Is there a formula for n/rad(n) Are there any known formulas for $n/\mathrm{rad}(n)$; summation or otherwise, that do not involve the Mobius function? Where may I find a list of formulas for this function of any type?
Here $\mathrm{rad}(n)$ is defined as the product of all (distinct) prime factors of $n$.
 A: A formula is give at OEIS:Radical of n, product of distinct prime factors of n:

Multiplicative with $a(p^{\alpha}) \,=\, p^{\alpha - 1}, \, $thus with $n \,=\, \prod_{i=1}^{\omega(n)} p_{i}^{\alpha_i}, \,$ we have
  $$
    \frac{n}{{\rm rad}(n)} = \prod_{i=1}^{\omega(n)} p_{i}^{\alpha_i - 1}, \, 
$$

where $\omega(n)$ is the number of distinct prime factors.
A: I know this is an old question but,
$n/rad(n)=\sum_{j=1}^{n}\lfloor{\frac{j^n}{n}}\rfloor-\lfloor{\frac{j^n-1}{n}}\rfloor $. This is because the two floor functions are just a characteristic equation that can only be $1$ or $0$ depending on whether or not $n|j^n$. The exponents in the prime factorization of $j^n$ are going to be larger than the exponents of the prime factorization of $n$, because $j$ is raised to the power of $n$. So, only the prime factors of $n$ and $j$ matter in whether or not $n|j^n$. Considering the factors, $n$ can only divide if $j$ is a multiple of $rad(n)$, so the sum is counting the multiples of $rad(n)\leq n$, but this is just $n/rad(n)$. This is probably exactly what you are looking for.
