OFFSET
2,1
COMMENTS
FORMULA
T(2n,k) = (n!)^2 * binomial(n+1,k) binomial(n-1,k-1);
T(2n+1,k) = n!*(n+1)!*binomial(n-1,k-1)*binomial(n+2,k) (n >= 1).
EXAMPLE
T(4,2) = 12 because we have 1234, 3214, 1432, 3412, 2134, 2314 and their reverses.
Triangle starts:
2;
6;
12, 12;
48, 72;
144, 432, 144;
720, 2880, 1440;
MAPLE
ae := proc (n, k) options operator, arrow: factorial(n)^2*binomial(n+1, k)*binomial(n-1, k-1) end proc: ao := proc (n, k) options operator, arrow: factorial(n)*factorial(n+1)*binomial(n-1, k-1)*binomial(n+2, k) end proc: T := proc (n, k) if `mod`(n, 2) = 0 then ae((1/2)*n, k) else ao((1/2)*n-1/2, k) end if end proc; for n to 12 do seq(T(n, k), k = 1 .. floor((1/2)*n)) end do; # yields sequence in triangular form
MATHEMATICA
T[n_, k_] := If[EvenQ[n], ((n/2)!)^2*Binomial[n/2+1, k]*Binomial[n/2-1, k-1], ((n-1)/2)!*((n-1)/2+1)!*Binomial[(n-1)/2-1, k-1]*Binomial[(n-1)/2+2, k]];
Table[T[n, k], {n, 2, 12}, {k, 1, Floor[n/2]}] // Flatten (* Jean-François Alcover, Sep 24 2024 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Dec 14 2008
STATUS
approved