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Search: id:A092287
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| A092287 |
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Product_{j=1..n} Product_{k=1..n} gcd(j,k). |
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+0
7
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| 1, 1, 2, 6, 96, 480, 414720, 2903040, 5945425920, 4334215495680, 277389791723520000, 3051287708958720000, 437332621360674939863040000, 5685324077688774218219520000, 15974941971638268369709427589120000, 982608696336737613503095822614528000000000
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Conjecture: Let p be a prime and let ordp(n,p) denote the exponent of the highest 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 the formula: ordp(a(n),p) = (floor(n/p))^2 + (floor(n/p^2))^2 + (floor(n/p^3))^2 + . . .. Compare this to the de Polignac-Legendre formula for the prime factorization of n!: ordp(n!,p) = floor(n/p) + floor(n/p^2) + floor(n/p^3) + . . .. This suggests that a(n) can be considered as generalization of n!. See A129453 for the analogue for a(n) of Pascal's triangle. See A129454 for the sequence defined as a triple product of gcd(i,j,k). - Peter Bala (pbala(AT)toucansurf.com), Apr 16 2007
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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FORMULA
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Also a(n) = product{k=1 to n} product{j=1 to n} LCM(1, 2, 3, ..., floor(min(n/k, n/j))).
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MAPLE
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f := n->mul(mul(gcd(j, k), k=1..n), j=1..n);
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CROSSREFS
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Cf. A018806, A090494.
Cf. A129365, A129439, A129453, A129454, A129455.
Sequence in context: A087277 A007188 A129364 this_sequence A035482 A007870 A081992
Adjacent sequences: A092284 A092285 A092286 this_sequence A092288 A092289 A092290
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), based on a suggestion from Leroy Quet, Feb 03 2004
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