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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 (list; table; graph; refs; listen; history; text; internal format)
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.
LINKS
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
Sequence in context: A176469 A141542 A364812 * A129455 A329322 A067924
KEYWORD
nonn,tabl
AUTHOR
Peter Bala, Apr 16 2007
STATUS
approved

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Last modified April 24 18:17 EDT 2024. Contains 371962 sequences. (Running on oeis4.)