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A230043 Numbers whose abundancy sigma(n)/n is a rational cube.

%I A230043 #51 Jul 09 2017 17:50:49
%S A230043 1,8232,32640,265825,3846879,6517665,14705145,16926000,31441920,
%T A230043 56471688,146475000,211421364,277368000,369022500,662518050,679568670,
%U A230043 968353620,2166699360,3091750900,3755367252,4122716598,6536970000,9740587500,10066738500,12423246290
%N A230043 Numbers whose abundancy sigma(n)/n is a rational cube.
%C A230043 All terms listed in the data section are deficient, but all 8-multiperfect numbers (which are abundant...) also belong to this sequence.
%C A230043 As with A230538, it is possible to find larger numbers with same ratio sigma(n)/n, in some cases using perfect numbers A000396 (see a230043.txt link). - _Michel Marcus_, Oct 30 2013
%C A230043 One motivation for this sequence lies in the fact that n*sigma(n) is a square (A069070) if and only if sigma(n)/n is a rational square. But this does not hold for higher powers: If sigma(n)/n = (p/q)^k then n*sigma(n) = (pq)^k (n/q^k)^2. - _M. F. Hasler_, Nov 02 2013
%C A230043 In his post to NMBRTHRY, Michiel Kosters gives a 233-digit number x such that sigma(x^3) is a cube. Actually this x^3 also belongs to the sequence, although there are no cubes in the current data. He has found many others such cubes that belong here, the smallest of which is 3590918978816938469301573291605^3, x having 31 digits, and x^3 92 digits. Is it possible to find the smallest such cube, or even a smaller one? - _Michel Marcus_, Jan 02 2014
%H A230043 Michel Marcus and Donovan Johnson, <a href="/A230043/b230043.txt">Table of n, a(n) for n = 1..33</a>
%H A230043 A. Flammenkamp, <a href="http://wwwhomes.uni-bielefeld.de/achim/mpn.html">The multiply perfect numbers page</a>
%H A230043 M. Kosters, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;63188f1d.1312">A solution to sigma(x^3)=y^3</a>, Post to NMBRTHRY, Dec 23 2013
%H A230043 Michel Marcus, <a href="/A230043/a230043.txt">Some other terms that belong to the sequence</a>
%e A230043 For n=8232, sigma(n)/n = 1000/343 = (10/7)^3.
%p A230043 isQcube := proc(r)
%p A230043     isA000578(numer(r)) and isA000578(denom(r)) ;
%p A230043 end proc: # see A000578 for isA000578()
%p A230043 isA230043 := proc(n)
%p A230043     abu := numtheory[sigma](n)/n ;
%p A230043     isQcube(abu) ;
%p A230043 end proc:
%p A230043 for n from 1 do
%p A230043     if isA230043(n) then
%p A230043         printf("%d,\n",n);
%p A230043     end if;
%p A230043 end do: # _R. J. Mathar_, Oct 08 2013
%o A230043 (PARI) is_A230043(n) = ispower(sigma(n)/n, 3);
%Y A230043 Cf. A069070 (abundancy is a square).
%K A230043 nonn
%O A230043 1,2
%A A230043 _Michel Marcus_, Oct 06 2013
%E A230043 a(11)-a(25) from _Donovan Johnson_, Oct 10 2013
%E A230043 a(26)-a(33) from _Donovan Johnson_, Dec 22 2013