A120883 (1/4)*number of lattice points with odd indices in a square lattice inside a circle around the origin with radius 2*n.
0, 1, 3, 8, 13, 20, 28, 39, 52, 64, 79, 96, 112, 135, 154, 179, 203, 228, 255, 281, 316, 349, 382, 416, 451, 494, 532, 573, 618, 661, 707, 756, 807, 859, 910, 963, 1015, 1076, 1137, 1198, 1256, 1321, 1386, 1452, 1523, 1594, 1667, 1737, 1808, 1889, 1965, 2046, 2123
Offset: 0
Keywords
Examples
a(3)=8 because the 8 lattice points in the first quadrant (x,y) = {(1,1), (1,3), (3,1), (1,5), (5,1), (3,3), (3,5), (5,3)} all satisfy x^2 + y^2 < (2*3)^2.
a(3)=8 because (1/2,1/2), (1/2,3/2), (1/2,5/2), (3/2,1/2), (3/2,3/2), (3/2,5/2), (5/2,1/2) and (5/2,3/2) all satisfy x^2 + y^2 <= n^2. - _Jon Perry_, Nov 04 2012
Crossrefs
Cf. A001182.
Formula
a(n) = Sum_{i=1..n} ceiling(sqrt(n^2 - (i - 1/2)^2) - 1/2). Proof outline: consider an integer grid (i,j), e.g., a pixel image. A positive half circle hull's boundary of radius n contains all points ('pixels') where (i - 1/2)^2 + (j - 1/2)^2 = n^2 => j = f(j) = sqrt(n^2 - (i - 1/2)^2) + 1/2. To obtain the number of elements of the hull's closure without nonpositive points, count by upper Riemann sums with interval length 1: (n) = A(n) = Sum_{i=1..n} (ceiling(f(i)) - 1). ('i=1' discards the (0,j) points and '-1' cancels the (i,0) points.) - Johannes Hoentsch, Feb 26 2019
Extensions
a(0) added by Jon Perry, Nov 04 2012
Comments