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A247455 Numbers k such that d(r,k) = 0 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {3*sqrt(2)}, and { } = fractional part.

%I A247455 #14 Sep 28 2014 09:10:57
%S A247455 1,8,9,10,11,15,21,25,29,38,42,48,51,54,57,58,59,62,64,66,70,72,78,81,
%T A247455 82,86,89,93,96,107,109,111,113,122,128,130,134,136,139,144,147,148,
%U A247455 149,151,153,161,162,165,169,173,181,182,183,187,191,195,200,202
%N A247455 Numbers k such that d(r,k) = 0 and d(s,k) = 0, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {3*sqrt(2)}, and { } = fractional part.
%C A247455 Every positive integer lies in exactly one of these: A247455, A247456, A247457, A247758.  Let s denote any of these; what can be said about lim(#s < n)/n, where (#s < n) represents the number of numbers in s that are < n?
%H A247455 Clark Kimberling, <a href="/A247455/b247455.txt">Table of n, a(n) for n = 1..1000</a>
%e A247455 {1*sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1,...
%e A247455 {3*sqrt(2)} has binary digits 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1,...
%e A247455 so that a(1) = 2 and a(2) = 8.
%t A247455 z = 400; r = FractionalPart[Sqrt[2]]; s = FractionalPart[3*Sqrt[2]];
%t A247455 u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t A247455 v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t A247455 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A247455 t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A247455 t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A247455 t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A247455 Flatten[Position[t1, 1]]  (* A247455 *)
%t A247455 Flatten[Position[t2, 1]]  (* A247456 *)
%t A247455 Flatten[Position[t3, 1]]  (* A247457 *)
%t A247455 Flatten[Position[t4, 1]]  (* A247458 *)
%Y A247455 Cf. A246356, A247456, A247457, A247458.
%K A247455 nonn,easy,base
%O A247455 1,2
%A A247455 _Clark Kimberling_, Sep 18 2014