A145893 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that j and p(j) are of opposite parities for k values of j (0<=k<=n).

1, 1, 0, 1, 0, 1, 2, 0, 4, 0, 4, 0, 16, 0, 4, 12, 0, 72, 0, 36, 0, 36, 0, 324, 0, 324, 0, 36, 144, 0, 1728, 0, 2592, 0, 576, 0, 576, 0, 9216, 0, 20736, 0, 9216, 0, 576, 2880, 0, 57600, 0, 172800, 0, 115200, 0, 14400, 0, 14400, 0, 360000, 0, 1440000, 0, 1440000, 0, 360000, 0, 14400

Offset

0,7

Comments

Mirror image of A145894.

Without the 0's it is the triangle of A145891.

Sum of entries in row n = n! = A000142(n).

Columns k=0,2,4,8,10,12 give: A010551, A226282, A226283, A226284, A226285, A226286. - Alois P. Heinz, May 29 2014

Links

Alois P. Heinz, Rows n = 0..140, flattened

Formula

T(2n,2k) = [n!*C(n,k)]^2; T(2n+1,2k) = n! (n+1)! C(n,k) C(n+1,k); elsewhere T(n,k)=0.

Example

T(3,2) = 4 because we have 132, 312, 213 and 231.
Triangle starts:
   1;
   1, 0;
   1, 0,   1;
   2, 0,   4, 0;
   4, 0,  16, 0,   4;
  12, 0,  72, 0,  36, 0;
  36, 0, 324, 0, 324, 0, 36;
  ...

Maple

T:=proc(n, k) if `mod`(n, 2) = 0 and `mod`(k, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n, (1/2)*k)^2 elif `mod`(n, 2) = 1 and `mod`(k, 2) = 0 then factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-1/2, (1/2)*k)*binomial((1/2)*n+1/2, (1/2)*k) else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form

Mathematica

T[n_, k_] := Which[EvenQ[n] && EvenQ[k], (n/2)!^2*Binomial[n/2, k/2]^2, OddQ[n] && EvenQ[k], (n/2-1/2)!*(n/2+1/2)!*Binomial[n/2-1/2, k/2] * Binomial[n/2+1/2, k/2], True, 0]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 20 2017, translated from Maple *)

Crossrefs

Cf. A000142, A145891, A145894.

Sequence in context: A035688 A364105 A046769 * A137561 A337605 A082024

Adjacent sequences: A145890 A145891 A145892 * A145894 A145895 A145896

Keyword

nonn,tabl

Author

Emeric Deutsch, Nov 30 2008

Status

approved