# Adjacent Pairs In Permutations Both Odd...
Leroy Quet
qqquet at mindspring.com
Tue Nov 5 01:38:09 CET 2002
UGGG!! I just checked this numerically with Mathematica, and it didn't
check out.
From Mathematica, I have the sequence (of the TOTAL number of occurrences
of adjacent elements both being odd in all of the permutations) starting
as:
0, 0, 4, 12, 144, 720, 8640, 60480, ...
What I am pretty sure of is (what I originally figured out {hopefully
correctly} & checks out numerically for the values above): If a(m) is the
m_th term,
a(2m) = (2m-1) a(2m-1).
My original formula works for m <= 4, which threw me off, since I only
checked my formula numerically by hand for m <= 4.
But I know not the recursive or closed formula for the sequence of
a(2m-1)s.
I'll maybe continue to work on this or simply give up. It seems like an
easy problem, but I keep wasting my time making errors.
Thanks,
Leroy Quet
-------
>Just sent this to sci.math:
>
>
>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)
>
>If so, then the number of times (over all permutations) that adjacent
>elements are odd would approach an average of once per permutation as m
>approaches infinity.
>
>--
>
>Now, the total number of occurrences of any elements being adjacent in
>all permutations of {1,2,3,...,m} is:
>(m-1) * m!
>
>
>We can get the same set of permutations by simply subtracting every
>element of each permutation from (m+1). If m is even, then the resulting
>permutations have elements with parity reversed from the original
>permutations.
>So the number of occurrences of EVEN elements being adjacent in all of
>the permutations of {1,2,3,...,m} is, for EVEN m, the same as for odd
>elements:
>
>2 * (m-1)! * floor((m-1)/2)
>
>= (m-1)! (m-2)
>
>So, if I am right, the total number of occurrences of an odd element
>being next to an even element (together with the number of occurrences of
>an even element being next to an odd element) is, for even m:
>
>(m-1) * m! - 2 * (m-1)! * (m-2) =
>
>(m-1)! (m^2 - 3m + 4)
>
>Half of this is for the number of occurrences of an odd being before an
>even.
>And (the other...) half is for the number of occurrences of an even being
>before
>an odd.
>
>So, the number of occurrences of two elements of identical-parity being
>adjacent over all permutations of {1,2,3,...m} is of O(m!).
>
>While the number of occurrences of two elements of opposite-parity being
>adjacent is of O(m!m).
>
>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?
>
>Thanks,
>Leroy Quet
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