A247356 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 = {sqrt(3)}, and { } = fractional part.

3, 5, 7, 16, 17, 19, 22, 23, 30, 32, 33, 41, 45, 49, 56, 61, 67, 74, 75, 76, 88, 90, 91, 98, 99, 101, 105, 108, 115, 116, 117, 120, 125, 131, 137, 138, 140, 141, 154, 158, 164, 167, 170, 172, 175, 178, 185, 188, 189, 193, 194, 199, 203, 221, 230, 231, 234

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.

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) = 3 and a(2) = 5.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1115

Crossrefs

Cf. A247454, A246356, A246357, A246358.

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 *)