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

franktaw at netscape.net franktaw at netscape.net
Wed Aug 5 06:23:32 CEST 2009

```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

```