cp's OEIS Frontend

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

%I A177916 #25 Dec 05 2024 17:50:36
%S A177916 1,3,5,15,21,39,55,57,105,111,155,165,195,205,219,273,285,327,399,465,
%T A177916 505,555,609,615,741,777,903,915,1095,1155,1255,1265,1365,1443,1515,
%U A177916 1533,1635,1705,1995,2067,2109,2145,2255,2265,2289,2373,2667,2715,2847
%N A177916 Numbers k such that k^3 divides 16^(k^2) - 1.
%C A177916 From _Robert Israel_, Apr 24 2015: (Start)
%C A177916 The only primes in the sequence are 3 and 5.
%C A177916 If m and n are in the sequence and are coprime, then m*n is in the sequence.
%C A177916 Are all members of the sequence divisible by 3 or 5? (End)
%H A177916 Robert Israel, <a href="/A177916/b177916.txt">Table of n, a(n) for n = 1..1000</a>
%p A177916 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
%t A177916 Select[Range[3000], IntegerQ[(16^(#^2) - 1) / #^3 ] &] (* _Vincenzo Librandi_, Apr 24 2015 *)
%t A177916 Join[{1},Select[Range[3000],PowerMod[16,#^2,#^3]==1&]] (* _Harvey P. Dale_, Dec 05 2024 *)
%o A177916 (Magma) [n: n in [1..2000] | Denominator((16^(n^2) - 1)/n^3) eq 1]; // _Vincenzo Librandi_, Apr 24 2015
%Y A177916 Cf. A129211, A129212, A177905, A127106, A177907, A127102, A177909, A177243.
%Y A177916 Cf. A177911, A177912, A177913, A177914, A177915, A177916, A177917, A177918, A177919, A177920.
%K A177916 nonn
%O A177916 1,2
%A A177916 _Alexander Adamchuk_, May 14 2010