Function with zeros plus/minus the primes While playing with Cohen's pari script prodeulerrat found a function.
For $s \in \mathbb{C}$ define
$$ f(s) = \prod_{p \text{ prime}} (1-\frac{s^2}{p^2})$$
The product converges everywhere, no poles and the zeros
are $\pm p$.
At integers one can tell if $f(n)=0$ via primality testing.
Cohen's script computes $f(s)$ in $O(|s|)$ and it
iterates over primes.

Q1 Is there an alternative way to compute $f(s)$?
Q2 An explicit series converging to $f(s)$?

$f(1)=1/\zeta(2)$.

Q3 Is there closed form for $f$ at integers?

Xray:

Complex plot:

 A: A standard method to improve the speed of convergence is to look for approximations, which can be explicitly evaluated. In this case one would take $f(s)\approx\zeta(2)^{-s^2}$. We have
$$
f(s)\zeta(2)^{s^2} = \exp\left(\sum_p s^2\log (1-\frac{1}{p^2}) + \log(1-\frac{s^2}{p^2})\right).
$$
Each summand is now of magnitude $\frac{s^4}{p^4}$, thus the speed of convergence has improved, since you have to add up primes substantially larger then $s$ anyway, unless you only look for very rough bounds.
If this is not enough, you can develop $\log(1-\frac{1}{p^2})$ and $\log(1-\frac{s^2}{p^2})$ into a Taylor series, and pull out another power of $\zeta$. The next term will probably give $\zeta(4)^{(s^4-s^2)/2}$, and the series should then converge like $\frac{s^6}{p^6}$.
At some point you will have to ask yourself whether the algebraic manipulations necessary to improve the convergence are worth the saving in computation time, the answer to this question of course depends on your application or interest.
A: For Q2 I'm not sure what you expect other than the obvious series which comes from multiplying out the product:
$$
f(s)=\sum_{n=1}^\infty \mu(n)\frac{s^{2\omega(n)}}{n^2}
$$
where $\omega(n)$ is the number of distinct prime divisors of $n$.
