Counting prime powers $p^{\frac{k}{t}} \left( t \in \mathbb{R}{+}, k \in \mathbb{N} \right)$ by changing $\rho$'s in $\psi(x)$? With $\rho=\beta+ \gamma \,i$ being a non-trivial zero of $\zeta(s)$, the logarithmic prime counting function is:
$$\psi(x) = x - \log(2\pi) - \frac12 \log\left(1- \frac{1}{x^2}\right)  -  \sum_{\rho} \left(\dfrac{x^{\beta+ \gamma \,i}}{\beta+ \gamma \,i}+\dfrac{x^{1-\beta- \gamma \,i}}{1-\beta- \gamma \,i}\right)$$
This function counts all prime powers of the type $p^k, k \in \mathbb{N}$ residing $\le x$. I wondered what would happen to the $\rho$'s when I would divide $k$ by a real number $t \in \mathbb{R}^{+}$. Numerical evidence suggests that the following formula holds for all $t$ (i.e. counting all prime powers of the form $p^{\frac{k}{t}})$:
$$\psi(x,t) = x - \log(2\pi) - \frac12 \log\left(1- \frac{1}{x^2}\right)  -  \sum_{\gamma} \left(\dfrac{x^{1-\frac{t}{2}+ t\,\gamma \,i}}{1-\frac{t}{2}+ t\,\gamma \,i}+\dfrac{x^{1-\frac{t}{2}- t\,\gamma \,i}}{1-\frac{t}{2}- t\,\gamma \,i}\right)$$
The graph below illustrates the point (for three values of $t$ and using the first 1000 $\rho$'s):
 
The adjusted real part $\beta = 1-\frac{t}{2}$ can now become any value $<1$. For instance: at $t=2$ then $\beta=0$ and $\psi(x,2)$ counts all primes, prime powers but now also its squared roots $\le x$.
Could $\psi(x,t)$ also be formally derived 'bottom-up' through Fourier analysis (and assuming the RH)?  
 A: Consider, for $\sigma>1$, the function
$$ \psi(x,t):=\frac{1}{2\pi i}\int_{\sigma-\infty}^{\sigma+\infty}\frac{x^s}{s}\cdot\frac{-\zeta'}{\zeta}\left(1+\frac{s-1}{t}\right)ds.$$
On the one hand, shifting the contour indefinitely to the left, the residue theorem yields
$$ \psi(x,t)=x-\frac{\zeta'}{\zeta}\left(1-\frac{1}{t}\right)-\sum_{n=1}^\infty \frac{x^{1-(2n+1)t}}{1-(2n+1)t}-\sum_\rho \frac{x^{1-t+t\rho}}{1-t+t\rho},$$
where the $\rho$-sum is over the zeros of $\zeta$ in the critical strip, with $\rho$ paired up with $1-\rho$. The second term comes from the residue at $s=0$, while the $n$-sum is the contribution of the $\zeta$-zeros at negative even integers $s=-2n$.
That is, the function $\psi(x,t)$ defined above is essentially the same as your $\psi(x,t)$ under the Riemann Hypothesis, except that the constant $\log(2\pi)$ and the tiny term $\frac12 \log\left(1- \frac{1}{x^2}\right)$ in your formula for $\psi(x,t)$ should be adjusted as given above for an exact match. On the other hand, by an inverse Mellin transform we obtain, for $x$ not an integer,
$$ \psi(x,t)=\sum_{n^{1/t}\leq x}\Lambda(n)n^{\frac{1}{t}-1},$$
where $\Lambda(n)$ is the von Mangoldt function supported on the prime powers $n=p^k$. (For $x$ an integer the same holds except that the term for $n=x^t$ must be halved.)
In particular, $\psi(x,t)$ jumps exactly at the numbers of the form $n^{1/t}=p^{k/t}$, and your observation has been explained rigorously.
P.S. Note that for $t=1$ we have
$$\frac{\zeta'}{\zeta}\left(1-\frac{1}{t}\right)=\log(2\pi)\qquad\text{and}\qquad \sum_{n=1}^\infty \frac{x^{1-(2n+1)t}}{1-(2n+1)t}=\frac12 \log\left(1- \frac{1}{x^2}\right).$$
