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Interested in the following function:

$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}=\sum_{n=1}^\infty \frac{\lambda_n}{n^s}, $$

where $\pi(n)$ is the prime counting function.

Was thinking about:

$$ \bar L(s,\chi)=\sum_{n=1}^\infty\frac{|\chi(n)|^2}{n^s}, $$

and

$$ L_k(s,\chi)=\sum\limits_{n=1}^\infty \frac{|\chi(n)|^{2k}}{n^s}. $$

However, I don't think these can yield non-periodic integer sequences in the numerator because $\exists k\in\Bbb Z^+: \chi(n)=\chi(n+k)\,\forall n. $

So to achieve non periodicity I settled on modifying the denominator, specifically changing $n$ to $\pi(n)$:

$$ \Psi(s)=\sum_{n=2}^\infty \frac{1}{\pi(n)^s}=1+\frac{1}{2^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{4^s}+... $$

and combining like terms:

$$ \Psi(s)=\sum_{n=1}^\infty \frac{\lambda_n}{n^s}=1+\frac{2}{2^s}+\frac{2}{3^s}+\frac{4}{4^s}+\frac{2}{5^s}+\frac{4}{6^s}+\frac{2}{7^s}+... $$

So far I've computed the non-periodic sequence, $\lambda_n$, to $34$ terms:

$\lambda_n=\{1,2,2,4,2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,10\}.$

Just found that $\lambda_n$ is sequence A001223 in the oeis; Prime gaps: differences between consecutive primes.

Questions:

Does this sum converge for all $Re(s)>1$?

Is there a closed form for the sums?

Can $\Psi(s)$ be analytically continued? If so, how?

Where does $\Psi(s)=0$?

Is there a Euler product for $\Psi(s)?$

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    $\begingroup$ It definitely converges for $Re(s)>1$ since $\pi(n)\geq\frac12\cdot\frac{n}{\log n}$ for all $n$ sufficiently large. In particular, the summand is $\ll \frac{(\log n)^s}{n^s}\leq\frac{1}{n^{s-\varepsilon}}$. $\endgroup$ – Clayton Apr 25 at 21:16
  • $\begingroup$ oh okay awesome $\endgroup$ – Ultradark Apr 25 at 21:17
  • $\begingroup$ Almost surely the other three questions do not have tidy answers. Although it definitely has a singularity at $s=1$ by Landau's theorem. $\endgroup$ – Greg Martin Apr 25 at 21:25
  • $\begingroup$ Yeah I figured they would be difficult questions, but definitely attainable $\endgroup$ – Ultradark Apr 25 at 21:26
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$$F(s)=\sum_{m=2}^\infty \pi(m)^{-s}=\sum_{n=1}^\infty n^{-s} (p_{n+1}-p_n)= \sum_{n=1}^\infty p_{n+1} (n^{-s}-(n+1)^{-s})\\ =\sum_{n=1}^\infty n \ln n (1+o(1)) (s n^{-s-1}+O(s(s+1)n^{-s-2})$$ so it converges and it is analytic for $\Re(s) > 1$ and as $s \to 1$, $F(s) \sim -s\zeta'(s) \sim \frac{1}{(s-1)^2}$.

For the analytic continuation under the RH $n=\pi(p_n) = Li(p_n)+O(p_n^{1/2+\epsilon})$ thus $p_n = Li^{-1}(n+O(n^{1/2+\epsilon}))=Li^{-1}(n)+O(n^{1/2+\epsilon})$ and $F(s)-s\sum_{n=1}^\infty n^{-s-1} Li^{-1}(n)$ is analytic for $\Re(s) > 1/2$.

So it is now a problem about the asymptotic expansion of $Li^{-1}$

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  • $\begingroup$ Thank you, could you elaborate on the last remark? $\endgroup$ – Ultradark Apr 25 at 22:14
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    $\begingroup$ @Ultradark You can start with a 1,2 or 3 terms asymptotic for $Li(x)$ to deduce a 3 terms asymptotic for its inverse (and show us what you get). The question is if the remainder is $O(x^a)$ with $a < 1$ or not $\endgroup$ – reuns Apr 25 at 22:20
  • $\begingroup$ why is the question if the remainder is $O(x^a)$ with $a<1$ $\endgroup$ – Ultradark Apr 28 at 0:21

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