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A246357 Numbers k such that d(r,k) = 0 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {sqrt(3)}, and { } = fractional part.

%I A246357 #13 Sep 26 2014 17:23:40
%S A246357 1,4,8,10,11,14,15,21,25,38,42,47,51,54,55,59,60,63,64,70,72,78,83,85,
%T A246357 86,92,100,107,109,119,121,128,134,136,147,148,150,153,157,162,168,
%U A246357 169,173,182,183,184,198,200,209,211,214,215,218,226,227,229,241
%N A246357 Numbers k such that d(r,k) = 0 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {sqrt(2)}, s = {sqrt(3)}, and { } = fractional part.
%C A246357 Every positive integer lies in exactly one of these: A246356, A246357, A246358, A247356.
%H A246357 Clark Kimberling, <a href="/A246357/b246357.txt">Table of n, a(n) for n = 1..1000</a>
%e A246357 {sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,...
%e A246357 {sqrt(3)} has binary digits 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0,..
%e A246357 so that a(1) = 1 and a(2) = 4.
%t A246357 z = 500; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[3]];
%t A246357 u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t A246357 v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t A246357 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A246357 t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A246357 t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A246357 t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A246357 Flatten[Position[t1, 1]]  (* A246356 *)
%t A246357 Flatten[Position[t2, 1]]  (* A246357 *)
%t A246357 Flatten[Position[t3, 1]]  (* A246358 *)
%t A246357 Flatten[Position[t4, 1]]  (* A247356 *)
%Y A246357 Cf. A247454, A246356, A246358, A247356.
%K A246357 nonn,easy,base
%O A246357 1,2
%A A246357 _Clark Kimberling_, Sep 17 2014