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%I #9 Jul 21 2017 10:44:14
%S 0,1,0,2,4,0,2,16,0,8,64,48,0,8,384,288,0,48,2880,1536,576,0,48,23040,
%T 12288,4608,0,384,208896,115200,30720,7680,0,384,2088960,1152000,
%U 307200,76800,0,3840,23193600,12533760,3456000,614400,115200,0,3840,278323200
%N Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} for which the number of j < ceiling(n/2) such that p(j) + p(n+1-j) = n+1 is equal to k (n>=1; 0<=k <=ceiling(n/2)).
%C For the permutation 31756284 of S_8 we have k=2 because p(2) + p(7) = 1+8 = 9 and p(3) + p(6) = 7+2 = 9; for the permutation 3214756 of S_7 we have k=2 because p(3) + p(5) = 1+7 = 8 and p(4) + p(4) = 4+4 = 8.
%C Row sums are the factorial numbers (A000142).
%C Row n contains 1 + ceiling(n/2)entries.
%C T(2n,n) = n!*2^n = A037223(2n) = number of centrosymmetric permutations in S[2n];
%C T(2n+1,n+1) = n!*2^n = A037223(2n+1) = number of centrosymmetric permutations in S[2n+1].
%C T(n,0) = A155518(n).
%C Sum_{k=0..ceiling(n/2)} k*T(n,k) = A155519(n).
%F T(2n,k) = n!*2^n*A055140(n,k);
%F T(2n-1,k) = (n-1)!*2^(n-1)*A055140(n,k);
%F here A055140(n,k) = A053871(n-k)*binomial(n,k), where g(n) = A053871(n) is defined by g(0)=1, g(1)=0, g(n) = 2(n-1)(g(n-1)+g(n-2)).
%e T(4,2)=8 because we have 1234, 4231, 1324, 4321, 2143, 3142, 2413 and 3412.
%e Triangle starts:
%e 0, 1;
%e 0, 2;
%e 4, 0, 2;
%e 16, 0, 8;
%e 64, 48, 0, 8;
%e 384, 288, 0, 48;
%p g[0] := 1: g[1] := 0: for n from 2 to 20 do g[n] := (2*(n-1))*(g[n-1]+g[n-2]) end do: T := proc (n, k) if `mod`(n, 2) = 0 then 2^((1/2)*n)*factorial((1/2)*n)*g[(1/2)*n-k]*binomial((1/2)*n, k) else 2^((1/2)*n-1/2)*factorial((1/2)*n-1/2)*g[(1/2)*n+1/2-k]*binomial((1/2)*n+1/2, k) end if end proc: for n to 12 do seq(T(n, k), k = 0 .. ceil((1/2)*n)) end do;
%Y Cf. A000142, A037223, A055140, A155518, A155519.
%K nonn,tabf
%O 1,4
%A _Emeric Deutsch_, Jan 26 2009
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