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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.

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    Here is a pointer: [generatingfunctionology](http://www.math.upenn.edu/~wilf/DownldGF.html).2012-05-08
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    Lagrange reversion, contour integration, and analysis of poles seem very much on the analytic side to me... and *gfology* explains them all. Nevermind. The other classic book in this area is *Analytic combinatorics*, by Flajolet and Sedgewick.2012-05-08
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    My last comment answered a comment by @Andy, now deleted. // On another note, there is a problem with the recursive relation on the coefficients $e_n$ since the only solution of the current one seems to be $e_n=0$ for every $n$.2012-05-08
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    @Didier Thank you. I deleted my comment because I was not precise, there is a part of analytical comb. in Wilfs book. Analytic combinatorics, is the book I started studying.2012-05-08
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    @Didier $e_n$: I made a mistake in the sum, the highest index is now changed to be $n-1$ not $n$2012-05-08
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    Which does not save your recursion.2012-05-08
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    Hm, I don't see any problem for n > 1 and $e_1 = 1$2012-05-08
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    Except that $e_n=(n-1)$times something yields $e_1=0$.2012-05-08
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    Ok I clarified it2012-05-08
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    A trivial observation: what you want is the dominant singularity of $xF$ instead of $F$ (because it's $0$), but it's ok since you'll get the asymptotics for $[x^n]F$ shifted by one. The dominant singularity of $xF$ is the singularity with smallest radius of $E$, $E'$ or $E''$. Using the recurrence we should be able to give a closed form for $E$.2012-05-08
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    $E$ in OEIS: http://oeis.org/A0006992012-05-08
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    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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