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A136514
Number of unit square lattice cells inside half-plane (two adjacent quadrants) of origin centered circle of radius n.
3
0, 2, 8, 16, 30, 44, 60, 82, 108, 138, 166, 196, 238, 278, 324, 366, 416, 468, 526, 588, 644, 714, 780, 848, 930, 1008, 1090, 1170, 1256, 1350, 1438, 1540, 1638, 1744, 1856, 1954, 2072, 2180, 2310, 2432, 2548, 2678, 2808, 2950, 3090, 3220, 3366, 3510, 3664
OFFSET
1,2
LINKS
FORMULA
Lim_{n -> oo} a(n)/(n^2) -> Pi/8.
a(n) = 2 * Sum_{k=1..n-1} floor(sqrt(n^2 - k^2)).
a(n) = A136513(2*n).
a(n) = 2*A001182(n). - R. J. Mathar, Jan 10 2008
EXAMPLE
a(2) = 2 because a circle centered at the origin and of radius 2 encloses (-1,1) and (1,1) in the upper half plane.
MATHEMATICA
Table[2*Sum[Floor[Sqrt[n^2 -k^2]], {k, n-1}], {n, 100}]
PROG
(Magma)
A136514:= func< n | n eq 1 select 0 else 2*(&+[Floor(Sqrt(n^2-j^2)): j in [1..n-1]]) >;
[A136514(n): n in [1..100]]; // G. C. Greubel, Jul 27 2023
(SageMath)
def A136514(n): return 2*sum(isqrt(n^2-k^2) for k in range(1, n))
[A136514(n) for n in range(1, 101)] # G. C. Greubel, Jul 27 2023
(PARI) a(n) = 2*sum(k=1, n-1, sqrtint(n^2-k^2)); \\ Michel Marcus, Jul 27 2023
CROSSREFS
Sequence in context: A137882 A194643 A327329 * A077071 A187216 A210729
KEYWORD
easy,nonn
AUTHOR
Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008
STATUS
approved