p is a prime $$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p) $$
$ (ab)_p\equiv ab(\mod p), 0<(ab)_p<p $

($\lim_{p\rightarrow \infty}\frac{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}}{p\ln^2p}=\frac{1}{2}$).