A237526 a(n) = number of unit squares in the first quadrant of a disk of radius sqrt(n) centered at the origin of a square lattice.

0, 0, 1, 1, 1, 3, 3, 3, 4, 4, 6, 6, 6, 8, 8, 8, 8, 10, 11, 11, 13, 13, 13, 13, 13, 15, 17, 17, 17, 19, 19, 19, 20, 20, 22, 22, 22, 24, 24, 24, 26, 28, 28, 28, 28, 30, 30, 30, 30, 30, 33, 33, 35, 37, 37, 37, 37, 37, 39, 39, 39, 41, 41, 41, 41, 45, 45, 45, 47, 47

Offset: 0

Author

L. Edson Jeffery, Feb 09 2014

Keywords

• nonn

Links

• Stefano Spezia, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000404, A001481, A057961, A232499.

Partial sums of A063725.

Programs

Mathematica

a[n_]:=Sum[Floor[Sqrt[n-k^2]],{k,Floor[Sqrt[n]]}]; Array[a,70,0] (* Stefano Spezia, Jul 19 2024 *)

PARI

A237526(n)=sum(k=1,sqrtint(n),sqrtint(n-k^2)) \\ M. F. Hasler, Feb 09 2014

Python

from math import isqrt
def A237526(n): return sum(isqrt(n-k**2) for k in range(1,isqrt(n)+1)) # Chai Wah Wu, Jul 18 2024

Formula

a(A000404(n)) = A232499(n).

a(n) = Sum_{k=1..floor(sqrt(n))} floor(sqrt(n-k^2)). - M. F. Hasler, Feb 09 2014

G.f.: (theta_3(x) - 1)^2/(4*(1 - x)), where theta_3() is the Jacobi theta function. - Ilya Gutkovskiy, Apr 17 2018

For n > 1, Pi*(n+2-sqrt(8n)) < a(n) < Pi*n. (This is trivial and can probably be improved by methods like Euler-Maclaurin and perhaps even a modification of the Dirichlet hyperbola method.) - Charles R Greathouse IV, Jul 17 2024