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 $\Psi(s)?$
$$F(s)=\sum_{m=2}^\infty \pi(m)^{-s}=\sum_{n=1}^\infty n^{-s} (p_{n+1}-p_n)= \sum_{n=1}^\infty p_{n+1} (n^{-s}-(n+1)^{-s})\\ =\sum_{n=1}^\infty n \ln n (1+o(1)) (s n^{-s-1}+O(s(s+1)n^{-s-2})$$ 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 $\Re(s) > 1/2$.