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%I #9 Sep 26 2014 21:12:06
%S 4,11,14,18,24,27,32,34,42,45,47,50,60,62,64,71,76,81,90,98,100,105,
%T 109,112,117,123,126,132,137,143,147,150,154,157,159,167,171,175,178,
%U 183,186,188,192,202,205,210,213,215,220,224,228,233,240,245,249,252
%N Numbers k such that d(r,k) = 1 and d(s,k) = 0, 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, A247522.
%H Clark Kimberling, <a href="/A247521/b247521.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) = 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 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, A247522.
%K nonn,easy,base
%O 1,1
%A _Clark Kimberling_, Sep 19 2014
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