5
$\begingroup$

If $\mathfrak{S}_n$ denotes the permutation group on $n$ letters, then $Inv(n)=\{\pi: \pi^2=1\}\subset\mathfrak{S}_n$ is the set of involutions or self-inverse permutations. The latter is enumerated by $$I_n=\sum_{k\geq0}\binom{n}{2k}\frac{(2k)!}{2^kk!}.$$ This is also the number of Young tableaux with $n$ cells; see OEIS A000085 for more.

Let's alter $I_n$ a little bit to write $$J_n=\sum_{k\geq0}\binom{n}{2k+1}\frac{(2k)!}{2^kk!}.$$ It is possible to prove that $J_{n+1}=\sum_{j=0}^nI_j$, simply a sum of involutions. Although this is okay, I really would like to ask:

Question. Are there other interesting (combinatorial or otherwise) interpretations of $J_n$?

Note. As an aside, observe that there is a "near-by" sequence $$a_n=\sum_{k\geq0}\binom{n}{2k+1}\frac{(2k+1)!}{2^kk!}$$ which is listed on OEIS A013989 as the number of fixed points in all involutions of $n+1$.

Improve this question
$\endgroup$
2
  • $\begingroup$ If $U: \mathbb{C}Y \to \mathbb{C}Y$ is the standard up operator on the vector space with basis indexed by the elements of Young's lattice (so $U[\mu] = \sum_{\mu \lessdot \lambda} [\lambda]$) then $I_n$ is the sum of the coefficients of $U^n[\varnothing]$ (i.e., the image of $U^n[\varnothing]$ under the map $[\lambda] \mapsto 1$ for all $\lambda$), while $J_{n}$ is the sum of the coefficients of $((1-U^n)/(1-U)) [\varnothing]$. But this is really just a restatement of what you have already observed. $\endgroup$
    – Sam Hopkins
    May 10, 2017 at 17:39
  • $\begingroup$ Interesting indeed. $\endgroup$
    – T. Amdeberhan
    May 11, 2017 at 0:14

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.