×

# Largest squarefree divisor

It is known that $$\displaystyle \lim_{k \to \infty} \left( \sum_i x_i^k \right)^{1/k} = \max_i x_i$$ for $$x_i \in \mathbb{R}^+$$. So, to extract the squarefree divisors of $$n$$ only, we make use of the Möbius function and take the following sum:

$\sum_{d|n} \mu^2(d)d^k = n^k \sum_{d|n} \mu^2(d)\left( \frac{n}{d} \right)^{-k} = n^k (\mu^2 * \text{Id}_{-k})(n)$

Taking the $$k$$th root and letting $$k \to \infty$$:

$\max_{d|n} \mu^2(d)d = \lim_{k \to \infty} \left( n^k (\mu^2 * \text{Id}_{-k})(n) \right)^{1/k} = n \lim_{k \to \infty} (\mu^2 * \text{Id}_{-k})(n)^{1/k}$

Note:

$$*$$ is the Dirichlet convolution, a binary operator defined on functions $$\mathbb{N} \longrightarrow \mathbb{C}$$ such that $$\displaystyle (f*g)(n) = \sum_{d|n} f(d)g(\frac{n}{d})$$.

$$\text{Id}_{k}(n) = n^k$$ for $$n \in \mathbb{N}$$, $$k \in \mathbb{C}$$.

Note by Jake Lai
2 years ago

Sort by:

Mind if you explain the first step? The

$$\displaystyle \lim_{k \to \infty} \left( \sum_i x_i^k \right)^{1/k} = \max_i x_i$$

?

- 2 years ago

It just means $$\displaystyle \lim_{k \to \infty} \left( \sum_{i=1}^n x_i^k \right)^{1/k} = \max_{i=1,2,\ldots ,n} x_i = \max \{ x_1,x_2, \ldots ,x_n \}$$.

- 2 years ago

No, as to why it is so. Also mind if you also define what $$\text{Id}$$ means? Thanks

- 2 years ago

This is a clean, nice proof. Also I've added clarifications to the note.

- 2 years ago