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If my understanding is correct, for "squarefree part" can be "squarefree kernel" in other cases, the generating Dirichlet series is $${\zeta(2s)\zeta(s-1)\over\zeta(2s-2)}=\prod_p\biggl(1+{p\over p^s}+{1\over p^{2s}}+{p\over p^{3s}}+\cdots\biggr)=\sum_n{\bar s(n)\over n^s}$$ alternating $1$ and $p$ as the coefficients, which is a $\zeta$ quotient as indicated. The residue at $s=2$ is $\zeta(4)/\zeta(2)={\pi^4/90\over\pi^2/6}={\pi^2\over 15}$, so that $$\sum_{n\le X} \bar s(n)\sim {\pi^2\over 15}X^2,$$ and then doing partial summation twice to get the $n^2$ in the denominator gives you the extra factor of $2$ in the denominator, with the result as you say.