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