A226181 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.

3, 5, 7, 11, 13, 17, 19, 23, 29, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 113, 131, 137, 139, 149, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 227, 233, 239, 257, 263, 269, 271, 281, 293, 311, 313, 317, 337, 347, 349

Offset: 1

Author

Lear Young, May 30 2013

Keywords

• nonn
• easy

Comments

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

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

Examples

			(41-1)/20 = 2. 20 is the period of the binary representation of 1/n, the odd part of 2 is 1.
		

Links

• Lear Young, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A007733, A136042.

Programs

Mathematica

Select[Prime[Range[2, 100]], # == 2^IntegerExponent[#, 2] &[(# - 1)/MultiplicativeOrder[2, #]] &] (* Peter J. C. Moses, May 28 2014 *)

PARI

is(n) = {
  m = valuation(n+1,2);
      k=(n+1)>>m;
      if(k!=1, for(i=0,(n-1)>>1,
        l=valuation(k+n,2);
        k=(k+n)>>l;
        m+=l;if(k==1,break)));
       ((n-1)/m)>>valuation((n-1)/m, 2)==1
       \\ m  equals znorder(Mod(2,n))
    }
forstep(i=3,1e3,2,if(is(i),print1(i, ", ")))
\\ Lear Young May 30 2013

PARI

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