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I am trying to calculate the following product

$$ \prod_{\substack{d\mid n \\ d>1}} \left( 1-\frac{\mu(d)}{\varphi(d)} \right)^{\varphi(d)} $$ where the functions $\varphi$ and $\mu$ are Euler's totient and Mobius functions, respectively. I am especially interested in finding explicit formula when $n$ is a square-free number. Any help will be appreciated.

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    $\begingroup$ It is not entirely clear to me exactly what you want; you ask for an explicit formula, yet the function is already given to us by a fairly explicit formula. Do you want to know if there is a "simpler" formula? Your function is not multiplicative, so standard tricks to simplify it do not work here. Alternatively, would information about the growth rate/average order/Dirichlet series of your function suffice? $\endgroup$
    – Daniel Loughran
    Commented Aug 18, 2014 at 16:40
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    $\begingroup$ One thing you might note is that your product is $0$ if $n$ is even. $\endgroup$
    – Robert Israel
    Commented Aug 18, 2014 at 17:47
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    $\begingroup$ If for instance $n=pq$, then the product is $(\frac{p-2}{p-1})^{p-1} (\frac{q-2}{q-1})^{q-1} (\frac{pq-p-q+2}{(p-1)(q-1)})^{(p-1)(q-1)}$ which has no evident cancellation or further factorisation, and only a small amount of like terms to collect. Doesn't seem like there is much else to be done here. $\endgroup$
    – Terry Tao
    Commented Aug 18, 2014 at 19:50
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    $\begingroup$ Asymptotics are more interesting; writing $(1 - \frac{\mu(d)}{\phi(d)})^{\phi(d)}$ as $\exp( \mu(d) - \frac{\mu(d)^2}{2\phi(d)} ) (1 + O(\frac{1}{\phi(d)^2} ))$, the product becomes $\exp( -\frac{1}{2} - \sum_{d|n} \frac{\mu(d)^2}{2\phi(d)}) ( 1 + O( \sum_{d|n; d>1} \frac{1}{\phi(d)^2} ))$ which can simplify to $\exp(-\frac{1}{2}-\frac{1}{2}\prod_{p|n} (1+\frac{1}{2(p-1)})) (1 + O( \sum_{p|n} \frac{1}{p^2} )))$. One can be a bit more accurate about the contribution of the small primes $p$ to the error term if one wants more precise asymptotics. $\endgroup$
    – Terry Tao
    Commented Aug 18, 2014 at 19:56
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    $\begingroup$ In view of the recent edits and the fact that it’s a bit odd for $\varphi(d)$ to appear in an exponent, I wonder whether that’s not a typo for $\mu(d)$ or better yet, $\mu(n/d)$. $\endgroup$
    – Emil Jeřábek
    Commented Aug 18, 2014 at 20:02

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