// Arup Guha
// 3/19/2026
// Solution to 2026 UCF HS Contest Problem D: Wormhole Wizardry

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

typedef long long ll;

int main() {

    // Read the input.
    ll n;
    cin >> n;

    /*** Always want to connect to lowest available. So split into two groups
         Everyone connected to 1, and then the rest of the pairs all spaced
         out by k. Here's the work...just do arithmetic series sums...
         (1,2), (1,3), (1,4), (1,k+1), (2, k+2), (3, k+3), ... (n-k, n)
         First set is k*2/2 + k(k+3)/2 = k*(k+5)/2
         second set 2 to n-k so sum is (2+n-k)/2*(n-k-1)
         third set is k+2 to n so (k+2+n)/2*(n-k-1).
    ***/

    // Try each k.
    for (ll k=1; k<=n-1; k++) {
        ll firstSet = k*(k+5)/2;
        ll lowSum = (2+n-k)*(n-k-1)/2;
        ll hiSum = (k+2+n)*(n-k-1)/2;
        cout << firstSet+lowSum+hiSum << endl;
    }

    return 0;
}