Nilpotency class of Frattini subgroup and group order

Suppose $$\psi(n)$$ denotes the minimal natural number $$k$$, such that there exists a finite group $$G$$, such that $$k = \max \{m \in \mathbb{N}| \exists \text{ prime } p, p^m | |G| \}$$, and $$\Phi(G)$$ has nilpotency class exactly $$n$$. Here $$\Phi$$ stands for Frattini subgroup. Is there some sort of a closed formula for $$\psi(n)$$?

What I currently know:

$$\psi(n) \geq n + 3$$ for $$n \geq 2$$

Proof:

Using the method from the answer to “If $|G|=p^3q^2$ then $\Phi(G)$ is cyclic for primes $p\neq q$.“, the statement becomes reduced to:

If $$G$$ is a finite group, such, that $$\Phi(G)$$ is a $$p$$-group of exact nilpotency class $$n \geq 2$$, then $$p^{n+3}| |G|$$. Here $$p$$ stands for an arbitrary prime number.

Now, suppose, $$p^{n+3}$$ does not divide $$|G|$$. Then, $$|\Phi(G)| | p^{n + 1}$$. That means, that $$|\Phi(G)| = p^{n + 1}$$, because all groups of order $$p^m$$, with $$m \leq n$$, have exact nilpotency class strictly less than $$n$$. Thus $$\Phi(G)$$ is a maximal class group. Thus it contains a non-abelian characteristic subgroup of order $$p^3$$ (which is the second element of its upper central series). And there we receive the contradiction with Lemma 1 from “The nilpotence class of the Frattini subgroup” by W.M. Hill and D.B. Parker, which states:

A non-abelian group of order $$p^3$$ can non occur as a normal subgroup contained in the Frattini subgroup of any finite group.

EDIT:

Actually that article proves an even more stronger result

$$\psi(n) \geq 2n + 1$$ for $$n \geq 2$$

(which I did not know before, because I found the full text of the article only today)

However, the question, whether this bound is sharp, or can be bettered, still remains.