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Let $S$ denote the set of natural numbers $m$ with the property that for all prime powers $p^k || m$ we have $k \equiv 1 \pmod{2}$.

What is the asymptotic density of $S$?

Note that $S$ contains all prime numbers and more generally, all square-free numbers, so that $\liminf_{X \rightarrow \infty} \frac{\# (S \cap [1,X])}{X} \geq \frac{6}{\pi^2}$.

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    $\begingroup$ You could try to compute the Dirichlet series $\sum_{n \in S} n^{-s}$. $\endgroup$ – François Brunault Oct 4 '18 at 14:05
  • $\begingroup$ Shouldn't it be something like zeta(3) times the square free density? Gerhard "Being Somewhat Thick About This" Paseman, 2018.10.04. $\endgroup$ – Gerhard Paseman Oct 4 '18 at 14:29
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The Dirichlet series $L (s)=\sum_{n \in S} n^{-s} $ has the Euler product expression $$L (s) = \prod_p \frac{1+p^{-s}-p^{-2s}}{1-p^{-2s}}.$$ So $S $ has analytic density $$\mu (S) = \operatorname{Res}_{s=1} L(s) = \lim_{s \to 1} \frac{L(s)}{\zeta(s)} = \prod_p \frac{1+1/p-1/p^2}{1+1/p}$$ as can also be seen by an informal argument. We can also rewrite this as $$ \mu(S) = \prod_p \bigl(1-\frac{1}{p(p+1)}\bigr) = \prod_p \frac{1-2/p^2+1/p^3}{1-1/p^2}$$ I don't expect this Euler product to have a nice expression since the roots of the Euler factor $1+X-X^2$ in the numerator are not roots of unity, so the product cannot be expressed, at least in an obvious way, in terms of values of the Riemann zeta function.

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    $\begingroup$ In fact the last expression you write can be written as $\prod_p \left(1 - \frac{2p-1}{p^3}\right)\left(1 - \frac{1}{p^2}\right)^{-1} = \zeta(2) C$, where $C = \prod_p \left(1 - \frac{2p-1}{p^3}\right)$ is known as the care-free constant. Very interesting! $\endgroup$ – Stanley Yao Xiao Oct 4 '18 at 16:16
  • $\begingroup$ @StanleyYaoXiao I didn't know this constant had a name. In fact, I don't know anything about these curious Euler products (are they periods? and so on). I also like the alternative expression $\mu(S) = \prod 1-1/(p (p+1)) $. $\endgroup$ – François Brunault Oct 4 '18 at 17:03
  • $\begingroup$ @StanleyYaoXiao's comment gives a numerical answer of .704441 $\endgroup$ – Matt F. Oct 4 '18 at 19:59
  • $\begingroup$ @MattF. Numerical evaluation of these constants is indeed not easy. There is a lot of interesting information at guests.mpim-bonn.mpg.de/moree/Moree.en.html They can get 1000 digits of accuracy in reasonable time (in 1999!). So for $\mu(S)=\prod 1-f/g$ with $f=1$ and $g=p(p+1)$ the numerical value can be found at guests.mpim-bonn.mpg.de/moree/Moree-details.en.shtml#t10-cfree $\endgroup$ – François Brunault Oct 4 '18 at 22:30
  • $\begingroup$ I don't understand the transition between the first display and the second where the Residue has a 1+1/p in the denominator. Is there a typo or a manipulation I am missing? Gerhard "Questionable Minds Wish To Know" Paseman, 2018.10.04. $\endgroup$ – Gerhard Paseman Oct 5 '18 at 2:04
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As I see it, the desired set is the disjoint union of (the rad(n) multiples of) squarefree numbers times n^2 as n ranges over all positive integers, which is like (zeta(3) + 1/64 + 3/512 + 2/729 + 1/1728 + 7/4096 + ...)/zeta(2) for the density.

Gerhard "Doing This On Minimal Caffeine" Paseman, 2018.10.04.

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    $\begingroup$ The terms listed come to .746882, which is bigger than the accepted answer. $\endgroup$ – Matt F. Oct 4 '18 at 20:02
  • $\begingroup$ I'll reconsider my thinking on this. I think it should be something like zeta(c)/zeta(2) though, where c is between 2 and 3 and zeta(c) stands for sum 1/(rad(n)n^2). Thanks for checking this answer. Does a straightforward computation tend to the accepted answer? Gerhard "Will Drink Some More Coffee" Paseman, 2018.10.04 $\endgroup$ – Gerhard Paseman Oct 4 '18 at 21:30
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    $\begingroup$ +1 for the caffeine stuff. (Yes, I know, I'm being unprofessional a likely in violation of a few minor MO rules.) $\endgroup$ – Yaakov Baruch Oct 5 '18 at 10:40

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