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In his paper, Daniel Grunberg shows a relationship between the Stirling Numbers of the first kind and the Harmonic numbers via series of partitions (see Equation 3.1 on Page 5 in the link above). If you are confused about what the summation is over the partitions of $r$, see the examples in the bottom of Page 5 in the link.

$$S_{r,1,n}=(-1)^r\sum_{\{r\}}\prod_{j=1}^l\frac{(-1)^{i_j}}{i_j!}\left(\frac{{H^{(r_j)}_n}}{r_j}\right)^{i_j}$$

It is shown that this equation can be inverted to give an expression for the Harmonic number (see end of Section 3 on Page 6).

$${H^{(r)}_n}=(-1)^rr\sum_{\{r\}}(-1)^{i_1+\space\cdots\space+i_l}\frac{(i_1+\cdots+{i_l}-1)!}{i_1!\cdots{i_l!}}S^{i_1}_{{r_1},1,n}\cdots S^{i_l}_{{r_l},1,n}$$

If we replace the generalized harmonic number $H^{(r)}_n$ with the prime zeta function $P(s)=\sum_{p\space\in\space \text{primes}}\frac{1}{p^s}$ while limiting $n$ to infinity, and now define the function

$$T_r(s)=\sum_{\{r\}}\prod_{j=1}^l\frac{(-1)^{i_j}}{i_j!}\left(\frac{{P(s\cdot r_j)}}{r_j}\right)^{i_j}$$

It's true then that,

$$P(rs)=r\sum_{\{r\}}(-1)^{i_1+\space\cdots\space+i_l}\frac{(i_1+\cdots+{i_l}-1)!}{i_1!\cdots{i_l!}}(T_{r_1}(s))^{i_1}\cdots(T_{r_l}(s))^{i_l}$$

It turns out that

$$\sum_{r=1}^\infty T_r(s)=\sum_{r=1}^\infty\sum_{\{r\}}\prod_{j=1}^l\frac{(-1)^{i_j}}{i_j!}\left(\frac{{P(s\cdot r_j)}}{r_j}\right)^{i_j}=\frac{1}{\zeta(s)}$$ where $\zeta(s)$ is the Riemann Zeta function.

Is there a way that we can invert this equation in a way similar to the one above to derive $P(s)$ or a simpler form involving $P(s)$ in terms of $\frac{1}{\zeta(s)}$?

It may be of interest to note that

$$\sum_{r=1}^\infty\frac{P(rs)}{r}=\sum_{r=1}^\infty\sum_{\{r\}}(-1)^{i_1+\space\cdots\space+i_l}\frac{(i_1+\cdots+{i_l}-1)!}{i_1!\cdots{i_l!}}(T_{r_1}(s))^{i_1}\cdots(T_{r_l}(s))^{i_l}=\ln\zeta(s)$$

Does this imply that there could be a derivative/anti-derivative relationship between $\frac{P(rs)}{r}$ and $T_r(s)$?

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