A145891 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k adjacent pairs of the form (odd,even) (0<=k<=floor(n/2)).

1, 1, 1, 1, 2, 4, 4, 16, 4, 12, 72, 36, 36, 324, 324, 36, 144, 1728, 2592, 576, 576, 9216, 20736, 9216, 576, 2880, 57600, 172800, 115200, 14400, 14400, 360000, 1440000, 1440000, 360000, 14400, 86400, 2592000, 12960000, 17280000, 6480000, 518400

Offset: 0

Author

Emeric Deutsch, Nov 30 2008

Keywords

• nonn
• tabf

Comments

Also number of permutations of {1,2,...,n} having k adjacent pairs of the form (even,odd). Example: T(3,1) = 4 because we have 123, 213, 231 and 321.

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

Mirror image of A134434.

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

Sum_{k>=0} k*T(n,k) = A077613(n).

Examples

			T(3,1) = 4 because we have 123, 132, 312 and 321.
Triangle starts:
   1;
   1;
   1,   1;
   2,   4;
   4,  16,   4;
  12,  72,  36;
  36, 324, 324, 36;
  ...
		

Crossrefs

Cf. A000142, A077613, A134434, A134435, A145892.

Programs

Maple

T:=proc(n,k) if `mod`(n, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n, k)^2 else factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-1/2, k)*binomial((1/2)*n+1/2, k) end if end proc: for n from 0 to 11 do seq(T(n,k), k =0..floor((1/2)*n)) end do; # yields sequence in triangular form

Mathematica

T[n_,k_]:=If[EvenQ[n],Floor[(n/2)!Binomial[n/2,k]]^2, ((n-1)/2)!((n+1)/2)!Binomial[(n-1)/2,k]Binomial[(n+1)/2,k]]; Table[T[n,k],{n,0,11},{k,0,Floor[n/2]}]//Flatten (* Stefano Spezia, Jul 12 2024 *)

Formula

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