OFFSET
0,3
COMMENTS
FORMULA
T(2n,k) = (n!)^2*C(n-1,k)*C(n+1,k+1); T(2n+1,k) = n!(n+1)! * C(n-1,k) * C(n+2,k+2).
EXAMPLE
T(4,1) = 12 because we have 1243, 1423, 1324, 1342, 3124, 3142, 2413, 4213, 2431, 4231, 3241 and 3421.
Triangle starts:
1;
1;
2;
6;
12, 12;
72, 48;
144, 432, 144;
1440, 2880, 720;
...
MAPLE
T:=proc(n, k) if `mod`(n, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n-1, k)*binomial((1/2)*n+1, k+1) else factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-3/2, k)*binomial((1/2)* n+3/2, k+2) end if end proc: 1; 1; for n from 2 to 12 do seq(T(n, k), k = 0 .. floor((1/2)*n)-1) end do; # yields sequence in triangular form
MATHEMATICA
T[n_, k_]:=If[EvenQ[n], ((n/2)!)^2Binomial[n/2-1, k]Binomial[n/2+1, k+1], ((n-1)/2)!((n+1)/2)!Binomial[(n-3)/2, k]Binomial[(n+3)/2, k+2]]; Join[{1, 1}, Flatten[Table[T[n, k], {n, 0, 12}, {k, 0, Floor[n/2]-1}]]] (* Stefano Spezia, Jul 12 2024 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Nov 30 2008
STATUS
approved