Questions about the prime counting function

[riemannian]

greetings . i have a couple of questions about the prime counting function .
when \pi _{0}(x) changes by 1, then it's logical to assume that it should happen at a prime argument . meaning :

\lim_{\xi\rightarrow 0}\pi _{0}(x+\xi )-\pi _{0}(x-\xi )=1

implies that x is a prime .
is this a true assumption ?

according to the literature, we can expand \pi_{0}(x) using the riemann R function .

R(x) = 1+\sum_{k=1}^{\infty}\frac{(ln x)^{k}}{k!k\zeta (k+1)}

\pi_{0}(x)= R(x)-\sum_{\rho} R(x^{\rho })-\frac{1}{lnx}-\frac{1}{\pi}tan^{-1}\left( \frac{\pi}{lnx}\right) = \sum_{k=1}^{\infty}\frac{(lnx)^{k}[1-\sum \rho^{k} ]}{k!k\zeta(k+1)}-\frac{1}{lnx}-\frac{1}{\pi}tan^{-1}\left( \frac{\pi}{lnx}\right)\rho being the nontrivial zeros of the zeta function .
is this correct ?? i mean , is the expansion correct ??

 

[riemannian]

i think i got my answer .

\Pi _{0}(x)=\sum_{n=1}^{\infty}\frac{1}{n}\pi_{0}(x^{1/n})

and \Pi_{0}(x) does change by 1 at primes .

now i am intrigued by the terms \sum \rho^{k} , it seems to me we don't need to know \rho themselves to evaluate \Pi_{0}(x) , we just need to evaluate \sum \rho^{k} . is this correct ??