A247458 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.

3, 5, 7, 13, 16, 17, 19, 23, 27, 30, 31, 32, 33, 34, 36, 39, 40, 43, 44, 46, 50, 53, 56, 61, 68, 73, 74, 75, 76, 80, 84, 87, 91, 94, 97, 99, 101, 103, 105, 114, 115, 116, 118, 120, 123, 124, 125, 127, 131, 132, 137, 140, 141, 142, 146, 154, 156, 158, 160

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) = 3 and a(2) = 5.
		

Links

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

Crossrefs

Cf. A247455, A247456, A247457.

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 *)