A136483 Number of unit square lattice cells inside quadrant of origin-centered circle of diameter n.

0, 0, 1, 1, 3, 4, 6, 8, 13, 15, 19, 22, 28, 30, 37, 41, 48, 54, 64, 69, 77, 83, 94, 98, 110, 119, 131, 139, 152, 162, 172, 183, 199, 208, 226, 234, 253, 263, 281, 294, 308, 322, 343, 357, 377, 390, 412, 424, 447, 465, 488, 504, 528, 545, 567, 585, 612, 628, 654

Offset: 1

Author

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

Keywords

• easy
• nonn

Examples

			a(5) = 3 because a circle of radius 5/2 in the first quadrant encloses (2,1), (1,1), (1,2).
		

Links

• Ivan Panchenko, Table of n, a(n) for n = 1..1000

Crossrefs

Alternating merge of A136484 and A001182.

Cf. A019683, A136485, A136513.

Programs

Magma

A136483:= func< n | n eq 1 select 0 else (&+[Floor(Sqrt((n/2)^2-j^2)): j in [1..Floor(n/2)]]) >;
[A136483(n): n in [1..100]]; // G. C. Greubel, Jul 28 2023

Mathematica

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

PARI

a(n) = sum(k=1, n\2, sqrtint((n/2)^2 - k^2)); \\ Michel Marcus, Jul 28 2023

SageMath

def A136483(n): return sum(isqrt((n/2)^2-j^2) for j in range(1,(n//2)+1))
[A136483(n) for n in range(1,101)] # G. C. Greubel, Jul 28 2023

Formula

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

Lim_{n -> oo} a(n)/(n^2) -> Pi/16 (A019683).

a(n) = (1/4) * A136485(n) = (1/2) * A136513(n).

a(n) = [x^(n^2)] (theta_3(x^4) - 1)^2 / (4 * (1 - x)). - Ilya Gutkovskiy, Nov 23 2021