distribution of prime powers Let $f(x)$ be the number of prime powers less than $x$
(by prime power I mean any number of the form $p^n$ such that $p$ is prime and $n \ge 1$.)  

Consider the limit: lim x->∞ f(x)/x
I would like to know the value of L
In particular, this article seems to suggest that the limit tend neither towards 0 nor 1 however I do not have the knowledge needed to understand the it:  

http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160270602/abstract

Thanks
Let's define $\Pi_n(x)$ as the number of prime powers $p^n$ with exponent $n$ that are less than $x$. Clearly $\Pi_1(x) = \Pi(x)$, the usual prime counting function.

Since $p^n &lt; x$ implies that $p &lt; x^{1/n}$, we have
$$
\Pi_n(x) = \Pi\left(x^{1/n}\right)\, .
$$
Your $f(x)$ is:
\begin{align}
f(x) &amp;= \sum_{n=1}^{\infty} \Pi_n(x)\\
&amp;= \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)\, .
\end{align}
For fixed $x$, $x^{1/n}$ will eventually go to 1 as $n\rightarrow\infty$, so this sum will actually terminate at some point. (Actually the sum need only go up to the point where $x^{1/n} &lt; 2$, i.e. $n_{max} = \left\lfloor \ln x / \ln 2 \right \rfloor$.)

Beyond that, at present, all I have are numerics (done in Mathematica) which loosely suggest that
$$
\lim_{x\rightarrow\infty} \frac{f(x)}{x} = 0\, ,
$$
although convergence seems to be very, very slow. (Perhaps going as $1/\ln(x)$?)

Edit:

Numerically, it looks like the limit
$$
\lim_{x\rightarrow\infty} \frac{f(x)}{x/\ln(x)}
$$
may exist, and have a value $\ge 1$. (I suspect the actual limit is 1 itself, but the convergence is very slow.)

Edit 2:

For those down-voting this, note that the original questioner asked me (in the comments above) to post the proof of $f(x) = \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)$. I'm just trying to be helpful. (Unless I've made an error here, in which case let me know.)
Let's define $\Pi_n(x)$ as the number of prime powers $p^n$ with exponent $n$ that are less than $x$. Clearly $\Pi_1(x) = \Pi(x)$, the usual prime counting function.

Since $p^n &lt; x$ implies that $p &lt; x^{1/n}$, we have
$$
\Pi_n(x) = \Pi\left(x^{1/n}\right)\, .
$$
Your $f(x)$ is:
\begin{align}
f(x) &amp;= \sum_{n=1}^{\infty} \Pi_n(x)\\
&amp;= \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)\, .
\end{align}
For fixed $x$, $x^{1/n}$ will eventually go to 1 as $n\rightarrow\infty$, so this sum will actually terminate at some point. (Actually the sum need only go up to the point where $x^{1/n} &lt; 2$, i.e. $n_{max} = \left\lfloor \ln x / \ln 2 \right \rfloor$.)

Beyond that, at present, all I have are numerics (done in Mathematica) which loosely suggest that
$$
\lim_{x\rightarrow\infty} \frac{f(x)}{x} = 0\, ,
$$
although convergence seems to be very, very slow. (Perhaps going as $1/\ln(x)$?)

Edit:

Numerically, it looks like the limit
$$
\lim_{x\rightarrow\infty} \frac{f(x)}{x/\ln(x)}
$$
may exist, and have a value $\ge 1$. (I suspect the actual limit is 1 itself, but the convergence is very slow.)

Edit 2:

For those down-voting this, note that the original questioner asked me (in the comments above) to post the proof of $f(x) = \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)$. I'm just trying to be helpful. (Unless I've made an error here, in which case let me know.)