莫比乌斯反演推演总结

鉴于本人水平有限就不证明了。

替换法则：$\sum_{n = 1}^{m}\sum_{d | n}^{} [\frac{n}{d}] \mu ([\frac{n}{d}]) =  \sum_{n = 1}^{m}\sum_{d | n}^{} d \mu (d)$

递推一：$\sum_{i = 1}^{n}\sum_{j = 1}^{n} lcm(a[i],a[j]) = \sum_{i = 1}^{n}\sum_{j = 1}^{n} a[i] * a[j] * \frac{1}{gcd(a[i],a[j])}$

$= \sum_{d = 1}^{max} \sum_{i = 1}^{n}\sum_{j = 1}^{n} (d | a[i]) * (d | a[j]) * a[i] * a[j] * \frac{1}{d} * [gcd(a[i],a[j]) = d] $

$=  \sum_{d = 1}^{max} \sum_{i = 1}^{n}\sum_{j = 1}^{n} (d | a[i]) * (d | a[j]) * a[i] * a[j] * \frac{1}{d} * [gcd(\frac{a[i]}{d},\frac{a[j]}{d}) = 1]$

$= \sum_{d = 1}^{max}* \frac{1}{d} \sum_{i = 1}^{n}\sum_{j = 1}^{n} (d | a[i]) * (d | a[j]) * a[i] * a[j] \sum_{t | \frac{a[i]}{d},t | \frac{a[j]}{d}}^{}\mu (d)$

$= \sum_{d = 1}^{max}* \frac{1}{d} \sum_{i = 1}^{n}\sum_{j = 1}^{n} (d | a[i]) * (d | a[j]) * a[i] * a[j] \sum_{t | d}^{}t * \mu (t)$

$= \sum_{d = 1}^{max} d \sum_{i = 1}^{n}\sum_{j = 1}^{n} (d | a[i]) * (d | a[j]) * \frac{a[i]}{d} * \frac{a[j]}{d} \sum_{t | d}^{}t * \mu (t)$