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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

long f(long x, long y, long z){
	return pow(x+y+z,2)-x*y-y*z-z*x;
}

int main(){
	long N;
	cin >> N;

	for(int i=1; i<=N; i++){
		long cnt=0;
		long max = sqrt(i);
		for(long x=1; x <= max; x++){
			for(long y=1; y<=x; y++){
				for(long z=1; z<=y; z++){
					if(f(x,y,z) == i){
						if(x==y){
							if(y==z){
								cnt++;
							}else{
								cnt=cnt+3;
							}
						}else if(y==z){
							if(z==x){
								cnt++;
							}else{
								cnt=cnt+3;
							}
						}else if(z==x){
							if(x==y){
								cnt++;
							}else{
								cnt=cnt+3;
							}
						}else{
							cnt=cnt+6;
						}
					}
				}
			}
		}
		cout << cnt << endl;
	}
}