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We should then reclaim the fact that when $z=0$ the series $G_q$ will be that of the prime numbers. The only non zero term will be when $r=1$ \begin{equation}  P_q=\frac{-1}{(z-1)^q}\sum_{r=1}^q \sum_{k=1}^r \bigg[ \frac{(-1)^{q-r}}{(q-r)!}P_{k+q-r}\prod_{m=1}^{q-r}(k+m-1) P_q=\frac{-1}{(-1)^q}\bigg[ \frac{(-1)^{q-1}}{(q-1)!}P_{q}\prod_{m=1}^{q-1}(m)  \bigg ]z^{r-1} ]  \end{equation}  Which is indeed reclaimed!