A129453 An analog of Pascal's triangle based on A092287. T(n,k) = A092287(n)/(A092287(n-k)*A092287(k)), 0 <= k <= n.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 16, 24, 16, 1, 1, 5, 40, 40, 5, 1, 1, 864, 2160, 11520, 2160, 864, 1, 1, 7, 3024, 5040, 5040, 3024, 7, 1, 1, 2048, 7168, 2064384, 645120, 2064384, 7168, 2048, 1, 1, 729, 746496, 1741824, 94058496, 94058496, 1741824, 746496, 729, 1

Offset: 0

Author

Peter Bala, Apr 16 2007

Keywords

• nonn
• tabl

Comments

It appears that the T(n,k) are always integers. This would follow from the conjectured prime factorization given in A092287. Calculation suggests that the binomial coefficients C(n,k) divide T(n,k) and indeed that T(n,k)/C(n,k) are perfect squares.

Examples

			Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1, 16, 24, 16,  1;
  1,  5, 40, 40,  5,  1;
		

Links

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

Crossrefs

Cf. A007318, A092287, A129454, A129455.

Programs

Magma

A092287:= func< n | n eq 0 select 1 else (&*[(&*[GCD(j,k): k in [1..n]]): j in [1..n]]) >;
A129453:= func< n,k | A092287(n)/(A092287(n-k)*A092287(k)) >;
[A129453(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 07 2024

Mathematica

A092287[n_]:= Product[GCD[j,k], {j,n}, {k,n}];
A129453[n_, k_]:= A092287[n]/(A092287[k]*A092287[n-k]);
Table[A129453[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2024 *)

SageMath

def A092287(n): return product(product( gcd(j,k) for k in range(1,n+1)) for j in range(1,n+1))
def A129453(n,k): return A092287(n)/(A092287(n-k)*A092287(k))
flatten([[A129453(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 07 2024

Formula

T(n, k) = (Product_{i=1..n} Product_{j=1..n} gcd(i,j)) / ( (Product_{i=1..n-k} Product_{j=1..n-k} gcd(i,j)) * ( Product_{i=1..k} Product_{j=1..k} gcd(i,j)) ), note that empty products equal to 1.

T(n, n-k) = T(n, k). - G. C. Greubel, Feb 07 2024