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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
3 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
"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
Your closed form is slightly off: $c_n=\frac1{n+1}\binom{2n}n$. – Brian M. Scott May 9 '12 at 3:27
If I am reading 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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