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%I #21 Feb 14 2024 17:28:09
%S 2,4,6,10,12,14,18,20,22,26,28,30,34,36,38,40,44,46,48,52,54,56,60,62,
%T 64,68,70,72,74,78,80,82,86,88,90,94,96,98,102,104,106,108,112,114,
%U 116,120,122,124,128,130,132,136,138,140,142,146,148,150,154,156,158,162
%N Lower Even Swappage of Upper Wythoff Sequence.
%C lim (1/n)*A141104(n) = 1 + tau.
%C Let S(n)=(1/2)*A141104(n). Is the complement of S equal to A004976?
%C 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
%H Vincent Russo and Loren Schwiebert, <a href="https://www.fq.math.ca/Papers1/49-2/RussoSchwiebert.pdf">Beatty Sequences, Fibonacci Numbers, and the Golden Ratio</a>, The Fibonacci Quarterly, Vol 49, Number 2, May 2011.
%H Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, <a href="https://arxiv.org/abs/2402.08331">Beatty Sequences for a Quadratic Irrational: Decidability and Applications</a>, arXiv:2402.08331 [math.NT], 2024. See p. 14.
%F 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.
%e Start with
%e a = (1,3,4,6,8,9,11,12,...) and b = (2,5,7,10,13,15,18,...).
%e After first swap,
%e a = (1,3,5,6,8,9,11,12,...) and b = (2,4,7,10,13,15,18,...).
%e After 2nd swap,
%e a = (1,3,5,7,8,9,11,12,...) and b = (2,4,6,10,13,15,18,...).
%Y Cf. A000201, A001950, A141105, A141106, A141107, A004976.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jun 02 2008, Aug 27 2008
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