OFFSET
1,1
COMMENTS
All the terms are divisible by 6 since if n == 1, 2, 4, or 5 (mod 6) then either n^3 - 1 or n^3 + 1 is divisible by 9, and if n == 3 (mod 6) then either n^3 - 1 or n^3 + 1 is divisible by 4. - Amiram Eldar, Oct 01 2019
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
6=2*3; 6^3-1=5*41; 6^3+1=7*31,..
MATHEMATICA
f[n_]:=SquareFreeQ[n]; lst={}; Do[If[f[n^3-1]&&f[n^3+1]&&f[n], AppendTo[lst, n]], {n, 3*6!}]; lst
PROG
(Magma) [k:k in [2..1400]| IsSquarefree(k) and IsSquarefree(k^3-1) and IsSquarefree(k^3+1)]; // Marius A. Burtea, Oct 01 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Joseph Stephan Orlovsky, Apr 14 2010
STATUS
approved