$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n\gcd (a_i,a_j)\gcd (i,j) \\ \sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}\gcd (a_{id},a_{jd})[\gcd (i,j)=1] \\ \sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}\mu (k)\sum _{i=1}^{\frac{n}{kd}}\sum _{j=1}^{\frac{n}{kd}}\gcd (a_{ikd},a_{jkd}) \\ T=kd \\ \sum _{T=1}^n\sum _{i=1}^{\frac{n}{T}}\sum _{j=1}^{\frac{n}{T}}\gcd (a_{iT},a_{jT})\sum _{d\mid T}d\mu (\frac{T}{d}) \\ \sum _{T=1}^n\varphi (T)\sum _{i=1}^{\frac{n}{T}}\sum _{j=1}^{\frac{n}{T}}\gcd (a_{iT},a_{jT}) \end{gathered}$$
我们考虑设$f(n,T)=\sum _{i=1}^{\frac{n}{T}}\sum _{j=1}^{\frac{n}{T}}\gcd (a_{iT},a_{jT})$，$g(x)$为$i\in [T,2T,\dots ,\frac{n}{T}T]$时$x$的出现次数。
$$\begin{gathered} f(n,T)=\sum _{i=1}^m\sum _{j=1}^{m}g(i)g(j)\gcd (i,j),(m=10^5) \\ \sum _{d=1}^{m}d\sum _{i=1}^{\frac{m}{d}}\sum _{j=1}^{\frac{m}{d}}g(id)g(jd)[\gcd (i,j)=1] \\ \sum _{d=1}^{m}d\sum _{k=1}^{\frac{n}{d}}\mu (k){\left(\sum _{i=1}^{\frac{m}{kd}}g(ikd)\right)}^2 \\ T=kd \\ \sum _{T=1}^m\varphi (T){\left(\sum _{i=1}^{\frac{m}{T}}g(iT)\right)}^2 \end{gathered}$$
考虑重新定义$g(n)$表示为是$n$的倍数的数字有多少个，则上式可以直接写成：
$$\sum _{T=1}^m\varphi (T)g(T{)}^2$$
由此我们可以在$O(n{\log }^{2}n)$的时间内完成这题。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10, mod = 1e9 + 7;int prime[N], phi[N], a[N], n, cnt;int sum[N], m;bool st[N];vector<int> fac[N];inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}void init() {phi[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;phi[i] = i - 1;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}for (int i = 1; i < N; i++) {for (int j = i; j < N; j += i) {fac[j].push_back(i);}}
}int f(int n, int T) {int ans = 0;for (int i = T; i <= n; i += T) {for (auto it : fac[a[i]]) {ans = sub(ans, 1ll * phi[it] * sum[it] % mod * sum[it] % mod);sum[it]++;ans = add(ans, 1ll * phi[it] * sum[it] % mod * sum[it] % mod);}}for (int i = T; i <= n; i += T) {for (auto it : fac[a[i]]) {sum[it]--;}}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);init();scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}int ans = 0;for (int T = 1; T <= n; T++) {ans = add(ans, 1ll * phi[T] * f(n, T) % mod);}printf("%d\n", ans);return 0;
}