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

Author

Clark Kimberling, Mar 01 2016

Keywords

• nonn
• easy

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).

Examples

			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.
		

Crossrefs

Cf. A269571, 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]] (* Davin Park, Nov 19 2016 *)