Let us look at a set of $n$ points in the plane ( $n \geq 5$ ) with the property that no four of the points are collinear.
Then , can we find out a closed expression for---
$f(n)=$the maximum number of points ( out of these $n$ points ) such that we can guarantee that no three of these $f(n)$ number of points are collinear .
(Obviously , $f(n)$ is taken over all possible legal configuration of the $n$ points)
This is a wide-open research problem, and we are very very far from finding a closed expression - we do not know the right order of magnitude.
Erdős would appear to be the one to have first posed this question, and the best known bounds are due to Füredi, who showed that $$ \Omega \left( \sqrt{n \log n} \right) = f(n) = o (n). $$ That is, $f(n)$ is at least of the order of magnitude of $\sqrt{n \log n}$, and $f(n)$ must be sublinear. As you can see, there is a multiplicative gap of almost $\sqrt{n}$ between the bounds. You can find his paper here, together with some extensions (where the $n$ points are allowed to have larger collinear sets) here.
Note that Füredi's upper bound uses the Hales-Jewett theorem, so it is truly an asymptotic result. Using the best-known bounds from the density Hales-Jewett theorem would give an upper bound of the form, for some constant $c > 0$, $$ f(n) \le \frac{cn}{\sqrt{ \log^* n }}, $$ where $\log^* n$ is the inverse of the tower function - it is the height of a tower of twos (iterative exponentiation) needed to get $n$.
This is an extremely slow-growing function; for example, $\log^* \left(2^{65536} \right) = 5$, and $\log^* \left ( 2^{ \left ( 2^{65536}\right) }\right) = 6$. Hence the upper bound is only significantly sublinear for unthinkably large values of $n$.
To finish, let me point out that $\log \log n$ goes to infinity much much much much faster than $\log^* n$, and of $\log \log n$ itself it was once famously said: