%I #11 Jun 14 2015 04:32:56
%S 2,8,9,89,96,204,224,296,541,576,1537,1701,4496,6124,6611,7685,7789,
%T 8381,8741,9025,12048,12105,12513,13711,15924,16160,17180,21486,21998,
%U 24657,26264,26354,29864,32477,43791,52518,53662,54018,56189,81281
%N Numbers whose anti-divisors sum to a perfect cube.
%C See A066272 for definition of anti-divisor.
%H Chai Wah Wu, <a href="/A109351/b109351.txt">Table of n, a(n) for n = 1..586</a>
%H Jon Perry, <a href="http://www.users.globalnet.co.uk/~perry/maths/antidivisor.htm">The Anti-Divisor</a>
%H Jon Perry, <a href="/A066272/a066272a.html">The Anti-divisor</a> [Cached copy]
%H Jon Perry, <a href="/A066272/a066272.html">The Anti-divisor: Even More Anti-Divisors</a> [Cached copy]
%e The anti-divisors of 89 = {2, 3, 59} sum to 64, a perfect cube, so 89 is in the sequence.
%t AntiDivisors[n_] := Union[Drop[Drop[Divisors[2*n-1], 1], -1], Map[2*n/#&, Drop[Select[Divisors[2*n], OddQ], 1]], Drop[Drop[Divisors[2*n+1], 1], -1]]; Do[s = Plus @@ AntiDivisors[n]; If[IntegerQ[s^(1/3)], Print[n]], {n, 2, 10^5}]
%o (Python)
%o from sympy import integer_nthroot
%o from sympy.ntheory.factor_ import antidivisors
%o A109351_list = [n for n in range(2,10**4) if integer_nthroot(sum(antidivisors(n)),3)[1]] # _Chai Wah Wu_, Jun 13 2015
%K nonn
%O 1,1
%A _Ryan Propper_, Aug 21 2005