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Let $a_n=[x^n]\mathrm{e}^{x+x^2/2}.$ How does one show that $$ a_n \sim\frac{1}{2\sqrt{\pi}} n^{-(n+1)/2}\mathrm{e}^{-n/2+\sqrt n -1/4}?$$

I'd also appreciate references illustrating relevant techniques.

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up vote 5 down vote accepted

The function $$ f(x) = e^{x+x^2/2} $$ is the exponential generating function for the number of involutions on finite sets.

An analytic derivation of the asymptotic formula $$ [x^n]f(x) = \frac{1}{2\sqrt{\pi}} n^{-(n+1)/2} e^{n/2+\sqrt{n}-1/4}\left(1+O\left(n^{-1/5}\right)\right) $$ can be found on pages 558-560 in Flajolet and Sedgewick's Analytic Combinatorics (freely available here). Flajolet also cites volume 3 of Knuth's The Art of Computer Programming, which he says contains a derivation of the bound through the use of the explicit formula $$ [x^n]f(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{1}{(n-2k)!2^k k!}. $$

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@Yiorgos, I like the regular typeface for $e$, no need to change it :) – Antonio Vargas Jan 28 '14 at 16:45
Ok - understood! – Yiorgos S. Smyrlis Jan 28 '14 at 16:46

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