Every simple group below are assumed non-abelian.
Let us call the class number $k(G)$ of a finite group $G$ the number of its conjugacy classes (also, the number of its irreducible complex representations, up to equivalence). Here is the computation of the order and the class number of the first finite simple groups:
gap> it:=SimpleGroupsIterator(10,10000);
<iterator>
gap> for G in it do Print([G,Order(G),NrConjugacyClasses(G)]); od;
[ A5, 60, 5 ][ PSL(2,7), 168, 6 ][ A6, 360, 7 ][ PSL(2,8), 504, 9 ][ PSL(2,11), 660, 8 ][ PSL(2,13), 1092, 9 ][ PSL(2,17), 2448, 11 ][ A7, 2520, 9 ][ PSL(2,19), 3420, 12 ][ PSL(2,16), 4080, 17 ][ PSL(3,3), 5616, 12 ][ PSU(3,3), 6048, 14 ][ PSL(2,23), 6072, 14 ][ PSL(2,25), 7800, 15 ][ M11, 7920, 10 ][ PSL(2,27), 9828, 16 ]
In the following picture, the points are $(k(G),|G|)$ with $G$ simple of order less than $10^8$.
My laptop can hardly compute such a picture for finite simple groups of order up to $10^9$ or $10^{10}$.
Question: Could you post your best such picture (computed yourself or found in the literature or on the web or elsewhere) which would be significatively better than above one?
Observation: Using NrConjugacyClassesPSL we can show that $k(\mathrm{PSL}(2,2^n)) = 2^n+1$, for $n \le 50$.
Bonus question: Is it true for all $n$?
Remark: The curve which appears in the middle of the picture should have equation $\frac{1}{2}(2r-6)(2r-5)(2r-4)$ and its points should exactly be given by the groups $\mathrm{PSL}(2,q)$ with $q$ odd prime power (see the bonus of this post for more details).