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 <iostream>
#include<time.h>
#include<string>
#include<cstring>
#include <cstdio>
#include<utility>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
#include<bitset>
#include<queue>
#include<stack>
#include<set>
#include <limits>
#include <climits>
#include <numeric>
#include<cassert>
#include<cctype>
#include<iterator>
#include<unordered_map>
using namespace std;
typedef long long ll;
typedef unsigned long long LL;
typedef vector<ll>VI;
typedef vector<bool>VB;
typedef pair<ll,ll>P;
typedef pair<double,double>P_D;
#define VV(T) vector<vector<T>>
#define PI 3.1415926535897932384626433832795
#define rep(i, n) for (ll i = 0; i < (ll)n; i++)
#define reprev(i, n) for (ll i = (ll)n - 1; i >= 0; i--)
#define bitrep(bit,A) for(int bit=A; ;bit=(bit-1)&A)
#define ALL(a)  (a).begin(),(a).end()
#define SZ(a) (ll)((a).size())
#define c_max(a, b) (((ll)a)>((ll)b)?(a):(b))
#define c_min(a,b) (((ll)a)<((ll)b)?(a):(b))
#define chmax(x,a) x=c_max(x,a)
#define chmin(x,a) x=c_min(x,a)
#define vmax(v) *max_element(ALL(v))
#define vmin(v) *min_element(ALL(v))
#define SORT(c) stable_sort(ALL(c))
#define $(x) {cout<<#x<<" = " <<(x)<<endl;}
#define fi first
#define se second
#define MAX 100100//5
#define MAX2 200100
#define MAX_6 1001001//6
#define MAX_7 10010010//7
#define SENTINEL 2000000000//9
#define NIL -1
const int MOD = 1000000007;
#define INF 1<<30
#define INFTY 1LL<<61
#define MAX_INT INT_MAX
#define Endl '\n'
#define greater greater<int>()
inline ll GCD(ll a,ll b){return b?GCD(b,a%b):a;}
inline ll lcm(ll a,ll b){return a*b/GCD(a,b);}
using Edge=pair<int,ll>;
using Graph=vector<vector<ll> >;//ll or Edge
using Edge=pair<int,ll>;
//const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
// g++ -o a a.cpp -Wall -lm -std=c++14

int main(){
    ll n;cin>>n;
    ll bottle[MAX];
    memset(bottle,0,sizeof(bottle));
    for(ll x=1;x<=n;x++){
        for(ll y=x;y<=n;y++){
            for(ll z=y;z<=n;z++){
                ll i=x*x+y*y+z*z+x*y+z*(x+y);
                if(i>n)break;
                if(z==x && y==x)bottle[i]++;
                else if(y==z || y==x)bottle[i]+=3;
                else bottle[i]+=6;
            }
        }
    }
    for(int i=1;i<=n;i++){
        cout<<bottle[i]<<endl;
    }
}