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
Examples
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.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
Programs
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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
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Mathematica
Table[2*Sum[Floor[Sqrt[(n/2)^2 -k^2]], {k,Floor[n/2]}], {n,100}] -
PARI
a(n) = 2*sum(k=1, n\2, sqrtint((n/2)^2-k^2)); \\ Michel Marcus, Jul 27 2023
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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
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
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