# 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.

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You have $\prod_p (1+p/p^s+1/p^{2s}+p/p^{3s}+\cdots)$, alternately $1$ and $p$ in coefficients, and likely can write this as a $\zeta$ expression. – Junkie Jun 8 '11 at 21:36
I guess it is $\zeta(2s)\zeta(s-1)/\zeta(2s-2)$. – Junkie Jun 8 '11 at 21:46

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.

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Thanks Gerry, this is just what I was looking for. – Granger Mar 20 '12 at 15:31

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.

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Junkie, by Perron's formula we have $$\sum_{n\le X} \bar s(n)=\int_{(3)}\frac{\zeta(2s)\zeta(s-1)}{\zeta(2s-2)}\frac{X^s}{s}\,ds\sim \frac{\pi^2}{30}X^2,$$ which is the limit conjectured by the OP. So there is no need to do any partial summation: the extra factor $2$ comes from the $s$ in the denominator of the integrand. – GH from MO Jun 8 '11 at 22:41
Ah yes, you are right, and really there should be more fretting about convergence. – Junkie Jun 8 '11 at 22:56
Thanks guys, but I was only asking if to anyones knowledge this had been considered before (hence reference request). It's sufficiently natural that I thought it must be classical? – Granger Jun 9 '11 at 9:35
This is a standard exercise in analytic number theory, regardless if it had been considered before. You got a solution, use it and enjoy it. It is possible that there is a more elementary proof, using convolutions and the hyperbola method. – GH from MO Jun 11 '11 at 4:20
@GH - I think it is standard etiquette (and saves space) to determine whether or not a result is already in print, before contributing a proof. Therefore whether a result has been considered before is important. I think you and Junkie misunderstood what I was asking. As it is, writing `this is a standard exercise' would have been a sufficient answer, so thank you. – Granger Jun 13 '11 at 16:50