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I have been unable to find anywhere a sharp bounding for the partial sum of the first $n$ square-free integers $\sum_{k=1}^{n} k$, where $k$ runs over the square-free integers. I guess it should be around $\frac{\pi^2}{6} \left(\frac{n^2+n}{2}\right)$, but I found no reference to check and I do not have a rigorous way to prove my guess, just the heuristic following from the partial sum of the first positive integers and the distribution of square-free numbers.

Any reference or proof of a sharp bound would be welcomed.

Thanks!

EDIT

After some trial an error experimentation, it seems that $$\sum_{k=1}^{n} k = \frac{\pi^2}{6}\left(\frac{n^2+n}{2}\right)-\frac{6}{\pi^2}n\frac{\log(n)}{\log\log(n)}+O\left(\frac{n}{\sqrt{\log\log(n)}}\right)$$ Here the graphical representation of the diferences between $\sum_{k=1}^{n} k$ and $\frac{\pi^2}{6}\left(\frac{n^2+n}{2}\right)-\frac{6}{\pi^2}n\frac{\log(n)}{\log\log(n)}$:

enter image description here

There is still a clear pattern in the differences function, so I believe that the theoretical formula can be improved even further.

Is there any theoretical explanation supporting this formula? It seems much better than the one obtained using abelian partial summation.

Juan Moreno
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  • Just to clarify : You search an approximation for the sum of the squarefree numbers (I assume $1$ is included) not exceeding $n$ ? Did you check your asumed expression for , say , $n=10^6$ and $n=10^7$ ? – Peter Oct 04 '23 at 12:47
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    The sum must be smaller than $\frac{n(n+1)}{2}$. Do you mean $\frac{6}{\pi^2}\cdot \frac{n(n+1)}{2}$ ? This seems in fact to work very well. – Peter Oct 04 '23 at 13:01
  • @Peter I mean what I said, the partial sum of the first $n$ square-free integers, not the sum of the square-free integers not exceeding $n$. In this case, the sum is greater than $\frac{n^2+n}{2}$. – Juan Moreno Oct 04 '23 at 16:52
  • The sum of the first $n$ squarefree numbers is tabulated at https://oeis.org/A173143 and the asymptotic $(\pi^2/12)n^2$ is given, but with no discussion or references. – Gerry Myerson Oct 06 '23 at 09:27

1 Answers1

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By Abelian partial summation,

$$ \sum_{n=1}^x n 1_{n \text{ square-free}} = \frac{6}{π^2} x^2 + O(x^{3/2}) - \int_1^x 1 \frac{6t}{π^2} dt + O(x^{3/2}) $$

Thus, your asymptotic is correct, and the error term has the order $x^{3/2}$.

EDIT: If you want to sum the first $m$ (changing letters here to avoid notational overload) square-free integers, then you may choose $x = π^2/6 m + O(\sqrt{m})$, so that there are $m$ square-free integers up to $x$, and insert this into the above asymptotic.

Cloudscape
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  • You can get a better error term if you use the Riemann hypothesis. – Cloudscape Oct 04 '23 at 13:17
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    Interesting trailing comment. Please consider adding an additional section to your answer where you demonstrate using the Reimann hypothesis to improve the error term (... from someone who has a moderate understanding of Reimann sums, but is totally ignorant of Abelian partial summation). – user2661923 Oct 04 '23 at 13:33
  • Basically, the Riemann hypothesis gives you a better order for the error term of the proportion of square-free integers. This you can then plug into the Abelian summation, about which you may read something here: https://en.wikipedia.org/wiki/Abel%27s_summation_formula – Cloudscape Oct 04 '23 at 14:00
  • To be honest , I only know from an improvement of the error term in the case of the prime numbers. Do you have a reference ? Also, you (as I as well) misunderstood the question (see comment above). – Peter Oct 04 '23 at 17:00
  • The estimate comes straight from Wikipedia ( https://en.wikipedia.org/wiki/Square-free_integer#Distribution ) – Cloudscape Oct 04 '23 at 17:04
  • @Cloudscape thanks for your answer! Any idea on why the theoretical formula I have posted seems to improve so much the error term? – Juan Moreno Oct 06 '23 at 09:16
  • @JuanMoreno By imitating the proof of Wikipedia, I managed to squeeze the error term down to x^(5/4)+ε under the Riemann hypothesis (using Abelian summation and estimates on the Mertens function to bound the error in the main term). I don't really know how to go beyond that. – Cloudscape Oct 06 '23 at 12:40
  • I had hoped that I could discover some cancellation in the process, but so far I've failed. – Cloudscape Oct 06 '23 at 12:40
  • @Cloudscape thanks for the effort and the interest! Lets see if we find a reasonable explanation of why my theoretical formula works so well, and if it can be improved even better – Juan Moreno Oct 06 '23 at 16:09