OFFSET
0,3
COMMENTS
Lim_{n->infinity} a(n)/n^2 = Pi/4.
a(n) gives the number of positive half-points (for example, 1/2 and 3/2) inside or on the circle of radius n. - Jon Perry, Nov 04 2012
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
EXAMPLE
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
KEYWORD
nonn
AUTHOR
Hugo Pfoertner, Jul 12 2006
EXTENSIONS
a(0) added by Jon Perry, Nov 04 2012
STATUS
approved