# 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
5 years, 10 months ago

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Mind if you explain the first step? The

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

?

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago