$\ln(n)$ can make a difference

We have $\zeta^2(s)=\sum_{n=1}^\infty \dfrac{\tau(n)}{n^s}$ This is easily found by dirichlet series . d.w.r.s $2\zeta(s)\zeta'(s)=-\sum_{n=1}^\infty \dfrac{\ln(n)\tau(n)}{n^s}$ put s=2 $\sum_{n=1}^\infty \dfrac{\ln(n)\tau(n)}{n^2}=-2\zeta(2)\zeta'(2)$ The popular closed form of $\zeta'(2)$ is $\dfrac{\pi^2}{6} (\gamma+\ln(2\pi)-12\ln(A))$ Putting that in we get $-\dfrac{\pi^4}{18}(1\gamma+\ln(2\pi)-12\ln(A))$ and $4+18+1+2-12=13$