Suppose one has a nested table of disitnct primes then the permutations of their products produce dupluicates at certain values.
For example, letting the primes $\{a,b\}=\{2, 3\}$, the products of their permutations over $2$ cycles gives $\{ 4,6,6,9 \}$, or $\{a^2, 2a b, b^2\}$.
Paying attention only to their expenent-type, we have $1\times a^2$ and $2\times ab$.
For $3$ distinct primes, we have $1\times a^3$ and $3\times a^2b$ and $6\times abc$.
These are simply the multinomial coefficients of the integer partitions of the number of primes in the set, the $4$th being $1,4,6,12,24$ for $a^4,a^3b,a^2b^2,a^2bc,abcd$ respectively, and so on.
However, if each subsequent term - eg, $a^2, b,c$ is used as the input instead of $a,b,c,d$, the result is $1,2,2,2$ for $a^4,a^3b,a^2b^2,a^2bc$. It is these partial sums that I am interested in. I can't seem to find a multinomial / partition formula for these values.
I include a few tables for reference:
Note
It is fairly easy to get the nest largest term for each (eg, the $6$ where $a^2bcd, \ a^2bcd$ meet), which is given by $(1-60/5!)(5!/10)$, where the $60$ is the coefficient in the column to the right, and the $10$ is the $2$nd coefficient given in this sequence ($M_2$ on table 24.2 in Abramowit's Handbook of mathematical functions), but can't get past this.