A247455 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 = {3*sqrt(2)}, and { } = fractional part.

1, 8, 9, 10, 11, 15, 21, 25, 29, 38, 42, 48, 51, 54, 57, 58, 59, 62, 64, 66, 70, 72, 78, 81, 82, 86, 89, 93, 96, 107, 109, 111, 113, 122, 128, 130, 134, 136, 139, 144, 147, 148, 149, 151, 153, 161, 162, 165, 169, 173, 181, 182, 183, 187, 191, 195, 200, 202

Offset: 1

Author

Clark Kimberling, Sep 18 2014

Keywords

• nonn
• easy
• base

Comments

Every positive integer lies in exactly one of these: A247455, A247456, A247457, A247758. Let s denote any of these; what can be said about lim(#s < n)/n, where (#s < n) represents the number of numbers in s that are < n?

Examples

			{1*sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1,...
{3*sqrt(2)} has binary digits 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1,...
so that a(1) = 2 and a(2) = 8.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A246356, A247456, A247457, A247458.

Programs

Mathematica

z = 400; r = FractionalPart[Sqrt[2]]; s = FractionalPart[3*Sqrt[2]];
u = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[r, 2, z]]
v = Flatten[{ConstantArray[0, -#[[2]]], #[[1]]}] &[RealDigits[s, 2, z]]
t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
Flatten[Position[t1, 1]]  (* A247455 *)
Flatten[Position[t2, 1]]  (* A247456 *)
Flatten[Position[t3, 1]]  (* A247457 *)
Flatten[Position[t4, 1]]  (* A247458 *)