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Is it possible to describe the set of sequences $a(n) = \mathcal{O}(1)$ such that $$\frac{F'}{F}(s) = \sum_{p^k} a(p^k) p^{-sk} \ln p \qquad (Re(s) > 1)$$ is the logarithmic derivative of a function meromorphic on the whole complex plane ?

The restriction that $\frac{F'}{F}$ is meromorphic with all its poles being of the form $\frac{\pm k}{s-\beta}$ seems to be strong enough for deducing something on $a(n)$ - if it helps assuming the (generalized) Riemann hypothesis.

If $F(s) = \frac{P(s)\prod_{k=1}^K L(n_k s-s_k,\pi_k)}{Q(s)\prod_{m=1}^M L(n_m s-s_m,\pi_m)}$, $L(s,\pi_.)$ are automorphic L-functions and $P(s),Q(s)$ are two entire Euler products, then $\frac{F'}{F}(s)$ fits the requirements. If $K,M \to \infty$ it should fit too assuming everything converges compactly. A part of the question is if you can find other classes of meromorphic Euler products ?

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  • $\begingroup$ "$F$ is meromorphic" is supposed to be the simplest possible relaxation of "$F$ has a functional equation" $\endgroup$
    – reuns
    Commented Aug 7, 2016 at 7:53
  • $\begingroup$ Clarification: do you really mean fully multiplicative, or just "weakly" multiplicative? Even in the "fully" multiplicative case, there are many elementary, natural Euler products that do not have a meromorphic continuation: "Estermann phenomenon"... Clarify? $\endgroup$
    – paul garrett
    Commented Aug 7, 2016 at 20:08
  • $\begingroup$ @paulgarrett I see it is not clear, I mean if you can find a general construction for $a(p^k)$ such that $\sum_{p^k} a(p^k) p^{-sk} \ln p$ is the logarithmic derivative of an Euler product $F(s) = \prod_p \sum_k b(k) p^{-sk}, Re(s) > 1$, whose analytic continuation is meromorphic (so any product/quotient of shifted L-function works, times a simples (entire or finite terms) Euler product, but I'd like to know if you can find some other and more illustrative examples, and ideally if you can find a constraint on $a$ and $F'/F$ poles) $\endgroup$
    – reuns
    Commented Aug 7, 2016 at 20:53
  • $\begingroup$ If the question is about producing Euler products with meromorphic continuation, I think automorphic L-functions provide the bulk of them... And there does not seem to be any "more elementary" prescription for the $a(p^n)$'s for a cuspform for $GL_2$. $\endgroup$
    – paul garrett
    Commented Aug 7, 2016 at 20:56
  • $\begingroup$ @paulgarrett let $A(x) = \sum_{p^k < x} a(p^k) \ln p = \sum_{n < x} a(n) \Lambda(n)$. My idea was trying something like $A(x) = \sum_{\beta} \pm \frac{x^{\beta}}{\beta}$ and the Riemann explicit formula gives $\Lambda(n) = 1 - \sum_{\rho} \frac{(n+1/2)^\rho-(n-1/2)^\rho}{\rho}$ so $A(x) = \sum_{n < x} a(n) -\sum_{\rho} \sum_{n < x} a(n)\frac{(n+1/2)^\rho-(n-1/2)^\rho}{\rho}$ using that $(n+1/2)^\rho-(n-1/2)^\rho$ looks like $n^{\rho -1}$ plus other terms I get $\sum_{\beta} \pm \frac{x^{\beta}}{\beta} \approx \sum_{n < x} a(n) -\sum_{\rho} \sum_{n < x} a(n) n^{\rho - 1}$ $\endgroup$
    – reuns
    Commented Aug 7, 2016 at 21:01

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