Let us take any odd natural $n$ and denote by $H(n)$ the number of ways how $n$ can be represented as the product of factors larger or equal than $3$, where the order of factors matters. For instance, if $n=27$, then $H(n)=4$, since

It is known that $H(n) < n^{\eta}$, where $\eta = 1.37779\dots$ is the unique positive real zero of $\left(1-\frac{1}{2^s}\right) \zeta(s)=2$, see Theorem 5 in [^{1}].
Our aim is to show that this upper bound is optimal.

**Lemma.**
For any $\varepsilon>0$ there exist infinitely many odd $n$ such that $H(n) > n^{\eta-\varepsilon}$.

*Proof.*
We will argue along the same lines as in Section 3 of [^{2}], where the similar optimality was obtained for all (even) numbers.
First, we notice that

Since this expression is a decreasing function of $s$, we take any $\varepsilon>0$ and find that

and there exists some odd $b$ such that

Recalling the monotonicity property and the definition of $\eta$, we can find some $\gamma \in (\eta-\varepsilon, \eta)$ such that

which we rewrite as

Let us now consider for each odd $m \in [3,b]$ and sufficiently large $t>0$ the function

Notice that

which yields, in particular, that

\begin{equation}\label{eq:summ} \left(\frac{\sum_m c_m}{t}\right)^{\sum_m c_m} \geq \left(\sum_{m=3,~ m ~\text{odd}}^{b} \frac{1}{m^{\gamma}}-\frac{b}{t}\right)^{\sum_m c_m} = \left(1-\frac{b}{t}\right)^{\sum_m c_m} \geq \left(1-\frac{b}{t}\right)^{t} = e^{-b} + o(1). \end{equation}

This inequality will be used slightly below.

Let us define

Since $m$ is odd, we see that $n$ is odd. Moreover, clearly, $H(n)$ is bounded from below by the number of multiset permutations

Let us estimate $v(n)$ by Stirling’s bounds $\sqrt{2\pi} n^{n+1/2} e^{-n} \leq n! \leq e n^{n+1/2} e^{-n}$:

First, recalling that $c_i \leq \frac{t}{i^\gamma}$ and using \eqref{eq:summ}, we get

Second, arguing exactly as in [^{2}], we deduce that

for some $c_b > 0$.

Combining the lattter two inequalities, we deduce that

for some $C>0$ and all sufficiently large $t$. The proof is complete.