Convergency radius of the generating series for A93637 Sequence A93637 of the OEIS (https://oeis.org/A093637) starting as $1,1,2,4,9,20,49,117,297,746,1947,\ldots$ is defined by
the coefficients $a_0,a_1,\ldots$ of the unique formal power series
defined by the equality
$$A(x)=\prod_{n=0}^\infty \frac{1}{1-a_nx^{n+1}}=\sum_{n=0}^\infty a_n x^n\ .$$
Experimentally, $a_{n+1}/a_n$ seems to converge to some a limit
roughly given by $2.96777$ suggesting a convergency radius
slightly larger than $1/3$ for $A(x)$.
Is there an easy argument ensuring that $A(x)$ has strictly positive convergency radius? Are there computable upper/lower bounds
for the convergency radius of $A(x)$?
 A: As noted in the OEIS entry, the sequence is bounded below by the number of unlabelled rooted trees, so the radius of convergence can be only equal or less. For unlabelled rooted trees, the radius of convergence is the reciprocal of Otter's constant $\rho=2.955765285{\ldots}$ (see A051491 for lots of digits). This is suspiciously close to Roland's numerical estimate so maybe the radius of convergence is the same.
Wikipedia kindly provides both sequences up to $n=1000$, which allows a numerical comparison. Let $a_n$ be the counts in Roland's question and $b_n$ be the number of unlabelled rooted trees. For $100\le n\le 1000$, the ratio $a_n/b_n$ is a very close match to
$$ 3.04 \times 1.00406^n.$$
This suggests that the radius of convergence of $\{a_n\}$ is a tiny bit less than $1/\rho$.  Established methods for asymptotically counting trees are likely to give precise results.
A: This is pretty simple, really. Note that we can obtain our power series in the following way. Define on (formal) power series with positive coefficients the transform
$$
T\sum_{k=0}^\infty b_kx^k=\text{Expansion of }\prod_{k=0}^\infty\frac 1{1-b_k x^{k+1}}.
$$
Start with $f_0(x)=1$ and iterate $f_n(x)=Tf_{n-1}(x)$. Then $f_n$ will have correct coefficients up to $a_n$.
Now we want to prove by induction that $f_n$ cannot be greater than $A$ on $[0,a]$.
Initially it is true for all $A\ge 1$. Now for $x\in[0,a]$ and $f_{n-1}(x)= \sum_{k=0}^\infty b_kx^k$,
$$
f_n(x)=\left[\prod_{k\ge 0}(1-b_kx^{k+1})\right]^{-1}\le \left[(1-\sum_{k\ge 0}b_kx^{k+1})\right]^{-1}
\\
=(1-xf_{n-1}(x))^{-1}\le (1-aA)^{-1}
$$
as long as  $aA<1$. Thus we have our upper bound if $(1-aA)^{-1}\le A$. We then just take $A=2$, $a=\frac 14$. This immediately proves that the convergence radius is at least $\frac 14$ because all approximations are uniformly bounded in $|z|\le 1/4$ (real positive $z$ gives the largest value with fixed $|z|$) and we have coefficient-wise convergence.
One can show an upper bound in the same way. Suppose that the convergence radius of the final series $f(x)=\sum_{k\ge 0}a_k x^k$ is greater than $u$. Then the function $f(x)$ is well-defined in $|z|\le u$ and we can write
$$
f(u)=\left[\prod_{k\ge 0}(1-a_ku^{k+1})\right]^{-1}\ge \exp\left[\sum_{k\ge 0}a_ku^{k+1}\right]=\exp[uf(u)]\,,
$$
so if $e^{uF}>F$ for all $F>0$, we have $f(u)>f(u)$, which is nonsense. The inequality is equivalent to $u^{-1}(uF)e^{-uF}<1$ and holds for $u>1/e$ since $\max_{y\ge 0}ye^{-y}=1/e$.
Thus the radius $r$ of convergence satisfies $\frac 14\le r\le \frac 1e$. One can easily improve these bounds by writing the iterations as $\sum_{k=0}^N a_kx^k+g_n(x)$ (with correct $a_k$ up to $k=N$), starting with $g_0(x)=0$, and using the same inequalities for the products in which the coefficients of $g_n$ participate but the corresponding equations for cutoffs would be impossible to solve algebraically though numerical solutions would be not too hard to obtain.
It may be a bit more interesting to discuss whether $f$ has an analytic continuation beyond the disk $|z|<r$. Of course, there is a huge blow up problem at $z=r$, but, say, what happens when $z\to-r$ doesn't look entirely obvious to me unless I miss something trivial.
