tesuto-Mobius

求 \begin{equation*}\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=k]\end{equation*} 的值.

莫比乌斯反演吧.

\begin{align*}
&=\sum_{i=1}^{\left\lfloor\frac n k\right\rfloor}\sum_{j=1}^{\left\lfloor\frac m k\right\rfloor}\sum_{d|\gcd(i,j)=1}\mu(d)\\
&=\sum_{i=1}^{\left\lfloor\frac n k\right\rfloor}\sum_{j=1}^{\left\lfloor\frac m k\right\rfloor}\sum_{d|\gcd(i,j)=1}\mu(d)\\
&=\sum_{i=1}^{\left\lfloor\frac n k\right\rfloor}\sum_{j=1}^{\left\lfloor\frac m k\right\rfloor}\sum_{d|i}\sum_{d|j}\mu(d)\\
&=\sum_{i=1}^{\left\lfloor\frac n k\right\rfloor}\sum_{d|i}\sum_{j=1}^{\left\lfloor\frac m k\right\rfloor}\sum_{d|j}\mu(d)\\
&=\sum_{d=1}^{\min\left(\left\lfloor\frac n k\right\rfloor,\left\lfloor\frac m k\right\rfloor\right)}\mu(d)\sum_{i=1}^{\left\lfloor\frac n k\right\rfloor}\sum_{d|i}\sum_{j=1}^{\left\lfloor\frac m k\right\rfloor}\sum_{d|j}1\\
&=\sum_{d=1}^{\min\left(\left\lfloor\frac n k\right\rfloor,\left\lfloor\frac m k\right\rfloor\right)}\mu(d)\sum_{i=1}^{\left\lfloor\frac n k\right\rfloor}\sum_{d|i}1\sum_{j=1}^{\left\lfloor\frac m k\right\rfloor}\sum_{d|j}1\\
&=\sum_{d=1}^{\min\left(\left\lfloor\frac n k \right\rfloor,\left\lfloor\frac m k\right\rfloor\right)}\mu(d)\left\lfloor\frac{\left\lfloor\frac n k\right\rfloor}d\right\rfloor\left\lfloor\frac{\left\lfloor\frac m k\right\rfloor}d\right\rfloor\\
&=\sum_{d=1}^{\min\left(\left\lfloor\frac n k \right\rfloor,\left\lfloor\frac m k\right\rfloor\right)}\mu(d)\left\lfloor\frac n{kd}\right\rfloor\left\lfloor\frac m{kd}\right\rfloor\\
\end{align*}