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A244428 Sum of divisors of n and product of divisors of n are both perfect cubes.

%I A244428 #24 Nov 26 2024 13:08:34
%S A244428 1,1164,8148,11596,12028,28128,32980,34144,34528,36244,38764,39916,
%T A244428 41164,41516,73200,75252,81172,84196,94023,100348,181948,182430,
%U A244428 192175,193380,193612,194044,195780,196896,200574,204180,208416,211620,214176,217668,220116,225696,230860,235716
%N A244428 Sum of divisors of n and product of divisors of n are both perfect cubes.
%C A244428 This is also the intersection of A020477 and A048944.
%C A244428 Numbers m such that sigma(m) is a cube and (m is a cube or number of divisors of m is a multiple of 3). - _Chai Wah Wu_, Mar 10 2016
%H A244428 Chai Wah Wu, <a href="/A244428/b244428.txt">Table of n, a(n) for n = 1..500</a>
%e A244428 The divisors of 1164 are {1, 2, 3, 4, 6, 12, 97, 194, 291, 388, 582, 1164}. 1*2*3*4*6*12*97*194*291*388*582*1164 = 2487241979165915136 = 1354896^3 = (1164^2)^3. 1+2+3+4+6+12+97+194+291+388+582+1164 = 2744 = 14^3. Thus, since both the sum of divisors and the product of divisors are perfect cubes, 1164 is a member of this sequence.
%t A244428 Select[Range[236000],AllTrue[{CubeRoot[DivisorSigma[1,#]],CubeRoot[Times@@Divisors[#]]},IntegerQ]&] (* _Harvey P. Dale_, Nov 26 2024 *)
%o A244428 (PARI) for(n=1,10^6,d=divisors(n);s=sum(i=1,#d,d[i]);p=prod(j=1,#d,d[j]);if(ispower(s,3)&&ispower(p,3),print1(n,", ")))
%o A244428 (Python)
%o A244428 from gmpy2 import iroot
%o A244428 from sympy import divisor_sigma
%o A244428 A244428_list = [i for i in range(1,10**4) if (iroot(i,3)[1] or not divisor_sigma(i,0) % 3) and iroot(int(divisor_sigma(i,1)),3)[1]] # _Chai Wah Wu_, Mar 10 2016
%Y A244428 Cf. A064710, A007955, A000203, A048944, A020477.
%K A244428 nonn
%O A244428 1,2
%A A244428 _Derek Orr_, Jun 27 2014