Old problem solution: https://blog.csdn.net/gmh77/article/details/99066792#commentBox
Before writing a little strange,This question does not reflect the sbSo again pushed
\(\because n=\sum_{d \mid n}{\varphi(d)}\)
\(\therefore \sum_{i=1}^{n}{f(i)}=\sum_{n}{\prod{{a_i}^{\left \lfloor \frac{p_i}{2} \right \rfloor}}}\)
\(=\sum_{n}{\sum_{\prod{{a_i}^{\left \lfloor \frac{p_i}{2} \right \rfloor}} \mid d}{\varphi(d)}}\)
\(=\sum_{d=1}^{\left \lfloor \sqrt{n} \right \rfloor}{\varphi(d)\sum_{n}{[d^2\mid n]}}\)$
\(=\sum_{d=1}^{\left \lfloor \sqrt{n} \right \rfloor}{\varphi(d)*\left \lfloor \frac{n}{d^2} \right \rfloor}\)