A194948 - OEIS

A194948

Numbers k such that sum of aliquot divisors of k, sigma(k) - k, is a cube.

1, 2, 3, 5, 7, 10, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 56, 59, 61, 67, 69, 71, 73, 76, 79, 83, 89, 97, 101, 103, 107, 109, 113, 122, 127, 131, 133, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

For prime numbers, the sum of their aliquot divisors is exactly 1 = 1^3.

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..2500 from Harvey P. Dale)

EXAMPLE

a(6) = 10, since the sum of aliquot divisors 1 + 2 + 5 = 8 = 2^3.

MAPLE

for n do s:=numtheory[sigma](n)-n; if root(s, 3)=trunc(root(s, 3)) then print(n); fi; od:

MATHEMATICA