Let
$$a(z) = \sum_{0<n} \frac{x^{n^2}}{(4n)!}$$
$$b(z) = \sum_{0<n} \frac{x^{n^2}}{(4n+3)!}$$
Both have a natural boundary at $|z| = 1$.
Let $c(z)$ be the series reversion of $b(z)$.
So $c(z)$ is the functional inverse of $b(z)$.
See : https://mathworld.wolfram.com/SeriesReversion.html
Does $f(z) = c(a(z))$ also have a natural boundary at $|z| = 1$ ?
Why is that ?
Notice $a$ and $b$ seem very similar, so you would expect $c(a(z))$ to at least converge in a larger radius ?
To imagine how this could perhaps work ,
Say $d(z)$ is a taylor series of the functional inverse of $a(z)$.
$$d(a(z)) = a(d(z)) = z$$
$$c(z) = a_0 + a_1 d(z) + a_2 d(z)^2 + a_3 d(z)^3 + ...$$
Then clearly
$$f(z) = c(a(z)) = a_0 + a_1 z + a_2 z^2 + a_3 z^3 + ...$$
And this could have any radius in principle, it just depends on the $a_n$.
Another thought.
How can a function converge outside its natural boundary ?
Well there are summability methods, lambert series etc.
But even if that fails the question above makes sense, just like
$a(d(z)) = z = b(c(z))$ everwhere since they are functional inverses.
Also sums like
$$ h(z) = \sum _{0<n} \frac{z^n}{(1 - z^n) n^4}$$
have a natural boundary, yet converge outside it as well.
Another example :
$$j(z) = a(z) - a(z^2/2)$$
$a(z)$ has a radius 1 and boundary natural boundary there.
$(z^2/2)$ has a radius $\sqrt 2$ and a natural boundary there.
However
$$j(2) = a(2) - a(4/2) = a(2) - a(2) = 0$$
So $j(z)$ has a taylor series with radius $1$ , natural boundary $1$, but $j(2) = 0$.
And there also ideas like $\frac{a(z)^2}{b(z)^2 + a(z) + 1}$ and such stuff.
Some people even consider extending these natural boundaries with "infinite numbers" or noncomplex numbers.
So many ideas exist.
Limits of truncated series is also an idea.