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Let $s=\sigma+it$ be a complex number and define the function:

$$F(s)=\sum_{k=2}^{\infty}\frac{p_\pi(k)}{k^s}$$

Where $p_\pi(k)$ is the number of unordered factorizations of $k$, corresponding to OEIS A001055

This is a Dirichlet series that converges for $\sigma>1$.

I know that finding the analytic continuation for any Dirichlet series is absolutely not trivial, and many of them require more knowledge about the nontrivial zeros of $\zeta(s)$. The function above is just an example of a function I am working with.

Is it possible to find the analytic continuation of the function above, with the known methods? If not, why?

Do we know any other analytic continuation for the Dirichlet series out of Dirichlet's L-Functions?

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  • $\begingroup$ Just to be clear $p_{\pi}(k)$ corresponds to OEIS sequence oeis.org/A001055? A001055 states "Dirichlet g.f.: Product_{k>=2} 1/(1 - 1/k^s)" which doesn't seem right to me. Is this correct? $\endgroup$ Aug 18, 2021 at 16:36
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    $\begingroup$ @StevenClark yeah, that's correct. I've also the proof of the last statement. $\endgroup$ Aug 18, 2021 at 18:19
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    $\begingroup$ Thanks. I'm not sure I'm interpreting your last question correctly, but there are many Dirichlet series that have analytic continuations that do not correspond to Dirichlet L-functions. For example I believe the Dirichlet series $\eta(s)=\sum\limits_{n=1}^\infty(-1)^{n+1}\ n^{-s}$, $\frac{1}{\zeta(s)}=\sum\limits_{n=1}^\infty\mu(n)\ n^{-s}$, $\log\zeta(s)=\sum\limits_{n=2}^{\infty } \frac{\Lambda(n)}{\log(n)}\ n^{-s}$, and $\frac{\zeta'(s)}{\zeta(s)}=-\sum\limits_{n=1}^\infty\Lambda(n)\ n^{-s}$ all have analytic continuations. $\endgroup$ Aug 18, 2021 at 19:05
  • $\begingroup$ @StevenClark very interesting. I also think that they all have analytic continuations. For example the Dirichlet eta function is itself an analytic continuation of $\zeta(s)$ for $\sigma>0$, so the analytic continuation of the zeta function over the whole complex plan should be an analytic continuation of the eta function too. I'll definitely do some other research. Thank you. $\endgroup$ Aug 18, 2021 at 19:18

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