )^{-1}$ must belong to the domain of analyticity in y of $T(t,1-p,ty)$ , which was studied in the previous section. Under these conditions for p, t, and z, the computation of the weight sequence generating function is as follows:

(3.6) $$ \begin{align} F_{\mathbf q (p,t) }(z) :&= \sum_{k\geq 1} q_k(p,t) z^k, \notag\\ &= \frac{1}{p} (pt)^{3/2} z^3 + \frac{1}{p} \sum_{l \geq 0} [y^l] T(1-p,t,ty) \sum_{k \geq 1} \binom{k+l-1}{k-1} (p t^3)^{k/2} z^k, \notag\\ &= \frac{1}{p} (pt)^{3/2} z^3 + \frac{1}{p} \sqrt{p t^3} z \sum_{l \geq 0} [y^l] T(1-p,t,ty) \, \left( 1 - \sqrt{p t^3} z\right)^{-l-1}, \notag \\ &= \frac{1}{p} (pt)^{3/2} z^3 + \frac{1}{p} \frac{\sqrt{pt^3}z}{1-\sqrt{pt^3}z} T \left( 1-p,t,\frac{t}{1-\sqrt{pt^3}z} \right)\!. \end{align} $$

We will need expressions for the pointed and unpointed disk generating functions associated with the weight sequence $\mathbf q (p,t)$ . For every $l \geq 0$ , let $\mathcal M^l$ denote the set of all rooted planar maps with root face of degree l (for $l=0$ , this set contains only the atomic map). The unpointed disk generating function is defined as follows for $|z|$ large enough:

$$\begin{align*}W_{\mathbf q (p,t)}(z) := \sum_{l \geq 0} \, \left( \sum_{\mathfrak m \in \mathcal M^l} \prod_{f \in F(\mathfrak m) \setminus \{ f_r \}} q_{\mathrm{deg} (f)}(p,t) \right) \, z^{-l-1}, \end{align*}$$

where we denote the root face of a planar map by $f_r$ . From our discussion establishing (3.2), we can compute the coefficients of these series. Indeed, from equation (3.2) and the fact that $q_l$ is $1/p$ times the generating series of triangulations with a boundary of perimeter l counted with a weight t per inner edge, a weight $\sqrt {pt}$ per boundary edge, and a weight $1-p$ per inner vertex adjacent to the boundary, we have

$$ \begin{align*} \sum_{\mathfrak m \in \mathcal M^l} \, \sum_{\mathfrak t \in \mathcal {T}^{\mathrm{perc}} \, : \, {\mathfrak C} (\mathfrak t) = \mathfrak m} t^{e(\mathfrak t)} p^{\mathrm v_{\bullet} (\mathfrak t)} (1-p)^{\mathrm v_{\circ} (\mathfrak t)} & = p \sqrt{p \,t}^{-l} \, q_l(p,t) \, \sum_{\mathfrak t \in \mathcal T_l} t^{e(\mathfrak t)} p^{\mathrm v_{\mathrm{out}}(\mathfrak t)}. \end{align*} $$

Comparing with the right-hand side of (3.2), this gives, for every $l \geq 1$ ,

(3.7) $$ \begin{align} \sum_{\mathfrak m \in \mathcal M^l} \prod_{f \in F(\mathfrak m) \setminus \{ f_r \}} q_{\mathrm{deg} (f)}(p,t) = \frac{1}{p\,\sqrt{p t}^l} \, [y^l] T(p,t,y), \end{align} $$

and thus

(3.8) $$ \begin{align} W_{\mathbf q (p,t) }(z) = \frac{1}{p\,z} \, T \left( p,t,\frac{t}{\sqrt{pt^3}z} \right)\!. \end{align} $$

The pointed disk generating function is defined similarly:

$$\begin{align*}W_{\mathbf q (p,t)}^{\bullet}(z) = \sum_{l \geq 0} \, \left( \sum_{\mathfrak m \in \mathcal M^l} |V(\mathfrak m)| \, \prod_{f \in F(\mathfrak m) \setminus \{ f_r \}} q_{\mathrm{deg} (f)}(p,t) \right) \, z^{-l-1}. \end{align*}$$

It has the following universal form: for $p\in (0,1)$ and $t \in (0,t_c]$ fixed, there exists real numbers $c_+(p,t)>2$ and $c_-(p,t) \in (-c_+(p,t) , c_+(p,t))$ (the lower bound is excluded since our maps are not bipartite) such that for $z \in \mathbb C \setminus [c_-(p,t) , c_+(p,t)]$ , one has

(3.9) $$ \begin{align} W_{\mathbf q (p,t) }^{\bullet} (z) = \frac{1}{\sqrt{(z-c_+(p,t))(z-c_-(p,t))}}. \end{align} $$

This expression, sometimes called the one-cut lemma, appears in several references. See, for example, [Reference Budd14, Proposition 12] or [Reference Borot, Bouttier and Guitter11, Section 6.1]. It is also established in these articles that the two disk generating functions W and $W^{\bullet }$ have the same domain of analyticity. Comparing our two expressions (3.8) and (3.9), we see that as a consequence

(3.10) $$ \begin{align} c_{\pm}(p,t) = \frac{1}{\sqrt{p \, t^3} \, y_{\pm} (p,t)}, \end{align} $$

where $ y_{\pm } (p,t)$ are the respective positive and negative singularities in y of the series $T(p,t,ty)$ defined in Lemma 2.2. Note that, as mentioned in the proof of Lemma 2.2, this directly implies that $y_-(p,t) < - y_+(p,t)$ .

5.1 Admissibility equations and volume-modified weight sequence

We review in this section additional background on Boltzmann planar maps that we will use in the proof of Theorem 1.2. We refer the reader to the references [Reference Bernardi, Curien and Miermont6, Reference Budd14, Reference Miermont25] for details.

For $p \in (0,1)$ and $t \in (0,t_c]$ , consider the two following bivariate power series in $(z_1,z_2)$ :

$$ \begin{align*} f^{\bullet} (p,t;z_1,z_2) &= \sum_{k,k' \geq 0} z_1^k z_2^{k'} \binom{2k+k' +1}{k+1,k,k'} q_{2+2k+k'} (p,t), \\ f^{\diamond} (p,t;z_1,z_2) &= \sum_{k,k' \geq 0} z_1^k z_2^{k'} \binom{2k+k'}{k,k,k'} q_{1+2k+k'} (p,t). \end{align*} $$

These two functions are linked with Boltzmann maps with weight sequence $(q_k(p,t_c))_{k\geq 1}$ by the Bouttier–Di Francesco–Guitter bijection [Reference Bouttier, Di Francesco and Guitter13] and will be instrumental in the remaining of this work. In our case, we can compute alternative expressions for these functions that will be more amenable to analysis (see Lemma 6.4).

Recall the parameters $c_+(p,t)$ and $c_-(p,t)$ defined in equation (3.10). Since the weight sequence $\mathbf q (p,t)$ is admissible, the two functions defined above are well defined at least in the domain $|z_1| \leq z^+(p,t)$ and $|z_2| \leq z^{\diamond } (p,t)$ where $z^+(p,t)$ and $z^{\diamond } (p,t)$ are positive real numbers defined by

(5.1) $$ \begin{align} c_{\pm}(p,t) = z^{\diamond}(p,t) \pm 2 \sqrt{z^+(p,t)}. \end{align} $$

Furthermore, from Proposition 4.2 and Lemma 4.4 of [Reference Bernardi, Curien and Miermont6], $(z^+(p,t),z^{\diamond }(p,t))$ is the minimal solution of the system of equations

(5.2) $$ \begin{align} \begin{cases} f^{\bullet}(p,t;z^+(p,t),z^{\diamond}(p,t)) = 1 - \frac{1}{z^+(p,t))},\\ f^{\diamond} (p,t;z^+(p,t),z^{\diamond}(p,t)) = z^{\diamond}(p,t). \end{cases} \end{align} $$

In addition, when $p=1/2$ and $t=t_c$ , the weight sequence is critical (see Theorem 1.1 of [Reference Bernardi, Curien and Miermont6]), and $(z^+(1/2,t_c),z^{\diamond }(1/2,t_c))$ is the unique solution of the system of equations (5.2) such that

(5.3) $$ \begin{align} \left( \partial_{z_2} + \sqrt{z_1} \partial_{z_1} \right) f^{\diamond} (p,t;z^+(p,t),z^{\diamond}(p,t)) = 1. \end{align} $$

For $g \in (0,1]$ , let $(z^+(p,t;g) , z^{\diamond } (p,t;g) )$ be the unique solution in $(0,z^+(p,t)] \times (0,z^{\diamond } (p,t)]$ of the system of equations

(5.4) $$ \begin{align} \begin{cases} f^{\bullet}(p,t;z^+(p,t;g),z^{\diamond}(p,t;g)) = 1 - \frac{g}{z^+(p,t;g))},\\ f^{\diamond} (p,t;z^+(p,t;g),z^{\diamond}(p,t;g)) = z^{\diamond}(p,t;g). \end{cases} \end{align} $$

Define

$$\begin{align*}c_{\pm}(p,t;g) = \frac{1}{\sqrt g} \left(z^{\diamond} (p,t;g) \pm 2 \sqrt{z^+ (p,t;g)} \right)\!. \end{align*}$$

From, e.g., [Reference Bernardi, Curien and Miermont6, equations (4.3) and (4.4)], the pointed disk generating function of Boltzmann maps with modified weight sequence $\mathbf q (p,t;g) := \left ( g^{(k -2)/2} \, q_{k} (p ,t) \right )_{k \geq 1}$ defined as in (3.7) is given by

(5.5) $$ \begin{align} W_{\mathbf q (p,t;g)}^{\bullet} (z) = \frac{1}{\sqrt{ (z- c_+(p,t;g)) (z-c_-(p,t;g))}}. \end{align} $$

5.2 Proof of Theorem 1.2

We start with the perimeter exponent as it is the easiest of the two. From equations (3.2) and (3.7), we can write

$$ \begin{align*} {\mathbb P}_n ^p \left( |V(\partial {\mathfrak C} ) | = k \right) &= \frac{[t^{3n}] \left(q_k(p,t) \cdot [z^{-(k+1)}] W_{\mathbf q (p,t)}(z) \right)}{[t^{3n}] \mathcal Z (p,t)}\\ &= \frac{1}{p} \frac{[t^{3n}] \left(\sqrt{pt^3}^{-k} q_k(p,t) \cdot [y^k] T(p,t,ty) \right)}{[t^{3n}] \mathcal Z (p,t)}. \end{align*} $$

Applying Lemmas 6.2 and 6.3 directly gives the limit of the previous display as $n \to \infty $ . Indeed, the series $\sqrt {pt^3}^{-k} q_k(p,t)$ and $[y^k] T(p,t,ty)$ have the same unique dominant singularity in $t^3$ and it is easy to get an asymptotic expansion at $t_c^3$ of the product at this singularity from the two separate expansions. By continuity for the local topology, this limit is the probability that $|V(\partial {\mathfrak C} ) | = k$ in the UIPT:

$$\begin{align*}{\mathbb P}_{\infty} ^p \left( |V(\partial {\mathfrak C} ) | = k \right) = \frac{1}{p} \left( \delta_k(p) \, t_c^k T_k(p,t_c) + \sqrt{pt_c^3}^{-k} q_k(p,t_c) \, \theta_k(p) \right)\!. \end{align*}$$

Using the asymptotics (6.7), (6.9), (6.11), and (6.13) derived in Section 6.3 gives that for $p=1/2$ ,

$$\begin{align*}{\mathbb P}_{\infty} ^{1/2} \left( |V(\partial {\mathfrak C} ) | = k \right) \underset{k \to \infty}{\sim} \kappa' \, k^{-4/3} \end{align*}$$

with

(5.6) $$ \begin{align} \kappa' = - 8 \, \frac{1}{\Gamma(4/3)} \left( \frac{8 \,3^{\frac{5}{6}}}{351}+\frac{2 \,3^{\frac{1}{3}}}{117} \right) \, \frac{3^{5/6}}{2 \Gamma(-2/3)} \simeq 0.454, \end{align} $$

proving the second statement of Theorem 1.2.

We now turn on the first statement of the theorem on the volume of the root cluster. Although we are only interested in the case $p=1/2$ , we start the proof with a generic $p \in (0,1)$ as it will be easier to follow. From Proposition 4.1, we have, for every $g \leq 1$ ,

$$ \begin{align*} \mathbb E_{\infty}^p &\left[ g^{|V({\mathfrak C})|} \right]\\ &= \sum_{\mathfrak m \in \mathcal M} \, g^{|V(\mathfrak m)|} \, \sum_{f \in F(\mathfrak m )} (p t^3_c)^{\mathrm{deg}(f)/2} \delta_{\mathrm{deg}(f)} (p) \, \prod_{f' \in F(\mathfrak m ) \setminus \{f\}} q_{\mathrm{deg}(f')} (p ,t_c). \end{align*} $$

Applying Euler’s formula then gives

$$ \begin{align*} \mathbb E_{\infty}^p \left[ g^{|V({\mathfrak C})|} \right] & = g \, \sum_{\mathfrak m \in \mathcal M} \, \sum_{f \in F(\mathfrak m )} (g \, p t^3_c)^{\mathrm{deg}(f)/2} \delta_{\mathrm{deg}(f)} (p) \\ &\hspace{3cm} \times \prod_{f' \in F(\mathfrak m ) \setminus \{f\}} g^{(\mathrm{deg}(f') -2)/2} \, q_{\mathrm{deg}(f')} (p ,t_c). \end{align*} $$

Using the exact same line of reasoning as in the proof of Theorem 1.1, we can compute this sum as the Hadamard product evaluated at $z=1$ of the functions $z \mapsto \Delta (p, \sqrt {g p t_c^3})$ and the function $z \mapsto \Phi (p,z)$ where $c_+(p,t_c)$ and $c_-(p,t_c)$ are replaced by the two constants $c_+(p,t_c;g)$ and $c_-(p,t_c;g)$ associated with the pointed disk generating function $\left ( (z- c_+(p,t_c;g)) (z-c_-(p,t_c;g) - z) \right )^{-1/2}$ of Boltzmann maps with weight sequence $\mathbf q (p,t_c;g) = \left ( g^{(k -2)/2} \, q_{k} (p ,t_c) \right )_{k \geq 1}$ introduced in Section 5.1. Indeed, the first function should be obvious, and the second comes from the antiderivative of cylinder generating functions associated with the weight sequence $\mathbf q (p,t_c;g)$ instead of the weight sequence $\mathbf q (p,t_c;1)$ , which has the same universal form. Of course, it remains to calculate these two constants, which are less explicit than their counterparts $c_+(p,t_c)$ and $c_-(p,t_c)$ . Nevertheless, after mimicking the part of the proof of Theorem 1.1 leading to (4.6), we arrive at the formula

(5.7) $$ \begin{align} \mathbb E_{\infty}^p \left[ g^{|V({\mathfrak C})|} \right] = \frac{g}{2 \pi} \int_{c_-(p,t_c;g)}^{c_+(p,t_c;g)} \frac{dz}{z} & \Delta \left( p, \sqrt{g \, p t^3_c} \, z \right) \left(z + \frac{c_+(p,t_c;g) + c_-(p,t_c;g)}{2} \right) \notag \\ & \times \sqrt{(c_+(p,t_c;g) - z)(z-c_-(p,t_c;g))}. \end{align} $$

We want to study the asymptotic behavior of the integral (5.7) when $g \to 1^-$ . To do so, we must first study the dependency in g of the two constants $c_{\pm }(p,t_c;g)$ . We do not have a simple formula for the unpointed or pointed disk generating function of the Boltzmann maps as was the previously the case. However, the two constants can be studied via the solution of the system of equation (5.4). Let us denote $z^+(g) =z^+(1/2,t_c;g)$ and $z^{\diamond } (g) =z^{\diamond } (1/2,t_c;g)$ the solution of this system. We can calculate an expansion as $g \to 1^-$ of these two quantities with Lemma 6.5. Indeed, the development of $f^{\diamond }$ gives

$$ \begin{align*} z^{\diamond} - z^{\diamond}(g) &= \partial_{z_1}f^{\diamond} (z^+,z^{\diamond}) \, \left( z^+ - z^+(g) \right) + \left(1 -\sqrt{z^+} \partial_{z_1}f^{\diamond} (z^+,z^{\diamond}) \right) \, \left( z^{\diamond} - z^{\diamond}(g) \right)\\ & \qquad - {\kappa^{\diamond}} \, \left( (z^+ - z^+(g)) + \sqrt{z^+} (z^{\diamond} - z^{\diamond}(g)) \right)^{7/6} \\ & \qquad + o \left( \left((z^+ - z^+(g)) + \sqrt{z^+} (z^{\diamond} - z^{\diamond}(g)) \right)^{7/6} \right)\!. \end{align*} $$

This yields in turn

$$ \begin{align*} z^+ -{\kern-1pt} z^+(g) &= {\kern-1pt}\sqrt{z^+} \, \left( z^{\diamond} {\kern-1pt}-{\kern-1pt} z^{\diamond}(g) \right) + \frac{\kappa^{\diamond}}{\partial_{z_1} f^{\diamond} (z^+,z^{\diamond})} \, \left( (z^+ {\kern-1pt}-{\kern-1pt} z^+(g)) + \sqrt{z^+} (z^{\diamond} - z^{\diamond}(g)) \right)^{7/6} \\ & \qquad + o \left({\kern-1pt} \left((z^+ - z^+(g)) + \sqrt{z^+} (z^{\diamond} {\kern-1pt}-{\kern-1pt} z^{\diamond}(g)) {\kern-1pt}\right)^{7/6} {\kern-1pt}\right)\!. \end{align*} $$

Therefore, we have

(5.8) $$ \begin{align} z^{\diamond} - z^{\diamond}(g) &= \frac{1}{\sqrt{z^+}} \, \left( z^+ - z^+(g) \right) \notag \\ & \qquad - \frac{\kappa^{\diamond}}{\sqrt{z^+} \, \partial_{z_1} f^{\diamond} (z^+,z^{\diamond})} \, \left(2(z^+ - z^+(g)) \right)^{7/6} + o \left( \left( z^+ - z^+(g) \right)^{7/6} \right)\!. \end{align} $$

Now, the development of $f^{\bullet }$ gives

$$ \begin{align*} \frac{1}{z^+} - \frac{g}{z^+(g)} &= - \frac{1}{(z^+)^2} \left( z^+ - z^+(g) \right) \\ & \qquad + \left( \frac{\kappa^{\diamond}}{\sqrt{z^+}} + \kappa^{\bullet} \right) \left(2(z^+ - z^+(g)) \right)^{7/6} + o \left( \left(2(z^+ - z^+(g)) \right)^{7/6} \right)\!. \end{align*} $$

Expanding the left-hand side of this identity then gives

$$ \begin{align*} \frac{1-g}{z^+} &+ \frac{1-g}{(z^+)^2} \left( z^+ - z^+(g) \right) \\ & = \left( \frac{\kappa^{\diamond}}{\sqrt{z^+}} + \kappa^{\bullet} \right) \left(2(z^+ - z^+(g)) \right)^{7/6} + o \left( \left(2(z^+ - z^+(g)) \right)^{7/6} \right)\!, \end{align*} $$

which in turns gives

$$ \begin{align*} z^+(g) = z^+ - \frac{1}{2} \, \left(\kappa^{\diamond} \sqrt{z^+} + \kappa^{\bullet} z^+ \right)^{-6/7} \left( 1- g \right)^{6/7} + o \left( \left( 1- g \right)^{6/7} \right)\!. \end{align*} $$

Plugging this expression in equation (5.8) gives

$$ \begin{align*} z^{\diamond}(g) = z^{\diamond} - \frac{1}{2 \sqrt{z^+}} \, \left(\kappa^{\diamond} \sqrt{z^+} + \kappa^{\bullet} z^+ \right)^{-6/7} \left( 1- g \right)^{6/7} + o \left( \left( 1- g \right)^{6/7} \right)\!. \end{align*} $$

We finally get an expansion for $c_{\pm }(g)$ :

$$ \begin{align*} c_+(g) & = z^{\diamond}(g) + 2 \sqrt{z^+(g)} \\ & = c_+(1) - \frac{1}{\sqrt{z^+}} \, \left(\kappa^{\diamond} \sqrt{z^+} + \kappa^{\bullet} z^+ \right)^{-6/7} \left( 1- g \right)^{6/7} + o \left( \left( 1- g \right)^{6/7} \right)\!,\\ c_-(g) & = z^{\diamond}(g) - 2 \sqrt{z^+(g)} \\ & = c_-(1) + o \left( \left( 1- g \right)^{6/7} \right)\!. \end{align*} $$

The change of variable $z = \phi (g,\xi ) := c_-(g) + \left ( c_+(g) - c_-(g) \right ) \xi $ in equation (5.7) gives

$$ \begin{align*} \mathbb E_{\infty}^p & \left[ g^{|V({\mathfrak C})|} \right]\\&= \frac{g (c_+(g) - c_-(g))^2}{2 \pi} \int_{0}^{1} d\xi \sqrt{\xi (1-\xi)} \frac{\Delta \left( 1/2, \sqrt{g \, t^3_c/2} \, \phi(g,\xi) \right)}{\phi(g,\xi)} \\& \hspace{4cm} \times \left(\phi(g,\xi) + \frac{c_+(g) + c_-(g)}{2} \right)\!, \\& = \frac{(c_+(1) - c_-(1))^2}{2 \pi} \int_{0}^{1} d\xi \sqrt{\xi (1-\xi)} \frac{\Delta \left( 1/2, \sqrt{g \, t^3_c/2} \, \phi(g,\xi) \right)}{\sqrt{g} \, \phi(g,\xi)} \\& \hspace{4cm} \times\left(\phi(1,\xi) + \frac{c_+(1) + c_-(1)}{2} \right) + \mathcal O \left( (1-g)^{6/7} \right)\!, \end{align*} $$

where we used the developments of $c_{\pm }(g)$ to obtain the second equality. In a similar fashion than in the proof of Lemma 6.5, the dominant singular term of the integral comes for the singularity at $g=1$ and $\xi =1$ of the term $\frac {\Delta \left ( 1/2, \sqrt {g \, t^3_c/2} \, \phi (g,\xi ) \right )}{\sqrt {g} \, \phi (g,\xi )}$ . The asymptotic expansion (6.10) established in Section 6.3 gives

$$\begin{align*}\frac{\Delta \left( 1/2 , z \right)}{z} \underset{z \to 1/2}{\sim} \left( \frac{16 \,3^{\frac{5}{6}}}{351}+\frac{4 \,3^{\frac{1}{3}}}{117} \right) \, \left( 1 - 2 z \right)^{-4/3}. \end{align*}$$

Applying the same techniques as in the proof of Lemma 6.5, we see that the main singular term of the expansion of $\mathbb E_{\infty }^p \left [ g^{|V({\mathfrak C})|} \right ]$ comes from the integral

$$ \begin{align*} \frac{\tilde \kappa}{\pi} \int_{0}^{1} d\xi & \sqrt{\xi (1-\xi)} \left( 1-2 \sqrt{g \, t^3_c/2} \, \phi(g,\xi) \right)^{-4/3} \\ &= \frac{\tilde \kappa}{8 \left(1 - 2 \sqrt{g \, t^3_c/2} c_-(g) \right)^{4/3}} \, _2F_1 \left(\frac{4}{3}, \frac{3}{2}; 3; \frac{2 \sqrt{g \, t^3_c/2} \left( c_+(g) - c_-(g)\right)}{1 - 2 \sqrt{g \, t^3_c/2} c_-(g)} \right) , \end{align*} $$

with

$$\begin{align*}\tilde \kappa = \frac{(c_+(1) - c_-(1))^2 \left(\phi(1,1) + \frac{c_+(1) + c_-(1)}{2} \right) \sqrt{t_c^3/2}}{2} \left( \frac{4 \,3^{\frac{5}{6}}}{351}+\frac{3^{\frac{1}{3}}}{117} \right)\!. \end{align*}$$

Using the singular expansion of the hypergeometric function at $1$ and the developments of $c_{\pm }(g)$ , we finally get

$$ \begin{align*} \mathbb E_{\infty}^p & \left[ g^{|V({\mathfrak C})|} \right] \\ &= 1 - \frac{\tilde \kappa}{8 \left(1 - 2 \sqrt{\, t^3_c/2} c_-(1) \right)^{4/3}} \, \frac{36 \sqrt{3} \, \Gamma \! \left(\frac{5}{6}\right) \, \Gamma \! \left(\frac{2}{3}\right) }{\pi^{\frac{3}{2}}} \\ & \qquad \left( \frac{2 \sqrt{t_c^3/2}}{1 - 2 \sqrt{t_c^3/2} c_-(1)} \frac{1}{\sqrt{z^+}} \, \left(\kappa^{\diamond} \sqrt{z^+} + \kappa^{\bullet} z^+ \right)^{-6/7} \left( 1- g \right)^{6/7} \right)^{1/6} + o \left( 1- g \right)^{1/7}. \end{align*} $$

This expansion is unfortunately not enough to extract the asymptotic behavior of the probabilities ${\mathbb P} _{\infty }^{1/2} \left ( | V({\mathfrak C}) | = n\right ) $ as $n \to \infty $ . Indeed, the generating series of these probabilities is $\mathbb E_{\infty }^p \left [ g^{|V({\mathfrak C})|} \right ]$ , but this function could have singularities of modulus $1$ other than $1$ contributing to the asymptotic. However, we do not have this problem for tail probabilities. For every n, denote $p_n = {\mathbb P} _{\infty }^{1/2} \left ( | V({\mathfrak C}) | \geq n\right ) $ . A simple computation gives

$$\begin{align*}\sum_{n \geq 0} p_n g^n = \frac{1-g \, \mathbb E_{\infty}^p \left[ g^{|V({\mathfrak C})|} \right] }{1-g} \underset{g \to 1^-}{\sim} \kappa_1 \, \left( 1- g \right)^{-6/7}, \end{align*}$$

with

$$ \begin{align*} \kappa_1 = \frac{\tilde \kappa}{8 \left(1 - 2 \sqrt{\, t^3_c/2} c_-(1) \right)^{4/3}} & \, \frac{36 \sqrt{3} \, \Gamma \! \left(\frac{5}{6}\right) \, \Gamma \! \left(\frac{2}{3}\right) }{\pi^{\frac{3}{2}}}\\ & \times \left( \frac{2 \sqrt{t_c^3/2}}{1 - 2 \sqrt{t_c^3/2} c_-(1)} \frac{1}{\sqrt{z^+}} \right)^{1/6} \, \left(\kappa^{\diamond} \sqrt{z^+} + \kappa^{\bullet} z^+ \right)^{-8/7}. \end{align*} $$

From there, a classical Tauberian theorem (see, e.g., Theorem VI.13 of [Reference Flajolet and Sedgewick19] and the following discussion) establishes the asymptotic behavior of $p_n$ given in the theorem with

(5.9) $$ \begin{align} \kappa = \frac{\kappa_1}{ \Gamma(8/7)} = \frac{63 \left(3^{\frac{1}{3}}+\frac{4 \,3^{\frac{5}{6}}}{3}\right) 3^{\frac{17}{21}} 7^{\frac{1}{7}} 137^{\frac{6}{7}} \Gamma \! \left(\frac{2}{3}\right)^{\frac{18}{7}} 2^{\frac{3}{14}} 5^{\frac{13}{14}}}{56,992 \pi^{\frac{12}{7}} \Gamma \! \left(\frac{1}{7}\right)} \simeq 0.278. \end{align} $$

6.1 Dependency in t of the weights

Recall that $V_c$ be the power series in z defined in (4.1) by $V_c(p,z) = V(1-p,U(t_c^3), 1/(1-z))$ , where V is defined in Lemma 2.2. Furthermore, recall that $\Delta (p,z)$ is the power series in z defined in (4.2) as

$$ \begin{align*}{} &\Delta(p,z) \notag\\ &= \hat \Delta (p,V_c(p,z)) \notag\\ & := 3 \, \frac {V_c \, (2 \sqrt 3 -3V_c) \, \left( 9 V_c^3 - 9 (\sqrt 3 + 1) V_c^2 + 3(3 + 2\sqrt 3) V_c - 2 (1-p) \sqrt 3 \right)} {(3(p - 1) + 2 \sqrt 3)\, \left( \sqrt 3 -3V_c \right)^3 \, \left(9 V_c^3 - 9 V_c^2 \sqrt 3 + 4 (1-p) \sqrt 3 \right)}\\ & = \sum_{k \geq 0} \delta_k(p) \, z^k. \end{align*} $$

Lemma 6.1 Fix $p \in (0,1)$ . The series $\Delta (p,z)$ is analytic in the domain

$$ \begin{align*} \mathbb C \setminus \left[ 1- \frac{1}{y_+(1-p,t_c)} ,+\infty \right) = \mathbb C \setminus \left[\frac{1}{y_+(p,t_c)} ,+\infty \right)\!. \end{align*} $$

Proof By definition, the series $z \mapsto V_c(p,z)$ is analytic on $\mathbb C \setminus \left [ 1- \frac {1}{y_+(1-p,t_c)} ,+\infty \right )$ . In addition, if z is in this domain, the denominator of $\hat \Delta $ does not vanish (see the discussion in the proof of Theorem 1.1 after equation (4.9)). The only statement left to prove is the equality

$$ \begin{align*} 1- \frac{1}{y_+(1-p,t_c)} = \frac{1}{y_+(p,t_c)}. \end{align*} $$

Since we have an explicit expression for $y_+(p,t_c)$ (again, see the discussion in the proof of Theorem 1.1 after equation (4.9)), it is a straightforward verification.

Lemma 6.2 Fix $p \in (0,1)$ . For every $k \geq 1$ , one has

$$ \begin{align*} \frac{[t^{3n}] q_{k} (p ,t) \sqrt{p t^3}^{-k}}{[t^{3n}] \mathcal Z ( p ,t)} \underset{n \to \infty}{\rightarrow} \delta_k(p). \end{align*} $$

Proof In the whole proof, $p\in (0,1)$ is fixed. All calculations are available in the Maple companion file [26]. We start by proving that, for every $k \geq 1$ , the series

$$\begin{align*}\tilde q_k(p,t) = \sqrt{p t^3}^{-k} \cdot \left(q_k(p,t) - (p t)^{3/2} \mathbf 1_{k=3}\right)\end{align*}$$

seen as a series in $t^3$ is algebraic and has a unique dominant singularity at $t_c^3$ . In view of (3.6), the generating series of these modified weights is given by

(6.1) $$ \begin{align} \tilde F(p,t,z) = \sum_{k\geq 1} \tilde q_k(p,t) \, z^{k-1} = \frac{1}{p} \, \frac{1}{1-z} T \left(1-p,t,\frac{t}{1-z}\right)\!. \end{align} $$

Injecting this into the algebraic equation verified by T, we get an algebraic equation verified by $\tilde F$ :

$$ \begin{align*} & \left(p \, \tilde F(p,t,z) - T(1-p,t,t) \right) \, \mathrm{Pol}_1 (p \, \tilde F(p,t,z),p,t^3,T(1-p,t,t),tT_1(p,t)) \\ & \qquad = z \cdot \mathrm{Pol}_2 (p \, \tilde F(p,t,z),p,z), \end{align*} $$

where $\mathrm {Pol}_1$ and $\mathrm {Pol}_2$ are explicit polynomials. The form of this equation has the following consequences. First, using Lemma 2.2, the series $\tilde F(p,t,z)$ is algebraic. Second, it is the unique solution of this equation with constant term in z equal to $T(1-p,t,t)/p$ , and we can compute its coefficients in z inductively. These coefficients are the modified weights $\tilde q_k(p,t)$ and their expressions are then rational fractions in p, $t^3$ , $tT_1(p,t)$ , and $T(1-p,t,t)$ , whose denominator (up to a factor p) are the kth power of

$$ \begin{align*} \mathrm{Pol}_1 (\tilde F(p,t,0)&,p,t^3,T(1-p,t,t),tT_1(p,t)) \\ & = \mathrm{Pol}_1 (T(1-p,t,t),p,t^3,T(1-p,t,t),tT_1(p,t)),\\ &= 3 t^3 (p-1) \, T^2(1-p,t,t) + p (p-1) + tT_1(t). \end{align*} $$

A quick glance at equation (2.1) with $y=1$ shows that the last display is the derivative of the algebraic equation verified by $T(1-p,t,t)$ with respect to T. Therefore, this quantity can only be $0$ at singularities of $T(1-p,t,t)$ , which leaves only $t^3_c$ according to Lemma 2.2. As a consequence, we just proved that the series $\tilde q (p,t)$ are all algebraic series in $t^3$ and all have a unique dominant singularity at $t_c^3$ .

Now that we know that $\tilde q_k(p,t)$ has a unique dominant singularity at $t_c^3$ , it will follow from the general form of Puiseux expansions of algebraic series near their singularities (see [Reference Flajolet and Sedgewick19, Theorem VII.7, p. 498]) that $[t^{3n}] \tilde q_k(p,t)$ has the same asymptotic behavior as $[t^{3n}] \mathcal Z (p,t)$ if we can prove that there exist two positive constants $c,c'$ , depending on k and p such that, for n large enough,

$$\begin{align*}c \cdot [t^{3n}] \mathcal Z (p,t) \leq [t^{3n}] \tilde q_k(p,t) \leq c' \cdot [t^{3n}] \mathcal Z (p,t). \end{align*}$$

The upper bound is easily obtained by putting a triangulation of the $1$ -gon with a white boundary vertex inside a cycle of k black vertices and summing over every possible necklace between the two. The lower bound is obtained similarly starting from cycle of k black vertices by putting a single white vertex inside the cycle and an edge joining this additional vertex to every boundary vertex, and putting an arbitrary triangulation with white boundary together with a matching necklace on the outside of the cycle. To sum up, we have proved that for every k, $\tilde q_k(p,t)$ seen as a series in $t^3$ has a unique dominant singularity at $t_c^3$ and that its asymptotic expansion at $t_c^3$ is of the form

(6.2) $$ \begin{align} \tilde q_k(p,t) = \tilde q_k(p,t_c) - \tilde a_k (p) \left(1 - \frac{t^3}{t^3_c} \right) + \tilde b_k(p) \left(1 - \frac{t^3}{t^3_c} \right)^{3/2} + o \left(1 - \frac{t^3}{t^3_c} \right)^{3/2}. \end{align} $$

This finishes the proof of the first statement (4.4) of the lemma, and we now have to identify the generating series of the numbers $\delta _k(p) = \tilde b_k(p) / \kappa (p)$ , where $\kappa (p)$ is the coefficient of the term $\left (1 - \frac {t^3}{t^3_c} \right )^{3/2}$ in the asymptotic expansion of $\mathcal Z(p,t)$ calculated in Proposition 3.1.

Using the rational parameterization of Lemma 2.2, we can find a rational parameterization for $\tilde F$ . Indeed, if $V = V(1-p,U,y)$ is the power series in $\mathbb Q[p,U] [\![ y ]\!] \subset \mathbb Q[p] [\![ t^3,y ]\!]$ defined in Lemma 2.2, we have

$$\begin{align*}\begin{aligned} \tilde F(p,t,z) &= \frac{1}{p} \hat y \left( 1-p,U(t^3),V \Big(1-p,U(t^3),1/(1-z) \Big) \right) \\ & \qquad \times \hat T\left( 1-p,U(t^3),V \Big( 1-p,U(t^3),1/(1-z) \Big) \right)\!, \end{aligned} \end{align*}$$

where $\hat y$ and $\hat T$ are rational fractions defined in Lemma 2.2. For z fixed such that $|1/(1-z)| < y_+(1-p,t_c)$ (which includes a neighborhood of $0$ for z since $y_+(1-p,t_c)> 1$ ), the series $\tilde F(p,t,z)$ seen as a series in $t^3$ has nonnegative coefficients and has radius of convergence $t^3_c$ . In addition, this implies that $(t^3,z ) \mapsto \tilde F(p,t,z)$ is analytic in the larger domain $\mathcal D(0,t^3_c) \times \mathcal D(0,y_+(1-p,t_c)/(y_+(1-p,t_c)-1))$ . We will produce an asymptotic expansion of $\tilde F(p,t,z)$ near $t^3_c$ using our rational parameterization. To do so, we start by computing the asymptotic expansion of $V \Big ( p,U(t^3),1/(1-z) \Big )$ near $t^3_c$ , with z fixed.

First, writing

$$\begin{align*}\begin{aligned} \frac{1}{1-z} &= \hat y \left( 1-p,U(t^3),V \Big(1-p,U(t^3),1/(1-z) \Big) \right)\\ & = \hat y \left( 1-p,U(t^3_c),V \Big(1-p,U(t^3_c),1/(1-z) \Big) \right)\!, \end{aligned} \end{align*}$$

we get an algebraic equation between $V_c^z {\kern-1pt}= V \Big (1-{\kern-1pt}p,U(t^3_c),1/(1-{\kern-1pt}z) \Big )$ , $V \Big (1-p,U(t^3), 1/(1-z) \Big )$ , and $U(t^3)$ . Plugging the asymptotic expansion (2.2) of $U(t^3)$ in this equation, we obtain an asymptotic expansion for $V \Big (p,U(t^3),1/(1-z) \Big )$ of the form:

$$ \begin{align*} V \Big(1-p,U(t^3),1/(1-z) \Big) & = V_c^z - a_1(V_c^z) \left(1 - \frac{t^3}{t^3_c} \right)^{1/2} + a_2(V_c^z) \left(1 - \frac{t^3}{t^3_c} \right)\\ & \qquad \qquad \qquad + a_3(V_c^z) \left(1 - \frac{t^3}{t^3_c} \right)^{3/2} + o \left(1 - \frac{t^3}{t^3_c} \right)^{3/2}, \end{align*} $$

where the $a_i$ ’s are explicit rational fractions whose expressions are given in the Maple companion file [26]. Injecting in turn the asymptotic expansions in $t^3$ of U and V in $\hat T$ and $\hat y$ , we find an asymptotic expansion for $\tilde F$ of the form:

$$\begin{align*}\tilde F (p,t,z) = \tilde F (p,t_c,z) + \tilde A(p,V_c^z) \left(1 - \frac{t^3}{t^3_c} \right) + \tilde B(p,V_c^z) \left(1 - \frac{t^3}{t^3_c} \right) + o \left(1 - \frac{t^3}{t^3_c} \right)^{3/2}, \end{align*}$$

where $\tilde A$ and $\tilde B$ are explicit rational functions, and are analytic on the disk $\mathcal D (0, y_+(1-p,t_c)/(y_+(1-p,t_c) - 1)$ (this is obvious from their expressions: $z \mapsto V_c^z$ is analytic in this region, and the poles of $\tilde A$ and $\tilde B$ fall outside it; see the Maple companion file [26]). Note that the error term in the previous expansion is a priori not uniform in z. To ensure that $\tilde A$ and $\tilde B$ are the respective generating series of the numbers $\tilde a_k$ and $\tilde b_k$ , we see that, as power series in $(z,t^3)$ , we have

$$\begin{align*}\tilde A(p,V_c^z) = \lim_{t \to t_c} \left( \tilde F (p,t,z) - \tilde F (p,t_c,z) \right) \cdot (1 - t^3/t_c^3)^{-1}, \end{align*}$$

and

$$\begin{align*}\tilde B(p,V_c^z) = \lim_{t \to t_c} \left( \tilde F (p,t,z) - \tilde F (p,t_c,z) - \tilde A(p,V_c^z) \left(1 - \frac{t^3}{t^3_c} \right) \right) \cdot (1 - t^3/t_c^3)^{-3/2}. \end{align*}$$

Combined with the analyticity properties of these series, this ensures that $\tilde A(p,V_c^z)$ and $\tilde B(p,V_c^z)$ are indeed the generating series of the numbers $\tilde a_k$ and $\tilde b_k$ .

Finally, the generating series of the numbers $\delta _k$ is then given by

$$\begin{align*}\Delta(p,z) = \frac{z}{\kappa(p) \, (1-z)} \, \tilde B(p,V_c^z) = \frac{\hat y \left( 1-p,U(t^3_c),V_c^z \right) - 1}{\kappa(p)} \, \tilde B(p,V_c^z), \end{align*}$$

giving the expression of the lemma. See the Maple file [26] for detailed computations.

Applying the same proof to the function $T(p,t,t y)$ instead of $\tilde F (p,t,z)$ defined in equation (6.1) allows to establish the asymptotic behavior in n of $[t^{3n}] t^k T_k(p,t)$ for every $k \geq 0$ . We do not reproduce the proof as it will be almost exactly the same as the proof of Lemma 6.2, with no additional difficulties but with the function

$$\begin{align*}\frac{1-p}{y} \, \tilde F \left(1-p, t , 1- \frac{1}{y} \right) \end{align*}$$

instead of $\tilde F (p,t,z)$ . The statement is as follows.

Lemma 6.3 Fix $p \in (0,1)$ and $k \geq 1$ . One has

(6.3) $$ \begin{align} \frac{[t^{3n}] t^k T_{k} (p ,t)}{[t^{3n}] \mathcal Z ( p ,t)} \underset{n \to \infty}{\rightarrow} \theta_k(p), \end{align} $$

where the generating series of the numbers $\theta _k(p)$ is given by

(6.4) $$ \begin{align} \Theta (p, y) := \sum_{k \geq 0} \theta_k(p) \, y^k = \frac{1-p}{y} \, \Delta \left( 1-p,1 - \frac 1 y \right)\!, \end{align} $$

which is analytic on $\mathbb C \setminus \left [y_+(p,t_c) , + \infty \right )$ .

6.2 BDFG functions

We start with an integral formula for the functions $f^{\bullet }$ and $f^{\diamond }$ .

Lemma 6.4 Fix $p \in (0,1)$ and $t \in (0,t_c]$ . Then, for $0< z_1 \leq z^+(p,t)$ and $0<z_2 \leq z^{\diamond } (p,t)$ , we have

$$ \begin{align*} f^{\bullet} & (p,t;z_1,z_2) \\ &= \sqrt{pt^3} \, 2 z_2 + \frac{t^3}{\pi} \int_{ \left(1 -\sqrt{pt^3} (z_2 - 2 \sqrt{z_1})\right)^{-1}} ^{ \left(1 -\sqrt{pt^3} (z_2 + 2 \sqrt{z_1})\right)^{-1}} {dz} \, \frac{1 - \sqrt{p t^3} z_2 - \frac{1}{z}}{2pt^3 \, z_1} \\ & \qquad \qquad \cdot \frac{T(1-p,t,t z)}{\sqrt{\left(1 - z \left(1- \sqrt{pt^3} (z_2 + 2 \sqrt{z_1}) \right) \right) \left(z \left(1 -\sqrt{pt^3} (z_2 - 2 \sqrt{z_1}) \right) -1 \right)}} , \\ f^{\diamond} & (p,t;z_1,z_2) \\ &= \sqrt{pt^3} \, (2z_2 +z_1^2) +\frac{\sqrt{t^3/p}}{\pi} \, \int_{ \left(1 -\sqrt{pt^3} (z_2 - 2 \sqrt{z_1})\right)^{-1}} ^{ \left(1 -\sqrt{pt^3} (z_2 + 2 \sqrt{z_1})\right)^{-1}} dz \\ & \qquad \qquad \cdot \frac{T(1-p,t,t z)}{\sqrt{\left(1 - z \left(1- \sqrt{pt^3} (z_2 + 2 \sqrt{z_1}) \right) \right) \left(z \left(1 -\sqrt{pt^3} (z_2 - 2 \sqrt{z_1}) \right) -1 \right)}}. \end{align*} $$

Proof Fix $p \in (0,1)$ and $t \in (0,t_c]$ . Replacing the weights by their expression (3.1) in the definitions of the two functions gives

$$ \begin{align*} f^{\bullet} & (p,t;z_1,z_2) \\ &= \frac{1}{p} 2 (pt)^{3/2} z_2 \\ & \quad + \frac{p t^3}{p} \sum_{k,k',l \geq 0} \binom{2k+k' +l +1}{k+1,k,k',l} (pt^3 \, z_1)^k \, (\sqrt{p t^3} \, z_2)^{k'} \, [y^l] T(1-p,t,ty), \\ f^{\diamond} & (p,t;z_1,z_2) \\ &= \frac{1}{p} (pt)^{3/2} (2z_2 +z_1^2) \\ & \quad + \frac{\sqrt{p t^3}}{p} \sum_{k,k',l \geq 0} \binom{2k+k' +l}{k,k,k',l} (pt^3 \, z_1)^k \, (\sqrt{p t^3} \, z_2)^{k'} \, [y^l] T(1-p,t,ty). \end{align*} $$

We can express these two functions as Hadamard products. Indeed, define the trivariate power series in $(z_1,z_2;z)$ :

$$ \begin{align*} h^{\bullet} & (z_1,z_2;z) \\ &:= \sum_{k,k',l \geq 0} \binom{2k+k' +l +1}{k+1,k,k',l} (pt^3 \, z_1)^k \, (\sqrt{p t^3} \, z_2)^{k'} \, z^l ,\\ &= \frac{1}{2pt^3 \, z_1} \left( \frac{1 - z- \sqrt{pt^3} z_2}{\sqrt{\left(1-z-\sqrt{pt^3} (z_2 + 2 \sqrt{z_1}) \right) \left(1-z-\sqrt{pt^3} (z_2 - 2 \sqrt{z_1}) \right)}} -1 \right)\!, \end{align*} $$

and

$$ \begin{align*} h^{\diamond}(z_1,z_2;z) &:= \sum_{k,k',l \geq 0} \binom{2k+k' +l}{k,k,k',l} (pt^3 \, z_1)^k \, (\sqrt{p t^3} \, z_2)^{k'} \, z^l ,\\ &= \frac{1 }{\sqrt{\left(1-z-\sqrt{pt^3} (z_2 + 2 \sqrt{z_1}) \right) \left(1-z-\sqrt{pt^3} (z_2 - 2 \sqrt{z_1}) \right)}}. \end{align*} $$

Then we have

$$ \begin{align*} f^{\bullet} (p,t;z_1,z_2) &= \sqrt{pt^3} \, 2 z_2 + t^3 \, T(1-p,t,t z) \odot h^{\bullet}(z_1,z_2;z) \vert_{z=1} , \\ f^{\diamond} (p,t;z_1,z_2) &= \sqrt{pt^3} \, (2z_2 +z_1^2) +\frac{\sqrt{p t^3}}{p} \, T(1-p,t,t z) \odot h^{\diamond} (z_1,z_2;z) \vert_{z=1}. \end{align*} $$

We can calculate these two Hadamard products as contour integrals in a similar fashion than in the proof of Theorem 1.1 where we established (4.8). For $(z_1,z_2) \in (0,z^+(p,t)] \times (0,z^{\diamond }(p,t)]$ , the functions $h^{\bullet }$ and $h^{\diamond }$ are analytic in the domain $|z| < 1 - \sqrt {p t^3} (z_2 + 2 \sqrt {z_1})$ that contains the domain $|z| < 1 - \sqrt {p t^3} c_+(p,t) = 1 - \frac {1}{y_+(p,t)}$ . This last domain is not empty since $y_+(p,t)>1$ from Lemma 2.2. Therefore, we can represent the Hadamard product as a contour integral similar to (4.8) if $ \left (1 - \frac {1}{y_+(p,t)} \right )^{-1} \leq y_+(1-p,t)$ . We can check that this is the case since we have explicit formulas for $y_+(p,t_c)$ and $y_+(p,t)$ is increasing in t (see the Maple file [26] for details). Therefore, for a contour $\gamma $ enclosing $0$ and a point in the interval $\left [ \left (1 - \frac {1}{y_+(p,t)} \right )^{-1} , y_+(1-p,t) \right ]$ , we have

$$ \begin{align*} f^{\bullet} (p,t;z_1,z_2) &= \sqrt{pt^3} \, 2 z_2 + \frac{t^3}{2i\pi} \oint_{\gamma} \frac{dz}{z} T(1-p,t,t z) \, h^{\bullet} (z_1,z_2,1/z) ,\\ f^{\diamond} (p;z_1,z_2) &= \sqrt{pt^3} \, (2z_2 +z_1^2) + \frac{\sqrt{t_c^3/p}}{2i\pi} \oint_{\gamma} \frac{dz}{z} T(1-p,t,t z) \, h^{\diamond} (z_1,z_2,1/z). \end{align*} $$

We obtain the expressions of the lemma after simplifications and taking contours converging to the cut $\left [ {\left (1 -\sqrt {pt^3} (z_2 - 2 \sqrt {z_1})\right )^{-1}} , { \left (1 -\sqrt {pt^3} (z_2 + 2 \sqrt {z_1})\right )^{-1}} \right ]$ .

When $p=1/2$ and $t=t_c$ , we can compute explicitly an asymptotic expansion of $f^{\bullet }$ and $f^{\diamond }$ at the point $(z^+(p,t),z^{\diamond }(p,t))$ that will be used in the proof of Theorem 1.2.

Lemma 6.5 Write $f^{\bullet }(z_1,z_2) = f^{\bullet }(1/2,t_c;z_1,z_2)$ , $f^{\diamond }(z_1,z_2) = f^{\diamond }(1/2,t_c;z_1,z_2)$ , $z^+ = z^+(1/2,t_c)$ , and $z^{\diamond } = z^{\diamond }(1/2,t_c)$ . Then $z_+ = \frac {27 \sqrt {3}}{32}$ and $z^{\diamond } = \frac {3^{1/4} \sqrt {2}}{4}$ and we have the following asymptotic expansions at $(z^+,z^{\diamond })^-$ :

$$ \begin{align*} f^{\bullet} & (z_1,z_2) \\&= 1 - \frac{1}{z^+} + \frac{1}{z^+} \left( \frac{1}{z^+} - \sqrt{z_+} \partial_{z_1} f^{\diamond}(z^+,z^{\diamond}) \right) \, \left( z_1 - z^+ \right) + \partial_{z_1} f^{\diamond}(z^+,z^{\diamond}) \, \left( z_2 - z^{\diamond} \right) \\& \qquad + \kappa^{\bullet} \, \left( (z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} + o \left( \left((z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \right)\!, \\f^{\diamond} & (z_1,z_2) \\&= z^{\diamond} + \partial_{z_1} f^{\diamond}(z^+,z^{\diamond}) \, \left( z_1 - z^+ \right) + \left( 1 - \sqrt{z^+} \partial_{z_1} f^{\diamond}(z^+,z^{\diamond}) \right) \, \left( z_2 - z^{\diamond} \right) \\ & \qquad + \kappa^{\diamond} \, \left( (z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} + o \left( \left((z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \right)\!, \end{align*} $$

where

$$ \begin{align*} \kappa^{\diamond} = \frac{4 \,2^{\frac{1}{6}} \Gamma \! \left(\frac{2}{3}\right)^{3} 3^{\frac{2}{3}} \sqrt{5}}{63 \pi^{2}} \quad \text{and} \quad \kappa^{\bullet} = \frac{512 \, 3^{\frac{1}{4}} \, \sqrt{2}}{81} \kappa^{\diamond}. \end{align*} $$

Proof The respective values of $z^+$ and $z^{\diamond }$ are computed from (3.10) and the explicit values of the singularities $y_+(1/2,t_c) = 2$ and $y_-(1/2,t_c) = -4$ of the function $y \mapsto T(1/2,t_c, t_cy)$ (see the Maple file [26] for details):

$$ \begin{align*} z^+ &= \frac{c_+(1/2,t_c) + c_-(1/2,t_c)}{2} = \frac{27 \sqrt{3}}{32} ,\\ z^{\diamond} &= \left(\frac{c_+(1/2,t_c) - c_-(1/2,t_c)}{4}\right)^2 = \frac{3^{3/4} \sqrt{2}}{4}. \end{align*} $$

Now, the change of variable

$$ \begin{align*} z &= \phi(z_1,z_2;\xi) \\ & := \frac{1}{1-\sqrt{pt^3} (z_2 - 2 \sqrt{z_1})} + \left(\frac{1}{1-\sqrt{pt^3} (z_2 + 2 \sqrt{z_1})} - \frac{1}{1-\sqrt{pt^3} (z_2 - 2 \sqrt{z_1})} \right) \xi \end{align*} $$

in the expressions of Lemma 6.4 gives

$$ \begin{align*} f^{\bullet} & (p,t;z_1,z_2) \\ &= \sqrt{pt^3} \, 2 z_2 + \frac{1}{2 p z_1} \frac{1}{\pi} \int_{ 0} ^{ 1} d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \, \frac{\phi(z_1,z_2;\xi) ( 1 - \sqrt{pt^3} z_2 ) -1}{\phi(z_1,z_2;\xi)} \\ & \hspace{6cm} \times T \left( 1-p,t, t \, \phi(z_1,z_2;\xi)\right)\!,\\ f^{\diamond} & (p,t;z_1,z_2) \\ &= \sqrt{pt^3} \, (2z_2 +z_1^2) + \sqrt{\frac{t^3}{p}} \frac{1}{\pi} \int_{0} ^{1} d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \, T \left( 1-p,t, t \, \phi(z_1,z_2;\xi) \right)\!. \end{align*} $$

We will see that, when $p=1/2$ and $t=t_c$ , the main term in the expansion of these functions stems from the singular behavior of T at $y_+ =2$ .

From Lemma 6.6 and the discussion that follows, we know that the function $y \mapsto T(1/2,t_c,t_c \, y)$ is analytic for $y\in [0,2)$ and has the following expansion as $y \to 2^-$ :

$$ \begin{align*} T& (1/2,t_c,t_c \, y) \\ & = \frac{\sqrt{3}}{2} - \frac{3^{5/6}}{2} \left( 1- \frac{y}{2} \right)^{2/3} + \frac{\sqrt{3}}{2} \left( 1- \frac{y}{2} \right) - 3^{1/6} \left( 1- \frac{y}{2} \right)^{4/3} + \mathcal O \left(\left( 1- \frac{y}{2} \right)^{5/3} \right)\!. \end{align*} $$

The function

$$\begin{align*}\varphi(y) := T(1/2,t_c,t_c \, y) - \left( \frac{\sqrt{3}}{2} - \frac{3^{5/6}}{2} \left( 1- \frac{y}{2} \right)^{2/3} + \frac{\sqrt{3}}{2} \left( 1- \frac{y}{2} \right) + 3^{1/6} \left( 1- \frac{y}{2} \right)^{4/3} \right) \end{align*}$$

is twice differentiable on $[0,2)$ . In addition, for $\xi \in (0,1)$ and if $(z_1,z_2) \in [0,z^+]\times [0,z^{\diamond }]$ , the quantity $\phi (z_1,z_2;\xi )$ varies in a subset of $(0,(1-1/y_+(1/2,t_c))^{-1}) = (0,2)$ . We then have

$$\begin{align*}\int_0^1 d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \left| \varphi' \left( \phi(z_1,z_2;\xi) \right) \right| < + \infty \end{align*}$$

and

$$\begin{align*}\int_0^1 d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \left| \varphi" \left( \phi(z_1,z_2;\xi) \right) \right| < + \infty. \end{align*}$$

As a consequence, the function

$$\begin{align*}\Phi (z_1 ,z_2) = \frac{1}{\pi} \int_0^1 d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \varphi \left( \phi(z_1,z_2;\xi) \right) \end{align*}$$

is twice differentiable on $[0,z^+]\times [0,z^{\diamond }]$ and has the following asymptotic expansion when $(z_1,z_2) \to (z^+,z^{\diamond })$ :

$$ \begin{align*} \Phi &(z_1,z_2) \\ &= \Phi(z^+, z^{\diamond}) + \nabla \Phi(z^+, z^{\diamond}) \cdot \left( z_1 -z^+, z_2 -z^{\diamond} \right) + \mathcal O \left( (z_1 -z^+)^2 +( z_2 -z^{\diamond})^2 \right)\!. \end{align*} $$

The singular parts of the expansions of $f^{\bullet }$ and $f^{\diamond }$ come from the singularities of the form $(1-y/2)^{\alpha }$ in the development of $T(1/2,t_c,t_c \, y)$ at $y=2$ for $\alpha \in \{ 2/3 , 4/3 \} $ (it is straightforward to check that the linear term $(1-y/2)$ contributes only to nonsingular parts in the expansion). Indeed, for $\alpha \in \{ 2/3 , 4/3 \}$ , set

$$ \begin{align*} I_{\alpha} &(z_1,z_2) \\ &= \frac{1}{\pi} \int_0^1 d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \left( 1 - \frac{\phi(z_1,z_2;\xi)}{2}\right)^{\alpha}, \\ & = \left(1- \frac{1}{2(1-\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1}))} \right)^{\alpha} \, \frac{1}{\pi} \int_0^1 d \xi \, \xi^{-1/2} (1-\xi)^{-1/2} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \cdot \left( 1 - \xi \frac{ \frac{1-\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1})}{1-\sqrt{t_c^3/2} (z_2 + 2 \sqrt{z_1})} -1 }{1-2\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1})} \right)^{\alpha}, \\ & = \left(1- \frac{1}{2(1-\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1}))} \right)^{\alpha} \, _2F_1 \left( - \alpha, \frac{1}{2};1; \frac{\frac{1-\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1}) }{1-\sqrt{t_c^3/2} (z_2 + 2 \sqrt{z_1})} -1 }{1-2\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1})} \right)\!, \end{align*} $$

where we used Euler’s integral representation of the hypergeometric function $_2F_1$ in the last line. This last equality is valid when $0< z_1 \leq z^+$ and $0<z_2 \leq z^{\diamond }$ since in this case $\sqrt {t_c^3/2} (z_2 + 2 \sqrt {z_1}) \leq 1/2$ and the variable in the hypergeometric function is in $(0,1]$ . Furthermore, using the values of $z^+$ and $z^{\diamond }$ , a simple computation done in the Maple file [26] gives

$$ \begin{align*} & \frac{\frac{1-\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1})}{1-\sqrt{t_c^3/2} (z_2 + 2 \sqrt{z_1})} -1}{1-2\sqrt{t_c^3/2} (z_2 - 2 \sqrt{z_1})} \\ & \hspace{1cm} = 1 - \frac{20 \sqrt 3}{81} \left( (z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right) + \mathcal O \left( (z^+ - z_1)^2 + (z^{\diamond} - z_2)^2 \right)\!. \end{align*} $$

Using the standard asymptotic development of hypergeometric functions at $1$ , we see that the first singular term in the development of $I_{2/3}$ is

$$\begin{align*}- \kappa \left( (z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \, \text{with} \, \kappa = \frac{18 \,2^{\frac{1}{3}} \, \Gamma \! \left(\frac{2}{3}\right)^{3}}{7 \pi^{2}} \, \left(\frac 3 5 \right)^{2/3} \, \left( \frac{20 \sqrt 3}{81} \right)^{7/6}. \end{align*}$$

The first singular term of $I_{4/3}(z_1,z_2)$ is similar, but with exponent $11/6$ instead of $7/6$ . This means that the first singular term in the development of $f^{\diamond }(1/2,z_1,z_2)$ is from $- \frac {3^{5/6}}{2} \, \sqrt {t_c^3/p} \, I_{2/3}(z_1,z_2)$ , and we have

$$ \begin{align*} f^{\diamond}(z_1,z_2) = f^{\diamond}(z^+,z^{\diamond}) &+ \nabla f^{\diamond}(z^+,z^{\diamond}) \cdot \left( z_1 - z^+ , z_2 - z^{\diamond} \right) \\ &+ \frac{3^{5/6}}{2} \, \sqrt{2 t_c^3} \, \kappa \, \left( (z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \\ &+ o \left( \left((z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \right)\!. \end{align*} $$

The statement for $f^{\diamond }$ follows using the fact that $f^{\diamond } ( z^+, z^{\diamond }) = z^{\diamond }$ and the criticality equation (5.3).

The expansion for $f^{\bullet } (1/2,z_1,z_2)$ is obtained similarly by replacing $\varphi $ by

$$ \begin{align*} \frac{1 - \sqrt{p t_c^3} z_2 - \frac{1}{y}}{2pt_c^3 \, z_1} T(1/2,t_c,t_c \, y) &- \frac{1 - \sqrt{t_c^3/2} z_2 - \frac{1}{2}}{t_c^3 \, z_1} \Bigg( \frac{\sqrt{3}}{2} - \frac{3^{5/6}}{2} \left( 1- \frac{y}{2} \right)^{2/3} \\ & + \frac{\sqrt{3}}{2} \left( 1- \frac{1}{2 t_c^3 z_1} \right) \left( 1- \frac{y}{2} \right) - 3^{1/6} \left( 1- \frac{y}{2} \right)^{4/3} \Bigg). \end{align*} $$

The first singular term in the development of $f^{\bullet }$ is then the one from $ - \frac {3^{5/6}}{2} \, \frac {1 - \sqrt {t_c^3/2} z^{\diamond } - \frac {1}{2}}{t_c^3 \, z^+} \frac {1}{z^+} I_{2/3}$ , and we have

$$ \begin{align*} f^{\bullet}(z_1,z_2) &= f^{\bullet}(z^+,z^{\diamond}) + \nabla f^{\bullet}(z^+,z^{\diamond}) \cdot \left( z_1 - z^+ , z_2 - z^{\diamond} \right) \\ & \qquad + \frac{3^{5/6}}{2} \, \frac{1 - 2\sqrt{t_c^3/2} z^{\diamond}}{2 t_c^3 \, (z^+)^2} \, \kappa \left( (z^+ - z_1) +\sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \\ & \qquad + o \left( \left( (z^+ - z_1) + \sqrt{z^+} (z^{\diamond} - z_2) \right)^{7/6} \right)\!. \end{align*} $$

The statement for $f^{\bullet }$ the follows from $f^{\bullet } (z^+,z^{\diamond }) = 1 - \frac {1}{z^+}$ and from the identities $\partial _{z_2} f^{\bullet } = \partial _{z_1} f^{\diamond }$ and $z_1 \partial _{z_1} f^{\bullet } + f^{\bullet } = \partial _{z_2} f^{\diamond }$ .

6.3 Perimeter asymptotics at criticality

We gather in this section asymptotics of several quantities appearing in this paper when $p=1/2$ and $t=t_c$ . They are all consequences of the following lemma characterizing the singularities of the function $y \mapsto V(1/2,U(t_c^3),y)$ defined in Lemma 2.2.

Lemma 6.6 The function $y \mapsto V(1/2,U(t_c^3),y)$ is analytic on $\mathbb C \setminus ( (-\infty , -4] \cup [2, +\infty ) )$ . In addition, it has the following asymptotic expansion in a slit neighborhood of $2$ :

(6.5) $$ \begin{align} V(1/2, U(t_c^3), y ) = \frac{\sqrt 3}{3} - \frac{1}{3^{\frac{1}{3}}} \left( 1 - \frac y 2 \right)^{\frac{1}{3}} +\frac{1}{3} \left( 1 - \frac y 2 \right) -\frac{1}{3^{\frac{4}{3}}} \left( 1 -\frac y 2 \right)^{\frac{4}{3}} + \mathcal O \left( \left( 1 - \frac y 2 \right)^{\frac{5}{3}} \right)\!. \end{align} $$

This lemma combined with the rational expressions in terms of V that we have allows to immediately compute asymptotic expansions. Indeed, the expressions $\hat T (1/2,U(t_c^3),V)$ and $\hat \Delta (1/2,V)$ defined, respectively, in Lemmas 2.2 and 6.2 are singular only when V is singular, and we can get an asymptotic expansion at their unique dominant singularity by plugging the expansion of V in their expression. As usual, calculations are available in the Maple companion file [26].

We get the following expansion for T:

(6.6) $$ \begin{align} T& (1/2,t_c,t_c \, y) \nonumber \\ & = \frac{\sqrt{3}}{2} - \frac{3^{5/6}}{2} \left( 1- \frac{y}{2} \right)^{2/3} + \frac{\sqrt{3}}{2} \left( 1- \frac{y}{2} \right) - 3^{1/6} \left( 1- \frac{y}{2} \right)^{4/3} + \mathcal O \left(\left( 1- \frac{y}{2} \right)^{5/3} \right)\!, \end{align} $$

and as a consequence,

(6.7) $$ \begin{align} t_c^k T_k(1/2,t_c) \underset{k \to \infty}{\sim} \frac{-3^{5/6}}{2 \Gamma(-2/3)} \, 2^k \, k^{-5/3}. \end{align} $$

Similarly, using the expression (6.1) gives the asymptotic expansion of the weights $\sqrt {t_c^3/2}^{-k} q_k(1/2,t_c)$ :

(6.8) $$ \begin{align} \tilde F(1/2,t_c,z) = 2\, \sqrt{3} - 2 \, 3^{5/6} \left( 1- 2 z \right)^{2/3} + + \mathcal O \left(\left( 1- 2z \right)^{5/3} \right)\!, \end{align} $$

and as a consequence,

(6.9) $$ \begin{align} \sqrt{t_c^3/2}^{-k} q_k(1/2,t_c) \underset{k \to \infty}{\sim} \frac{-2 \, 3^{5/6}}{ \Gamma(-2/3)} \, 2^{-k} \, k^{-5/3}. \end{align} $$

Note that these two singular expansions and the corresponding asymptotics were established in [Reference Bernardi, Curien and Miermont6] using different and more involved techniques. The rational parameterization that we have simplifies this analysis a lot.

Using $\hat \Delta $ , we get the following expansion:

(6.10) $$ \begin{align} \Delta \left( 1/2 , z \right) \underset{z \to 1/2}{\sim} \left( \frac{32 \,3^{\frac{5}{6}}}{351}+\frac{8 \,3^{\frac{1}{3}}}{117} \right) \, \left( 1 - 2 z \right)^{-4/3}, \end{align} $$

giving

(6.11) $$ \begin{align} \delta_k(1/2) \underset{k \to \infty}{\sim} \frac{1}{\Gamma(4/3)} \left( \frac{32 \,3^{\frac{5}{6}}}{351}+\frac{8 \,3^{\frac{1}{3}}}{117} \right) 2^{-k} \, k^{1/3}. \end{align} $$

Finally, using the expression (6.4), we get

(6.12) $$ \begin{align} \Theta \left( 1/2 , y \right) \underset{y \to 2}{\sim} \left( \frac{8 \,3^{\frac{5}{6}}}{351}+\frac{2 \,3^{\frac{1}{3}}}{117} \right) \, \left( 1 - y/2 \right)^{-4/3}, \end{align} $$

giving

(6.13) $$ \begin{align} \theta_k(1/2) \underset{k \to \infty}{\sim} \frac{1}{\Gamma(4/3)} \left( \frac{8 \,3^{\frac{5}{6}}}{351}+\frac{2 \,3^{\frac{1}{3}}}{117} \right) 2^{k} \, k^{1/3}. \end{align} $$