A136513 Number of unit square lattice cells inside half-plane (two adjacent quadrants) of origin centered circle of diameter n.

0, 0, 2, 2, 6, 8, 12, 16, 26, 30, 38, 44, 56, 60, 74, 82, 96, 108, 128, 138, 154, 166, 188, 196, 220, 238, 262, 278, 304, 324, 344, 366, 398, 416, 452, 468, 506, 526, 562, 588, 616, 644, 686, 714, 754, 780, 824, 848, 894, 930, 976, 1008, 1056, 1090, 1134, 1170

Offset: 1

Author

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

Keywords

• easy
• nonn

Examples

			a(3) = 2 because a circle centered at the origin and of radius 3/2 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

Alternating merge of A136514 and A136515.

Cf. A136483, A136485.

Programs

Magma

A136513:= func< n | n eq 1 select 0 else 2*(&+[Floor(Sqrt((n/2)^2-j^2)): j in [1..Floor(n/2)]]) >;
[A136513(n): n in [1..100]]; // G. C. Greubel, Jul 27 2023

Mathematica

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

PARI

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

SageMath

def A136513(n): return 2*sum(isqrt((n/2)^2-k^2) for k in range(1,(n//2)+1))
[A136513(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..floor(n/2)} floor(sqrt((n/2)^2 - k^2)).

a(n) = 2 * A136483(n).

a(n) = (1/2) * A136485(n).

a(n) = [x^(n^2)] (theta_3(x^4) - 1)^2 / (2 * (1 - x)). - Ilya Gutkovskiy, Nov 24 2021