忘了

\[s(i)=\sum_{i=1}^{n} f(i) \\g(1)\times s(n)=\sum_{i=1}^n (f\times g)(i)-\sum_{i=2}^{n}g(i)\times s(\lfloor \frac ni\rfloor) \]

\[\sum_{i=1}^n \sum_{j=1}^n i \times j \times \gcd(i,j) \\=\sum_{d=1}^n \sum_{i=1}^{\lfloor \frac nd \rfloor} \sum_{j=1} ^ {\lfloor \frac nd \rfloor} i \times j \times d^3 \times [\gcd(i,j)=1] \\=\sum_{d=1}^n \sum_{i=1}^{\lfloor \frac nd \rfloor} \sum_{j=1}^{\lfloor \frac nd \rfloor} i \times j \times d^3 \times \sum_{k|\gcd(i,j)} \mu(k) \\=\sum_{d=1}^n d^3 \times \sum_{k=1}^{\lfloor \frac n d \rfloor} (\frac {\lfloor \frac n {d \times k} \rfloor\times(\lfloor \frac n {d \times k} \rfloor+1)} 2 )^2 \times k^2 \mu(k) \\ p=d \times k \\=\sum_{p=1}^n p^2 \phi(p)\times (\frac {\lfloor \frac n p \rfloor\times(\lfloor \frac n p \rfloor+1)} 2 )^2 \]

\[f(i)=i^2\times \phi(i) \\s(i)=\sum_{i=1}^{n} f(i) \\g(1)\times s(n)=\sum_{i=1}^n (f\times g)(i)-\sum_{i=2}^{n}g(i)\times s(\lfloor \frac ni\rfloor) \]

posted @ 2026-02-03 15:53  Sgt_Dante  阅读(1)  评论(0)    收藏  举报