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;
#define mem(a,b) memset(a,b,sizeof a) ;
typedef pair<int,int> pii ;
#define vi vector<int>
#define vit vector<int> :: iterator 
#define all(ans) ans.begin(),ans.end()
//#define sz(a) int(a.size())
#define ub upper_bound
#define lb lower_bound
#define create(Name,N,M,val) vector<vector<int> > Name(N,vector<int>(M,val))
#define ull unsigned long long
#define ll long long 
#define db double 
#define ldb long double
#define fl float
#define rep(i,a,b) for(int i=a;i<b;i++)
#define lrep(i,a,b) for(ll i=a;i<b;i++)
//#define all(c) c.begin(),c.end()
#define tr(c,it) for(typeof(c.begin()) it=c.begin();it!=c.end();it++)
#define mpr make_pair
#define pb push_back
#define fr first
#define sn second
#define bpp(x) __builtin_popcountll(x)
const db PI = 3.141592653589793238462643383;
const long long INF = 1e9+7;
const long long mod = 998244353;
int getID(char c)
{
	if(c == ' ') return 0;
	return c - 'a' + 1 ;
}
/*
bool ok(string &s ,int l ,int r)
{
while(s[l]==s[r]) l++,r--;
return l>=r; 
}
int prime[]={2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89,97,101,103};
int gcd(int a, int b)
{
while(b)
{
a%=b;
swap(a,b);
}
return a;
}
*/
ll pw(ll a,ll b)
{
ll res=1;
while(b)
{
if(b&1) res=(res*a)%INF;
a=(a*a)%INF;
b>>=1;
}
return res;
}
/*
bool isFib(int n)
{
int x,y;
x=sqrt(5*n*n+4); y=sqrt(5*n*n-4);
return ((x*x)==(5*n*n+4))||((y*y)==(5*n*n-4));
}
int ExtU(int a,int b,int &x,int &y)
{
if(a==0)
{
x=0; y=1; return b;
}
int x1,y1;
int d=ExtU(b%a,a,x1,y1);
x=y1-(b/a)*x1;
y=x1;
return d;
}
ll factors(ll n)
{
ll ct=0ll;
for(ll i=1ll;i*i<=n;i++)
{
if(n%i==0)
{
if(i*i==n)
ct++;
else
ct+=2ll;
}
return ct;
}
} 
vector<int>factorize(int n)
{
vi ans;
while(n%2==0)
{
ans.pb(2); n/=2;
}
for(int i=3;i<=sqrt(n);i+=2)
{
while(n%i==0)
{
ans.pb(i); n/=i;
}
}
if(n>2)
ans.pb(n);
return ans;
}
int mnpr[500003]={0};
void sprime(int n)
{
for(int i=2;i*i<=n;i++)
{
if(mnpr[i]==0)
{
for(int j=i*i;j<=n;j+=i)
{
if(mnpr[j]==0)
mnpr[j]=i;
}
}
}
for(int i=1;i<=n;i++)
if(mnpr[i]==0)
mnpr[i]=i;
}
int mxpr(int n)
{
int mx; 
while(n&1==0)
{
mx=2; n>>=1;
}
for(int i=3;i*i<=n;i+=2)
{
while(n%i==0)
{
mx=i; n/=i;
}
}
if(n>2)
mx=n;
return mx;
}
bool seg[100005]={true};
void ss(int r,int l)
{
rep(i,0,r-l+2)
seg[i]=true;
for(int i=2;i*i<=r;i++)
{
for(int j=max(i*i,((l+i-1)/i)*i);j<=r;j+=i)
seg[j-l]=false;
}
}
int phi(int n)
{
int res=n;
for(int i=2;i*i<=n;i++)
{
if(n%i==0)
{
while(n%i==0)
n/=i;
res-=res/i;
}
}
if(n>1)
res-=res/n;
return res;
}
int ncr(int n,int r)
{
int C[r+1];
mem(C,0);
C[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=min(i,r);j;j--)
{
C[j]=(C[j]+C[j-1])%INF;
}
}
return C[r];
}
int bitzero(int n)
{
int ct=0;
while(n)
{
ct+=!(n&1);
n>>=1;
}
return ct;
}
string twocom(string &s)
{
int i;
for(i=s.length()-1;i>=0;i--)
if(s[i]=='1')
break;
if(i==-1)
{
return '1'+s;
}
for(int k=i-1;k>=0;k--)
{
if(s[k]=='1')
s[k]='0';
else
s[k]='1';
}
return s;
}
void sl()
{
int n;
goto e;
cin>>n;
e:
;
cout<<n;
return ;
}
*/
/*
void solv()
{
string s; cin>>s;
reverse(all(s));
int a[s.size()+1];
a[0]=0;
int ct[2019]={0};
ct[0]=1;
int ml=1;
rep(i,0,s.size())
{
a[i+1]=(a[i]+(s[i-1]-'0')*ml)%2019;
ml=10*ml % 2019;
ct[a[i+1]]++;
}
int ans=0;
rep(i,0,2019)
{
int x=ct[i]*(ct[i]-1)/2;
ans+=x;
}
cout<<ans;
return ;
}
*/
/*
struct sg
{
int st,end;
};
struct cmp
{
	bool operator () ( sg a, sg b )
	{
	int l1=a.end-a.st+1;
	int l2=b.end-b.st+1;
	return (l1<l2)||(l1==l2 && (a.st>b.st) );
	}
};

auto compare = [&](auto a, auto b){
return a.first<b.first || (a.first==b.first && a.second<b.second);
};
sort(bridges.begin(), bridges.end(), compare);

int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
*/
/*
int n,m,curtime=0; 
vi v[15];
bool vis[15] ;
int par[15] , dis[15] , arp[15] , lw[15];
vi brg[15];
void ini()
{
	curtime=0;
	rep(i,1,n+1)
	{
	v[i].clear();
	vis[i]=0;
	arp[i]=0;
	dis[i]=0;
	par[i]=-1;
	lw[i]=1e8;
	}
}
void dfs(int u) 
{
	vis[u]=1; dis[u]=lw[u]= (++curtime);
	int chd=0;
	rep(i,0,v[u].size())
	{
		/* if(par==v[u][i])
		continue; *
		if(!vis[v[u][i]])
		{
			chd++; 
			par[v[u][i]]=u;
			dfs(v[u][i]);
			lw[u]=min(lw[u],lw[v[u][i]]);
			if(par[u]==-1 && chd>1)
			arp[u]=1;
			if(par[u]!=-1 && lw[v[u][i]]>=dis[u]) // Means No back edge
			arp[u]=1;
			if(lw[v[u][i]]>dis[u])
			brg[u].pb(v[u][i]);
		}
		else if((v[u][i])!=par[u])
		{
		lw[u]=min(lw[u],dis[v[u][i]]);
		}
	}
}
*/
void solve()
{
int num; cin>>num;
rep(n,1,num+1)
{
int ans=0;
int x; x=(int)(sqrt(n))+1;
rep(i,1,x)
{
rep(j,i,x)
{
rep(k,j,x)
{
if( (i*i + j*j + k*k + i*k + i*j + j*k) == n)
{
if(i==j && j==k && i==k)
ans++;
else if(i==j || j==k || i==k)
ans+=3;
else
ans+=6;
}
}
}
}
cout<<ans<<"\n";
}
return;
}
int main()
{
ios_base::sync_with_stdio(0);
	cin.tie(0);
	int t=1;
//	cin>>t;
while(t--)
{
solve();
}
return 0;
}

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