Reference requested for $\lim_{n \rightarrow \infty} \frac{\sum_{i=1}^{n} \bar{s}(i)}{n^2} = \frac{\pi^2}{30}$ While analysing the average runtime of an algorithm, I came across the following identity, and would like to know if anybody knows of any references for it? 
For $i \in \mathbb{N}$, let $\bar{s}(i)$ denote the square-free part of $i$, eg., $\bar{s}(12) = 3$ (and $\bar{s}(1)=1$). Then
$$
\lim_{n \rightarrow \infty} \frac{\sum_{i=1}^{n} \bar{s}(i)}{n^2} = \frac{\pi^2}{30}.
$$
Many thanks.
 A: 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 by Perron's formula
$$\sum_{n\le X} \bar s(n)={1\over 2\pi i}\int_{(\sigma)}{\zeta(2s)\zeta(s-1)\over\zeta(2s-2)}{X^s ds\over s} \sim {\zeta(4)\over 2\zeta(2)}X^2={\pi^4/90\over2\pi^2/6}X^2={\pi^2\over 30}X^2,$$
with usual conditions about convergence in vertical strips, which are OK here. Dividing by $X^2$ gives the desired limit.
A: The question asked for a reference, not a proof. A reference is Karl Greger, Square divisors and square-free numbers, Mathematics Magazine 51, No. 4 (Sept. 1978), 211-219. I make no claim that the result was unknown before Greger's paper. 
