Let $I\_k$ denote the [enumeration of involutions](http://mathworld.wolfram.com/PermutationInvolution.html) among permutations in $\mathfrak{S}\_k$. I always enjoy these numbers. Of course, here is yet another cute experimental finding for which I ask validity. 



> 
> **Question.** Let $n!!=1!2!\cdots n!$. Is the following true?
>  $$\det\left[I\_{i+j}\right]\_{i,j=0}^n=n!!$$
> 
> 
> 



a Hankel matrix of involution numbers
As in arXiv:0902.1650 it suffices to show that $a(n,0)=I\_n$ if $a(n,j)$ satisfies $a(n,j)=a(n-1,j-1)+a(n-1,j)+(j+1)a(n-1,j+1)$ with $a(n,-1)=0$ and $a(0,j)=[j=0]$.
But it is easily verified that $a(n,j)=\binom{n}{j}I\_{n-j},$ because then the above recursion reduces to $I\_n=I\_{n-1}+(n-1)I\_{n-2}.$


**Edit**: Another proof along the same lines follows immediately from formula (2.7) in [this paper](https://arxiv.org/pdf/1406.2356.pdf) entitled *Involutions and their progenies*:
$\sum\_k\binom{n}{k}I\_{n-k} \binom{m}{k}I\_{m-k} k!=I\_{m+n}.$


