A247457 Numbers k such that d(r,k) = 1 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.

2, 18, 22, 26, 35, 41, 45, 49, 65, 67, 71, 77, 79, 88, 90, 95, 98, 108, 110, 112, 117, 126, 133, 135, 138, 143, 145, 152, 155, 172, 175, 188, 194, 196, 203, 208, 210, 212, 221, 223, 230, 234, 239, 243, 260, 262, 268, 278, 292, 294, 296, 299, 310, 312, 319

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, A247458.

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.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A247455, A247456, 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 *)