|
%I #40 Sep 14 2022 15:26:18
%S 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,
%T 11,17,9,9,22,34,6,15,13,29,23,22,29,45,26,19,51,14,24,39,28,39,18,37,
%U 57,17,38,41,15,68,32,24,66,42,10,50,27,10,53,72,25,26
%N Lexicographically earliest sequence of nonnegative integers such that no three points (i,a(i)), (j,a(j)), (n,a(n)) are collinear.
%C (a(n)-a(j))/(n-j) <> (a(j)-a(i))/(j-i) 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.
%H Alois P. Heinz, <a href="/A236266/b236266.txt">Table of n, a(n) for n = 0..20000</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) = A236335(n+1) - 1. - _Alois P. Heinz_, Jan 23 2014
%e 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.
%p a:= proc(n) option remember; local i, j, k, ok;
%p for k from 0 do ok:=true;
%p for j from n-1 to 1 by -1 while ok do
%p for i from j-1 to 0 by -1 while ok do
%p ok:= (n-j)*(a(j)-a(i))<>(j-i)*(k-a(j)) od
%p od; if ok then return k fi
%p od
%p end:
%p seq(a(n), n=0..60);
%t a[0] = a[1] = 0; a[n_] := a[n] = Module[{i, j, k, ok}, For[k = 0, True, k++, ok = True; For[j = n-1, ok && j >= 1, j--, For[i = j-1, ok && i >= 0, i--, ok = (n-j)*(a[j]-a[i]) != (j-i)*(k-a[j])]]; If[ok, Return[k]]]];
%t Table[a[n], {n, 0, 70}] (* _Jean-François Alcover_, Jun 16 2018, after _Alois P. Heinz_ *)
%Y Cf. A005836, A179040, A231334, A236335, A255708, A255709.
%K nonn,look
%O 0,5
%A _Alois P. Heinz_, Jan 21 2014
|
