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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. Have you? –  Arturo Magidin May 9 '12 at 3:23
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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. –  leslie townes May 9 '12 at 3:25
    
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. –  Andy May 9 '12 at 3:25
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"of course I did this"- and you should have mentioned that in your question to begin with. :) –  J. M. May 9 '12 at 3:27
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If I am reading 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) –  leslie townes May 9 '12 at 3:34
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