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 Python (3.8.2) (judge verdict: Wrong Answer)

import numpy as np
import math

def func(x, y, z):
    return x ** 2 + y ** 2 + z ** 2 + x*y + y*z + z*x

def count_xyz(n):
    counts = np.zeros(n + 1, dtype=int)
    for x in range(1, math.ceil(math.sqrt(n)) + 1):
        if func(x, x, x) > n:
                break
        for y in range(x, math.ceil(math.sqrt(n)) + 1):
            if func(x, y, y) > n:
                    break
            for z in range(y, math.ceil(math.sqrt(n)) + 1):
                if func(x, y, z) > n:
                    break
                if len({x, y, z}) == 1:
                    counts[func(x, y ,z)] += 1
                elif  len({x, y, z}) == 2:
                    counts[func(x, y ,z)] += 3
                else:
                    counts[func(x, y ,z)] += 6
    
    return counts


def main():
    N = int(input())
    counts = count_xyz(N)
    for i in range(1, N):
        print(counts[i])

if __name__ == "__main__":
    main()