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%I #13 Feb 14 2024 17:28:03
%S 1,5,7,9,13,15,17,19,23,25,27,31,33,35,39,41,43,47,49,51,53,57,59,61,
%T 65,67,69,73,75,77,81,83,85,89,91,93,95,99,101,103,107,109,111,115,
%U 117,119,123,125,127,129,133,135,137,141,143,145,149,151,153,157,159,161,163
%N Lower Odd Swappage of Upper Wythoff Sequence.
%C 1. lim (1/n)*A141106(n) = 1 + tau.
%C 2. Let S(n)=(1/2)*(1+A141106(n)). Is the complement of S equal to A035487?
%C #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
%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. 15.
%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 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.
%e Start with
%e a = (1,3,4,6,8,9,11,12,...) and b = (2,5,7,10,13,15,18,...).
%e After 1st swap,
%e a = (2,3,4,6,8,9,11,12,...) and b = (1,5,7,10,13,15,18,...).
%e After 2nd swap,
%e a = (2,3,4,5,6,9,11,12,...) and b = (1,5,7,9,13,15,18,...).
%Y Cf. A000201, A001950, A141104, A141105, A141107, A004976.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jun 02 2008