samples/texts/2952501/page_20.md

To simplify notation, let us note in this proof $\rho = 8$ and $\alpha = 4$. The property $B_r^\bullet(\bar{T}_n^{(p)}) = \Delta$ holds if and only if $\mathcal{T}_n^{(p)}$ is obtained from $\Delta$ by gluing to the top boundary³ an arbitrary triangulation with a semi-simple alternating boundary of length $2q$, and with $n-N$ black triangles, and if the distinguished vertex is chosen among the inner vertices of the glued triangulation. Thus:

$$\mathbb{P}(B_r^{\bullet }({\bar{T}}_n^{(p)})=\mathrm{\Delta })=\frac{B_{n-N,q}}{B_{n,p}}\cdot \frac{\mathrm{\#}\text{inner vertices in glued triangulation}}{\mathrm{\#}\text{inner vertices in total triangulation}}\mathrm{.}(5.3)\backslash mathbb\{P\}\; (B\_r^\backslash bullet\; (\backslash bar\{T\}\_n^\{(p)\})\; =\; \backslash Delta)\; =\; \backslash frac\{B\_\{n-N,q\}\}\{B\_\{n,p\}\}\; \backslash cdot\; \backslash frac\{\backslash \#\backslash text\{inner\; vertices\; in\; glued\; triangulation\}\}\{\backslash \#\backslash text\{inner\; vertices\; in\; total\; triangulation\}\}.\; \backslash quad\; (5.3)$$

Therefore:

$$\underset{n\to \mathrm{\infty }}{\lim }\mathbb{P}(B_r^{\bullet }({\bar{T}}_n^{(p)})=\mathrm{\Delta })=\frac{C(q)}{C(p)}{\rho }^{-N}\mathrm{.}(5.4)\backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash mathbb\{P\}\; (B\_r^\backslash bullet\; (\backslash bar\{T\}\_n^\{(p)\})\; =\; \backslash Delta)\; =\; \backslash frac\{C(q)\}\{C(p)\}\; \backslash rho^\{-N\}.\; \backslash quad\; (5.4)$$

As we have:

$$N=\mathrm{\mid }\mathcal{M}(\mathrm{\Delta })\mathrm{\mid }-p+\sum _{v\in {\mathcal{F}}^{\ast }}N(M_v)=\sum _{1\le i\le q}\mathrm{\mid }{\tau }_i\mathrm{\mid }-p+\sum _{v\in {\mathcal{F}}^{\ast }}N(M_v)=q+\sum _{v\in {\mathcal{F}}^{\ast }}(c_v+N(M_v))-p,N\; =\; |\backslash mathcal\{M\}(\backslash Delta)|\; -\; p\; +\; \backslash sum\_\{v\; \backslash in\; \backslash mathcal\{F\}^*\}\; N(M\_v)\; =\; \backslash sum\_\{1\; \backslash le\; i\; \backslash le\; q\}\; |\backslash tau\_i|\; -\; p\; +\; \backslash sum\_\{v\; \backslash in\; \backslash mathcal\{F\}^*\}\; N(M\_v)\; =\; q\; +\; \backslash sum\_\{v\; \backslash in\; \backslash mathcal\{F\}^*\}\; (c\_v\; +\; N(M\_v))\; -\; p,$$

we get:

$$\underset{n\to \mathrm{\infty }}{\lim }\mathbb{P}(B_r^{\bullet }({\bar{T}}_n^{(p)})=\mathrm{\Delta })=\frac{{\rho }^{-q}C(q)}{{\rho }^{-p}C(p)}\prod _{v\in {\mathcal{F}}^{\ast }}{\rho }^{-c_v}{\rho }^{-N(M_v)}\mathrm{.}\backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash mathbb\{P\}\; (B\_r^\backslash bullet\; (\backslash bar\{T\}\_n^\{(p)\})\; =\; \backslash Delta)\; =\; \backslash frac\{\backslash rho^\{-q\}\; C(q)\}\{\backslash rho^\{-p\}\; C(p)\}\; \backslash prod\_\{v\; \backslash in\; \backslash mathcal\{F\}^*\}\; \backslash rho^\{-c\_v\}\; \backslash rho^\{-N(M\_v)\}.$$

Now, since $\sum_{v \in \mathcal{F}^*} (c_v - 1) = p - q$, we can multiply the right-hand side by $(\alpha\rho)^{p-q-\sum_{v \in \mathcal{F}^*} (c_v-1)}$, which yields:

$$\underset{n\to \mathrm{\infty }}{\lim }\mathbb{P}(B_r^{\bullet }({\bar{T}}_n^{(p)})=\mathrm{\Delta })=\frac{{\alpha }^{-q}C(q)}{{\alpha }^{-p}C(p)}\prod _{v\in {\mathcal{F}}^{\ast }}{\rho }^{-1}{\alpha }^{-c_v+1}{\rho }^{-N(M_v)},\backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash mathbb\{P\}\; (B\_r^\backslash bullet\; (\backslash bar\{T\}\_n^\{(p)\})\; =\; \backslash Delta)\; =\; \backslash frac\{\backslash alpha^\{-q\}\; C(q)\}\{\backslash alpha^\{-p\}\; C(p)\}\; \backslash prod\_\{v\; \backslash in\; \backslash mathcal\{F\}^*\}\; \backslash rho^\{-1\}\; \backslash alpha^\{-c\_v+1\}\; \backslash rho^\{-N(M\_v)\},$$

that is:

$$\underset{n\to \mathrm{\infty }}{\lim }\mathbb{P}(B_r^{\bullet }({\bar{T}}_n^{(p)})=\mathrm{\Delta })=\frac{{\alpha }^{-q}C(q)}{{\alpha }^{-p}C(p)}\prod _{v\in {\mathcal{F}}^{\ast }}\theta (c_v)\frac{{\rho }^{-N(M_v)}}{Z(c_v+1)},\backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash mathbb\{P\}\; (B\_r^\backslash bullet\; (\backslash bar\{T\}\_n^\{(p)\})\; =\; \backslash Delta)\; =\; \backslash frac\{\backslash alpha^\{-q\}\; C(q)\}\{\backslash alpha^\{-p\}\; C(p)\}\; \backslash prod\_\{v\; \backslash in\; \backslash mathcal\{F\}^*\}\; \backslash theta(c\_v)\; \backslash frac\{\backslash rho^\{-N(M\_v)\}\}\{Z(c\_v\; +\; 1)\},$$

for $\theta(k) = \rho^{-1}\alpha^{-k+1}Z(k+1)$. $\square$

Let us give a few properties of $\theta$ that will be useful in the sequel. These properties are obtained from the analytic combinatorial work in [9], rather than explicit enumeration as was the case for usual triangulations in [13].

First, the asymptotics of Z give:

$$\theta (k){\sim }_k\frac{1}{2}\sqrt{\frac{3}{\pi }}{k}^{-5\mathrm{/}2}\mathrm{.}(5.5)\backslash theta(k)\; \backslash sim\_k\; \backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash frac\{3\}\{\backslash pi\}\}\; k^\{-5/2\}.\; \backslash quad\; (5.5)$$

Moreover, $\theta$ has the following generating function $g_\theta$:

$$g_{\theta }(x)=\sum _{k=0}^{\mathrm{\infty }}\theta (k)x^k=1-\frac{3}{{(\sqrt{\frac{4-x}{1-x}}+1)}^2-1}\mathrm{\forall }x\in [0,1]\mathrm{.}(5.6)g\_\backslash theta(x)\; =\; \backslash sum\_\{k=0\}^\{\backslash infty\}\; \backslash theta(k)x^k\; =\; 1\; -\; \backslash frac\{3\}\{\backslash left(\backslash sqrt\{\backslash frac\{4-x\}\{1-x\}\}\; +\; 1\backslash right)^2\; -\; 1\}\; \backslash quad\; \backslash forall\; x\; \backslash in\; [0,\; 1].\; \backslash quad\; (5.6)$$

³Note that, as $\Delta$ is rooted, we can fix an arbitrary rule to determine where to glue the root of the other triangulation.