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A246356 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 = {sqrt(3)}, and { } = fractional part.

%I A246356 #15 Sep 26 2014 17:23:30
%S A246356 6,9,12,20,24,28,29,37,48,52,57,58,62,66,69,81,82,89,93,96,102,104,
%T A246356 106,111,113,122,129,130,139,144,149,151,159,161,163,165,166,177,179,
%U A246356 181,186,187,190,191,195,201,202,204,217,219,220,222,225,228,232,233
%N A246356 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 = {sqrt(3)}, and { } = fractional part.
%C A246356 Every positive integer lies in exactly one of these: A246356, A246357, A246358, A247356.  Let s denote any of these; is lim(#s < n)/n = 1/4, where (#s < n) represents the number of numbers in s that are < n?
%H A246356 Clark Kimberling, <a href="/A246356/b246356.txt">Table of n, a(n) for n = 1..1000</a>
%e A246356 {sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,...
%e A246356 {sqrt(3)} has binary digits 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0,..
%e A246356 so that a(1) = 6.
%t A246356 z = 500; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[3]];
%t A246356 u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t A246356 v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t A246356 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A246356 t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A246356 t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A246356 t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A246356 Flatten[Position[t1, 1]]  (* A246356 *)
%t A246356 Flatten[Position[t2, 1]]  (* A246357 *)
%t A246356 Flatten[Position[t3, 1]]  (* A246358 *)
%t A246356 Flatten[Position[t4, 1]]  (* A247356 *)
%Y A246356 Cf. A247454, A246357, A246358, A247356.
%K A246356 nonn,easy,base
%O A246356 1,1
%A A246356 _Clark Kimberling_, Sep 17 2014