分析：这题都过了2000了，应该很简单。。写这篇只是为了凑篇数= =
假设在第i级的时候开方过后的数为i∗t[i]，t[i]是第i级的系数。那么
(3t[3])2−(2t[2])≡0(mod2)
(4t[4])2−(3t[3])≡0(mod3)
(5t[5])2−(4t[4])≡0(mod4)
…
(3t[3])2≡0(mod2)
(4t[4])2≡0(mod3)
(5t[5])2≡0(mod4)
显然，最小的情况应该就是t[i]=i−1, 化简一下公式，在i的情况下应该是i∗i∗(i−1)−(i−2)，注意i=2时特殊情况，应该为2。
/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define offcin ios::sync_with_stdio(false)
#define sigma_size 26
#define lson l,m,v<<1
#define rson m+1,r,v<<1|1
#define slch v<<1
#define srch v<<1|1
#define sgetmid int m = (l+r)>>1
#define LL long long
#define ull unsigned long long
#define mem(x,v) memset(x,v,sizeof(x))
#define lowbit(x) (x&-x)
#define bits(a) __builtin_popcount(a)
#define mk make_pair
#define pb push_back
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const LL INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const LL mod = 1e9+7;
const int maxmat = 10;
const ull BASE = 31;
/*****************************************************/
int main(int argc, char const *argv[]) {
int N;
cin>>N;
for (int i = 2; i <= N + 1; i ++) {
if (i == 2) puts("2");
else printf("%I64d\n", 1LL * i * i * (i -1) - 1LL * (i - 2));
}
return 0;
}