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%I #21 Apr 02 2024 12:07:20
%S 0,1,2,16,60,540,3024,32256,241920,3024000,28512000,410572800,
%T 4670265600,76281004800,1017080064000,18598035456000,284549942476800,
%U 5762136335155200,99527809425408000,2211729098342400000,42575785143091200000,1030334000462807040000
%N Number of leading even entries in all permutations of {1,2,...,n}.
%F a(n) = Sum_{k=0..floor(n/2)} k*A152664(n,k).
%F a(2n+1) = n(2n+1)!/(n+2);
%F a(2n) = n(2n)!/(n+1).
%F D-finite with recurrence 2*(n+3)*a(n) +(-5*n-8)*a(n-1) +(-2*n^3-2*n^2-n-4)*a(n-2) +(n-2)*(3*n^2-3*n+2)*a(n-3) +(n-3)*(n-2)^2*a(n-4)=0. - _R. J. Mathar_, Jul 26 2022
%e The permutation 4,6,2,1,5,3 begins with three even numbers, so would contribute 3 to a(6).
%e a(3)=2 because in the permutations 123, 132, 213, 231, 312, 321 we have 0+0+1+1+0+0 = 2 leading odd entries.
%e a(45) = 16: Here are the permutations of 1234, 24 in all:
%e 1(234) total 6, no. of initial even terms = 0
%e 3(124) ditto
%e 21(34) total 2, no. of initial even terms 1*2 = 2
%e 23(14) ditto
%e 24(13) total 2, no. of initial even terms 2 twice = 4
%e Subtotal from 2*** is 2+2+4 = 8
%e Subtotal from 4*** is also 2+2+4 = 8
%e Total a(4) = 8+8 = 16.
%p ao := proc (n) options operator, arrow; n*factorial(2*n+1)/(n+2) end proc: ae := proc (n) options operator, arrow; n*factorial(2*n)/(n+1) end proc: a := proc (n) if `mod`(n, 2) = 1 then ao((1/2)*n-1/2) else ae((1/2)*n) end if end proc; seq(a(n), n = 1 .. 20);
%t a[n_] := If[OddQ[n], (n-1)*n!/(n+3), n*n!/(n+2)];
%t Table[a[n], {n, 1, 20}] (* _Jean-François Alcover_, Apr 29 2023 *)
%Y Cf. A152662, A152663, A152664.
%K nonn
%O 1,3
%A _Emeric Deutsch_, Dec 13 2008
%E Examples expanded by _N. J. A. Sloane_, Sep 09 2019
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