A236335 - OEIS

%I #36 Sep 14 2022 15:25:26

%S 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 12,18,10,10,23,35,7,16,14,30,24,23,30,46,27,20,52,15,25,40,29,40,19,

%U 38,58,18,39,42,16,69,33,25,67,43,11,51,28,11,54,73,26,27

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

%C An integer can't appear more than twice in the sequence, which means the sequence tends to infinity.

%C An increasing version of this sequence is A236336.

%H Grant Garcia, <a href="/A236335/b236335.txt">Table of n, a(n) for n = 1..10000</a>

%H 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 a(n) = A236266(n-1) + 1. - _Alois P. Heinz_, Jan 23 2014

%e 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 b[1] = 1;

%t b[n_] := b[n] =

%t   Min[Complement[Range[100],

%t     Select[Flatten[

%t       Table[b[k] + (n - k) (b[j] - b[k])/(j - k), {k, n - 2}, {j,

%t         k + 1, n - 1}]], IntegerQ[#] &]]]

%t Table[b[k], {k, 70}]

%Y Cf. A229037, A185256, A231334, A236266, A236336, A300002.

%K nonn

%O 1,3

%A _Tanya Khovanova_, Jan 22 2014