A145891 - OEIS

%I #13 Jul 12 2024 16:00:17

%S 1,1,1,1,2,4,4,16,4,12,72,36,36,324,324,36,144,1728,2592,576,576,9216,

%T 20736,9216,576,2880,57600,172800,115200,14400,14400,360000,1440000,

%U 1440000,360000,14400,86400,2592000,12960000,17280000,6480000,518400

%N 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)).

%C 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.

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

%C Mirror image of A134434.

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

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

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

%e T(3,1) = 4 because we have 123, 132, 312 and 321.

%e Triangle starts:

%e    1;

%e    1;

%e    1,   1;

%e    2,   4;

%e    4,  16,   4;

%e   12,  72,  36;

%e   36, 324, 324, 36;

%e   ...

%p 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

%t 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 *)

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

%K nonn,tabf

%O 0,5

%A _Emeric Deutsch_, Nov 30 2008