OFFSET
1,1
COMMENTS
lim (1/n)*A141104(n) = 1 + tau.
This question has an affirmative answer, as proved by Russo and Schwiebert in the link below. It can also be proved using the Walnut theorem-prover, using synchronized Fibonacci automata for the two sequences. 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
Vincent Russo and Loren Schwiebert, Beatty Sequences, Fibonacci Numbers, and the Golden Ratio, The Fibonacci Quarterly, Vol 49, Number 2, May 2011.
Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, Beatty Sequences for a Quadratic Irrational: Decidability and Applications, arXiv:2402.08331 [math.NT], 2024. See p. 14.
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 odd 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. A141104 is the sequence obtained by thus swapping all odds 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 = (1,3,5,6,8,9,11,12,...) and b = (2,4,7,10,13,15,18,...).
After 2nd swap,
a = (1,3,5,7,8,9,11,12,...) and b = (2,4,6,10,13,15,18,...).
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jun 02 2008, Aug 27 2008
STATUS
approved