OFFSET
1,4
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).
FORMULA
EXAMPLE
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;
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;
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Jan 26 2009
STATUS
approved