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From this question here:

Moreover, if multiplicative function $\mathrm{core}(n)$ is defined to map positive integers "$n$" to square-free numbers by reducing the exponents in the prime power representation modulo $2$, or in formula: $$ \mathrm{core}(p^e) = p^{e\mod 2}, $$ with $\mathrm{core}(1) =1$. Or equivalently, $$\mathrm{core}(p^{2k+1})=p,\mathrm{core}(p^{2k})=1$$ Since the Dirichlet $\mathrm{core}$ generating function is: \begin{align} \begin{split} C(s)&=\sum_{n\ge 1}\frac{\mathrm{core}(n)}{n^s} =\dots=\frac{\zeta(2s)\zeta(s-1)}{\zeta(2s-2)}. \end{split} \end{align}

Is it possible to use Möbius inversion How to use an inverse Mellin transform on the last equation to get $\mathrm{core}(n)$?

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draks ...
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    Möbius inversion doesn't relate a Dirichlet series $\sum_{n=1}^\infty c(n)n^{-s}$ to its sequence of coefficients $c_n$; it relates a divisor sum $\sum_{d\mid n} f(d)$ to $f(n)$. If you want to extract $c_n$ from a Dirichlet series $\sum_{n=1}^\infty c(n)n^{-s}$, it would be an inverse Mellin transform: $$ c_n = \frac1{2\pi i} \int_{a-i\infty}^{a+i\infty} \bigg( \sum_{n=1}^\infty \frac{c(n)}{n^s} \bigg) \frac{n^s-(n-1)^s}s ,ds. $$ – Greg Martin Feb 10 '16 at 07:27
  • @draks : the Mellin transform is nothing more than a slight modification of the Laplace transform, so yes as Greg said if it converges somewhere then we can find the function back : https://en.wikipedia.org/wiki/Perron%27s_formula $\implies$ read a course on the Laplace transform – reuns Feb 10 '16 at 07:56
  • @user1952009: So it boils down to the following: Given $\frac{\zeta(2s)\zeta(s-1)}{\zeta(2s-2)}=\sum_{n=1}^{\infty} \frac{\mathrm{core}(n)}{n^{s}}$ I have to calculate $\displaystyle \mathrm{core}(x) =\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{\zeta(2z)\zeta(z-1)}{\zeta(2z-2)} \frac{x^{z}}{z} dz $, correct? – draks ... Feb 10 '16 at 15:24
  • @draks : $c > 2$, no it will give you $\sum_{n \le x} core(n)$. by the way it is a contour (complex) integral : replace $z$ by $c+i\omega$ and $dz$ by $id\omega$ it will be a normal (real) integral $\int_{-\infty}^\infty$. and if you want to understand why all that, read a course on the Fourier/Laplace transform – reuns Feb 10 '16 at 17:03
  • @user1952009, ah right, misread Perrons formula. I've read a bit on contour integrals in Edward's book on Riemann's Zeta Function. Why is $c>2$? – draks ... Feb 10 '16 at 20:27

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