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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. 1
 1, 2, 4, 7, 11, 16, 22, 28, 35, 44, 53, 63, 74, 86 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS This is the complementary problem to A240114: a(n) + A240114(n) = n(n+1)/2. 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). LINKS Ed Wynn, A comparison of encodings for cardinality constraints in a SAT solver, arXiv:1810.12975 [cs.LO], 2018. EXAMPLE 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:         O             O             O        O ,           . ,           . .       , . O         , O .         . O .      . O , .       O . , O       . O O . 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. CROSSREFS Cf. A227116, A240114, A319158. Sequence in context: A126613 A024224 A025727 * A025702 A025717 A025726 Adjacent sequences:  A319156 A319157 A319158 * A319160 A319161 A319162 KEYWORD nonn,more,hard AUTHOR Ed Wynn, Sep 12 2018 STATUS approved

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Last modified December 3 02:27 EST 2020. Contains 338898 sequences. (Running on oeis4.)