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Number of permutations of [2n+1] beginning with exactly n ascents.
3

%I #6 Sep 09 2015 12:09:13

%S 1,2,15,168,2520,47520,1081080,28828800,882161280,30474662400,

%T 1173274502400,49819040409600,2313026876160000,116576554558464000,

%U 6338850154116480000,369890550169620480000,23056510960573009920000,1529010726859052236800000

%N Number of permutations of [2n+1] beginning with exactly n ascents.

%H Alois P. Heinz, <a href="/A262035/b262035.txt">Table of n, a(n) for n = 0..365</a>

%F E.g.f.: (1-2*x)/(4*sqrt(1-4*x)*x^2)+(2*x^2-1)/(4*x^2).

%F a(n) = 2*(n+1)*(2*n+1)*a(n-1)/(n+2) for n>1, a(n) = n+1 for n<=1.

%F a(n) = (2*n+1)!/(n+1)! - floor((2*n+2)!/((n+1)!*(n+1)*(n+2)*2)).

%F a(n) = A262033(2n+1) - A262034(2n+1).

%e a(0) = 1: 1.

%e a(1) = 2: 132, 231.

%e a(2) = 15: 12435, 12534, 12543, 13425, 13524, 13542, 14523, 14532, 23415, 23514, 23541, 24513, 24531, 34512, 34521.

%p a:= proc(n) option remember; `if`(n<2, n+1,

%p 2*(n+1)*(2*n+1)*a(n-1)/(n+2))

%p end:

%p seq(a(n), n=0..20);

%Y Cf. A262033, A262034.

%K nonn

%O 0,2

%A _Alois P. Heinz_, Sep 08 2015