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

1, 1164, 8148, 11596, 12028, 28128, 32980, 34144, 34528, 36244, 38764, 39916, 41164, 41516, 73200, 75252, 81172, 84196, 94023, 100348, 181948, 182430, 192175, 193380, 193612, 194044, 195780, 196896, 200574, 204180, 208416, 211620, 214176, 217668, 220116, 225696, 230860, 235716

Offset: 1

Author

Derek Orr, Jun 27 2014

Keywords

• nonn

Comments

This is also the intersection of A020477 and A048944.

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

Examples

			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.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A064710, A007955, A000203, A048944, A020477.

Programs

Mathematica

Select[Range[236000],AllTrue[{CubeRoot[DivisorSigma[1,#]],CubeRoot[Times@@Divisors[#]]},IntegerQ]&] (* Harvey P. Dale, Nov 26 2024 *)

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,", ")))

Python

from gmpy2 import iroot
from sympy import divisor_sigma
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