# Adjacent Pairs In Permutations Both Odd...

[Leroy Quet]

I will repost this here:
Erick Bryce Wong posted to sci.math:

>
>Leroy Quet <qqquet at mindspring.com> wrote:
>>Consider all of the m! permutations of {1,2,3,4,...,m}.
>>
>>I THINK (I think that I think) that
>>
>>the TOTAL number of occurrences of adjacent elements both being odd in
>>all of the permutations is:
>>
>>2 * (m-1)! * floor((m-1)/2)
>> [snip]
>>Therefore, IF true, the number of opposite-parity pairs grows much
>>faster than the number of same-parity pairs.
>>
>>But this seems counter-intuitive to me. Is there an easy way to see
>>this intuitively as fact, if, in fact, it is indeed fact?
>
>I get ceil(m/2) * ceil(m/2-1) * (m-1)!, which certainly jibes better
>with both of our intutions, since it corresponds to about 1/4 of the
>total number (m-1)*m! of adjacent positions over all permutations.
>
> -- Erick

This agrees, at least, with my formula:

a(2m) = (2m-1) * a(2m-1).

I don't know how he got the a(2m-1)'s, however.

Thanks,
Leroy Quet