Riemann Zeta function with the prime counting function in place of $n$

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.

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

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 $$\Psi(s)?$$

• 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 $\ll \frac{(\log n)^s}{n^s}\leq\frac{1}{n^{s-\varepsilon}}$. – Clayton Apr 25 at 21:16
• oh okay awesome – Ultradark Apr 25 at 21:17
• Almost surely the other three questions do not have tidy answers. Although it definitely has a singularity at $s=1$ by Landau's theorem. – Greg Martin Apr 25 at 21:25
• Yeah I figured they would be difficult questions, but definitely attainable – Ultradark Apr 25 at 21:26

$$F(s)=\sum_{m=2}^\infty \pi(m)^{-s}=\sum_{n=1}^\infty n^{-s} (p_{n+1}-p_n)= \sum_{n=1}^\infty p_{n+1} (n^{-s}-(n+1)^{-s})\\ =\sum_{n=1}^\infty n \ln n (1+o(1)) (s n^{-s-1}+O(s(s+1)n^{-s-2})$$ 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 $$Li^{-1}$$
• @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 – reuns Apr 25 at 22:20
• why is the question if the remainder is $O(x^a)$ with $a<1$ – Ultradark Apr 28 at 0:21