1,2

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).

Ed Wynn, A comparison of encodings for cardinality constraints in a SAT solver, arXiv:1810.12975 [cs.LO], 2018.

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.

Cf. A227116, A240114, A319158.

Sequence in context: A126613 A024224 A025727 * A025702 A025717 A348575

Adjacent sequences: A319156 A319157 A319158 * A319160 A319161 A319162

nonn,more,hard

Ed Wynn, Sep 12 2018

approved