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A160826 Improvement of A125852 over A053416, A053479 and A053417 1


%S 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,

%T 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,

%U 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

%N Improvement of A125852 over A053416, A053479 and A053417

%C 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)?

%C 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).

%C 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?

%H H. v. Eitzen, <a href="/A160826/b160826.txt">Table of n, a(n) for n=1..1000</a>

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

%e 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.

%e a(9) = A125852(9) - max(A053416(9),A053479(9),A053417(9)) = 77 - max(73,69,76) = 1.

%K nonn

%O 1,10

%A _Hagen von Eitzen_, May 27 2009

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