For positive integers $n\in\mathbb{N}^*$, the radical of $n$ is defined as $\text{rad}(n):=\prod_{p\vert n}p$ where the product extends over the prime divisors of $n$. It can also be thought of as the greatest square-free divisor of $n$. We will also denote $\text{corad}(n):=\frac{n}{\text{rad}(n)}\in\mathbb{N}^*$ for simplicity. With this in mind, I thought of defining recursively $\text{rad}_0(n):=\text{rad}(n)$ and $\text{corad}_{0}(n):=\text{corad}(n)$ and, if $\text{rad}_{k-1}(n)$ and $\text{corad}_{k-1}(n)$ are well defined, then $\text{rad}_k(n):=\text{rad}(\text{corad}_{k-1}(n))$ and $\text{corad}_k(n):=\text{corad}(\text{corad}_{k-1}(n))$. This might seem arbitrary but, in fact, it leads to an interesting decomposition of $n$ since, by noticing that $n=\text{rad}(n)\text{corad}(n)$ by definition, we have that $n=\prod_{k=0}^\infty \text{rad}_k(n)$ since, eventually, $\text{rad}_k(n)$ and $\text{corad}_k(n)$ will become constant $=1$. I'll be refering to this as the radical factorization of $n$ since, for every $n\in\mathbb{N}^*$, it exists and is unique. My question then is: are there any results in the literature related to this topic? That is, the topic of decomposing positive integers into square-free integers in a unique manner?
A simple fact that is quite easy to prove is the following. Reminded by the prime factorization of positive integers, I thought of defining a function analogous to the prime omega functions as $\varpi(n):=|\{\text{rad}_k(n)\neq1:k\geq0\}|=\sum_{\text{rad}_k(n)\neq1}1$ which counts the number of terms $\neq1$ that appear in the radical factorization of $n$. Then, it turns out that $\varpi(n)=\max\{\nu_p(n):p\in\mathbb{P}\}$ where $\nu_p(n)\in\mathbb{N}$ is the greatest integer such that $p^{\nu_p(n)}$ divides $n$, i.e. the exponent of $p$ in the prime factorization of $n$. This is easy to see as the corad function subtracts $1$ from each exponent $\neq0$ of the prime factorization of $n$.
I have also thought of another way of defining square-free factorizations of the integers (that may seem more interesting) inspired by the Wikipedia article on square-free numbers. Given $n=\prod_{p_i\vert n}p_i^{e_i}$, we define its square residue as $\text{sqR}(n):=\prod_{e_i \text{odd}}p_i$ and its square part $\text{sqP}(n):=\sqrt{\frac{n}{\text{sqR(n)}}}$. Notice that the second one is well defined since every exponent of the prime factorization $n/\text{sqR}(n)$ is even. In fact, $n/\text{sqR}(n)=\text{sqP}(n)^2$ is the greatest perfect square dividing $n$. This appears naturally when simplifying square roots of integers as $\sqrt{n}=\text{sqP}(n)\sqrt{\text{sqR}(n)}$ is the optimal simplification. Analogously, we can define by recursion $\text{sqR}_0(n):=\text{sqR}(n)$, $\text{sqP}_0(n):=\text{sqP}(n)$, $\text{sqR}_k(n):=\text{sqR}(\text{sqP}_{k-1}(n))$ and $\text{sqP}_k(n):=\text{sqP}(\text{sqP}_{k-1}(n))$. Then, by observing that $n=\text{sqR}(n)\text{sqP}(n)^2$, we get the much more interesting factorization $n=\prod_{k=0}^\infty\text{sqR}_k(n)^{2^k}$ which we may call the square residue factorization. The length of the square residue factorization of $n$ clearly grows much slower than the previous one. In fact, by making use of the weak inequality $\text{sqP}(n)\leq \sqrt{n}$, one can easily see that the length must be no longer than $1+\left\lfloor\log_2\log_2 n\right\rfloor$.