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Search: id:A061701
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| A061701 |
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Smallest number m such that GCD of d(m^2) and d(m) is 2n+1. |
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+0
4
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| 1, 12, 4608, 1728, 1260, 509607936, 2985984, 144, 56358560858112, 5159780352, 302400, 6232805962420322304, 1587600, 900900, 201226394483583074212773888, 15407021574586368, 248832
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Comment from Dean Hickerson (dean(AT)math.ucdavis.edu), Jun 23, 2001: a(n) exists for every n. In other words, every positive odd integer k is equal to the GCD of d(m^2) and d(m) for some m. To see this, let m = 2^(k^2 - 1) * 3^((k-1)/2). Then d(m) = k^2 * (k+1)/2 and d(m^2) = (2 k^2 - 1) * k. Both of these are divisible by k, and (8k-4) d(m) - (2k+1) d(m^2) = k, so the GCD is k.
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FORMULA
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a(n) = Min[m : GCD[d(m^2), d(m)] = 2n+1]
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EXAMPLE
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For n = 7, GCD[d(20736),d(144)] = GCD[45,15] = 15 = 2*7+1.
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CROSSREFS
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Cf. A000005, A000290, A048691.
Sequence in context: A096732 A127233 A009094 this_sequence A134821 A013508 A003793
Adjacent sequences: A061698 A061699 A061700 this_sequence A061702 A061703 A061704
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KEYWORD
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nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Jun 18 2001
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EXTENSIONS
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More terms from David Wasserman (wasserma(AT)spawar.navy.mil), Jun 20 2002
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