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

1,3

Links

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

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

Example

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.

Mathematica

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

Prog

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

Crossrefs

Alternating merge of A136514 and A136515.

Cf. A136483, A136485.

Sequence in context: A320067 A248823 A284616 * A365662 A214932 A322132

Adjacent sequences: A136510 A136511 A136512 * A136514 A136515 A136516

Keyword

easy,nonn

Author

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

Status

approved