Permutations And Their Inverses

[Leroy Quet]

Permutations with no cycles of length 1 or 2.  By usual methods,
the EGF is exp(-x-x^2/2)/(1-x)

0, 0, 2, 6, 24, 160, 1140, 8988, 80864, 809856

Maybe also a(0) = 1.

Identified as
A038205	  Number of derangements of n where minimal cycle size is 3.

Someone please correct the title, it should be "at least 3".  I'm going
to bed (2:10am).  Good night.

Brendan.

* Leroy Quet <qq-quet at mindspring.com> [070602 01:56]:
> I am wondering if this sequence is in the EIS. If it isn't, could someone 
> please submit it?
> 
> a(n) = the number of permutations {b(k)} of {1,2,3,...n} where b(m) does 
> not equal c(m) for all m, 1 <=m <=n, where {c(k)} is the inverse 
> permutation of {b(k)}.
> 
> 
> For example, with the particular permutation of the first 4 positive 
> integers, (2,4,1,3), we have the inverse permutation (3,1,4,2). Since 2 
> doesn't = 3, 4 doesn't = 1, 1 doesn't = 4, and 3 doesn't = 2, this 
> permutation and its inverse are counted.
> 
> Obviously, every acceptable permutation is a derangement, but not the 
> other way around.
> 
> For example, the permutation (4,3,2,1) is a derangement, but its inverse 
> is also (4,3,2,1).
> 
> I get that the sequence {a(k)} begins: 0,0,2,6
> 
> I did a search for "0,2,6 permutations inverse" on the OEIS, and nothing 
> came up.
> 
> Could someone please calculate {a(k)}, and submit it if it isn't already 
> in the database?
> 
> Thanks,
> Leroy Quet



Brendan McKay wrote:
>Permutations with no cycles of length 1 or 2.  By usual methods,
>the EGF is exp(-x-x^2/2)/(1-x)
>
>0, 0, 2, 6, 24, 160, 1140, 8988, 80864, 809856
>
>Maybe also a(0) = 1.
>
>Identified as
>A038205	  Number of derangements of n where minimal cycle size is 3.
>
>Someone please correct the title, it should be "at least 3".  I'm going
>to bed (2:10am).  Good night.


Of course! Believe it or not, I know what a cycle is. I just prefer not 
to think in terms of them for some reason.

Oh, well, sorry for the dumb question.

Leroy Quet