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A272062 Positive numbers k such that k^2 - 1 divides 8^k - 1.

%I A272062 #32 Sep 08 2022 08:46:16
%S A272062 2,4,8,10,16,22,36,40,64,96,100,196,210,256,280,316,456,560,820,1200,
%T A272062 1236,1296,1360,1408,1600,1870,2380,2556,3516,3616,4096,4200,4356,
%U A272062 5656,6112,6256,6480,8008,8688,10192,10356,11440,11952,12160,13728,14950,16192,17020,19432,21880,22036
%N A272062 Positive numbers k such that k^2 - 1 divides 8^k - 1.
%C A272062 From _Robert Israel_, Jun 08 2018: (Start)
%C A272062 All terms are even.
%C A272062 Are 2, 8 and 560 the only terms == 2 (mod 6)?  There are no others up to 3*10^9. (End)
%H A272062 Robert Israel, <a href="/A272062/b272062.txt">Table of n, a(n) for n = 1..2559</a>
%e A272062 a(1) = 2 because (8^2 - 1)/(2^2 - 1) = 21.
%p A272062 A272062:=n->`if`((8^n-1) mod (n^2-1) = 0, n, NULL): seq(A272062(n), n=2..5*10^4); # _Wesley Ivan Hurt_, Apr 21 2016
%t A272062 Select[Range[2, 22100], Divisible[8^# - 1, #^2 - 1] &] (* _Michael De Vlieger_, Apr 19 2016 *)
%o A272062 (Magma) [0] cat [n: n in [2..30000] | Denominator((8^n-1)/(n^2-1)) eq 1];
%o A272062 (PARI) is(n)=Mod(8,n^2-1)^n==1 \\ _Charles R Greathouse IV_, Apr 19 2016
%Y A272062 Cf. positive numbers n such that n^2 - 1 divides (2^k)^n - 1: A247219 (k=1), A271842 (k=2), this sequence (k=3).
%K A272062 nonn
%O A272062 1,1
%A A272062 _Juri-Stepan Gerasimov_, Apr 19 2016