Special cases
For $n = p_i p_2 \cdots p_k$, (all with exponent $1$), the multiplicative partitions $m_1, m_2,\dots, m_{k+1}$ are given by the coefficients of the expanded series $$f(x,k)=\dfrac{1-(k-1) x}{\prod _{n=1}^k (1-n x)}$$ where the $m$th partition of $p_1 p_2 \cdots p_k$ is give by the $(k-m+2)$th coefficient.
f[k_] := (1 - (k - 1) x)/Product[1 - n x, {n, 1, k}]
g[a_, b_] := CoefficientList[Series[f@b, {x, 0, b + a}], x][[a - b + 2]]
h[n_] := g[n, #] & /@ Range[n + 1]
h[#] & /@ Range@6
gives
\begin{array}{c}
1 & 1\\
1 & 2 & 1\\
1 & 4 & 4 & 1\\
1 & 8 & 13 & 7 & 1\\
1 & 16 & 40 & 35 & 11 & 1\\
1 & 32 & 121 & 155 & 80 & 16 & 1\\
\end{array}
eg $30$ has partitions of length $m_1=1,\ m_2=4,\ m_3=4,\ m_4=1$:
$m_1=1$:
$m_2=4$:
- $1 \times 30$
- $2 \times 15$
- $3 \times 10$
- $5 \times 6$
$m_3=4$:
- $1 \times 2 \times 15$
- $1 \times 3 \times 10$
- $1 \times 5 \times 6$
- $2 \times 3 \times 5$
$m_4=1$:
- $1 \times 2 \times 3 \times 5$
The multiplicative partitions $m_1,m_2,\dots,m_{\left\lfloor \sqrt{2 (k+1)}+1/2\right\rfloor}$ for $p^k$ (where $p$ is prime) are given by the coefficients of the expanded series $$f_1(x,k)=\dfrac{1}{\prod _{n=1}^k (1-x^n)}$$ where the $m$th partition is give by the $(m+ 1 - k(k- 1)/2)$th coefficient.
f1[k_] := 1/Product[(1 - x^n), {n, 1, k}]
g1[a_, b_] := CoefficientList[Series[f1@b, {x, 0, b + a}], x][[1 - b (b - 1)/2 + a]]
h1[n_] := g1[n, #] & /@ Range@Floor[Sqrt[2 (n + 1)] + 1/2]
h1[#] & /@ Range@6
gives
\begin{array}{c}
1 & 1\\
1 & 1\\
1 & 2 & 1\\
1 & 2 & 1\\
1 & 3 & 2\\
1 & 3 & 3 & 1\\
\end{array}
eg $512$ has partitions of length $m_1=1,\ m_2=5,\ m_3=7,\ m_4=3$.
$m_1=1$:
$m_2=5$:
- $1 \times 512$
- $2 \times 256$
- $4 \times 128$
- $8 \times 64$
- $16 \times 32$
$m_3=7$:
- $1 \times 2 \times 256$
- $1 \times 4 \times 128$
- $1 \times 8 \times 64$
- $1 \times 16 \times 32$
- $2 \times 4 \times 64$
- $2 \times 8 \times 32$
- $4 \times 8 \times 16$
$m_4=3$:
- $1 \times 2 \times 4 \times 64$
- $1 \times 2 \times 8 \times 32$
- $1 \times 4 \times 8 \times 16$
It would be nice to find a generalised solution for any $p_i^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, but not sure whether this is easiliy attainable.
There are no extra numbers there. The numbers that you mentioned are the numbers from the set $\{0,1,2,\ldots,23\}$ wich are simultaneously congruent with $1$, $3$, $5$, or $7$ modulo $8(=2^3)$ and congruent with $1$ or $2$ modulo $3$. Take $19$, for instance, and note that $19\equiv3\pmod8$ and that $19\equiv1\pmod3$.
Best Answer
There is a nice randomized algorithm for finding the roots when $m=p$ is prime, it goes roughly as follows: first compute $$G(x) = \operatorname{gcd}\bigl( x^p-x,F(x)\bigr)$$ then $G(x)$ is the product of all the roots of $F(x)$ that are in $\mathbb Z/p$. Now for several random values of $a$ compute $$ \operatorname{gcd}\bigl( (x+a)^{(p-1)/2}-1,G(x)\bigr)$$ this will probably result on a non trivial factor of $G$, because it separates roots $\alpha$ for which $x+a$ is a quadratic residue from those which are not. You can repeat with the resulting factors until all the factors have degree one. This algorithm is very fast. It is described here.