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Given

$E' = (E^2 + E - x)/2xE$
$xF = E^3 E' + 2xE^3 E'' + E^2 - x^2$

where $E = \sum_{n > 0}{e_n x^n}$
with $e_n = (n-1) \sum^{n-1}_{i = 1}{e_i e_{n-i}}$ for $n > 1$ and $e_1 = 1$

I am interested in finding the singularity of $F$ with smallest modulus, when interpreting $F$ as a function in the complex plane.

I just started studying analytic combinatorics by my self but my calculus knowledge is a bit rusty, so any pointers would be appreciated.

  • 0
    Another observation: the dominant singularity of $x^kF$ is the one of $E$, because of the differential equation for $E'$ (and also we have $E'' = (xE + x^2 - E^3 - E^4)/(4x^2 E^3)$). (I added a constant $k$ to get rid of the $x^k$ in the denominator).2012-05-09

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The relevant OEIS page, mentioned by @Diego in the comments, links to Analytic Combinatorics of Chord Diagrams by P. Flajolet and M. Noy, which refers to P.R. Stein and C.J. Everett, On a Class of Linked Diagrams II. Asymptotics, Discrete Mathematics 21 1978, 309-318, to assert that $ e_n\sim\frac1{\mathrm e}\frac{(2n)!}{2^nn!}. $ Stirling's approximation yields $ e_n\sim\frac{\sqrt2}{\mathrm e}\left(\frac{2n}{\mathrm e}\right)^n, $ hence the radius of convergence of the series $E(x)=\sum\limits_{n\geqslant0}e_nx^n$ is zero. The same is true for $F(x)$.