cp's OEIS Frontend

A247519 Numbers k such that d(r,k) = 0 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.

%I A247519 #14 Dec 03 2014 00:57:02
%S A247519 3,9,10,23,31,44,49,56,57,58,59,70,75,80,84,85,86,87,88,89,93,94,95,
%T A247519 96,97,104,116,119,120,121,122,128,129,130,131,134,135,136,139,140,
%U A247519 141,142,145,146,149,166,173,174,177,182,185,190,191,199,200,201,208
%N A247519 Numbers k such that d(r,k) = 0 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 A247519 Every positive integer lies in exactly one of these: A247519, A247520, A247521, A247522.  Prefixing the binary digits for r by 1 gives the binary digits for s.
%H A247519 Clark Kimberling, <a href="/A247519/b247519.txt">Table of n, a(n) for n = 1..1000</a>
%e A247519 r has binary digits 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, ...
%e A247519 s has binary digits 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, ...
%e A247519 so that a(1) = 3.
%t A247519 z = 400; r1 = GoldenRatio; r = FractionalPart[r1]; s = FractionalPart[r1/2];
%t A247519 u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t A247519 v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t A247519 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A247519 t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A247519 t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A247519 t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A247519 Flatten[Position[t1, 1]] (* A247519 *)
%t A247519 Flatten[Position[t2, 1]] (* A247520 *)
%t A247519 Flatten[Position[t3, 1]] (* A247521 *)
%t A247519 Flatten[Position[t4, 1]] (* A247522 *)
%Y A247519 Cf. A247520, A247521, A247522.
%K A247519 nonn,easy,base
%O A247519 1,1
%A A247519 _Clark Kimberling_, Sep 19 2014