I'm solving the following formula.
$\sum_{0 \le m \le n/2} \frac{1}{(n-2m)! m!}$
For me it is never simplified further. Is there anyone to give me an hint?
I tried :
Let $r = 2m$.
$$ \begin{aligned} \sum_{0 \le m \le n/2} \frac{1}{(n-2m)! m!} &= \sum_{0 \le r \le n} \frac{n!}{(n-r)! (r/2)!} \frac{1}{n!} \\[10pt] &= \sum_{0 \le r \le n} \frac{n!(r/2)!}{(n-r)! (r/2)!(r/2)!} \frac{1}{n!} \\[10pt] &= \sum_{0 \le r \le n} \frac{n!(r/2)! 2^r}{(n-r)! r!} \frac{1}{n!}\\[10pt] &= \sum_{0 \le r \le n} \binom{n}{r} (r/2)! 2^r \frac{1}{n!}\\[10pt] \end{aligned} $$
It looks like it doesn't have a closed form. The sequence
$$a_n=\sum_{0\leq m\leq n/2} \frac{n!}{(n-2m)!m!}$$
exists on OEIS here, which lists come interesting properties, but gives no closed form, as well as the recurrence relation
$$a_n=a_{n-1}+2(n-1)a_{n-2}.$$