A269570 - OEIS

%I #10 Dec 07 2016 18:48:13

%S 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,

%T 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,

%U 6,2,3,8,2,7,2,5,5,2,2,4,3,1,6,11,4

%N Binary fractility of n.

%C 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}.

%C 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).

%e D(1/7) = (3,3,3,3, ... )

%e D(2/7) = (2,3,3,3, ... )

%e D(3/7) = (2,1,2,1,2,1,2,1, ... )

%e D(4/7) = (1,3,3,3, ... )

%e D(5/7) = (1,2,1,2,1,2, ... )

%e D(6/7) = (1,1,2,1,2,1, ... )

%e There are 2 distinct periods: (3) and (1,2), so that a(7) = 2.

%t 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 *)

%Y Cf. A269571, A269572.

%K nonn,easy

%O 2,5

%A _Clark Kimberling_, Mar 01 2016