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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
int n, k, l, r;
int ans;
signed main() {
cin >> n >> k;
ans = n * k; // 看题解的推导公式
for (l = 1; l <= n; l = r + 1) { // 枚举左端点,每次跳着走，下次的位置就是本次r的位置+1
if (k / l == 0) break; // 1、当k/l=0的时候，比如3/4=0,3/5=0,3/6=0...,以后就都是0，不用再往后算了，无贡献
r = min(k / (k / l), n); // 2、注意右端点和n取个min，>n没有贡献
ans -= (k / l) * (l + r) * (r - l + 1) / 2;
// 等差数列求和：首项:l,末项:r.项数:(r-l+1),根据公式推导，k/l = 块内值
}
cout << ans << endl;
