Provoking involutions further

Question

Let $\mathfrak{S}_n$ denote the permutation group, and $I_0(n)=\sum_{j\geq0}\binom{n}{2j}\frac{(2j)!}{2^jj!}$ stand for involutions see A000085 for more interpretations. There is also these numbers $I_1(n)=\sum_{j\geq0}\binom{n}jI_0(j)I_0(n-j)$ described in A000898 by several means.

Let me introduce the numbers $I_2(n)=\sum_{j\geq0}\binom{n}jI_0(j)^2I_0(n-j)^2$. I was able to verify the exponential generating function $$\sum_{n\geq0}I_2(n)\frac{x^n}{n!}=\frac1{1-x^2}e^{\frac{2x}{1-x}}.$$ However, it is desirable to know:

Question. Is there a combinatorial meaning to the numbers $I_2(n)$?

Remark. Of course, it is also interesting if one can provide any other context where $I_2(n)$ appears.

Answer 1

The generating function for involutions with respect to the number of fixed points is given by an evaluation of Hermite polynomials . The bilinear generating function of Hermite polynomials is given by "Mehler's formula": $$\sum_{n\geq 0}\frac{t^n H_n(x)H_n(y)}{2^nn!}=\frac{1}{\sqrt{1-t^2}}\exp\left(\frac{t^2(x^2+y^2)-2txy}{t^2-1}\right).$$

Foata wrote the paper "A combinatorial proof of the Mehler formula" where he gives a combinatorial interpretation and bijective proof of the formula. Mehler's formula (after an appropriate specialization) is technically related to the square root of your generating function. Your generating function also has a similar refinement and it's what Zeilberger calls "the heterosexual version of Mehler's formula". To see your identity in Zeilberger's set up you need to set $x=y=1$ and $s=t$ in his main identity.

Answer 2

Define a standard bitableau of size $n$ to be a pair $(P_1, P_2)$ of standard tableaux of total size $n$ such that each of the integers $1,\dotsc, n$ occurs exactly once in either tableau.

Then $I_2(n)$ is the number of pairs of standard bitableaux $((P_1, P_2), (Q_1, Q_2))$ of size $n$ such that $P_1$ has the same content as $Q_1$. In fact, the $j$th summand in the sum defining $I_2(n)$ is the number of such pairs where $P_1$ and $Q_1$ have $j$ cells, and the same content.