About A066289
Thomas Baruchel
thomas.baruchel at laposte.net
Sat Oct 4 17:23:26 CEST 2003
Hi, I found more terms for
%I A066289
%S A066289 1,6,120,672,30240,32760,31998395520
%N A066289 Numbers n such that Mod[DivisorSigma[2k-1,n],n]=0 holds presumably for all k; i.e. all odd-power-sum of divisors of n are divisible by n.
%F A066289 DivisorSigma[2k-1,n]/n] is integer for all k=1,2,3,..,200,..
%e A066289 Tested for each n and k<200. Otherwise the proof for all k seems laborious, since number of divisors of terms of sequence rapidly increases:{1,4,16,24,96,96,2304}. True subset of multiple 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.
these terms are:
* 796928461056000
* 212517062615531520
* 680489641226538823680000
* 13297004660164711617331200000
But I don't know if they come exactly after the last current term,
and i don't know either if no term is missing between them.
Is there something (greatness ???) that would make you think they
come exactly after the last current term with no missing term
between them ?
Regards,
--
« nous devons agir comme si la chose qui peut-être ne sera pas devait
être » (Kant, Métaphysique des moeurs, doctrine du droit, II conclusion)
Thomas Baruchel <baruchel at laposte.net>
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