%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.
%A _Ed Wynn_, Sep 12 2018