A304035 a(n) is the number of lattice points inside a square bounded by the lines x=-n/sqrt(2), x=n/sqrt(2), y=-n/sqrt(2), y=n/sqrt(2).

1, 9, 25, 25, 49, 81, 81, 121, 169, 225, 225, 289, 361, 361, 441, 529, 625, 625, 729, 841, 841, 961, 1089, 1089, 1225, 1369, 1521, 1521, 1681, 1849, 1849, 2025, 2209, 2401, 2401, 2601, 2809, 2809, 3025, 3249, 3249, 3481, 3721, 3969, 3969, 4225, 4489, 4489, 4761, 5041, 5329, 5329, 5625, 5929, 5929

Offset: 1

Author

Kirill Ustyantsev, May 05 2018

Keywords

• nonn

Comments

If we calculate the first difference of this sequence and then substitute nonzero numbers as 1, we get exactly A080764.

If we include boundary points of the squares we get same sequence (obviously).

Duplicates appear at 4, 7, 11, 14, 18, 21, 24, 28, 31, 35, 38, 41, 45, 48, 52, 55 (= A083051 ?). - Robert G. Wilson v, Jun 20 2018

Crossrefs

Cf. A080764, A049472, A051132, A303642.

Programs

PARI

a(n) = sum(x=-n, n, sum(y=-n, n, ((2*x^2 < n^2) && (2*y^2 < n^2)))); \\ Michel Marcus, May 22 2018

Python

import math
for n in range (1, 100):
 count=0
 for x in range (-n, n):
  for y in range (-n, n):
   if ((2*x*x < n*n) and (2*y*y < n*n)):
    count=count+1
 print(count)

Formula

a(n) = A051132(n) - A303642(n).