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A136513
Number of unit square lattice cells inside half-plane (two adjacent quadrants) of origin centered circle of diameter n.
5
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
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.
Sequence in context: A320067 A248823 A284616 * A365662 A214932 A322132
KEYWORD
easy,nonn
AUTHOR
Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008
STATUS
approved