Problem:
Score : 300 points

Problem Statement

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions:

1 \leq x,y,z

x^2 + y^2 + z^2 + xy + yz + zx = n

Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

Constraints

All values in input are integers.

1 \leq N \leq 10^4

Input

Input is given from Standard Input in the following format:

N

Output

Print N lines. The i-th line should contain the value f(i).

Sample Input 1

20

Sample Output 1

0
0
0
0
0
1
0
0
0
0
3
0
0
0
0
0
3
3
0
0

For n=6, only (1,1,1) satisfies both of the conditions. Thus, f(6) = 1.

For n=11, three triples, (1,1,2), (1,2,1), and (2,1,1), satisfy both of the conditions. Thus, f(6) = 3.

For n=17, three triples, (1,2,2), (2,1,2), and (2,2,1), satisfy both of the conditions. Thus, f(17) = 3.

For n=18, three triples, (1,1,3), (1,3,1), and (3,1,1), satisfy both of the conditions. Thus, f(18) = 3.

Solution in PyPy3 (7.3.0) (judge verdict: Accepted)

from math import sqrt
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline

N = int(input())

def solv_quadratic_equation(a, b, c):
    D2 = b**2 - 4*a*c
    if D2<0:
        return 0.1, 0.1
    D = sqrt(D2)
    x_1 = (-b + D) / (2 * a)
    x_2 = (-b - D) / (2 * a)

    return x_1, x_2

for n in range(1,N+1):
    ans=0
    if n < 6:
        print(0)
        continue
    for x in range(1,int(sqrt(n-3))+1):
        for y in range(1,int(sqrt(n-3))+1):
            z1,z2=solv_quadratic_equation(1,x+y, x**2+y**2+x*y-n)
            if z1.is_integer() and z1>=1:
                ans+=1
            if z2.is_integer() and z2>=1:
                ans+=1

    print(ans)