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A129454
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a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k).
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5
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1, 1, 1, 2, 6, 1536, 7680, 8806025134080, 61642175938560, 2168841254587541957294161920, 7562281854741110985626291951024209920, 1362299589723309231779453337910253309054734620740812800000000
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OFFSET
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0,4
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COMMENTS
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Conjecture: Let p be a prime and let ordp(n,p) denote the largest power of p which divides n. For example, ordp(48,2)=4 since 48 = 3*(2^4). Then we conjecture that the prime factorization of a(n) is given by ordp(a(n),p)=(floor(n/p))^3 + (floor(n/p^2))^3 + (floor(n/p^3))^3 + . . .. Compare with the comments in A092287.
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LINKS
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FORMULA
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a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k), for n > 2, otherwise a(n) = 1.
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MATHEMATICA
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A129454[n_]:= Product[GCD[j, k, m], {j, n-1}, {k, n-1}, {m, n-1}];
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PROG
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(Magma)
A129454:= func< n | n le 1 select 1 else (&*[(&*[(&*[GCD(GCD(j, k), m): k in [1..n-1]]): j in [1..n-1]]): m in [1..n-1]]) >;
(SageMath)
def A129454(n): return product(product(product(gcd(gcd(j, k), m) for k in range(1, n)) for j in range(1, n)) for m in range(1, n))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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