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%I #25 Feb 14 2024 17:27:45
%S 1,3,5,8,10,11,14,16,18,21,22,24,27,29,31,32,35,37,39,42,43,45,48,50,
%T 52,55,56,58,60,63,65,66,69,71,73,76,77,79,82,84,86,87,90,92,94,97,99,
%U 100,103,105,107,110,111,113,115,118,120,121,124,126,128,131,132,134,137,139,141,144,145,147,149,152,154,155,158,160,162,165,166,168,171,173,175,176
%N a(n) = n + floor(ns/r) + floor(nt/r) with r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2.
%C This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
%C a(n) = n+[ns/r]+[nt/r],
%C b(n) = n+[nr/s]+[nt/s],
%C c(n) = n+[nr/t]+[ns/t], where []=floor.
%C With r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2 gives a=A189377, b=A189378, c=A189379.
%C (Conjecture) These are the numbers n such that (n+1)-sections of the Fibonacci word contain "000" (the commoner bit) but not "111" (the rarer bit). - _Don Reble_, Apr 07 2021
%C Conjecture proved April 8 2021, using the Walnut theorem prover. - _Jeffrey Shallit_, Apr 09 2021
%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 pp. 11-12.
%t r=2; s=(-1+5^(1/2))/2; t=(1+5^(1/2))/2;
%t a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t Table[a[n], {n, 1, 120}] (*A189377*)
%t Table[b[n], {n, 1, 120}] (*A189378*)
%t Table[c[n], {n, 1, 120}] (*A189379*)
%Y Cf. A189378, A189379.
%K nonn
%O 1,2
%A _Clark Kimberling_, Apr 20 2011
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