A269571 Numbers having binary fractility 1.

2, 3, 4, 5, 8, 11, 13, 16, 19, 29, 32, 37, 53, 59, 61, 64, 67, 83, 101, 107, 128, 131, 139, 149, 163, 173, 179, 181, 197, 211, 227, 256, 269, 293, 317, 347, 349, 373, 379, 389, 419, 421, 443, 461, 467, 491, 509, 512, 523, 541, 547, 557, 563, 587, 613, 619

Offset: 1

Author

Clark Kimberling, Mar 01 2016

Keywords

• nonn

Comments

For each x in (0,1], let 1/2^p(1) + 1/2^p(2) + ... be the infinite binary representation of x. Let d(1) = p(1) and d(i) = p(i) - p(i-1) for i >=2. Call (d(i)) the powerdifference sequence of x, and denote it by D(x). Call m/n and u/v equivalent if every period of D(m/n) is a period of D(u/v). Define the binary fractility of n to be the number of distinct equivalence classes of {m/n: 0 < m < n}.

Examples

			D(1/5) = (3,1,3,1,3,1,3,1,...)
D(2/5) = (2,1,3,1,3,1,3,1,...)
D(3/5) = (1,3,1,3,1,3,1,3,...)
D(4/5) = (1,1,3,1,3,1,3,1,...).
This shows that all m/5, for 0<m<5 are equivalent to 1/5, so that there is only 1 equivalence class.
		

Links

• Robert Price, Table of n, a(n) for n = 1..76

Crossrefs

Cf. A269570, A269572.

Programs

Mathematica

A269570[n_] := CountDistinct[With[{l = NestWhileList[
        Rescale[#, {1/2^(Floor[-Log[2, #]] + 1),
           1/2^(Floor[-Log[2, #]])}] &, #, UnsameQ, All]},
     Min@l[[First@First@Position[l, Last@l] ;;]]] & /@
   Range[1/n, 1 - 1/n, 1/n]] (* from Davin Park, Nov 19 2016 *)
Select[Range[1000], A269570[#] == 1 &] (* Robert Price, Sep 20 2019 *)

Extensions

Corrected Offset by Robert Price, Sep 20 2019

a(37)-a(56) from Robert Price, Sep 20 2019