A136514 Number of unit square lattice cells inside half-plane (two adjacent quadrants) of origin centered circle of radius n.

0, 2, 8, 16, 30, 44, 60, 82, 108, 138, 166, 196, 238, 278, 324, 366, 416, 468, 526, 588, 644, 714, 780, 848, 930, 1008, 1090, 1170, 1256, 1350, 1438, 1540, 1638, 1744, 1856, 1954, 2072, 2180, 2310, 2432, 2548, 2678, 2808, 2950, 3090, 3220, 3366, 3510, 3664

Offset: 1

Author

Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008

Keywords

• easy
• nonn

Examples

			a(2) = 2 because a circle centered at the origin and of radius 2 encloses (-1,1) and (1,1) in the upper half plane.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001182, A136513.

Programs

Magma

A136514:= func< n | n eq 1 select 0 else 2*(&+[Floor(Sqrt(n^2-j^2)): j in [1..n-1]]) >;
[A136514(n): n in [1..100]]; // G. C. Greubel, Jul 27 2023

Mathematica

Table[2*Sum[Floor[Sqrt[n^2 -k^2]], {k,n-1}], {n,100}]

PARI

a(n) = 2*sum(k=1, n-1, sqrtint(n^2-k^2)); \\ Michel Marcus, Jul 27 2023

SageMath

def A136514(n): return 2*sum(isqrt(n^2-k^2) for k in range(1,n))
[A136514(n) for n in range(1,101)] # G. C. Greubel, Jul 27 2023

Formula

Lim_{n -> oo} a(n)/(n^2) -> Pi/8.

a(n) = 2 * Sum_{k=1..n-1} floor(sqrt(n^2 - k^2)).

a(n) = A136513(2*n).

a(n) = 2*A001182(n). - R. J. Mathar, Jan 10 2008