Let $S(z)$ be the OGF of bracketings. Apply the standard Lagrangean framework to derive an asymptotic expression for $[z^n]S(z)$.
Remark: Apparently the topic "saddle-point method for Lagrangean forms" is not included in the Flajolet & Sedgewick book, so I wrote down the theorem below, it is basically (C).
From the Flajolet & Sedgewick book (p.81) we know that for $S(z)$ holds
$$S(z) = \frac{1+z-\sqrt{z^2-6z+1}}{4} \quad \text{ and } \quad S(z) = z + \frac{S(z)^2}{1-S(z)}.$$
However, I do not see how to apply the saddle-point method for Lagrangean forms here. Could you please give me a hint?
We consider the Lagrangean framework $$f(z) = z \phi(f(z)),$$ where both $\phi$ and $f$ are formal power series. By the Lagrange inversion formula we have for $n \ge 1$ that $$[z^n]f(z) = n^{-1}[z^{n-1}]\phi(z)^n. \qquad \text{(A)}$$
We assume the following conditions:
- $\phi(0) \ne 0$
- (Nonnegativity) $a_j \le 0 \quad (j \ge 0)$,
- (Aperiodicity) $\gcd\{j : a_j > 0\} = 1$;
- (Sub-criticality) $\phi$ is analytic in $\lvert z \rvert < R, 0 < R \le \infty$;
- (Sub-criticality) $r\phi^\prime(r) = \phi(r)$ for some $r \in (0,R)$
Under these assumptions we obtain by the saddle-point estimation for large powers (see the excerpt from the Flajolet & Sedgewick book, p. 586,587,588, below) with $A(z) = 1, B(z) = \phi(z)$ and $\lambda n = n-1$ the asymptotic estimate
$$[z^{n-1}]\phi(z)^n \sim \frac{\phi(r)^n}{r^{n-1}\sqrt{2 \pi n b(r)}}, \qquad \text{(B)}$$
where
$$b(r) = \frac{r \phi^\prime(r)}{\phi(r)} + \frac{r^2 \phi^{\prime\prime}(r)}{\phi(r)} - \left( \frac{r \phi^\prime(r)}{\phi(r)}\right)^2 = \frac{r^2 \phi^{\prime\prime}(r)}{\phi(r)},$$
where the last equality is due to $r \phi^\prime(r) = \phi(r)$. Thus (B) has the form
$$[z^{n-1}]\phi(z)^n \sim \frac{\phi(r)^n}{r^n\sqrt{2 \pi \phi^{\prime\prime}(r)/ \phi(r)}} n^{-\frac{1}{2}} = \frac{1}{r^n\sqrt{2 \pi \phi^{\prime\prime}(r)/ \phi(r)}} n^{-\frac{1}{2}} \left( \frac{\phi(r)}{r} \right)^n.$$
Thus, by (A) we have
$$[z^n]f(z) \sim \sqrt{\frac{\phi(r)}{2\pi \phi^{\prime\prime}(r)}} n^{-\frac{3}{2}} \left( \frac{\phi(r)}{r} \right)^n. \qquad \text{(C)}$$
