Number of prime number? Let $G_n=\{1,2,3,\ldots, n\}$ be the set of natural number. 
Thus for each $G_i$ had order/number of elements $i$.
Let $p_i$ denote the number of primes in $G_i$.
Define a function $f:\mathbb N\rightarrow \mathbb Q$ by $f(i)=\frac{p_i}{i}$. That was, $f$ was a function of the "density" of prime number in $G_i$. (I took the idea of distribution from statistics. Notice this was not a distribution, but just a description of how the prime number was separated.) 
Could we obtain such a function $f$ ? If not, could we obtain a continuous function $g:\mathbb R^+\rightarrow \mathbb R$  such that it's the least function, meaning there was no other similar function $g_2(x)<g(x)$ for all $x$, that fits $f$? 
(Meaning $g(x)$ close to $f(x)$ but $g(x) \geq f(x)$ for all $x\in\mathbb N$)
 A: According to Prime Number Theorem, the number of prime numbers below some $N$ becomes $\frac N {\log(N)}$ as $N$ becomes large. Your function is simply the number of primes divided by $N$, and therefore your $f(i)$ would approach $\frac1{\log(i)}$ as $i$ becomes large.
A: There is a need to strictly define "close" and "similar" here. I know the way in which you think about your question so you do not need to define those terms because of me.
It could be hard to find your continuous $f$ that is as simple as possible and that at the same time takes those values at positive integers because we do not have a nice formula for $n$-th prime number.
Also, you know that to every continuous function you can assign an infinite number of other continuous functions that are as close as you want them to be to that function so you probably want that your $f$ is also differentiable, probably as many times as we want.
You want an exact expression for prime-counting function defined all over positive reals and that does not look easy enough to obtain, at present time we have just an approximations and it is a question when we will have the thing that you want.
