|
%I #17 Feb 10 2021 01:09:38
%S 1,7,73,487,511,2593,3409,18151,35551,39367,71119,80191,97687,189289,
%T 209953,248857,262657,275569,379081,472393,497833,561337,683809,
%U 1262791,1299079,1325023,1469671,1838599,2653567,2873791,3306751,5191687,5853943,7131151,8839537,9093553,15326569,19171729
%N Numbers k such that 2^(3^j) == 1 (mod k) for some j.
%C Numbers k such that the multiplicative order of 2 mod k is a power of 3.
%C If x and y are coprime members of the sequence, then x*y is in the sequence.
%C All divisors of a member of the sequence are in the sequence.
%C All prime-power divisors of 2^(3^k)-1 are in the sequence. In particular, the sequence contains infinitely many primes. - _Robert Israel_, Sep 02 2018
%H Robert Israel, <a href="/A318688/b318688.txt">Table of n, a(n) for n = 1..82</a>
%e a(3) = 73 is in the sequence because the multiplicative order of 2 mod 73 is 9 which is a power of 3.
%p N:= 10^6: # to get all terms <= N
%p Res:= NULL:
%p p:= 5:
%p do
%p p:= nextprime(p);
%p if p > N then break fi;
%p q:= 1: t:= 2:
%p while q < p-1 do
%p q:= 3*q;
%p t:= t^3 mod p;
%p if t = 1 then
%p Res:= Res, p;
%p v:= 1;
%p while 2 &^ t mod (p^(v+1)) = 1 do v:= v+1 od:
%p V[p]:= v;
%p break
%p fi
%p od
%p od:
%p S:= {1}:
%p for p in Res do
%p S:= `union`(S, seq(map(`*`,select(`<=`,S,N/p^i),p^i),i=1..V[p]))
%p od:
%p sort(convert(S, list));
%Y Cf. A000244, A002326
%K nonn
%O 1,2
%A _Robert Israel_, Aug 31 2018
|
