I need to prove that the generating function of $a_n = \sum\limits_{k = 0}^{\lfloor\frac{n}{2}\rfloor}{n \choose 2k}\dfrac{(2k)!}{k!2^k}$ is $e^{x+x^2/2}$.
I tried it like this. I know that $e^x=\sum_{i=0}^{\infty}\dfrac{x^i}{i!}$
So $e^{x+x^2/2}=\sum_{i=0}^{\infty}\dfrac{(x(1+\dfrac{x}{2}))^i}{i!}=\sum_{i=0}^{\infty}\dfrac{x^i\sum_{j=0}^i{i\choose j}\left(\dfrac{x}{2}\right)^j}{i!}$
Let's get the coefficient next to $x^k$. $\dfrac{x^i{i\choose j}\left(\dfrac{x}{2}\right)^j}{i!}$ will have an $x$ in $k$'th power, when $i+j=k$ and $j\leq i \leftrightarrow j\leq k-j \leftrightarrow 2j\leq k$ Let's put $i=k-j$ , then $\dfrac{x^{k}{k-j\choose j}}{(k-j)! 2^j}$
An I'm stuck...
Think of $e^{x+x^2/2}$ as $e^x\cdot e^{x^2/2}$: the series is
$$\left(\sum_{n\ge 0}\frac1{n!}\left(\frac{x^2}2\right)^n\right)\left(\sum_{n\ge 0}\frac{x^n}{n!}\right)=\left(\sum_{n\ge 0}\frac{x^{2n}}{2^nn!}\right)\left(\sum_{n\ge 0}\frac{x^n}{n!}\right)\;,$$
in which the $x^n$ term is
$$\begin{align*} \sum_{k=0}^{\lfloor n/2\rfloor}\left(\frac1{2^kk!}\cdot\frac1{(n-2k)!}\right)x^n&=\sum_{k=0}^{\lfloor n/2\rfloor}\frac{n!}{2^kk!(n-2k)!}\left(\frac{x^n}{n!}\right)\\\\ &=\sum_{k=0}^{\lfloor n/2\rfloor}\left(\frac{n!}{(2k)!(n-2k)!}\cdot\frac{(2k)!}{2^kk!}\cdot\frac{x^n}{n!}\right)\\\\ &=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\frac{(2k)!}{2^kk!}\left(\frac{x^n}{n!}\right)\\\\ &=a_n\left(\frac{x^n}{n!}\right)\;, \end{align*}$$
so $e^{x+x^2/2}$ is the exponential generating function of $\langle a_n:n\in\Bbb N\rangle$.