%I #9 Sep 26 2014 21:12:16

%S 1,5,6,7,12,15,16,19,20,21,25,28,29,35,36,37,38,39,40,51,52,53,54,65,

%T 66,67,68,72,73,77,78,82,91,101,102,106,107,110,113,114,124,151,152,

%U 155,160,161,162,163,164,168,169,179,180,193,194,195,196,197,203

%N Numbers k such that d(r,k) = 1 and d(s,k) = 1, 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 these: A247519, A247520, A247521.

%H Clark Kimberling, <a href="/A247522/b247522.txt">Table of n, a(n) for n = 1..1000</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) = 5.

%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 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];

%t t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];

%t t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];

%t t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];

%t Flatten[Position[t1, 1]] (* A247519 *)

%t Flatten[Position[t2, 1]] (* A247520 *)

%t Flatten[Position[t3, 1]] (* A247521 *)

%t Flatten[Position[t4, 1]] (* A247522 *)

%Y Cf. A247519, A247520, A247521.

%K nonn,easy,base

%O 1,2

%A _Clark Kimberling_, Sep 19 2014