I came across this paper from 2016 and it struck me because I tried this exact same thing in 2015 explained in this question. I know now that I was wrong because of a flaw in my thinking (I don't know where the error in the paper is in Mladen's work, but I know there has to be a mistake so if you can spot it that would be terrific). However, I still find the result, $$P(s)\approx1-\sqrt{\frac 2{\zeta(2^0s)}-\sqrt{\frac 2{\zeta(2^1s)}-\sqrt{\frac 2{\zeta(2^2s)}-\cdots-\sqrt{\lim\limits_{n \to \infty}\frac 2{\zeta(2^ns)}-1}}}}$$ approximating the prime zeta function, to be quite interesting (Mladen's equation excludes the minus one at the right end of the nested radical).

How good exactly is this approximation?

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    $\begingroup$ If $G(s) = 1 -F(s)$ where $F(s)$ is your nested radical function, then $G(2s) = \frac{2}{\zeta(s)} - G(s)^2 $. I would be amazed if this was a good approximation to $P(s)$ because it would imply $P$ almost satisfies this absurd looking functional equation. $\endgroup$ – user335907 Jun 14 '17 at 23:16
  • $\begingroup$ Actually it is saying $G(2s) = 1-\frac{2}{\zeta(s)} + (1-G(s))^2$ and yes, it seems to be an approximate functional equation @james.nixon $\endgroup$ – tyobrien Jun 15 '17 at 5:36

You're right, the result must be wrong. If the proof of the main result, $(1 − P(s))^2 =2/\zeta(s)− 1 + P(2s),$ were correct, it would be valid for all $s>1,$ not just integers. That's impossible, since the LHS grows to infinity as $s\rightarrow1+,$ while the RHS $\rightarrow-1+P(2).$ That's because $\zeta(s)$ has a pole at $s=1,$ and $P(s)$ a logarithmic singularity.
Numerically, the difference between both sides is small for large $s,$ predictably, as both approach $1$. For $s=2,$ it's $\approx0.007185545379369163,$ not even a satisfactory approximation. For $s=1.5,$ it's $\approx0.08228197886084833$, for $s=1.25,$ it's $\approx0.4091107555146472$.

  • $\begingroup$ (If $F(s) = \sum_{n=1}^\infty a_n n^{-s}$ is a non-identically zero convergent Dirichlet series, then as $\Re(s) \to \infty$ : $F(s) \sim a_m m^{-s}$ where $a_m$ is the first non-zero coefficient. Thus $F(k) \ne 0$ for $k$ large enough) $\endgroup$ – reuns Jun 15 '17 at 13:11
  • $\begingroup$ I agree, for small $s$ it is not a good approximation, but I was hoping we could derive some sort of asymptotic expression describing the quality of the approximation. $\endgroup$ – tyobrien Jun 15 '17 at 14:33
  • $\begingroup$ @Ty O'Brien It's the well known formula connecting $P(s)$ with $\ln\zeta(n\,s).$ Since $\zeta$ is approaching $1$ quickly for large real arguments, you don't need many values of $\mu(n),$ even a fixed table would suffice. That's not Project Euler, yet! :) $\endgroup$ – user436658 Jun 15 '17 at 14:33
  • $\begingroup$ @TyO'Brien You mean $\Re(s)$ small .... For $\Re(s)$ large the asymptotic expansion of $\sum_{n=1}^\infty a_n n^{-s}$ is $\sum_{n=1}^{N-1} a_n n^{-s}+\mathcal{O}(N^{-s})$ $\endgroup$ – reuns Jun 15 '17 at 15:16

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