A066289 - OEIS

A066289

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.

1, 6, 120, 672, 30240, 32760, 31998395520, 796928461056000, 212517062615531520, 680489641226538823680000, 13297004660164711617331200000, 1534736870451951230417633280000, 6070066569710805693016339910206758877366156437562171488352958895095808000000000

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OFFSET

1,2

COMMENTS

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, ...}.

Tested for each k and j <= 1000. - Thomas Baruchel, Oct 10 2003

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.

LINKS

Table of n, a(n) for n=1..13.

FORMULA

DivisorSigma(2*j-1, k)/k is an integer for all j = 1, 2, 3, ..., 200, ...

CROSSREFS

Cf. A066135, A066284, A007691, A066290.

Sequence in context: A054776 A076231 A076234 * A351862 A170917 A115678

Adjacent sequences: A066286 A066287 A066288 * A066290 A066291 A066292

KEYWORD

nonn

AUTHOR

Labos Elemer, Dec 12 2001

EXTENSIONS

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

STATUS

approved

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