# [seqfan] Re: Permutation Ascents

zak seidov zakseidov at yahoo.com
Sat Aug 28 14:37:52 CEST 2010

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--- On Sat, 8/28/10, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:

> From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>
> Subject: [seqfan]  Permutation Ascents
> To: seqfan at seqfan.eu
> Date: Saturday, August 28, 2010, 7:41 AM
> Take the permutation of (1,2,3,4):
> (2,4,1,3)
> and its inverse permutation:
> (3,1,4,2).
>
> Notice that the number of ascents ((p(k+1)>p(k))
> of (2,4,1,3) is 2,
> but the number of ascents of the inverse permutation is 1.
>
> 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.
>
This is given by A025585  	 	 Central Eulerian numbers A(2n-1, n).

> Let a(n) = the number of permutations P of (1,2,3,..., n)
> where the number of ascents of P do not equal the number of
> ascents of the inverse permutation of P.
> (And where the number of descents does not match either, of
> course.)
>
> Is {a(n)} already in the OEIS?

This given by n!-A025585(n).

>
> I am most interested in the sequence {a(n)/n!}.
> Does it approach zero? What is it asymptotical towards?
>
> And maybe there is a closed form for {a(n)}? (Involving,
> say, Eulerian numbers.)
>
>
> Thanks,
> Leroy Quet
>
> [ ( [ ([( [ ( ([[o0Oo0Ooo0Oo(0)oO0ooO0oO0o]]) ) ] )]) ] )
> ]
>
>
>
>
>
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