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 A269570 Binary fractility of n. 14
 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 (list; graph; refs; listen; history; text; internal format)
 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 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: A096369 A332289 A102297 * A243759 A098398 A306211 Adjacent sequences:  A269567 A269568 A269569 * A269571 A269572 A269573 KEYWORD nonn,easy AUTHOR Clark Kimberling, Mar 01 2016 STATUS approved

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Last modified December 2 14:09 EST 2021. Contains 349445 sequences. (Running on oeis4.)