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A319159 Given an equilateral triangular grid with side n, containing n(n+1)/2 points, a(n) is the minimal number of points to be selected, such that any equilateral triangle of points will include at least one of the selection.

%I A319159 #23 Aug 08 2020 11:11:12
%S A319159 1,2,4,7,11,16,22,28,35,44,53,63,74,86
%N A319159 Given an equilateral triangular grid with side n, containing n(n+1)/2 points, a(n) is the minimal number of points to be selected, such that any equilateral triangle of points will include at least one of the selection.
%C A319159 This is the complementary problem to A240114: a(n) + A240114(n) = n(n+1)/2.
%C A319159 This is the same problem as A227116 and A319158, except that here the triangles may have any orientation.  Due to the additional requirements, a(n) >= A227116(n) >= A319158(n).
%H A319159 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.
%e A319159 For n=4, this sequence has the same value a(4)=4 as A227116 and A319158, but if we look at the three solutions to those sequences (unique up to symmetry), representing selected points by O:
%e A319159         O             O             O
%e A319159        O ,           . ,           . .
%e A319159       , . O         , O .         . O .
%e A319159      . O , .       O . , O       . O O .
%e A319159 We see that only the last of these is a solution here -- the others have rotated triangles not including any selected point (for example, as shown with commas).  The last selection is therefore the unique solution (up to symmetry) for a(4)=4.
%Y A319159 Cf. A227116, A240114, A319158.
%K A319159 nonn,more,hard
%O A319159 1,2
%A A319159 _Ed Wynn_, Sep 12 2018