A141107 Upper Odd Swappage of Upper Wythoff Sequence.

3, 5, 7, 11, 13, 15, 19, 21, 23, 27, 29, 31, 35, 37, 39, 41, 45, 47, 49, 53, 55, 57, 61, 63, 65, 69, 71, 73, 75, 79, 81, 83, 87, 89, 91, 95, 97, 99, 103, 105, 107, 109, 113, 115, 117, 121, 123, 125, 129, 131, 133, 137, 139, 141, 143, 147, 149, 151, 155, 157, 159, 163

Offset

1,1

Comments

1. lim (1/n)*A141107(n) = 1 + tau

2. Let S(n)=(1/2)*(1+A141107(n)). Is the complement of S equal to A004976?

3. Is A141107 = 1 + A141104?

Both #2 and #3 are true. They can be proved with the Walnut theorem-prover, using the synchronized Fibonacci automaton for the sequences A141104 and A141107. These automata take n and y as input, in Fibonacci (Zeckendorf) representation, and accept iff y = a(n) for the respective sequence. - Jeffrey Shallit, Jan 27 2024

Links

Table of n, a(n) for n=1..62.

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 greatest number in a such that after swapping b(n) and a(m), the resulting new a and b are both increasing. A141107 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 first swap,
a = (1,2,4,6,8,9,11,12,...) and b = (3,5,7,10,13,15,18,...).
After 2nd swap,
a = (1,2,4,6,8,9,10,12,...) and b = (3,5,7,11,13,15,18,...).

Crossrefs

Cf. A000201, A001950, A141104, A141105, A141106, A004976.

Sequence in context: A062086 A283680 A005239 * A047484 A036991 A165887

Adjacent sequences: A141104 A141105 A141106 * A141108 A141109 A141110

Keyword

nonn

Author

Clark Kimberling, Jun 02 2008

Status

approved