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A066289
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Numbers k such that k divides DivisorSigma(2*j-1, k) for all j; i.e., all odd-power-sums of divisors of k are divisible by k.
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5
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1, 6, 120, 672, 30240, 32760, 31998395520, 796928461056000, 212517062615531520, 680489641226538823680000, 13297004660164711617331200000, 1534736870451951230417633280000, 6070066569710805693016339910206758877366156437562171488352958895095808000000000
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OFFSET
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1,2
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COMMENTS
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Tested for each k and j < 200. Otherwise the proof for all j seems laborious, since the number of divisors of terms of sequence rapidly increases: {1, 4, 16, 24, 96, 96, 2304, ...}.
The given terms have been tested for all j. - Don Reble, Nov 03 2003
This is a proper subset of the multiply perfect numbers A007691. E.g., 8128 from A007691 is not here because its remainder at Sigma[odd,8128]/8128 division is 0 or 896 depending on odd exponent.
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LINKS
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FORMULA
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DivisorSigma(2*j-1, k)/k is an integer for all j = 1, 2, 3, ..., 200, ...
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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The following numbers belong to the sequence, but there may be missing terms in between: 796928461056000 (also belongs to A046060); 212517062615531520 (also belongs to A046060); 680489641226538823680000 (also belongs to A046061); 13297004660164711617331200000 (also belongs to A046061). - Thomas Baruchel, Oct 10 2003
Extended to 13 confirmed terms by Don Reble, Nov 04 2003. There is a question whether there are other members below a(13). However, there are none in Achim's list of multiperfect numbers (see A007691); Richard C. Schroeppel has suggested that that list is complete to 10^70 - if so, a(1..12) are correct; as for a(13), Rich says there's only "an epsilon chance that some undiscovered MPFN lies in the gap." So it is very likely to be correct. - Don Reble
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STATUS
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approved
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