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I'm interested in a sequence of numbers whose ordinary generating function obeys the equation: $$F(z) = 1-z^2+z(F(z))^3.$$

Is there some (relatively simple) way to get a good upper bound on the corresponding sequence of the type $O(A^n)$? By "good" upper bound I mean at least $A < 6.75$ (which is what one obtains for the analogue without the $z^2$ term in the equation).

After some googling I'm vaguely aware of the Lagrange inversion formula which can be used when the $z^2$ term is not there and also of the notion that "singularities in the generating function determine exponential growth".

But the notion of singularity is very blurry to me (never did any complex analysis). My best guess is that the smallest modulus singularity is at about $0.15559 + 0i$, yielding a bound of $O(0.15559^{-n}) = O(6.43^n)$, but I don't really have an idea of what I'm doing so I could be way off.

Tassle
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    If $P(w) = w - (1 - z^2 + z w^3)$ and $\sum_{n \geq 0} a_n z^n$ is the root of $P$ which is regular at $z = 0$, the radius of convergence will be the distance to the nearest non-zero root of the discriminant of $P$. Therefore $a_n = O((A + \epsilon)^n)$, where $A$ is the largest positive root of $4 z^5 - 27 z^4 + 54 z^2 - 27$.

    We can use the Lagrange inversion theorem, write $a_n$ as an integral and apply the steepest descent method to obtain $$a_n \sim \frac {z_0} {2 \sqrt \pi n^{3/2}} A^n,$$ where $z_0$ is the smallest positive root of $8748 z^{10} - 6561 z^8 - 20250 z^4 + 8539$.

     – Maxim Jul 28 '20 at 21:27
  • @Maxim Thank you Maxim! Would you mind providing a hint or source as to why the radius of convergence is a root of the discriminant of P? I this true for any cubic P (eg. if I change $z^2$ to $z^3$) or only this particular one? (Sorry if this is trivial) – Tassle Jul 29 '20 at 08:13
  • Nevermind, I think I figured it out. Thanks again! – Tassle Jul 29 '20 at 10:20

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