OFFSET
1,2
COMMENTS
From Robert Israel, Apr 24 2015: (Start)
The only primes in the sequence are 3 and 5.
If m and n are in the sequence and are coprime, then m*n is in the sequence.
Are all members of the sequence divisible by 3 or 5? (End)
LINKS
Robert Israel, Table of n, a(n) for n = 1..1000
MAPLE
A177916:=n->`if`((16&^(n^2)-1) mod n^3 = 0, n, NULL): seq(A177916(n), n=1..1000); # Wesley Ivan Hurt, Apr 23 2015
MATHEMATICA
Select[Range[3000], IntegerQ[(16^(#^2) - 1) / #^3 ] &] (* Vincenzo Librandi, Apr 24 2015 *)
Join[{1}, Select[Range[3000], PowerMod[16, #^2, #^3]==1&]] (* Harvey P. Dale, Dec 05 2024 *)
PROG
(Magma) [n: n in [1..2000] | Denominator((16^(n^2) - 1)/n^3) eq 1]; // Vincenzo Librandi, Apr 24 2015
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Alexander Adamchuk, May 14 2010
STATUS
approved