A246356 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(3)}, and { } = fractional part.

6, 9, 12, 20, 24, 28, 29, 37, 48, 52, 57, 58, 62, 66, 69, 81, 82, 89, 93, 96, 102, 104, 106, 111, 113, 122, 129, 130, 139, 144, 149, 151, 159, 161, 163, 165, 166, 177, 179, 181, 186, 187, 190, 191, 195, 201, 202, 204, 217, 219, 220, 222, 225, 228, 232, 233

Offset: 1

Author

Clark Kimberling, Sep 17 2014

Keywords

• nonn
• easy
• base

Comments

Every positive integer lies in exactly one of these: A246356, A246357, A246358, A247356. Let s denote any of these; is lim(#s < n)/n = 1/4, where (#s < n) represents the number of numbers in s that are < n?

Examples

			{sqrt(2)} has binary digits 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1,...
{sqrt(3)} has binary digits 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0,..
so that a(1) = 6.
		

Links

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

Crossrefs

Cf. A247454, A246357, A246358, A247356.

Programs

Mathematica

z = 500; r = FractionalPart[Sqrt[2]]; s = FractionalPart[Sqrt[3]];
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]]  (* A246356 *)
Flatten[Position[t2, 1]]  (* A246357 *)
Flatten[Position[t3, 1]]  (* A246358 *)
Flatten[Position[t4, 1]]  (* A247356 *)