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n+[nr/t]+[ns/t]; r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2.
3

%I #19 Feb 14 2024 17:27:52

%S 2,4,7,9,12,15,17,20,23,25,28,30,33,36,38,41,44,46,49,51,54,57,59,62,

%T 64,67,70,72,75,78,80,83,85,88,91,93,96,98,101,104,106,109,112,114,

%U 117,119,122,125,127,130,133,135,138,140,143,146,148,151,153,156,159,161,164,167,169,172,174,177,180,182,185,187,190,193,195,198,201,203,206,208,211,214,216,219

%N n+[nr/t]+[ns/t]; r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2.

%C (Conjecture) These are the numbers n such that (n+1)-sections of the Fibonacci word contain both "000" and "111". - _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 (See A189377.)

%Y Cf. A189377, A189378.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 20 2011