A145894 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 the same parity for k values of j (0<=k<=n).

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

Offset

0,8

Comments

Mirror image of A145893.

Without the 0's, it is the triangle of A134434.

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

T(n,n) = A010551(n).

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

Example

T(3,1) = 4 because we have 132, 312, 213 and 231.
Triangle starts:
1;
0,   1;
1,   0,   1;
0,   4,   0,  2;
4,   0,  16,  0,   4;
0,  36,   0, 72,   0, 12;
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) = 1 then 2*factorial((1/2)*n+1/2)^2*binomial((1/2)*n-1/2, (1/2)*k-1/2)^2/(k+1) 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] && OddQ[k], (2*(n/2+1/2)!^2*Binomial[n/2-1/2, k/2-1/2]^2)/(k+1), True, 0]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 19 2017, translated from Maple *)

Crossrefs

Cf. A000142, A134434, A145893, A010551.

Sequence in context: A323982 A279415 A201399 * A366889 A359589 A021881

Adjacent sequences: A145891 A145892 A145893 * A145895 A145896 A145897

Keyword

nonn,tabl

Author

Emeric Deutsch, Nov 30 2008

Status

approved