A175332 Numbers whose binary expansion is of the form 11+0*.

3, 6, 7, 12, 14, 15, 24, 28, 30, 31, 48, 56, 60, 62, 63, 96, 112, 120, 124, 126, 127, 192, 224, 240, 248, 252, 254, 255, 384, 448, 480, 496, 504, 508, 510, 511, 768, 896, 960, 992, 1008, 1016, 1020, 1022, 1023, 1536, 1792, 1920, 1984, 2016, 2032, 2040, 2044

Offset: 1

Author

Joerg Arndt, Apr 12 2010

Keywords

• easy
• nonn

Comments

Also numbers n such that the set (2^j)%n consists only of the powers of two.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to binary expansion of n.

Crossrefs

Cf. A023758, A007088, A211705.

Programs

Haskell

import Data.Set (singleton, deleteFindMin, insert)
a175332 n = a175332_list !! (n-1)
a175332_list = f $ singleton 3 where
  f s = x : f (if even x then insert z s' else insert z $ insert (z+1) s')
        where z = 2*x; (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Sep 24 2014

PARI

is_11p0s(n)=
{ /* Return whether binary expansion has form 11+0* */
  local(b);
  if ( n<3, return(0) );
  b = binary( n/(2^valuation(n,2) ) );
  if ( #b<2, return(0) );
  for (j=1,#b, if(b[j]==0, return(0) ) );
  return(1);
}
for (n=1,2100, if (is_11p0s(n), print1(n,", ") ) ); /* show terms */

Python

def a_next(a_n): t = a_n + (a_n & 1); return t | (t >> 1)
a_n = 3; a = []
for i in range(53): a.append(a_n); a_n = a_next(a_n) # Falk Hüffner, Feb 21 2022

Formula

Sum_{n>=1} 1/a(n) = -2 + A211705. - Amiram Eldar, Feb 26 2022