%I #10 Sep 26 2014 21:14:27

%S 2,16,17,18,22,26,30,31,32,33,34,35,39,40,43,44,45,49,67,73,74,75,76,

%T 79,87,90,94,97,98,114,115,116,117,123,124,125,126,131,132,137,140,

%U 141,142,145,154,155,170,171,174,175,188,192,193,196,205,206,207,212

%N 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(8)}, and { } = fractional part.

%C Every positive integer lies in exactly one of these: A247631, A247632, A247633, A247634.

%H Clark Kimberling, <a href="/A247634/b247634.txt">Table of n, a(n) for n = 1..1122</a>

%e r has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, ...

%e s has binary digits 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, ...

%e so that a(1) = 1 and a(2) = 4.

%t z = 400; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[8]];

%t u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]

%t v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]

%t t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];

%t t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];

%t t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];

%t t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];

%t Flatten[Position[t1, 1]] (* A247631 *)

%t Flatten[Position[t2, 1]] (* A247632 *)

%t Flatten[Position[t3, 1]] (* A247633 *)

%t Flatten[Position[t4, 1]] (* A247634 *)

%Y Cf. A247631, A247632, A247633.

%K nonn,easy,base

%O 1,1

%A _Clark Kimberling_, Sep 23 2014