|
%I #19 Feb 14 2024 17:27:59
%S 2,6,8,10,14,16,18,20,24,26,28,32,34,36,40,42,44,48,50,52,54,58,60,62,
%T 66,68,70,74,76,78,82,84,86,90,92,94,96,100,102,104,108,110,112,116,
%U 118,120,124,126,128,130,134,136,138,142,144,146,150,152,154,158,160,162
%N Upper Even Swappage of Upper Wythoff Sequence.
%C 1. lim (1/n)*A141105(n) = 1 + tau.
%C 2. Let S(n)=(1/2)*A141105(n). Is the complement of S equal to A035487?
%C 3. Is A141105 = 1+A141106?
%C Both 2 and 3 are true. These can be verified with the Walnut theorem-prover, using a synchronized Fibonacci automaton for the sequence. This automaton accepts the Fibonacci (Zeckendorf) representation of n and x in parallel if and only if x = A141105(n). - _Jeffrey Shallit_, Jan 27 2024
%H Clark Kimberling, <a href="/A141105/b141105.txt">Table of n, a(n) for n = 1..10000</a>
%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 odd 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. A141105 is the sequence obtained by thus swapping all odds out of A001950.
%F a(n) = round(n*tau^2), with tau = (1+sqrt(5))/2 (the golden ratio A001622).
%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 = (1,3,4,5,8,9,11,12,...) and b = (2,6,7,10,13,15,18,...).
%e After 2nd swap,
%e a = (1,3,4,5,7,9,11,12,...) and b = (2,6,8,10,13,15,18,...).
%t r = GoldenRatio^2; f[x_] := Floor[x];
%t g[x_] := If[Mod[f[x], 2] == 0, f[x], f[x] + 1];
%t t = Table[g[r*n], {n, 1, 100}]
%t (* _Clark Kimberling_, Nov 03 2012 *)
%Y Cf. A000201, A001622, A001950, A141104, A141106, A141107, A004976.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jun 02 2008
|
