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Numbers k such that d(r,k) = d(s,k), where d(x,k) = k-th binary digit of x, r = {golden ratio}, s = {(golden ratio)/2}, and { } = fractional part.
3

%I #16 May 05 2018 08:14:46

%S 1,3,5,6,7,9,10,12,15,16,19,20,21,23,25,28,29,31,35,36,37,38,39,40,44,

%T 49,51,52,53,54,56,57,58,59,65,66,67,68,70,72,73,75,77,78,80,82,84,85,

%U 86,87,88,89,91,93,94,95,96,97,101,102,104,106,107,110

%N Numbers k such that d(r,k) = d(s,k), where d(x,k) = k-th binary digit of x, r = {golden ratio}, s = {(golden ratio)/2}, and { } = fractional part.

%C Every positive integer lies in exactly one of the sequences A247423 and A247524.

%H Clark Kimberling, <a href="/A247523/b247523.txt">Table of n, a(n) for n = 1..2000</a>

%e r has binary digits 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, ...

%e s has binary digits 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, ...

%e so that a(1) = 1 and a(2) = 3.

%t z = 400; r1 = GoldenRatio; r = FractionalPart[r1]; s = FractionalPart[r1/2];

%t u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]];

%t v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]];

%t t = Table[If[u[[n]] == v[[n]], 1, 0], {n, 1, z}];

%t Flatten[Position[t, 1]] (* A247523 *)

%t Flatten[Position[t, 0]] (* A247524 *)

%Y Cf. A247524, A247519.

%K nonn,easy,base

%O 1,2

%A _Clark Kimberling_, Sep 19 2014