OFFSET
1,2
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
PROG
(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)
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Kirill Ustyantsev, May 05 2018
STATUS
approved