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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.

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2 Answers 2

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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.

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  • $\begingroup$ I see Zeilberger's combinatorial formula specializes to give the required solution. Thanks. However, to say "Mehler formula is related to the the square root of your generating function" may not be of direct help. $\endgroup$ Apr 17, 2017 at 15:18
  • $\begingroup$ @T.Amdeberhan that was just meant for history/context. The combinatorial set up in Zeilberger 's paper is precisely an answer to the question. $\endgroup$ Apr 17, 2017 at 15:22
  • $\begingroup$ Now, I understand and appreciate your rendition. Thanks. $\endgroup$ Apr 17, 2017 at 15:23
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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.

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