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

1,5

Links

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

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

Example

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

Mathematica

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

Prog

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

Crossrefs

Alternating merge of A136484 and A001182.

Cf. A019683, A136485, A136513.

Sequence in context: A139041 A367081 A244607 * A240494 A280250 A356081

Adjacent sequences: A136480 A136481 A136482 * A136484 A136485 A136486

Keyword

easy,nonn

Author

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

Status

approved