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: Time Limit Exceeded)

import itertools
import math
N = int(input())
n = int(math.sqrt(N/6))+2
atai = []
for num in range(1,n+1):
  atai += [num]
  
xyz = list(itertools.combinations_with_replacement(atai,3))
for num in range(1,N+1):
  count = 0
  test = 0
  for i in range(len(xyz)):
    test = (xyz[i][0]*xyz[i][0]+xyz[i][1]*xyz[i][1]+xyz[i][2]*xyz[i][2]+xyz[i][0]*xyz[i][1]+xyz[i][0]*xyz[i][2]+xyz[i][1]*xyz[i][2])
    if(test == num):
      if(xyz[i][0]==xyz[i][1]==xyz[i][2]):
        count += 1
      elif(xyz[i][0]==xyz[i][1]) or (xyz[i][0]==xyz[i][2]) or (xyz[i][1]==xyz[i][2]):
        count += 3
      else:
        count += 6
  print(count)