∑ i = 1 n i g c d ( i , n ) \sum_{i = 1} ^{n} \frac{i}{gcd(i, n)} i=1∑ngcd(i,n)i
= ∑ d ∣ n ∑ i = 1 n i d ( g c d ( i , d ) = = d ) = \sum _{d \mid n} \sum_{i = 1} ^{n} \frac{i}{d} (gcd(i, d) == d) =d∣n∑i=1∑ndi(gcd(i,d)==d)
= ∑ d ∣ n ∑ i = 1 n d i ( g c d ( i , d ) = = 1 ) = \sum_{d\mid n} \sum_{i = 1} ^{\frac{n}{d}} i(gcd(i, d) == 1) =d∣n∑i=1∑dni(gcd(i,d)==1)
= ∑ d ∣ n n d ∗ ϕ ( n d ) 2 = \sum_{d \mid n} \frac{\frac{n}{d} * \phi(\frac{n}{d})} {2} =d∣n∑2dn∗ϕ(dn)
结果就出来了,我们只要仿照埃氏筛法,统计一遍答案就行了。
/*
Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include
#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;
inline ll read() {
ll x = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return x * f;
}
const int N = 5e6 + 10;
int phi[N], n;
bool st[N];
vector<int> prime;
ll sum[N], ans[N];
void init() {
st[0] = st[1] = 1;
phi[1] = 1, sum[1] = 1;//注意把sum[1]置为1,特判。
for(int i = 2; i < N; i++) {
if(!st[i]) {
prime.pb(i);
phi[i] = i - 1;
sum[i] = 1ll * i * phi[i] / 2;
}
for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {
st[i * prime[j]] = 1;
if(i % prime[j]) {
phi[i * prime[j]] = phi[i] * (prime[j] - 1);
sum[i * prime[j]] = 1ll * phi[i * prime[j]] * (i * prime[j]) / 2;
}
else {
phi[i * prime[j]] = phi[i] * prime[j];
sum[i * prime[j]] = 1ll * phi[i * prime[j]] * (i * prime[j]) / 2;
break;
}
}
}
for(int i = 1; i < N; i++) {
for(int j = i; j < N; j += i) {
ans[j] += sum[i];
}
}
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
init();
n = read();
for(int i = 1; i <= n; i++)
cout << ans[i] << "\n";
return 0;
}