Swapping the series and product in the zeta function euler product equality I was thinking that we have a well known function satisfying the following relation:
$$ \sum_{n\in \Bbb N} h_n(x) =\prod_{\text{p prime}} j_p(x) $$
which is of course the Riemann zeta function, $\zeta(s).$ This function is expressible as a product of primes.
I wondered if there could exist a reverse sort of function:
$$ \sum_{\text{p}~\text{prime}}f_p(x)=\prod_{n\in \Bbb N} g_n(x) $$
where we need the sum over primes equal to some product function.

Does a function like this exist? How would you construct it?

 A: Given any two function $\;h_n(x)\;$ and $\;j_p(x)\;$ with $p$ prime
and where
$$ \sum_{n\in \Bbb N} h_n(x) =\prod_{p\text{ prime}} j_p(x), \tag1 $$
then define another two functions $\;f_p(x)\;$ and $\;g_n(x)\;$ with
$$ h_n(x) :=f_{p_n}(x),\quad \text{and} \quad g_n(x) :=j_{p_n}(x) \tag2 $$ where $\;p_n\;$ is the $n$-th prime. Then
$$ \sum_{p\text{ prime}}f_p(x) = 
   \sum_{n\in\Bbb N} h_n(x) \quad \text{ and } \quad
  \prod_{n\in\Bbb N} g_n(x) =\prod_{p\text{ prime}}j_p(x). \tag3 $$
Thus, equation $(1)$ is equivalent to
$$ \sum_{p\text{ prime}}f_p(x)=\prod_{n\in \Bbb N} g_n(x) \tag4 $$
because they both equate the same sum to the same product.

The standard example of equation $(1)$ is the Riemann $\;\zeta\;$ function:
$$ \zeta(x) = \sum_{n\in \Bbb N} n^{-x} = \prod_{p\text{ prime}}
 \frac1{1-p^{-x}}, \tag5 $$
where $\;h_n(x) = n^{-x}\;$ and $\;j_p(x) = (1-p^{-x})^{-1}.\;$
Define the inverse function whose domain is all primes by
$\; q(p_n) := n.\;$ The equivalent to equation $(5)$ is
$$ \zeta(x) = \sum_{p\text{ prime}} q(p)^{-x} = \prod_{n\in \Bbb N}
 \frac1{1-p_n^{-x}}. \tag6 $$
