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A262034 Number of permutations of [n] beginning with at least ceiling(n/2) ascents.

%I A262034 #18 Dec 04 2022 08:32:43
%S A262034 1,0,1,1,4,5,30,42,336,504,5040,7920,95040,154440,2162160,3603600,
%T A262034 57657600,98017920,1764322560,3047466240,60949324800,106661318400,
%U A262034 2346549004800,4151586700800,99638080819200,177925144320000,4626053752320000,8326896754176000
%N A262034 Number of permutations of [n] beginning with at least ceiling(n/2) ascents.
%H A262034 Alois P. Heinz, <a href="/A262034/b262034.txt">Table of n, a(n) for n = 0..733</a>
%F A262034 E.g.f.: (exp(x^2)*(x+1)-(x^4/2+x^2+x+1))/x^3.
%F A262034 a(n) = 2*((n^2-1)*a(n-2)-a(n-1))/(n+3) for n>3, a(0)=a(2)=a(3)=1, a(1)=0.
%F A262034 a(n) = n!/(n/2+1)! if n even, a(n) = floor(C(n+1,(n+1)/2)/(n+3)*((n-1)/2)!) if n odd.
%F A262034 a(2n) = A262033(2n) = A001761(n).
%F A262034 a(2n+1) = A102693(n+1).
%F A262034 Sum_{n>=2} 1/a(n) = (39*exp(1/4)*sqrt(Pi)*erf(1/2) - 6)/16, where erf is the error function. - _Amiram Eldar_, Dec 04 2022
%e A262034 a(4) = 4: 1234, 1243, 1342, 2341.
%e A262034 a(5) = 5: 12345, 12354, 12453, 13452, 23451.
%p A262034 a:= proc(n) option remember; `if`(n<4, [1, 0, 1$2][n+1],
%p A262034       2*((n^2-1)*a(n-2)-a(n-1))/(n+3))
%p A262034     end:
%p A262034 seq(a(n), n=0..30);
%t A262034 np=Rest[With[{nn=30},CoefficientList[Series[(Exp[x^2](x+1)-x^4/2+x^2+x+1)/ x^3,{x,0,nn}],x] Range[0,nn]!]//Quiet];Join[{1},np] (* _Harvey P. Dale_, May 18 2019 *)
%Y A262034 Cf. A001761, A102693, A262033, A262035.
%K A262034 nonn
%O A262034 0,5
%A A262034 _Alois P. Heinz_, Sep 08 2015