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A129453
An analog of Pascal's triangle based on A092287. T(n,k) = A092287(n)/(A092287(n-k)*A092287(k)), 0 <= k <= n.
3
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
0,5
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.
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
EXAMPLE
Triangle starts:
1;
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 16, 24, 16, 1;
1, 5, 40, 40, 5, 1;
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Peter Bala, Apr 16 2007
STATUS
approved