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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

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 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 November 15 01:15 EST 2019. Contains 329142 sequences. (Running on oeis4.)