When I tested this in *Mathematica*, I had expected it to say it did not converge. However, I got this:

$$\prod_{n=1}^\infty n^{\mu(n)}=\frac{1}{4 \pi ^2}$$

**Note: this is the reciprocal of (3)** zeta-regularized product over all primes.

This indicates there is a slight skew toward the square-free numbers that have an odd number of factors.

I haven't been able to find this in the literature, but my guess is that it is already known.

Can someone point me in the right direction?

**Edit: We have been told it is a bug.** *Mathematica V.10* does not produce a symbolic answer.

**Edit 2** We can make this work by creating a function $f(k)$ that returns the $k$-th square-free number (in the constructive? order). We can find a few ideas from A019565.

$$\prod_{n=1}^{\infty}f(n)^{\mu(f(n))}$$

$$\sum_{n=1}^\infty \mu(f(n))$$

At the steps when $n>2$ is a power of $2$, sum$=0$ and prod$=1$.

Also, at those steps, $f(n)$ is the primorial with $p_k$ as the greatest factor, with $k=\frac{\log(n)}{\log(2)}$

There is a revised question over here. It is a variation to Edit 2.