# [seqfan] Re: please advise: count of decompositions of permutatations of n into products of involution pairs

Meeussen Wouter (bkarnd) wouter.meeussen at vandemoortele.com
Wed Aug 5 10:09:06 CEST 2009

```Franklin,

you are right.
Thanks for catching this.
My calculation disregarded decompositions among the 2-cycles, exactly as
you showed.

W.

-----Original Message-----
From: seqfan-bounces at list.seqfan.eu
[mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
franktaw at netscape.net
Sent: woensdag 5 augustus 2009 6:24
To: seqfan at list.seqfan.eu
permutatations of n into products of involution pairs

I don't think this is correct, unless I misunderstood something.  For
n=2, the permutation (1 2) can be factored as (1 2) * (1)(2) or (1)(2)
* (1 2); and (1)(2) as (1)(2) * (1)(2) or (1 2) * (1 2), which I think
would make a(2) = 4.

In fact, it seems to me that a(n) is just the square of the number of
involutions of n, since every ordered pair of involutions generates some
permutation.

So what is it you're calculating?

-----Original Message-----
From: wouter meeussen <wouter.meeussen at pandora.be> ...

Any permutation can be written as a product of a pair of involutions,
and often in many ways.
...

About ten years later, it came to me that the set of permutations of n
generates more than n! couples of involutions as its decompositions (no,
not prime, not unique).

Count of which is, teasingly to n=36: (offset 1)

1, 2, 10, 46, 326, 2476, 22492, 222860, 2539756, 31101976, 424163576,
6183804232, 98022462760, 1653222913616, 29862161016976, 570997442386576,
11573674977168272, 247136949802012960, 5552950221534095776,
130868386599399370976, 3227555172686193633376, 83164157620475009758912,
2232922569081925813952960, 62421208989699816969363136

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