%I
%S 1,3,5,9,11,13,15,19,21,25,27,29,33,35,37,39,45,53,55,57,59,61,63,65,
%T 67,69,75,77,81,83,87,91,95,99,101,105,107,111,115,117,121,125,131,
%U 133,135,139,141,143,145,147,149,159,163,165,169,171,173,175,177,179,181
%N Numbers k such that the multiplicative order of 2 modulo k is psi(k), psi = A002322.
%C Numbers k such that the multiplicative order of 2 modulo k is at its maximum possible value.
%C Numbers k such that the binary expansion of 1/k has period psi(n).
%C Numbers k such that A002326((k1)/2) = A002322(k).
%C This is a generalization of A167791, so A167791 is a proper subsequence.
%C Write k as k = Product_{i=1..t} (p_i)^(e_i) where p_i are distinct primes. If (p_i)^(e_i) belongs to A167791 (and thus here) for 1 <= i <= t, then k is also here, but the converse is not true. In fact, this sequence has terms such that none of (p_i)^(e_i) belongs to A167791, the smallest of which is 301 = 7*43. The multiplicative order of 2 modulo 7 and 43 are 3 (< psi(7) = 6) and 14 (< psi(43) = 42), so the multiplicative order of 2 modulo 301 is lcm(3, 14) = 42 = psi(301).
%H Robert Israel, <a href="/A319009/b319009.txt">Table of n, a(n) for n = 1..10000</a>
%e The multiplicative order of 2 modulo 15 is 4 = A002322(15), so 15 is a term.
%e The multiplicative order of 2 modulo 21 is 6 = A002322(21), so 21 is a term.
%e The multiplicative order of 2 modulo 51 is 8, but A002322(51) = 16, so 51 is not a term.
%p select(n > numtheory:order(2,n)=numtheory:lambda(n), [seq(i,i=1..1000,2)]); # _Robert Israel_, Sep 12 2018
%o (PARI) forstep(n=1, 200, 2, if(znorder(Mod(2, n))==lcm(znstar(n)[2]), print1(n, ", ")))
%Y Cf. A001122, A002322, A002326, A167791.
%K nonn
%O 1,2
%A _Jianing Song_, Sep 07 2018
