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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int sq = 0;


        for (int a=1;a<=n;a++) {
            long ans = 0;
            sq = (int)Math.sqrt(n) + 22;
            // 配列
            for(int x=1;x<=sq;x++){
                for(int y=x;x+y<=sq;y++){
                    for(int z=y;x+y+z<=sq;z++){
                        if(x*(x+y+z) + y*(y+z) + z*z == a) {
                            if(x!=y && x!=z && y!=z) {
                                ans+=6;
                            } else if(x==y && x==z) {
                                ans++;
                            } else {
                                ans+=3;
                            }
                        }
                    }
                }
            }
            System.out.println(ans);
        }
    }
}