Consider the subset $S \subseteq \mathbb{Z}$ given by:
$$S = \{ 2^i 3^j : i,j \ge 0 \}$$
Define the sequence $(a_k)$ to be the elements of $S$ in increasing order (with the standard order on $\mathbb{Z}$), and further define:
$$ [a_n]! = \prod_{k = 1}^n a_k$$
I am interested to know if this generalisation of the factorial satisfies the analogous fundamental property:
$$\frac{[a_{m+n}]!}{[a_m]![a_n]!} \in S$$
and more generally, if this relation would hold given $S$ generated as the multiplicative span of any finite set of primes.
This seems very plausible, but I haven't found it to be immediate. It is obvious if $S$ is generated by only one prime. It may not hold if $S$ is generated by infinitely many primes, as the case $S = \{1,3,5,7,9,...\}$ demonstrates.
I am aware of the Bhargava factorial corresponding to sets $S \subseteq \mathbb{Z}$, an important recently-discovered generalisation of the factorial which satisfies the property above. Unfortunately, the Bhargava factorial of $S$ is not equal to the function I have given above. But it's possible that this factorial could be realised as a Bhargava factorial of some other subset of $T \subseteq \mathbb{Z}$.
For reference, the first terms of the sequence $(a_k)$ for $S = \{2^i 3^j\}$ are $1,2,3,4,6,8,9,12,16,18,24,...$.
This gives:
$$[a_0]! = 1$$ $$[a_1]! = 1$$ $$[a_2]! = 2$$ $$[a_3]! = 6$$ $$[a_4]! = 24$$ $$[a_5]! = 144$$