A227116 Given an equilateral triangular grid with side n, containing n(n+1)/2 points, a(n) is the minimal number of points to be removed from the grid, so that, if 3 of the remaining points are chosen, they do not form an equilateral triangle with sides parallel to the grid.

0, 1, 2, 4, 7, 9, 14, 18, 23, 29, 36, 44, 52, 61, 71

Offset: 1

Author

Heinrich Ludwig, Jul 01 2013

Keywords

• nonn
• hard
• more

Comments

This is the complementary problem to A227308.

Numbers found by an exhaustive computational search for all solutions (see history).

Examples

			n = 11: at least a(11) = 36 points (.) out of the 66 have to be removed, leaving 30 (X) behind:
              .
             X X
            X . X
           X . . X
          X . . . X
         X . . . . X
        . X X . X X .
       . X . X X . X .
      . . X X . X X . .
     X . . . . . . . . X
    . X X X . . . X X X .
There is no equilateral subtriangle with all vertices = X and sides parallel to the whole triangle.
		

Links

• Heinrich Ludwig, Illustration of a(2)..a(15)
• Ed Wynn, A comparison of encodings for cardinality constraints in a SAT solver, arXiv:1810.12975 [cs.LO], 2018.

Crossrefs

Cf. A227308, A152125, A227133

Formula

a(n) + A227308(n) = n(n+1)/2.

Extensions

Added a(12), a(13), Heinrich Ludwig, Sep 02 2013

Added a(14), Giovanni Resta, Sep 19 2013

a(15) from Heinrich Ludwig, Oct 27 2013