A174451 Triangle T(n, k, q) = n!*(n+1)!*q^k/((n-k)!(n-k+1)!) if floor(n/2) > k-1, otherwise n!*(n+1)!*q^(n-k)/(k!*(k+1)!) for q = 3, read by rows.

1, 1, 1, 1, 18, 1, 1, 36, 36, 1, 1, 60, 2160, 60, 1, 1, 90, 5400, 5400, 90, 1, 1, 126, 11340, 680400, 11340, 126, 1, 1, 168, 21168, 1905120, 1905120, 21168, 168, 1, 1, 216, 36288, 4572288, 411505920, 4572288, 36288, 216, 1, 1, 270, 58320, 9797760, 1234517760, 1234517760, 9797760, 58320, 270, 1

Offset: 0

Author

Roger L. Bagula, Mar 20 2010

Keywords

• nonn
• tabl
• easy

Examples

			Triangle begins as:
  1;
  1,   1;
  1,  18,     1;
  1,  36,    36,       1;
  1,  60,  2160,      60,          1;
  1,  90,  5400,    5400,         90,          1;
  1, 126, 11340,  680400,      11340,        126,       1;
  1, 168, 21168, 1905120,    1905120,      21168,     168,     1;
  1, 216, 36288, 4572288,  411505920,    4572288,   36288,   216,   1;
  1, 270, 58320, 9797760, 1234517760, 1234517760, 9797760, 58320, 270, 1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Cf. A174449 (q=1), A174450 (q=2), this sequence (q=3).

Cf. A000108.

Programs

Magma

F:= Factorial; // T = A174451
T:= func< n,k,q | Floor(n/2) gt k-1 select F(n)*F(n+1)*q^k/(F(n-k)*F(n-k+1)) else F(n)*F(n+1)*q^(n-k)/(F(k)*F(k+1)) >;
[T(n,k,3): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 29 2021

Mathematica

T[n_, k_, q_]:= If[Floor[n/2]>k-1, n!*(n+1)!*q^k/((n-k)!*(n-k+1)!), n!*(n+1)!*q^(n-k)/(k!*(k+1)!)];
Table[T[n, k, 3], {n,0,12}, {k,0,n}]//Flatten

Sage

f=factorial
def A174451(n,k,q):
    if ((n//2)>k-1): return f(n)*f(n+1)*q^k/(f(n-k)*f(n-k+1))
    else: return f(n)*f(n+1)*q^(n-k)/(f(k)*f(k+1))
flatten([[A174451(n,k,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Nov 29 2021

Formula

T(n, k, q) = n!*(n+1)!*q^k/((n-k)!(n-k+1)!) if floor(n/2) > k-1, otherwise n!*(n+1)!*q^(n-k)/(k!*(k+1)!) for q = 3.

T(n, n-k, q) = T(n, k, q).

From G. C. Greubel, Nov 29 2021: (Start)

T(2*n, n, q) = q^n*(2*n+1)!*Catalan(n) for q = 3.

T(n, k, q) = binomial(n, k)*binomial(n+1, k+1) * ( k!*(k+1)!*q^k/(n-k+1) if (floor(n/2) >= k), otherwise ((n-k)!)^2*q^(n-k) ), for q = 3. (End)

Extensions

Edited by G. C. Greubel, Nov 29 2021