tsy’s number

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n\frac{\varphi (i)\varphi (j^2)\varphi (k^3)}{\varphi (i)\varphi (j)\varphi (k)}\varphi (gcd(i,j,k)) \\ 我们假定j=p_1^{k_1}{p}_2^{k_2}{p}_3^{p_3}\dots \dots p_n^{k_n}，有\varphi (j)=p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)p_3^{k_3-1}(p_3-1)\dots \dots p_n^{k_n-1}(p_n-1)， \\ \varphi (j^2)=p_1^{2k_1-1}(p_1-1)p_2^{2k_2-1}(p_2-1)p_3^{2k_3-1}(p_3-1)\dots \dots p_n^{2k_n-1}(p_n-1) \\ \frac{\varphi (j^2)}{\varphi (j)}=p_1^{k_1}{p}_2^{k_2}{p}_3^{p_3}\dots \dots p_n^{k_n}=j \\ 同样的，我们不难推出\frac{\varphi (k^3)}{\varphi (k)}=k^2 \\ 对上式化简： \\ =\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^{n}j{k}^2\varphi (gcd(i,j,k)) \\ =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^{n}j{k}^2\varphi (d)(gcd(i,j,k)=d) \\ =\sum _{d=1}^n{d}^3\varphi (d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}\sum _{k=1}^{\frac{n}{d}}j{k}^2(gcd(i,j,k)=1) \\ =\sum _{d=1}^n{d}^3\varphi (d)\sum _{t=1}^{\frac{n}{d}}\mu (t)t^3\sum _{i=1}^{\frac{n}{dt}}\sum _{j=1}^{\frac{n}{dt}}\sum _{k=1}^{\frac{n}{dt}}j{k}^2 \\ 设f(n)=\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^{n}j{k}^2，T=dt，再次对上式化简： \\ =\sum _{T=1}^{n}f(\frac{n}{T})T^3\sum _{d\mid T}\varphi (d)\mu (\frac{T}{d}) \\ 我们假设g(n)=n^3\sum _{d\mid n}\varphi (d)\mu (\frac{n}{d}) \end{gathered}$$

$$\begin{gathered} 显然n^3是较好解决的，所以我们考虑后面的式子\sum _{d\mid n}\varphi (d)\mu (\frac{n}{d}) \\ 写成迪利克雷卷积的形式g(n)=(\varphi \ast \mu )(n)，显然g(n)是一个积性函数 \\ n=1,g(1)=\varphi (1)\mu (1)=1 \\ n=p,g(p)=\varphi (1)\mu (p)+\varphi (p)\mu (1)=1\ast (-1)+(p-1)\ast 1=p-2 \\ n=p^{k}d，由于p^k与d互质，所以有g(n)=g(p^k)g(d) \\ 所以我们重点考虑如何求g(p^k) \\ g(p^k)=\sum _{i=0}^k\varphi (p^i)\mu (p^{k-i}) \\ =\sum _{i=0}^k\varphi (p^{k-i})\mu (p^i) \\ 显然对于i\ge 2时\mu (p^i)=0,所以 \\ g(p^k)=\sum _{i=0}^1\varphi (p^{k-i})\mu (p^i)=\varphi (p^k)-\varphi (p^{k-1}) \\ 这里预处理就结束了，接下来就能快乐的数论分块了。 \end{gathered}$$

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#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 | 1using 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 f = 1, x = 0;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 f * x;
}const int N = 1e7 + 10, mod = 1 << 30, inv3 = 715827883;int prime[N], cnt;ll f[N], g[N], n;bool st[N];ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}void init() {g[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;g[i] = i - 2;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {int num = 1, temp = i;while(temp % prime[j] == 0) {temp /= prime[j], num++;}g[i * prime[j]] = quick_pow(prime[j], num - 2) * ((1ll * prime[j] * prime[j] % mod - 2 * prime[j] % mod + 1 + mod) % mod) % mod * g[temp] % mod;break;}g[i * prime[j]] = g[i] * g[prime[j]];}}for(int i = 1; i < N; i++) {f[i] = ((1ll * i * (1 + i) / 2 % mod) * i % mod) * ((1ll * (i + 1) * i / 2) % mod * (2 * i + 1) % mod * inv3 % mod) % mod;g[i] = (1ll * i * i % mod * i % mod * g[i] % mod + g[i - 1]) % mod;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T = read();while(T--) {n = read();ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * f[n / l] * ((g[r] - g[l - 1] + mod) % mod) % mod) % mod;}printf("%lld\n", ans);}return 0;
}