[seqfan] Re: Permutation Ascents
Douglas McNeil
mcneil at hku.hk
Sat Aug 28 15:19:44 CEST 2010
LQ:
>> It seems (I have no proof) that the vast majority of
>> permutations of any
>> (1,2,3,..., n)
>> each have the same number of ascents (and the same number
>> of descents) as their inverse permutation.
>>
ZS:
> This is given by A025585 Central Eulerian numbers A(2n-1, n).
Are you sure, Zak? I don't see how the elements of A025585 can make
sense here, they're too big. Bigger than n!, so I don't think they
can be counting these permutations. If I understand the question
correctly, empirically I find
sage: descents_differ = lambda p: len(p.descents()) !=
len(p.inverse().descents())
sage: a = [len(filter(descents_differ, Permutations(n))) for n in range(1, 10)]
sage: a
[0, 0, 0, 2, 24, 228, 2088, 19732, 197890]
and a(9) is the first case for which a(n) > b(n). Switching to
statistical methods to estimate the a(n)/n! fraction:
10 0.585400000000000
20 0.750800000000000
50 0.855000000000000
100 0.901100000000000
200 0.934900000000000
500 0.956900000000000
1000 0.967800000000000
I think it asymptotes to 1. But I may be missing something.
Doug
--
Department of Earth Sciences
University of Hong Kong
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