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, 2, 4, 7, 11, 16, 22, 28, 35, 44, 53, 63, 74, 86

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

Table of n, a(n) for n=1..14.

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 A348575

Adjacent sequences: A319156 A319157 A319158 * A319160 A319161 A319162

Keyword

nonn,more,hard

Author

Ed Wynn, Sep 12 2018

Status

approved