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A236335 Lexicographically earliest sequence of positive integers whose graph has no three collinear points.

%I A236335 #36 Sep 14 2022 15:25:26
%S A236335 1,1,2,2,5,4,9,3,3,6,8,5,6,9,17,4,8,15,13,24,17,13,26,32,14,7,12,29,
%T A236335 12,18,10,10,23,35,7,16,14,30,24,23,30,46,27,20,52,15,25,40,29,40,19,
%U A236335 38,58,18,39,42,16,69,33,25,67,43,11,51,28,11,54,73,26,27
%N A236335 Lexicographically earliest sequence of positive integers whose graph has no three collinear points.
%C A236335 An integer can't appear more than twice in the sequence, which means the sequence tends to infinity.
%C A236335 An increasing version of this sequence is A236336.
%H A236335 Grant Garcia, <a href="/A236335/b236335.txt">Table of n, a(n) for n = 1..10000</a>
%H A236335 Dániel T. Nagy, Zoltán Lóránt Nagy, and Russ Woodroofe, <a href="https://arxiv.org/abs/2209.01447">The extensible No-Three-In-Line problem</a>, arXiv:2209.01447 [math.CO], 2022.
%F A236335 a(n) = A236266(n-1) + 1. - _Alois P. Heinz_, Jan 23 2014
%e A236335 Consider a(5). The previous terms are 1,1,2,2. The value of a(5) can't be 1 because points (1,1),(2,1),(5,1) (corresponding to values a(1), a(2), a(5)) are on the same line: y=1. Points (3,2),(4,2),(5,2) are on the same line y=2, so a(5) can't be 2. Points (1,1),(3,2),(5,3) are on the same line: y=x/2+1/2, so a(5) can't be 3. Points (2,1),(3,2),(5,4) are on the same line: y=x-1, so a(5) can't be 4. Thus a(5)=5.
%t A236335 b[1] = 1;
%t A236335 b[n_] := b[n] =
%t A236335   Min[Complement[Range[100],
%t A236335     Select[Flatten[
%t A236335       Table[b[k] + (n - k) (b[j] - b[k])/(j - k), {k, n - 2}, {j,
%t A236335         k + 1, n - 1}]], IntegerQ[#] &]]]
%t A236335 Table[b[k], {k, 70}]
%Y A236335 Cf. A229037, A185256, A231334, A236266, A236336, A300002.
%K A236335 nonn
%O A236335 1,3
%A A236335 _Tanya Khovanova_, Jan 22 2014