Preliminaries.
Apart from this first definition, what follows can be skipped for the reader in a hurry.
Definition 1. Let's define a function $\Phi$ as such:
$$\begin{matrix} \Phi\colon&\mathbb N_{\geqslant 1} &\to &\mathbb N[X] \\ &n=\prod_{i} {p_i}^{\alpha_i} &\mapsto &\sum_i \alpha_i X^{i-1},\end{matrix}$$
where $(p_i)=(2,3,5,\ldots)$ is the ordered list of prime numbers and $\prod_i {p_i}^{\alpha_i}$ is the prime decomposition of $n$.
Examples 1.
$\Phi(12)=X+2$
$\Phi(3\,920)=2X^3+X^2+4$ since $3\,920=2^4\cdot 5\cdot 7^2$
Remark 1. We have $\Phi(1)=0$ and $\Phi(mn)=\Phi(m)+\Phi(n)$.
Remark 2. The application $\Phi$ is bijective, we will denote by $\Phi^{-1}$ its inverse.
More context.
Definition 2. Let's the following function $\Psi$:
$$\begin{matrix} \Psi\colon&\mathbb N_{\geqslant 1} &\to &\mathbb N \\ &n &\mapsto &\displaystyle\sum_{k=0}^{\sqrt n}\Phi^{-1}(\Phi^{(k)}(n)),\end{matrix}$$
where $\Phi^{(k)}$ is the $k$-th derivative of $\Phi\in \mathbb Z[X]$.
Example 2.
$\Psi(10)=4$
$\Psi(25)=124$
$\Psi(30)=54$
Let's plot the first values of $\Psi$:
Remark 4. We can show that $\Psi(n)\to \infty$ since for all $n$, $\Psi(n)\geqslant \log n/\log 2$.
The question.
I am interested in the values which make this function $\Psi$ "explode".
Let's comprehend what I mean by "explode" by looking at those graphs:
We already saw that $\Psi(25)$ was quite bigger than the other $\Psi(n)$ for $n$ close to $25$, now we see that $49$, $98$, $121$ and $169$ are such values, where $\Psi(n)$ gets significantely big for those values.
Appart from $98$, which is $49\times 2$, all of those values seem to be equal to $p^2$ where $p$ is a prime number.
Which is why I make the following assumption:
Conjecture 1. $\Psi(n)$ gets significantly bigger than any $\Psi(k)$ for $k\leqslant n$ is, and only if, $n$ is the square of a prime number.
So my questions are the following:
Do you have any leads on how to prove conjecture 1?
I fail to make more tests than that since the numbers are getting huge, feel free to share your work.
Can we find some kind of a formula for $\Psi(n)?$
Can we bound $\Psi$ in a better way than remark 5?
I proved the following remark, where $\pi(n)$ denote the number of primer numbers less or equals to $n$ and $\lfloor.\rfloor$ is the floor function, but I am not really satisfied with this bound.
Remark 5. We have for all $n$:
$$\Psi(n) \leqslant \lfloor(\pi(n)+1)\log(\pi(n)+1)+\log(\log(\pi(n)+1))+13\rfloor!.$$