A247522 Numbers k such that d(r,k) = 1 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.

1, 5, 6, 7, 12, 15, 16, 19, 20, 21, 25, 28, 29, 35, 36, 37, 38, 39, 40, 51, 52, 53, 54, 65, 66, 67, 68, 72, 73, 77, 78, 82, 91, 101, 102, 106, 107, 110, 113, 114, 124, 151, 152, 155, 160, 161, 162, 163, 164, 168, 169, 179, 180, 193, 194, 195, 196, 197, 203

Offset

1,2

Comments

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

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) = 1 and a(2) = 5.

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, A247520, A247521.

Sequence in context: A022566 A047320 A327301 * A011761 A106745 A165776

Adjacent sequences: A247519 A247520 A247521 * A247523 A247524 A247525

Keyword

nonn,easy,base

Author

Clark Kimberling, Sep 19 2014

Status

approved