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

1,1

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}.

Links

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

Example

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.

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 *)

Crossrefs

Cf. A269570, A269572.

Sequence in context: A186041 A322531 A302591 * A281493 A181341 A174181

Adjacent sequences: A269568 A269569 A269570 * A269572 A269573 A269574

Keyword

nonn

Author

Clark Kimberling, Mar 01 2016

Extensions

Corrected Offset by Robert Price, Sep 20 2019

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

Status

approved