A141106 Lower Odd Swappage of Upper Wythoff Sequence.

1, 5, 7, 9, 13, 15, 17, 19, 23, 25, 27, 31, 33, 35, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 65, 67, 69, 73, 75, 77, 81, 83, 85, 89, 91, 93, 95, 99, 101, 103, 107, 109, 111, 115, 117, 119, 123, 125, 127, 129, 133, 135, 137, 141, 143, 145, 149, 151, 153, 157, 159, 161, 163

Offset

1,2

Comments

1. lim (1/n)*A141106(n) = 1 + tau.

2. Let S(n)=(1/2)*(1+A141106(n)). Is the complement of S equal to A035487?

#2 is true. This can be proved using a synchronized automaton for A035487 and A141106. These automata take the Fibonacci (Zeckendorf) representations of n and y in parallel, and accept if and only if y = a(n). - Jeffrey Shallit, Jan 27 2024

Links

Table of n, a(n) for n=1..63.

Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, Beatty Sequences for a Quadratic Irrational: Decidability and Applications, arXiv:2402.08331 [math.NT], 2024. See p. 15.

Formula

Let a = (1,3,4,6,8,9,11,12,...) = A000201 = lower Wythoff sequence; let b = (2,5,7,10,13,15,18,...) = A001950 = upper Wythoff sequence. For each even b(n), let a(m) be the least number in a such that after swapping b(n) and a(m), the resulting new a and b are both increasing. A141106 is the sequence obtained by thus swapping all evens out of A001950.

Example

Start with
a = (1,3,4,6,8,9,11,12,...) and b = (2,5,7,10,13,15,18,...).
After 1st swap,
a = (2,3,4,6,8,9,11,12,...) and b = (1,5,7,10,13,15,18,...).
After 2nd swap,
a = (2,3,4,5,6,9,11,12,...) and b = (1,5,7,9,13,15,18,...).

Crossrefs

Cf. A000201, A001950, A141104, A141105, A141107, A004976.

Sequence in context: A200268 A251627 A128161 * A047478 A048974 A089193

Adjacent sequences: A141103 A141104 A141105 * A141107 A141108 A141109

Keyword

nonn

Author

Clark Kimberling, Jun 02 2008

Status

approved