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

EXAMPLE

CROSSREFS

KEYWORD

nonn

AUTHOR

Hagen von Eitzen, May 27 2009

STATUS

approved