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

1,1

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

Links

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

Example

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

Mathematica

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

Prog

(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

Crossrefs

Cf. A007733, A136042.

Sequence in context: A002556 A130101 A130057 * A363286 A120637 A278454

Adjacent sequences: A226178 A226179 A226180 * A226182 A226183 A226184

Keyword

nonn,easy

Author

Lear Young, May 30 2013

Status

approved