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
Keywords
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
Programs
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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,", ")))
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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
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