

A236266


Lexicographically earliest sequence of nonnegative integers such that no three points (i,a(i)), (j,a(j)), (n,a(n)) are collinear.


7



0, 0, 1, 1, 4, 3, 8, 2, 2, 5, 7, 4, 5, 8, 16, 3, 7, 14, 12, 23, 16, 12, 25, 31, 13, 6, 11, 28, 11, 17, 9, 9, 22, 34, 6, 15, 13, 29, 23, 22, 29, 45, 26, 19, 51, 14, 24, 39, 28, 39, 18, 37, 57, 17, 38, 41, 15, 68, 32, 24, 66, 42, 10, 50, 27, 10, 53, 72, 25, 26
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OFFSET

0,5


COMMENTS

(a(n)a(j))/(nj) <> (a(j)a(i))/(ji) for all 0<=i<j<n. No value occurs more than twice. Each triangle with (distinct) vertices (i,a(i)), (j,a(j)), (n,a(n)) has area larger than zero.


LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..20000
Dániel T. Nagy, Zoltán Lóránt Nagy, and Russ Woodroofe, The extensible NoThreeInLine problem, arXiv:2209.01447 [math.CO], 2022.


FORMULA

a(n) = A236335(n+1)  1.  Alois P. Heinz, Jan 23 2014


EXAMPLE

For n=4 the value of a(n) cannot be less than 4 because otherwise we would have a set of three collinear points, {(0,0),(1,0),(4,0)} or {(2,1),(3,1),(4,1)} or {(0,0),(2,1),(4,2)} or {(1,0),(2,1),(4,3)}. Thus a(4) = 4 is the first value that is in accordance with the constraints.


MAPLE

a:= proc(n) option remember; local i, j, k, ok;
for k from 0 do ok:=true;
for j from n1 to 1 by 1 while ok do
for i from j1 to 0 by 1 while ok do
ok:= (nj)*(a(j)a(i))<>(ji)*(ka(j)) od
od; if ok then return k fi
od
end:
seq(a(n), n=0..60);


MATHEMATICA

a[0] = a[1] = 0; a[n_] := a[n] = Module[{i, j, k, ok}, For[k = 0, True, k++, ok = True; For[j = n1, ok && j >= 1, j, For[i = j1, ok && i >= 0, i, ok = (nj)*(a[j]a[i]) != (ji)*(ka[j])]]; If[ok, Return[k]]]];
Table[a[n], {n, 0, 70}] (* JeanFrançois Alcover, Jun 16 2018, after Alois P. Heinz *)


CROSSREFS

Cf. A005836, A179040, A231334, A236335, A255708, A255709.
Sequence in context: A199077 A345133 A016703 * A099289 A231892 A065628
Adjacent sequences: A236263 A236264 A236265 * A236267 A236268 A236269


KEYWORD

nonn,look


AUTHOR

Alois P. Heinz, Jan 21 2014


STATUS

approved



