[seqfan] Permutation Ascents

[Leroy Quet]

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

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