A242118 Number of unit squares that intersect the circumference of a circle of radius n centered at (0,0).

0, 4, 12, 20, 28, 28, 44, 52, 60, 68, 68, 84, 92, 92, 108, 108, 124, 124, 140, 148, 148, 164, 172, 180, 188, 180, 196, 212, 220, 220, 228, 244, 252, 260, 260, 268, 284, 284, 300, 300, 308, 316, 332, 340, 348, 348, 364, 372, 380, 388, 380

Offset: 0

Author

Kival Ngaokrajang, May 05 2014

Keywords

• nonn

Comments

For the points that form the Pythagorean triple (for example see illustration n = 5, on the first quadrant at coordinate (4,3) and (3,4)), the transit of circumference occurs exactly at the corners, therefore there are no additional intersecting squares on the upper or lower rows (diagonally NE & SW directions) of such points.

If the center of the circle is instead chosen at the middle of a square grid centered at (1/2,0), the sequence will be 2*A004767(n-1).

Links

• Kival Ngaokrajang, Illustration of initial terms

Crossrefs

Cf. A009003, A004767.

Programs

Python

a = lambda n: sum(4 for x in range(n) for y in range(n)
                    if x**2 + y**2 < n**2 and (x+1)**2 + (y+1)**2 > n**2)

Python

from sympy import factorint
def a(n):
    r = 1
    for p, e in factorint(n).items():
        if p%4 == 1: r *= 2*e + 1
    return 8*n - 4*r if n > 0 else 0

Formula

a(n) = 4*Sum{k=1..n} ceiling(sqrt(n^2 - (k-1)^2)) - floor(sqrt(n^2 - k^2)). - Orson R. L. Peters, Jan 30 2017

a(n) = 8*n - A046109(n) for n > 0. - conjectured by Orson R. L. Peters, Jan 30 2017, proved by Andrey Zabolotskiy, Jan 31 2017

Extensions

Terms corrected by Orson R. L. Peters, Jan 30 2017