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A246358 Numbers k such that d(r,k) = 1 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 A246358 #13 Sep 26 2014 17:23:49
%S A246358 2,13,18,26,27,31,34,35,36,39,40,43,44,46,50,53,65,68,71,73,77,79,80,
%T A246358 84,87,94,95,97,103,110,112,114,118,123,124,126,127,132,133,135,142,
%U A246358 143,145,146,152,155,156,160,171,174,176,180,192,196,197,205,206
%N A246358 Numbers k such that d(r,k) = 1 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 A246358 Every positive integer lies in exactly one of these: A246356, A246357, A246358, A247356.
%H A246358 Clark Kimberling, <a href="/A246358/b246358.txt">Table of n, a(n) for n = 1..1000</a>
%e A246358 {sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,...
%e A246358 {sqrt(3)} has binary digits 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0,..
%e A246358 so that a(1) = 2 and a(2) = 13.
%t A246358 z = 500; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[3]];
%t A246358 u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t A246358 v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t A246358 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A246358 t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A246358 t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A246358 t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A246358 Flatten[Position[t1, 1]]  (* A246356 *)
%t A246358 Flatten[Position[t2, 1]]  (* A246357 *)
%t A246358 Flatten[Position[t3, 1]]  (* A246358 *)
%t A246358 Flatten[Position[t4, 1]]  (* A247356 *)
%Y A246358 Cf. A247454, A246356, A246357, A247356.
%K A246358 nonn,easy,base
%O A246358 1,1
%A A246358 _Clark Kimberling_, Sep 17 2014