A141107 - OEIS

%I #12 Feb 14 2024 17:28:06

%S 3,5,7,11,13,15,19,21,23,27,29,31,35,37,39,41,45,47,49,53,55,57,61,63,

%T 65,69,71,73,75,79,81,83,87,89,91,95,97,99,103,105,107,109,113,115,

%U 117,121,123,125,129,131,133,137,139,141,143,147,149,151,155,157,159,163

%N Upper Odd Swappage of Upper Wythoff Sequence.

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

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

%C 3. Is A141107 = 1 + A141104?

%C 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

%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 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.

%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 = (1,2,4,6,8,9,11,12,...) and b = (3,5,7,10,13,15,18,...).

%e After 2nd swap,

%e a = (1,2,4,6,8,9,10,12,...) and b = (3,5,7,11,13,15,18,...).

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

%K nonn

%O 1,1

%A _Clark Kimberling_, Jun 02 2008