https://www.luogu.com/article/vmzsatr0

直接就 $O(n^{2/3})$ 了

打表搜OEIS得：https://oeis.org/search?q=1+4+7+13+16+25+28+38+44+53&language=english&go=Search

注意到文字：

a(n) = Sum_{k=1..n} A000005(k)*floor(n/k). - Benoit Cloitre, Apr 19 2007
To compute a(n) for huge n (see A180365) in sublinear use a(n) = 3*Sum_{i=1..n3} A006218(n/i) - Sum_{j=1..n3} floor(n/(i*j)) + n3^3, where n3 = floor(n^(1/3)). - Andrew Lelechenko, Apr 15 2011

按照这里做应该也是 $O(n^{2/3})$，不过常数原大于上面做法，略小于官方题解。