login
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).
8

%I #19 Jul 12 2024 15:17:58

%S 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,

%T 144,0,1728,0,2592,0,576,0,576,0,9216,0,20736,0,9216,0,576,2880,0,

%U 57600,0,172800,0,115200,0,14400,0,14400,0,360000,0,1440000,0,1440000,0,360000,0,14400

%N 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).

%C Mirror image of A145894.

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

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

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

%H Alois P. Heinz, <a href="/A145893/b145893.txt">Rows n = 0..140, flattened</a>

%F 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.

%e T(3,2) = 4 because we have 132, 312, 213 and 231.

%e Triangle starts:

%e 1;

%e 1, 0;

%e 1, 0, 1;

%e 2, 0, 4, 0;

%e 4, 0, 16, 0, 4;

%e 12, 0, 72, 0, 36, 0;

%e 36, 0, 324, 0, 324, 0, 36;

%e ...

%p 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

%t 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 *)

%Y Cf. A000142, A145891, A145894.

%K nonn,tabl

%O 0,7

%A _Emeric Deutsch_, Nov 30 2008