%I #12 Sep 28 2014 09:11:21
%S 3,5,7,13,16,17,19,23,27,30,31,32,33,34,36,39,40,43,44,46,50,53,56,61,
%T 68,73,74,75,76,80,84,87,91,94,97,99,101,103,105,114,115,116,118,120,
%U 123,124,125,127,131,132,137,140,141,142,146,154,156,158,160
%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 = {3*sqrt(2)}, and { } = fractional part.
%C Every positive integer lies in exactly one of these: A247455, A247456, A247457, A247458.
%H Clark Kimberling, <a href="/A247458/b247458.txt">Table of n, a(n) for n = 1..1000</a>
%e {1*sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1,...
%e {3*sqrt(2)} has binary digits 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1,...
%e so that a(1) = 3 and a(2) = 5.
%t z = 400; r = FractionalPart[Sqrt[2]]; s = FractionalPart[3*Sqrt[2]];
%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]] (* A247455 *)
%t Flatten[Position[t2, 1]] (* A247456 *)
%t Flatten[Position[t3, 1]] (* A247457 *)
%t Flatten[Position[t4, 1]] (* A247458 *)
%Y Cf. A247455, A247456, A247457.
%K nonn,easy,base
%O 1,1
%A _Clark Kimberling_, Sep 18 2014
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