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

1,10

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?

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))

Example

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.

Crossrefs

Sequence in context: A049060 A092462 A256357 * A057149 A087499 A100046

Adjacent sequences: A160823 A160824 A160825 * A160827 A160828 A160829

Keyword

nonn

Author

Hagen von Eitzen, May 27 2009

Status

approved