I asked this question in SEM but I got no answer, so I'm trying my luck here.

Let the Dirichlet series $\phi(s)=\sum_{n\ge 1}\frac{a(n)}{n^s}$ be absolutely convergent for $\Re(s)>1$ and extend to a meromorphic function on the half plane $\Re(s)>1/2$ with only pole at $s=1.$ write $$\phi(s)=\sum_{\text{prime}}\frac{a(p)}{p^s}+\sum_{\text{non-prime}}\frac{a(n)}{n^s}$$ If we suppose that $\sum_{\text{prime}}\frac{a(p)}{p^s}$ converges absolutely for $\Re(s)>1$ and has a pole at $s=1.$ We can deduce that the series $\sum_{\text{prime}}\frac{a(p)}{p^s}$ can be continued analytically to a meromorphic function in the half plane $\Re(s)>1/2$ with only pole at $s=1$ ?

  • Assuming the Riemann hypothesis is false, let $a_n = \ln n$, and you have a counter-example.

  • Assuming the Riemann hypothesis is true, let $a_p = \ln p + p^{-1/4}\ln p, a_{p+1} = \ln(p+1) - p^{-1/4}\ln p, a_n = \ln n$ otherwise, and you get a counter-example.

  • $\begingroup$ Thanks for your answer !!! I am really interested to the case where the $a(n)$ are Fourier coefficients of a modular forms... $\endgroup$ – Adam Aug 24 '16 at 15:09
  • $\begingroup$ @Adam the simplest counter-example is probably $a_n = \ln n + (-1)^n n^{-1/4} \ln n$. And if you meant $a(n)/\ln n$ are coefs of a modular form (otherwise $\sum_n a_n n^{-s}$ has no pole at $s=1$, or $\sum_p a_p p^{-s}$ is not meromorphic at $s=1$) by the Hecke operator theory it means $\phi(s) = \sum_{m=0}^M \sum_n b_n^m n^{-s} \ln n$ where $ b_n^m$ are multiplicative (and there is a Riemann hypothesis for each $\sum_n b_n^m n^{-s}$), so it should really be equivalent with GRH $\endgroup$ – reuns Aug 24 '16 at 15:22
  • 1
    $\begingroup$ @Adam, I think you must add it to your question, and you should specify if you're interested to modular forms of integral wieght or half-ntegral weight or Maass wave forms ...! $\endgroup$ – Med Aug 24 '16 at 15:28
  • 1
    $\begingroup$ @Adam it changes completely the question, and I can't answer if you meant non integral weight modular forms $\endgroup$ – reuns Aug 24 '16 at 15:43
  • $\begingroup$ @Med what would be the answer in the case of modular forms ? $\endgroup$ – reuns Aug 24 '16 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.