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%I #10 Sep 26 2014 21:32:58
%S 1,4,6,12,15,21,25,29,38,42,48,52,55,60,64,66,70,72,78,83,86,89,93,96,
%T 100,102,104,107,109,111,113,119,122,130,134,136,139,144,151,153,157,
%U 159,163,166,169,173,177,179,184,187,191,195,198,202,204,209,211
%N 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(8)}, and { } = fractional part.
%C Every positive integer lies in exactly one of these: A247631, A247632, A247633, A247634.
%H Clark Kimberling, <a href="/A247632/b247632.txt">Table of n, a(n) for n = 1..1097</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, A247633, A247634.
%K nonn,easy,base
%O 1,2
%A _Clark Kimberling_, Sep 23 2014
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