A247520 Numbers k such that d(r,k) = 0 and d(s,k) = 1, where d(x,k) = k-th binary digit of x, r = {golden ratio}, s = {(golden ratio)/2}, and { } = fractional part.

2, 8, 13, 17, 22, 26, 30, 33, 41, 43, 46, 48, 55, 61, 63, 69, 74, 79, 83, 92, 99, 103, 108, 111, 115, 118, 125, 127, 133, 138, 144, 148, 153, 156, 158, 165, 170, 172, 176, 181, 184, 187, 189, 198, 204, 207, 212, 214, 216, 221, 227, 229, 235, 242, 248, 250

Offset

1,1

Comments

Every positive integer lies in exactly one of these: A247519, A247520, A247521, A247522.

Links

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

Example

r has binary digits 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, ...
s has binary digits 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, ...
so that a(1) = 2.

Mathematica

z = 400; r1 = GoldenRatio; r = FractionalPart[r1]; s = FractionalPart[r1/2];
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]] (* A247519 *)
Flatten[Position[t2, 1]] (* A247520 *)
Flatten[Position[t3, 1]] (* A247521 *)
Flatten[Position[t4, 1]] (* A247522 *)

Crossrefs

Cf. A247519, A247521, A247522.

Sequence in context: A030389 A136738 A288227 * A304801 A184519 A156245

Adjacent sequences: A247517 A247518 A247519 * A247521 A247522 A247523

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 19 2014

Status

approved