A269570 Binary fractility of n.

1, 1, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 3, 4, 1, 2, 3, 1, 2, 5, 2, 2, 2, 2, 2, 3, 3, 1, 5, 6, 1, 4, 3, 5, 3, 1, 2, 4, 2, 2, 6, 3, 2, 7, 3, 2, 2, 4, 3, 7, 2, 1, 4, 4, 3, 4, 2, 1, 5, 1, 7, 12, 1, 6, 5, 1, 3, 5, 6, 2, 3, 8, 2, 7, 2, 5, 5, 2, 2, 4, 3, 1, 6, 11, 4

Offset

2,5

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}.

An equivalent definition of equivalence follows. Let S(x,h) denote the h-th partial sum of the infinite binary sum for x. Then m/n and u/v are equivalent if there exist integers p, h > 0, k > 0 such that (2^p)(m/n - S(m/n,h)) = u/v - S(u/v,k).

Links

Table of n, a(n) for n=2..86.

Example

D(1/7) = (3,3,3,3, ... )
D(2/7) = (2,3,3,3, ... )
D(3/7) = (2,1,2,1,2,1,2,1, ... )
D(4/7) = (1,3,3,3, ... )
D(5/7) = (1,2,1,2,1,2, ... )
D(6/7) = (1,1,2,1,2,1, ... )
There are 2 distinct periods: (3) and (1,2), so that a(7) = 2.

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]] (* Davin Park, Nov 19 2016 *)

Crossrefs

Cf. A269571, A269572.

Sequence in context: A332289 A365096 A102297 * A243759 A098398 A306211

Adjacent sequences: A269567 A269568 A269569 * A269571 A269572 A269573

Keyword

nonn,easy

Author

Clark Kimberling, Mar 01 2016

Status

approved