Tough Combinatorics Question (possibly related to Stirling numbers)

We can select $i\geq0$ persons to be used once in ${N\choose i}$ ways, and then can select $j\geq0$ persons to be used twice in ${N-i\choose j}$ ways. Necessarily $i+j\leq N$.

Assume that the $i$, resp. $j$, persons have been selected. We then produce a clone of each of the $j$ persons and have before us $i+2j$ persons, $j$ of them clones. These $i+2j$ persons can be linearly arranged in $(i+2j)!$ ways, but we have to divide this number by $2^j$ since we cannot distinguish between a real person and its clone.

It follows that the total number of admissible arrangements comes to $$\sum_{i=0}^N\sum_{j=0}^{N-i}{N!\over i!\,j!\,(N-i-j)!}\ {(i+2j)!\over 2^j}\ .$$ Maybe this expression can be simplified somewhat.

Hint: The first few values of the formula \begin{align*} a_N=\sum_{i=0}^N\sum_{j=0}^{N-i}{N!\over i!\,j!\,(N-i-j)!}\ {(i+2j)!\over 2^j} \end{align*} stated in the answer of @ChristianBlatter are

$$\begin{array}{rr} N&a_N\\ \hline 0&1\\ 1&3\\ 2&19\\ 3&271\\ 4&7\,365\\ 5&326\,011\\ 6&21\,295\,783\\ 7&1\,924\,223\,799\\ 8&229\,714\,292\,041\\ 9&385\,007\,742\,568\,755\\ 10&6\,630\,796\,801\,779\,771\\ 11&1\,527\,863\,209\,528\,564\,063\\ 12&420\,814\,980\,652\,048\,751\,629\\ 13&136\,526\,522\,051\,229\,388\,285\,611\\ \vdots&\vdots \end{array}$$ They are stored as A003011 in OEIS giving the number of permutations of up to $n$ kinds of objects, where each kind of object can occur at most two times.

Since there is no simpler formula stated an essential simplification of the formula is not plausible.