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A120883 (1/4)*number of lattice points with odd indices in a square lattice inside a circle around the origin with radius 2*n. 1

%I

%S 0,1,3,8,13,20,28,39,52,64,79,96,112,135,154,179,203,228,255,281,316,

%T 349,382,416,451,494,532,573,618,661,707,756,807,859,910,963,1015,

%U 1076,1137,1198,1256,1321,1386,1452,1523,1594,1667,1737,1808,1889,1965,2046,2123

%N (1/4)*number of lattice points with odd indices in a square lattice inside a circle around the origin with radius 2*n.

%C Lim_{n->infinity} a(n)/n^2 = Pi/4.

%C 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

%F 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 non-positive 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

%e 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.

%e 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

%o (JavaScript)

%o for (i=0;i<50;i++) {

%o c=0;

%o for (a=1/2;a<=i;a++)

%o for (b=1/2;b<=i;b++)

%o if (Math.pow(a,2)+Math.pow(b,2)<=Math.pow(i,2)) c++;

%o document.write(c+", ");

%o }

%Y Cf. A001182.

%K nonn

%O 0,3

%A _Hugo Pfoertner_, Jul 12 2006

%E a(0) added by _Jon Perry_, Nov 04 2012

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Last modified July 20 15:59 EDT 2019. Contains 325185 sequences. (Running on oeis4.)