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A160826 Improvement of A125852 over A053416, A053479 and A053417 1
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified December 16 09:34 EST 2018. Contains 318160 sequences. (Running on oeis4.)