Why the clique numbers of almost all graphs are concentrated at two values? I read that for fixed $0\leq p \leq 1$, the clique number of $G(n, p)$ is concentrated at $d$ and $d+1$, where $d$~$2log_{\frac{1}{p}}n$.
The key point is that $d$ is the maximum integer with $\binom{n}{d}p^{\binom{d}{2}}\geq ln(n)$.
Is there any simple proof?
 A: I believe that Paul Hudford's answer thought that the question is about the chromatic number, whereas the question is about the clique number of the random graph. Another inaccuracy with Paul's answer (although it has nothing to do with the question asked) is that in the paper being referred to, Heckel proves anti-concentration for intervals of length $\sim n^{1/4}$ (not $\sim n^{1/2}$), but recently in joint work with Riordan, Heckel announced a proof of an optimal bound of the form $\sim n^{1/2}$: https://www.math.princeton.edu/events/non-concentration-chromatic-number-random-graph-2019-11-06t200000
To answer the original question, it is important to realize that when a statement is made about almost all graphs, the object in question is $G(n,1/2)$, as opposed to $G(n,p)$ where $p$ is a parameter that shrinks with $n$. When $p$ is some other value that is constant, usually the situation is similar.
Something that is true is the following: $G(n,1/2)$ with high probability has its clique number to be one of two special values, and these values are near $2\log(n)$. The intuition is as follows:
Let $X_k$ be the random variable counting the number of $k$-cliques in $G(n,1/2)$. If $\mathbb{E}(X_k)\to 0$, by Markov's inequality, $G$ almost surely does not contain a $k$-clique. If $\mathbb{E}(X_k)\to \infty$ on the other hand, calculating the variance will reveal that almost surely $G$ almost surely contains a $k$-clique. What needs to be true for two-point-concentration to hold is that $\mathbb{E}[X_k]$ has a phase shift that occurs around $k\sim2\log(n)$.
You can calculate that the ratio $\frac{\mathbb{E}[X_{k+1}]}{\mathbb{E}[X_{k}]}=o(1)$ when $k$ is close to $2\log(n)$. What this means is that $\mathbb{E}[X_k]$ cannot stay some constant value for more than a couple of values of $k$. Say for example that $\mathbb{E}[X_k]\sim 10$ for some special value of $k$ (near $2\log(n)$). Then, the result about the ratio I'm referring to shows that $\mathbb{E}[X_{k+1}]\to 0$ and $\mathbb{E}[X_{k-1}]\to \infty$, which already gives a three-point-concentration result.
For more details, I suggest Zhao's notes available here (pages 34-35): https://yufeizhao.com/pm/probmethod_notes.pdf
