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;
typedef long long ll;
int go(int x){
	int cont=0;
	for(int m=2;m*m<=(2*(x-2));m++){
		for(int n=2;n*n<=(2*(x-2));n++){
			int t=(2*x)-(m*m)-(n*n);
			if(t>=4){
				int y=sqrt(t);
				if(t==y*y){
					int a=(m+y-n)/2;
					int b=(m+n-y)/2;
					int c=(n+y-m)/2;
					int l=(a*a)+(b*b)+(c*c)+(a*b)+(a*c)+(b*c);
				    if(l==x){
						if(a>=1 && b>=1 && c>=1) cont++;
					}
				}
			}
		}
	}
	return cont;
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n; cin>>n; 
	for(int i=1;i<=n;i++){
		cout<<go(i)<<"\n";
	}
    return 0;
}