A136484 Number of unit square lattice cells inside quadrant of origin centered circle of diameter 2n+1.

0, 1, 3, 6, 13, 19, 28, 37, 48, 64, 77, 94, 110, 131, 152, 172, 199, 226, 253, 281, 308, 343, 377, 412, 447, 488, 528, 567, 612, 654, 703, 750, 796, 847, 902, 957, 1013, 1068, 1129, 1187, 1252, 1313, 1378, 1446, 1511, 1582, 1650, 1725, 1800, 1877, 1955, 2034

Offset: 0

Author

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

Keywords

• easy
• nonn

Comments

Number of unit square lattice cells inside quadrant of origin centered circle of radius n+1/2.

Examples

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A019683, A136483, A136486, A136515.

Programs

Magma

A136484:= func< n | n eq 0 select 0 else (&+[Floor(Sqrt((n+1/2)^2-j^2)): j in [1..n]]) >;
[A136484(n): n in [0..100]]; // G. C. Greubel, Jul 29 2023

Mathematica

Table[Sum[Floor[Sqrt[(n+1/2)^2 - k^2]], {k,n}], {n,0,100}]

SageMath

def A136484(n): return sum(floor(sqrt((n+1/2)^2-k^2)) for k in range(1, n+1))
[A136484(n) for n in range(101)] # G. C. Greubel, Jul 29 2023

Formula

a(n) = Sum_{k=1..n} floor(sqrt((n+1/2)^2 - k^2)).

a(n) = (1/2) * A136515(n).

a(n) = (1/4) * A136486(n).

a(n) = A136483(2*n+1).

Lim_{n -> oo} a(n)/(n^2) -> Pi/16 (A019683).