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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.

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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*}

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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 $F(2) F(3) G(5) G(7)$.

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