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We have two conjectured generalizations of the question asked at a Hankel matrix of involution numbers by Tewodros Amdeberhan. Let $n!!=1!\,2!\cdots n!$.

Conjecture 1. Let $I_k$ denote the number of involutions in the symmetric group $\mathfrak{S}_k$. Then the Smith normal form of the matrix $[I_{i+j}]_{i,j=0}^n$ has diagonal entries $0!, 1!, \dots, n!$,

Conjecture 2. Let $J_k=k!\sum_{\lambda\vdash k}s_\lambda$, where $s_\lambda$ is a Schur function. Set $J_0=1$. When the symmetric function $\det[J_{i+j}]_{i,j=0}^n$ is expanded in terms of power sums, then every coefficient is an integer divisible by $n!!$. (Tewodros' question is equivalent to the coefficient of $p_1^{n(n+1)}$ being equal to $n!!$.)

Is there a nice formula or combinatorial interpretation of the coefficients in Conjecture 2?

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  • $\begingroup$ This is a really cool extension of the discussion on involution. Glad to see it. $\endgroup$
    – T. Amdeberhan
    Commented Feb 8, 2017 at 17:28

1 Answer 1

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On Conjecture 1: As remarked by Johann Cigler in the linked question (and shown in the references linked there) the matrix $M:=\left[I_{i+j}\right]_{i\ge0\atop j\ge0}$ diagonalises as $M=U^TDU$ with $D:=\operatorname{diag(k!)}$, and $U$ an upper triangular integer coefficients matrix with unit diagonal elements, hence with integer coefficient inverse: so this is also the SNF of $M$.

For example:

$$ \left[ \begin {array}{ccccc} 1&1&2&4&10\\ 1&2&4&10& 26\\ 2&4&10&26&76\\ 4&10&26&76&232 \\ 10&26&76&232&764\end {array} \right] =$$$$= \left[ \begin {array}{ccccc} 1&&&&\\ 1&1&&& \\ 2&2&1&&\\ 4&6&3&1& \\ 10&16&12&4&1\end {array} \right] \left[ \begin {array}{ccccc} 1& & & & \\ &1& & & \\ & &2& & \\ & & &6& \\ & & & &24\end {array} \right] \left[ \begin {array}{ccccc} 1&1&2&4&10\\ &1&2&6& 16\\ & &1&3&12\\ & & &1&4 \\ & & & &1\end {array} \right] $$

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  • $\begingroup$ This diagonalisation of $M$ is precisely the same as in my answer to the linked Amdeberhan’s question. $\endgroup$
    – Johann Cigler
    Commented Feb 9, 2017 at 15:57
  • $\begingroup$ (Actually I can't see the diagonalisation there, but yes, there is a uniqueness result :) ) $\endgroup$
    – Pietro Majer
    Commented Feb 9, 2017 at 16:43
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    $\begingroup$ $\sum_k\binom{n}{k}I_{n-k} \binom{m}{k}I_{m-k} k!=I_{m+n} $ is equivalent with the fact that if $A$ is the matrix with entries $\binom{i}{j}I_{i-j}$ and $D$ is the diagonal matrix with elements $i!$ then $M=ADA^T.$ $\endgroup$
    – Johann Cigler
    Commented Feb 9, 2017 at 16:59
  • $\begingroup$ The identity Johann Cigler mentioned is equation (2.7) in his link: arxiv.org/pdf/1406.2356.pdf $\endgroup$
    – T. Amdeberhan
    Commented Feb 9, 2017 at 18:34
  • $\begingroup$ ok thank you guys $\endgroup$
    – Pietro Majer
    Commented Feb 9, 2017 at 19:05

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