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: Accepted)

n = int(input())

num = range(1, int(n ** 0.5) + 1)
from itertools import combinations_with_replacement
table = [0] * n
for i in combinations_with_replacement(num, 3):
    #print(i)
    index = (i[0] + i[1] + i[2]) ** 2 - (i[0] * i[1] + i[0] * i[2] + i[1] * i[2])

    if (i[0] + i[1] + i[2]) ** 2 - (i[0] * i[1] + i[0] * i[2] + i[1] * i[2]) > n:
        continue
    if i[0] == i[1] == i[2]:
        table[index - 1] += 1
    elif i[0] != i[1] != i[2]:
        table[index - 1] += 6
    else:
        table[index - 1] += 3
for i in range(n):
    print(table[i])