TITLE: Asymptotic growth rate of coefficients of generating function
QUESTION [10 upvotes]: how to calculate the asymptotic growth rate of coefficients generating function $T(z)$ satisfied this identity
$T(z)=z+\frac{T(z)^3}{6}+\frac{T(z^2)T(z)}{2}+\frac{T(z^3)}{3}$

REPLY [3 votes]: Some remarks about bounding the coefficients of $T(z)$ from above and from below. 
I assume $T_0=0$ in $T(z)=\sum_{k=0}^\infty T_k z^k$, to which it corresponds a unique formal series solution of the functional equation for $T(z)$, e.g. obtained by iteration of the composition operator on the RHS. Let's first prove that even order coefficients vanish, and odd order coefficients increase. 
The Ansatz $T(z):=zS(z^2)$ produces this equation for $S(z):=\sum_{k=0}^\infty S_k z^k$:
$$S(z)=1+\frac{1}{6}zS(z)^3 +\frac{1}{2}zS(z^2)S(z)+ \frac{1}{3}zS(z^3)\ .$$
The latter equation also have a unique formal series solution, which proves that $T$ has indeed the above form $zS(z^2)$, that is, all even order coefficients of $T$ vanish. Since $S$ can be obtained by iteration of the composition operator on the RHS, starting e.g. by $1$, it is clear that $S_k\ge0$ for all $k$, and since $S_0=1$  we also have $[z]^k S(z)^3\ge 3 [z^k]S(z)$ and $  [z^k]S(z^2)  S(z)\ge [z^k]S(z)$. Thus $$S(z)\ge 1 + z S(z) 
 \ ,$$
(coefficient-wise), proving  that $S$ has increasing coefficients. Having $S(z)$ positive, increasing coefficients also implies, again from the equation, 
$$6S(z)\le 6 + z S(z)^3 + 3zS(z)^2+2zS(z)                      \ ,$$ 
(coefficient-wise). This implies that $S$ is dominated coefficient-wise by the power series solution $y(z)$ of $6y=6+z(y^3+3y^2+y)$. By uniqueness, $y$ is the expansion at $z=0$ of the local inverse of $$h(y)=\frac{6(y-1)}{y(y+1)(y+2)}$$ at $y=1$. This proves that $S$ has a positive radius of convergence $\rho$.
A bound from above for the  coefficients of $S$ can be deduced from 
$$6S(z)\ge 6 + z S(z)^3 + 3zS(z) +2z                       \ ,$$ 
(coefficient-wise), which implies that the coefficients of $S(z)$ are larger than the coefficients of the local inverse at $y=1$ of
$$g(y)=\frac{6(y-1)}{y^3+3y+2}\ .$$
Putting $y=x+1$, by the Lagrange inversion applied to $h(x+1)$ and $g(x+1)$ one gets, for $k\ge1$
$$\frac{1}{k}[x^{k-1}](1+x+ x^2/2 +x^3/6)^k \le S_k \le \frac{1}{k}[x^{k-1}](1+ 11x/6 +x^2 + x^3/6)^k$$
etc.