A123689 Number of points in a square lattice covered by a circle of diameter n if the center of the circle is chosen such that the circle covers the minimum possible number of lattice points.

0, 2, 4, 10, 16, 26, 32, 46, 60, 74, 88, 108, 124, 146, 172, 194, 216, 248, 276, 308

Offset: 1

Author

Hugo Pfoertner, Oct 09 2006

Keywords

• more
• nonn

Comments

a(n)<=min(A053411(n),A053414(n),A053415(n)).

Using brute force computation and a step size of 1/1000 (though 1/200 suffices), the [conjectured] terms a(21) to a(40) would be: 332, 374, 408, 438, 484, 522, 560, 608, 648, 698, 740, 794, 846, 894, 952, 1006, 1060, 1124, 1184, 1248. - Jean-François Alcover, Jan 08 2018

Examples

			a(1)=0: Circle with diameter 1 with center (0.5,0.5) covers no lattice points; a(2)=2: Circle with diameter 2 with center (0,eps) covers 2 lattice points;
a(3)=4: Circle with diameter 3 with center (0.5,0.5) covers 4 lattice points.
		

Links

• Hugo Pfoertner, Minimal number of points in the square lattice covered by circular disks. Illustrations.

Crossrefs

Cf. A123690, A053411, A053414, A053415, A122224, A291259.

The corresponding sequences for the hexagonal lattice and the honeycomb net are A125851 and A127405, respectively.

Programs

Mathematica

dx = 1/200; y0 = 0; (* To speed up computation, the step size dx is experimentally adjusted and the circle center is taken on the x-axis. *)
cnt[pts_, ctr_, r_] := Count[pts, pt_ /; Norm[pt - ctr] <= r];
a[n_] := Module[{r, pts, innerCnt, an, center}, r = n/2; pts = Select[ Flatten[ Table[{x, y}, {x, -r - 1, r + 1}, {y, -r - 1, r + 1}], 1], r - 1 <= Norm[#] <= r + 1 &]; innerCnt = Sum[If[Norm[{x, y}] < r - 1, 1, 0], {x, -r - 1, r + 1}, {y, -r - 1, r + 1}]; {an, center} = Table[{innerCnt + cnt[pts, {x, y0}, r], {x, y0}}, {x, -1/2, 1/2, dx}] // Sort // First; Print["a(", n, ") = ", an, ", center = ", center // InputForm]; an];
Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Jan 08 2018 *)