A177916 Numbers k such that k^3 divides 16^(k^2) - 1.

1, 3, 5, 15, 21, 39, 55, 57, 105, 111, 155, 165, 195, 205, 219, 273, 285, 327, 399, 465, 505, 555, 609, 615, 741, 777, 903, 915, 1095, 1155, 1255, 1265, 1365, 1443, 1515, 1533, 1635, 1705, 1995, 2067, 2109, 2145, 2255, 2265, 2289, 2373, 2667, 2715, 2847

Offset: 1

Author

Alexander Adamchuk, May 14 2010

Keywords

• nonn

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

Crossrefs

Cf. A129211, A129212, A177905, A127106, A177907, A127102, A177909, A177243.

Cf. A177911, A177912, A177913, A177914, A177915, A177916, A177917, A177918, A177919, A177920.

Programs

Magma

[n: n in [1..2000] | Denominator((16^(n^2) - 1)/n^3) eq 1]; // Vincenzo Librandi, Apr 24 2015

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 *)