Asymptotics of the e.g.f. $\exp(2x+x^2/2)$ (OEIS A005425) The sequence A005425$(n)$ on OEIS counting the number of partial involutions on $\{1, \dots, n\}$ has $\sum_{n \geq 0} (a_n / n!) x^n = \exp(2x + x^2/2)$ as its e.g.f.
How does one show that
$$
a_n \sim \exp(2 \sqrt n - n/2 - 1) \cdot n^{n/2} / \sqrt 2 ?
$$

I know that $a_n = \sum_k \binom{n}{k} b_k$ where $b_k$ is the sequence A000085 on OEIS with e.g.f $\exp(x + x^2/2)$.
In the post Asymptotic Analysis of coefficients of $\mathrm{e}^{x+x^2/2}$ a reference to pp.558-560 of Flajolet and Sedgewick's Analytic Combinatorics has been given.
However the book only covers the asymptotic analysis of $(b_n)$,
and I am not sure if the technique can be applied to $(a_n)$.
 A: The coefficient $a_n$ is given by $(d^{n} f/{dx}^n) (0)$ for $f(x) = e^{2x +x^2/2}$. From to Cauchy's integral formula, we have (with any $R>0$)
$$
a_n = \frac{n!}{2\pi i} \oint_{|z|=R} \frac{f(z)}{z^{n+1}} dz
 = \frac{n!}{2\pi i} \oint_{|z|=R} \exp\Bigl[2z + z^2/2 -(n+1) \log z \Bigr] dz \,.
$$
The integrand has stationary values at
$$ z_\pm = -1 \pm \sqrt{n+2} = \pm \sqrt{n} +O(1)\,.$$
Using the method of steepest decent, we thus obtain ($z = z_+ + i x$)
$$ a_n \sim \frac{n!}{2\pi}  e^{2 z_+ + z_+^2/2 -(n+1) \log z_+} \int e^{-x^2 (1+ O(n^{-1/2})) + O(x^3)} dx \,.  $$
Your result should follow by this and the Stirling approximation.
A: In $A005425$, there is another definition given by Groux Roland in year $2011$
$$a_n=\frac 1{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}} \,(x+2)^n\,dx$$ which gives
$$a_{2n}=\frac{2^n }{\sqrt{\pi }} \,\Gamma
   \left(n+\frac{1}{2}\right)\,\, _1F_1\left(-n;\frac{1}{2};-2\right)$$
$$a_{2n+1}=\frac{2^{n+2}  }{\sqrt{\pi }}\,\Gamma
   \left(n+\frac{3}{2}\right)\,\, _1F_1\left(-n;\frac{3}{2};-2\right)$$
Now, you need to combine Stirling approximation and the expansion of Kummer hypergeometric function for large negative $n$.
The simpler is probably to use the relation with generalized Laguerre polynomial (surprizingly not mentioned in the $OEIS$ page) since
$$\, _1F_1\left(-n;\frac{1}{2};-2\right)=\frac{L_n^{-\frac{1}{2}}(-2)}{\binom{n-\frac{1}{2}}{n}}$$
$$\, _1F_1\left(-n;\frac{3}{2};-2\right)=\frac{L_n^{\frac{1}{2}}(-2)}{\binom{n+\frac{1}{2}}{n}}$$ which make
$$a_{2n}=2^n \,n!\,L_n^{-\frac{1}{2}}(-2)$$
$$a_{2n+1}=2^{n+1} \,n!\,L_n^{\frac{1}{2}}(-2)$$
Using the asymptotics of Laguerre polynomials for negative argument
$$a_{2n}= 2^n\,n!\,\frac{e^{2 \sqrt{2(n+1)}-1}}{2 \sqrt{\pi } \sqrt{n+1}}$$
$$a_{2n+1}= 2^{n+1}\,n!\,\frac{e^{2 \sqrt{2(n+1)}-1}}{2 \sqrt{2 \pi }}$$
Now, using Stirling approximation and Taylor series
$$\log(a_{2n})=n \log \left(\frac{2n}{e}\right)+2 \sqrt{2n} -\frac{2+\log
   (2)}{2} + \sqrt{\frac{2}{n}}+O\left(\frac{1}{n}\right)$$
$$\log(a_{2n+1})=n \log \left(\frac{2n}{e}\right)+2 \sqrt{2n}-\frac{2-\log (n)}{2}+ \sqrt{\frac{2}{n}}+O\left(\frac{1}{n}\right)$$
which are better asymptotics.
Now, using $t=e^{\log(t)}$, you will get your formula and even more.
A: By Cauchy's formula
$$
a_n  = \frac{{n!}}{{2\pi {\rm i}}}\oint_{\left| z \right| = \sqrt n } {\frac{{\exp (2z + z^2 /2)}}{{z^{n + 1} }}{\rm d}z} ,
$$
where we chose the radius of the circle to be $\sqrt{n}$ (it can be any number for the function is entire). The substitution $z = \sqrt n {\rm e}^{{\rm i}\theta }$ yileds
$$
a_n  = \frac{{n!}}{{2\pi }}\frac{1}{{n^{n/2} }}\int_{ - \pi }^\pi  {\exp (n({\rm e}^{2{\rm i}\theta } /2 - {\rm i}\theta ))\exp (2\sqrt n {\rm e}^{{\rm i}\theta } ){\rm d}\theta } .
$$
The relevant saddle point of ${\rm e}^{2{\rm i}\theta } /2 - {\rm i}\theta$ is at $\theta=0$, whence, by the saddle point method,
\begin{align*}
a_n & \sim \frac{{n!}}{{2\pi }}\frac{1}{{n^{n/2} }}\int_{ - \pi }^\pi  {\exp (n(1/2 - \theta ^2 ))\exp (2\sqrt n ){\rm d}\theta } 
\\ &
 = \frac{{n!}}{{2\pi }}\frac{1}{{n^{n/2} }}\exp (n/2 + 2\sqrt n )\int_{ - \pi }^\pi  {\exp ( - \theta ^2 n){\rm d}\theta } 
\\ &
 \sim \frac{{n!}}{{2\pi }}\frac{1}{{n^{n/2} }}\exp (n/2 + 2\sqrt n )\int_{ - \infty }^{ + \infty } {\exp ( - \theta ^2 n){\rm d}\theta } 
\\ &
 = \frac{{n!}}{{2\pi }}\frac{1}{{n^{n/2} }}\exp (n/2 + 2\sqrt n )\frac{1}{{\sqrt n }}\int_{ - \infty }^{ + \infty } {\exp ( - t^2 ){\rm d}t} 
\\ &
 = \frac{{n!}}{{2\sqrt {\pi n} }}\frac{1}{{n^{n/2} }}\exp (n/2 + 2\sqrt n ),
\end{align*}
as $n\to +\infty$. Finally, with the aid of Stirling's formula,
$$
a_n  \sim \frac{{n^{n/2} }}{{\sqrt 2 }}\exp ( - n/2 + 2\sqrt n )
$$
as $n\to +\infty$.
