|
%I #9 Sep 26 2014 21:32:43
%S 8,9,10,11,14,20,24,28,37,47,51,54,57,58,59,62,63,69,81,82,85,92,106,
%T 121,128,129,147,148,149,150,161,162,165,168,181,182,183,186,190,200,
%U 201,214,217,218,219,225,226,227,228,232,236,241,245,248,249,258
%N Numbers k such that d(r,k) = 0 and d(s,k) = 0, 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. Deleting the initial 1 from the representation of r gives the representation of s.
%H Clark Kimberling, <a href="/A247631/b247631.txt">Table of n, a(n) for n = 1..1181</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) = 8 and a(2) = 9.
%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. A247632, A247633, A247634, A247519.
%K nonn,easy,base
%O 1,1
%A _Clark Kimberling_, Sep 23 2014
|
