[seqfan] Re: Permutation Ascents
Ron Hardin
rhhardin at att.net
Sat Aug 28 15:15:14 CEST 2010
I get for a(1..13)
0 0 0 2 24 228 2088 19732 197890 2131060 24729108 309027560 4148115048
subject to finding some kind of bug.
rhhardin at mindspring.com
rhhardin at att.net (either)
----- Original Message ----
> From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>
> To: seqfan at seqfan.eu
> Sent: Sat, August 28, 2010 7:41:52 AM
> Subject: [seqfan] Permutation Ascents
>
> 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.
>
> 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?
>
> 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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