%I #23 Jun 28 2021 13:42:59
%S 2,4,8,11,13,14,17,18,22,24,26,27,30,32,33,34,41,42,43,45,46,47,48,50,
%T 55,60,61,62,63,64,69,71,74,76,79,81,83,90,92,98,99,100,103,105,108,
%U 109,111,112,115,117,118,123,125,126,127,132,133,137,138,143
%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="/A247524/b247524.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) = 2 and a(2) = 4.
%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 *)
%t Module[{nn=150,gr,g2},gr=Rest[RealDigits[GoldenRatio,2,nn+1][[1]]];g2 = RealDigits[ GoldenRatio/2,2,nn][[1]];Position[Thread[{gr,g2}],_?(#[[1]] != #[[2]]&),1,Heads->False]]//Flatten (* _Harvey P. Dale_, Jun 28 2021 *)
%Y Cf. A247523.
%K nonn,easy,base
%O 1,1
%A _Clark Kimberling_, Sep 19 2014