%I
%S 1,2,4,7,11,16,22,28,35,44,53,63,74,86
%N 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 This is the complementary problem to A240114: a(n) + A240114(n) = n(n+1)/2.
%C 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 Ed Wynn, <a href="https://arxiv.org/1810.12975">A comparison of encodings for cardinality constraints in a SAT solver</a>, arXiv:1810.12975 [cs.LO], 2018.
%e 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 O O O
%e O , . , . .
%e , . O , O . . O .
%e . O , . O . , O . O O .
%e 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 Cf. A227116, A240114, A319158.
%K nonn,more,hard
%O 1,2
%A _Ed Wynn_, Sep 12 2018
