Among several possible approaches we can view this problem as a
straightforward application of the Polya Enumeration Theorem and very
similar to the problem discussed at this MSE
link.
Recall the recurrence by Lovasz for the cycle index $Z(S_n)$ of
the multiset operator $\mathfrak{M}_{=n}$ on $n$ slots, which is
$$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l})
\quad\text{where}\quad
Z(S_0) = 1.$$
Suppose the prime factorization of $n$ is given by
$$n=\prod_p p^v.$$
Applying PET it now follows almost by inspection that the desired
count of multiplicative partitions into $k$ factors is given by the
term
$$F(n, k) = \left[\prod_p X_p^v\right]
Z(S_k)\left(-1 + \prod_p \frac{1}{1-X_p}\right)$$
where the square bracket denotes coefficient extraction
of formal power series.
We are interested in
$$G(n) = \sum_{k=1}^{\Omega(n)} F(n, k).$$
Now recall the generating function of the cycle index of the symmetric
group which is
$$H(z) = \sum_{q\ge 0} Z(S_q) z^q =
\exp\left(a_1z + a_2\frac{z^2}{2} + a_3\frac{z^3}{3}+\cdots\right)
= \exp\left(\sum_{l\ge 1} a_l \frac{z^l}{l}\right).$$
This gives for $G(n)$
$$G(n) = \left[\prod_p X_p^v\right]
\exp\left(\sum_{l\ge 1} \frac{1}{l}
\left(-1 + \prod_p \frac{1}{1-X_p^l}\right)\right).$$
Recalling $v_p(n)$ which is the exponent of the maximum power of $p$
that divides $n$ we finally obtain
$$G(n) = \left[\prod_p X_p^v\right]
\exp\left(\sum_{l=1}^{\max_p v_p(n)} \frac{1}{l}
\left(-1 + \prod_p \frac{1}{1-X_p^l}\right)\right).$$
This formula has a certain aesthetic value. Implemented in Maple it
will produce the following sequence:
$$1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 5, 1, 4, 1, 4,
\\ 2, 2, 1, 7, 2, 2, 3, 4, 1, 5, 1, 7, 2, 2, 2, 9, 1, 2, 2, 7,
\\ 1, 5, 1, 4, 4, 2, 1, 12, \ldots$$
which points us to OEIS A001055 where
additional information awaits, including simple recurrences for
practical computation of this function.
The Maple code for the above formula is as follows (here we have
replaced the prime number index on the variable $X$ by a positional
index on the variable $Y$):
with(numtheory);
facts :=
proc(n)
option remember;
local f, gf, p, res;
f := op(2, ifactors(n));
gf := exp(add(1/l*
(-1+mul(1/(1-Y[p]^l), p=1..nops(f))),
l=1..max(map(p->p[2], f))));
res := gf;
for p to nops(f) do
res := coeftayl(res, Y[p]=0, f[p][2]);
od;
res;
end;
Best Answer
You are almost right. As already observed in the comments, note that, calling $k_1, k_2, k_3... $ the exponents of each prime number in the factorization, if your final product $(k_1+1)(k_2+1)(k_3+1)... $ is odd, then this means that all exponents are even and so the number is a perfect square. In this case, to find the total nunber of ways to express the number as a product of two factors (regardless of whether these two factors are equal or different) you would have to add $ 1$ to your final product and divide to 2, since the symmetric distributions of the prime numbers in two factors $a \cdot b $ include one where $a=b $. For example, for $36= 2^2 \cdot 3^2$, we have $3 \cdot 3=9$ "total" divisors ($1, 2, 3, 4, 6, 9, 12, 18, 36$) but $5$ ways to express $36$ as a product of two factors ($36 \cdot 1, 18 \cdot 2, 12 \cdot 3, 9 \cdot 4, 6 \cdot 6$).
So if the question talks about two "different" factors, you have to subtract 1 and divide to 2 (in this example we have to exclude the product $6 \cdot 6$).
In contrast, if the final product is even, then the number is not a perfect square and so you only have to divide to 2.