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I am partly repeating myself here again. Is this a correct relation between the Riemann zeta function and the Prime zeta function?

$$ \zeta (s) = \sum\limits_{n=1}^{\infty}\frac{1}{n^{s}}$$

$\displaystyle \log(\zeta (s)) = \sum\limits_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$ where $\displaystyle a_{1} = 0\;$, $\displaystyle a_{n} = \frac{\Lambda(n)}{\log(n)}$ and $\Lambda(n)$ is the Mangoldt function defined by $\displaystyle \Lambda(n) = \log(p)$ if $n = p^{k}$ for $p$ a prime.

$\displaystyle ω_{1}(\log(\zeta (s))) = \sum\limits_{n=1}^{\infty}\frac{\log(a_{n})}{n^{s}},$ where $ω_{1}(\;)$ is an operation and $\displaystyle \log(a_n) = 0$ if $a_n = 0$ $$ω_{1}(\log(\zeta (s)))\zeta (s) = \sum\limits_{n=1}^{\infty}\frac{b_{n}}{n^{s}}$$

$\displaystyle ω_{2}(ω_{1}(\log(\zeta (s)))\zeta (s)) = \sum\limits_{n=1}^{\infty}\frac{e^{b_{n}}}{n^{s}}\;$ where $ω_{2}(\;)$ is another operation and $\displaystyle e^{b_{n}}=\exp(b_{n})$

$$\sum\limits_{p\;\text{prime}} \frac1{p^s} = \log(ω_2(ω_1(\log(\zeta (s)))\zeta (s)))$$

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Can you explain in more detail this "operation" $\omega_1$. Are $\omega_1$ and $\omega_2$ the same thing? What precisely is the definition. – Eric Naslund May 28 '11 at 21:59
@Eric: I was puzzled by the same thing, but then I thought that this might be analytic number theorists' slang. I read it as follows: view the the Dirichlet series $\sum\limits_{n=1}^{\infty} \dfrac{a_n}{n^s} = D(s,(a_n))$ as a function of the sequence $(a_n)$. Then $\omega_1 D(s,(a_n))$ is the Dirichlet series $D(s,(\log{a_n}))$ and $\omega_2 D(s,(b_n)) = D(s, (e^{b_n}))$. I'm not sure whether this is well-defined if one thinks of $D(s,(a_n))$ as an actual function of $s$ without very strong hypotheses, but the "operation" seems to make sense this way. – t.b. May 28 '11 at 22:22 gives a close-formed expression for ζ in terms of P and vice-versa. Both are simple but neither are Dirichlet series. This is far from an answer to Mats' question but maybe a useful hint to someone. – Dan Brumleve May 29 '11 at 5:14
@Eric: I should have written unknown operations $ω_{1}$ and $ω_{2}$. – Mats Granvik May 29 '11 at 8:48

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