A160826 Improvement of A125852 over A053416, A053479 and A053417.

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 4, 3, 0, 0, 0, 1, 0, 0, 0, 0, 2, 4, 5, 1, 3, 1, 0, 3, 2, 3, 4, 3, 4, 5, 6, 9, 4, 3, 0, 1, 0, 0, 0, 2, 4, 3, 4, 5, 10, 14, 3, 6, 0, 7, 0, 4, 5, 1, 8, 6, 0, 4, 7, 8, 6, 5, 11, 5, 9, 12, 12, 4, 0, 11, 7, 12, 0, 3, 1, 0, 1, 5, 0, 6, 2, 10, 11, 25, 17, 3, 2, 0, 9, 0, 12, 5, 0, 4, 2

Offset: 1

Author

Hagen von Eitzen, May 27 2009

Keywords

• nonn

Comments

How many more lattice points of a hexagonal lattice can be covered by placing a disk of diameter n at an optimal center instead of one of the three obvious centers (a lattice point, midpoint between two lattice points, barycenter of a fundamental triangle)?

The first difference occurs at n=9, when a diameter 9 disc around e.g. (1/2, 4*sqrt(5)) covers more lattice points than one around (0,0) or (1/2,0) or (1/2,sqrt(3)/6).

Clearly a(n) = O(n) as all "extra" points have norm approximately n^2/4 if the optimal center is chosen near (0,0). Does a(n)/n converge? Are there only finitely many n with a(n)=0?

Examples

			For diameters n=2,4,6,8 a disc around (0,0) and for n=1,3,5,7 a disc around(1/2,0) happens to be optimal (covers as many points as possible); therefore a(1)=a(2)=...=a(8)=0.
a(9) = A125852(9) - max(A053416(9),A053479(9),A053417(9)) = 77 - max(73,69,76) = 1.
		

Links

• H. v. Eitzen, Table of n, a(n) for n=1..1000

Formula

a(n) = A125852(n) - max(A053416(n),A053479(n),A053417(n))