# Are $p$ and $q$ independent if my probability is $\sum_{n=1}^{\infty}\frac{1}{(pqn)!}$?

While checking for statistical independence of two events $$A_p$$ and $$A_q$$, for distinct primes $$p$$ and $$q$$, the sum $$\sum_{n=1}^{\infty}\frac{1}{(pqn)!}$$ appeared.

Can this sum be written in the form $$F(p)G(q)$$?

I can try to write the sum as a generalized hypergeometric function, e.g., $$_0F_{pq-1}(; b_1,...,b_n; 1)$$, but I cannot see how that would help check for independence.

If the sum is $$S(pq)$$, then the lack of any such factorization follows just from checking that $$S(6)/S(3) \ne S(10)/S(5)$$.
It's worth remarking that there is a simpler expression for $$S(pq)$$, since \begin{align*} S(pq) + 1 = \sum_{n=0}^\infty \frac1{(pqn)!} &= \sum_{\substack{m\ge 0 \\ pq \mid m}} \frac1{m!} \\ &= \sum_{m\ge0} \frac1{m!} \cdot \frac1{pq} \sum_{k=1}^{pq} e^{2\pi i km/pq} \\ &= \frac1{pq} \sum_{k=1}^{pq} \sum_{m\ge0} \frac1{m!}e^{2\pi i km/pq} = \frac1{pq} \sum_{k=1}^{pq} e^{e^{2\pi i k/pq}}. \end{align*}
Let $$S(p,q) = \sum_{n \ge 1} \frac1{(pqn)!}$$. Then by far the largest term of $$S(p,q)$$ is the first term: $$S(p,q) \approx \frac1{(pq)!}$$.
Therefore $$S(2,5) S(3,7) \ne S(2,7) S(3,5)$$ because $$\frac1{10!} \cdot \frac1{21!}$$ is over a thousand times smaller than $$\frac1{14!} \cdot \frac1{15!}$$.
However, if there were a factorization $$S(p,q) = F(p) G(q)$$, then both of the above would have to be equal to $$F(2) F(3) G(5) G(7)$$.