OFFSET
1,1
COMMENTS
1. lim (1/n)*A141105(n) = 1 + tau.
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
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
Luke Schaeffer, Jeffrey Shallit, and Stefan Zorcic, Beatty Sequences for a Quadratic Irrational: Decidability and Applications, arXiv:2402.08331 [math.NT], 2024. See p. 15.
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 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.
a(n) = round(n*tau^2), with tau = (1+sqrt(5))/2 (the golden ratio A001622).
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 = (1,3,4,5,8,9,11,12,...) and b = (2,6,7,10,13,15,18,...).
After 2nd swap,
a = (1,3,4,5,7,9,11,12,...) and b = (2,6,8,10,13,15,18,...).
MATHEMATICA
r = GoldenRatio^2; f[x_] := Floor[x];
g[x_] := If[Mod[f[x], 2] == 0, f[x], f[x] + 1];
t = Table[g[r*n], {n, 1, 100}]
(* Clark Kimberling, Nov 03 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jun 02 2008
STATUS
approved