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A227116 Given an equilateral triangular grid with side n, containing n(n+1)/2 points, a(n) is the minimal number of points to be removed from the grid, so that, if 3 of the remaining points are chosen, they do not form an equilateral triangle with sides parallel to the grid.

%I A227116 #66 Jul 07 2023 14:55:46
%S A227116 0,1,2,4,7,9,14,18,23,29,36,44,52,61,71
%N A227116 Given an equilateral triangular grid with side n, containing n(n+1)/2 points, a(n) is the minimal number of points to be removed from the grid, so that, if 3 of the remaining points are chosen, they do not form an equilateral triangle with sides parallel to the grid.
%C A227116 This is the complementary problem to A227308.
%C A227116 Numbers found by an exhaustive computational search for all solutions (see history).
%H A227116 Heinrich Ludwig, <a href="/A227116/a227116.png">Illustration of a(2)..a(15)</a>
%H A227116 Ed Wynn, <a href="https://arxiv.org/abs/1810.12975">A comparison of encodings for cardinality constraints in a SAT solver</a>, arXiv:1810.12975 [cs.LO], 2018.
%F A227116 a(n) + A227308(n) = n(n+1)/2.
%e A227116 n = 11: at least a(11) = 36 points (.) out of the 66 have to be removed, leaving 30 (X) behind:
%e A227116               .
%e A227116              X X
%e A227116             X . X
%e A227116            X . . X
%e A227116           X . . . X
%e A227116          X . . . . X
%e A227116         . X X . X X .
%e A227116        . X . X X . X .
%e A227116       . . X X . X X . .
%e A227116      X . . . . . . . . X
%e A227116     . X X X . . . X X X .
%e A227116 There is no equilateral subtriangle with all vertices = X and sides parallel to the whole triangle.
%Y A227116 Cf. A227308, A152125, A227133
%K A227116 nonn,hard,more
%O A227116 1,3
%A A227116 _Heinrich Ludwig_, Jul 01 2013
%E A227116 Added a(12), a(13), _Heinrich Ludwig_, Sep 02 2013
%E A227116 Added a(14), _Giovanni Resta_, Sep 19 2013
%E A227116 a(15) from _Heinrich Ludwig_, Oct 27 2013