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Let $\mu$ denote the Moebius function. What is a combinatorial interpretation of the following integer, \begin{align} \prod_{d \mid n} d!^{\,\mu(n/d)}, \end{align} where the product is taken over divisors of $n$? Does it have a simpler representation in terms of known functions? Note: The Online Encyclopedia of Integer Sequences does not have an entry containing the corresponding sequence.

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You can take the log and use Mobius inversion to show that (if $f$ is your function,) then $n! = \prod_{d|n} f(d)$. That lets you compute $f(n)$ recursively as $$\frac{n!}{\prod_{d|n, d<n} f(d)}$$ –  Thomas Andrews Jul 27 '12 at 17:40
@ThomasAndrews, that formula doesn't get any closer to a combinatorial interpretation. –  vonbrand Mar 11 at 2:19
If it did, do you think I might have posted it as an answer? @vonbrand –  Thomas Andrews Mar 11 at 3:18
Well, the later formula suggest one approach. If we can find a faithful action of a group of size $\prod_{d |n, d<n} f(d)$ on $S_{n}$, then that formula will count the orbits of that action. Now, I don't know what that group or action might be, but.... –  coolpapa May 26 at 7:48

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