%I #31 May 29 2014 13:53:40
%S 3,5,7,11,13,17,19,23,29,37,41,47,53,59,61,67,71,73,79,83,89,97,101,
%T 103,107,113,131,137,139,149,163,167,173,179,181,191,193,197,199,211,
%U 227,233,239,257,263,269,271,281,293,311,313,317,337,347,349
%N Primes p such that p-1 divided by the period of the binary expansion of 1/p equals 2^x for some nonnegative integer x.
%C Equivalently, p-1 divided by the period of the decimal expansion of 1/p equals 2^x for some nonnegative integer x. Composite numbers satisfying this condition are given in A243050. - _Lear Young_, May 30 2013
%C Let pi_1(x) and pi(x) be the numbers of primes of this sequence and all primes not exceeding x, respectively. Then, for x>=3, p_1(x)/pi(x) >= C_Artin = 0.37395581... Numerical results suggest that it is likely lim pi_1(x)/pi(x) = 2*C_Artin. - _Peter J. C. Moses_ and _Vladimir Shevelev_, May 29 2014
%H Lear Young, <a href="/A226181/b226181.txt">Table of n, a(n) for n = 1..1000</a>
%e (41-1)/20 = 2. 20 is the period of the binary representation of 1/n, the odd part of 2 is 1.
%t Select[Prime[Range[2, 100]], # == 2^IntegerExponent[#, 2] &[(# - 1)/MultiplicativeOrder[2, #]] &] (* _Peter J. C. Moses_, May 28 2014 *)
%o (PARI)
%o is(n) = {
%o m = valuation(n+1,2);
%o k=(n+1)>>m;
%o if(k!=1, for(i=0,(n-1)>>1,
%o l=valuation(k+n,2);
%o k=(k+n)>>l;
%o m+=l;if(k==1,break)));
%o ((n-1)/m)>>valuation((n-1)/m, 2)==1
%o \\ m equals znorder(Mod(2,n))
%o }
%o forstep(i=3,1e3,2,if(is(i),print1(i, ", ")))
%o \\ _Lear Young_ May 30 2013
%o (PARI)
%o forstep(i=1,1e3,2,j = (i-1)/znorder(Mod(2,i));if(j>>valuation(j, 2)==1,print1(i, ", "))) \\ _Lear Young_ May 31 2013
%Y Cf. A007733, A136042.
%K nonn,easy
%O 1,1
%A _Lear Young_, May 30 2013