Periodic Antisymmetric Matrices counted by Hermitte Numbers

[wouter meeussen]

Hi Gottfried,

sorry for the long interuption. Busy at work.
I was just pointing out that these matrices form a combinatorial construct that is counted by the
(unsigned) Hermite Numbers. Somewhat rare I think.
formula: 2^(k/2) Mod[k+1,2] k!/k!!
They seem to be equivalent to all permutations of 2n with the restriction that all cycles have
length two (pairings): example:
{1,5},{2,7},{3,4},{6,8} stands for the matrix

0  0  0  0  x  0  0  0
0  0  0  0  0  0  x  0
0  0  0  x  0  0  0  0
0  0 -x  0  0  0  0  0
-x 0  0  0  0  0  0  0
0  0  0  0  0  0  0  x
0 -x  0  0  0  0  0  0
0  0  0  0  0 -x  0  0

and now each +x can be choosen as +1 or -1. Hence the 2^(k/2).

The crux is to grok why all these and only these (antisymmetric) matrices are periodic, and why with
period 4.

W.

----- Original Message ----- 
From: "Gottfried Helms" <Annette.Warlich at t-online.de>
To: "wouter meeussen" <wouter.meeussen at pandora.be>; <Seqfan at ext.jussieu.fr>
Sent: Monday, December 05, 2005 11:24 AM
Subject: Re: Periodic Antisymmetric Matrices counted by Hermitte Numbers


Am 03.12.2005 23:08 schrieb wouter meeussen:
> Antisymmetric: Transpose(A)= -A, so zero diagonals
> Periodic: some matrix power = Identity Matrix
> A067994= Hermitte Numbers= 1,0,2,0,12,0,120,0,1680, ...
>
> Their periods are all 4.
> Strong law of small numbers checked upto n=8.
> Who says this is self-evident?
>
> W.
>
>
>
>
>
A
0    2^0
2    2^1 *1
4    2^2 *1*3
6    2^3 *1*3*5
8    2^4 *1*3*5*7
10   2^5 *1*3*5*7*9 --- but this seems very regular.

What was your expectation about periods=4 ?
Is it something with phi()-function or order of cyclic subgroups?
Then the reason for some irregularity may be, that 9 isn't prime.

Gottfried Helms