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The $n$-th Catalan number $c_n$ has the closed form $\frac1{n+1}\binom{2n}{n}$ and follows the recursion $c_n = \sum\limits_{i = 0}^{n-1} c_{n-1-i}c_i$

I am interested in the quantity $e_n$ which follows the recursion $e_n= (n-1) \sum\limits_{i = 1}^{n-1}{e_i e_{n-i}}$ for $n > 1$, with $e_1 = 1$.

I am wondering if it is possible to approximate $e_n$ using $c_n$?

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    For this kind of question, the first thing you should should try is to look it up in the On-Line Encyclopedia of Integer Sequences, [OEIS](http://http://oeis.org/). Have you?2012-05-09
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    http://oeis.org/A000699 may be of interest. Since there aren't any nice formulas in the large number of references there, it's a hint that there might not be any simple nice formulas.2012-05-09
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    Thank you, yes I am familiar with most of the literature and I am also not too optimistic to get an exact formula, but i was hoping more for some simple approximation. I don't need too much precision.2012-05-09
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    "of course I did this"- and you should have mentioned that in your question to begin with. :)2012-05-09
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    Your closed form is slightly off: $c_n=\frac1{n+1}\binom{2n}n$.2012-05-09
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    If I am reading http://arxiv.org/abs/hep-th/9912093 right, they guess that $e_n \approx .4 \cdot 2^{n-1} \Gamma(n + \frac{1}{2})$ (see the first paragraph of section 5 and their Table 2)2012-05-09
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    @leslie townes Thank you very much for the interesting link. If I understand it correctly, I am afraid they justify it empirically.2012-05-09
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    @Andy yeah, it is a guess and not a theorem. But having a good guess often makes it possible to prove a theorem (it wouldn't surprise me if the limiting value in their Table 2 is $\frac{1}{\sqrt{2\pi}} = .3989\dots$, which leads to the guess that $e_n \approx \frac{1}{\sqrt{2\pi}} 2^{n-1} \Gamma(n + \frac{1}{2})$, with my $e_n \approx a_n$ meaning that $e_n/a_n \to 1$ as $n \to \infty$).2012-05-09
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    For who might attempt to prove that guess, it may help to clean that guess up a bit: using Stirling-type stuff, it is equivalent to the guess that $e_n \approx \frac{1}{2} (\frac{2n}{e})^n$2012-05-09
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    btw: it is known that $lim_{n\to\infty} e_n / (2n-1)!! = 1/e$2012-05-09

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