A155517 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)).

0, 1, 0, 2, 4, 0, 2, 16, 0, 8, 64, 48, 0, 8, 384, 288, 0, 48, 2880, 1536, 576, 0, 48, 23040, 12288, 4608, 0, 384, 208896, 115200, 30720, 7680, 0, 384, 2088960, 1152000, 307200, 76800, 0, 3840, 23193600, 12533760, 3456000, 614400, 115200, 0, 3840, 278323200

Offset: 1

Author

Emeric Deutsch, Jan 26 2009

Keywords

• nonn
• tabf

Comments

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.

Row sums are the factorial numbers (A000142).

Row n contains 1 + ceiling(n/2)entries.

T(2n,n) = n!*2^n = A037223(2n) = number of centrosymmetric permutations in S[2n];

T(2n+1,n+1) = n!*2^n = A037223(2n+1) = number of centrosymmetric permutations in S[2n+1].

T(n,0) = A155518(n).

Sum_{k=0..ceiling(n/2)} k*T(n,k) = A155519(n).

Examples

			T(4,2)=8 because we have 1234, 4231, 1324, 4321, 2143, 3142, 2413 and 3412.
Triangle starts:
    0,   1;
    0,   2;
    4,   0,   2;
   16,   0,   8;
   64,  48,   0,   8;
  384, 288,   0,  48;
		

Crossrefs

Cf. A000142, A037223, A055140, A155518, A155519.

Programs

Maple

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;

Formula

T(2n,k) = n!*2^n*A055140(n,k);

T(2n-1,k) = (n-1)!*2^(n-1)*A055140(n,k);

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)).