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A247356 Numbers k such that d(r,k) = 1 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 A247356 #13 Sep 26 2014 17:24:07
%S A247356 3,5,7,16,17,19,22,23,30,32,33,41,45,49,56,61,67,74,75,76,88,90,91,98,
%T A247356 99,101,105,108,115,116,117,120,125,131,137,138,140,141,154,158,164,
%U A247356 167,170,172,175,178,185,188,189,193,194,199,203,221,230,231,234
%N A247356 Numbers k such that d(r,k) = 1 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 A247356 Every positive integer lies in exactly one of these: A246356, A246357, A246358, A247356.
%H A247356 Clark Kimberling, <a href="/A247356/b247356.txt">Table of n, a(n) for n = 1..1115</a>
%e A247356 {sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,...
%e A247356 {sqrt(3)} has binary digits 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0,..
%e A247356 so that a(1) = 3 and a(2) = 5.
%t A247356 z = 500; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[3]];
%t A247356 u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
%t A247356 v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
%t A247356 t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A247356 t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A247356 t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t A247356 t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t A247356 Flatten[Position[t1, 1]]  (* A246356 *)
%t A247356 Flatten[Position[t2, 1]]  (* A246357 *)
%t A247356 Flatten[Position[t3, 1]]  (* A246358 *)
%t A247356 Flatten[Position[t4, 1]]  (* A247356 *)
%Y A247356 Cf. A247454, A246356, A246357, A246358.
%K A247356 nonn,easy,base
%O A247356 1,1
%A A247356 _Clark Kimberling_, Sep 17 2014