TITLE: Are there any Combinatoric proofs of Bertrand's postulate?
QUESTION [14 upvotes]: I feel like there must exist a combinatoric proof of a theorem like: There is a prime between $n$ and $2n$, or $p$ and $p^2$ or anything similar to this stronger than there is a prime between $p$ and $(\prod_p p) + 1$ (Euclid's theorem).
I was trying to prove one by the Sieve on this grid
    1     2     3     4 ... p
  p+1   p+2   p+3   p+4 ... 2p
 2p+1  2p+2  2p+3  2p+4 ... 3p
...........................
...........................
........................... p^2
p^2+1 p^2+2 p^2+3 p^2+4 ...

but it did not work.
Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes?

REPLY [10 votes]: Edited: Here is a very relevant link on Robin Chapman's page.  It is the combinatorial proof which Erdos originally used to prove Bertrands postulate.  (Mentioned in Pete L. Clark's comment)
The PDF is very similar to the proofs I mentioned here, (and here)  and is based entirely on the primes dividing the central binomial coefficient.
I will leave in the proof regarding a lower bound for $\pi(x)$, since it is still relevant.

What follows is a proof Chebyshev's Theorem based on the primes dividing the central binomial coefficient.  First, two well known facts:
Fact 1: Let $r(n,p)$ denote the highest power of a prime $p$ that divides $n!$. Then $$r(n,p)=\left\lfloor\frac{N}{p}\right\rfloor+\left\lfloor\frac{2N}{p^{2}}\right\rfloor+\left\lfloor\frac{2N}{p^{3}}\right\rfloor+\cdots$$
Fact 2: For every integer $N\geq1$ we have that $$\frac{4^{N}}{2N}\leq\binom{2N}{N}.$$
Theorem: We have that $$\pi(x)\geq\ln(2)\frac{x}{\ln x}-2.$$
Proof. Consider $\binom{2N}{N}.$ For each prime $p$, let $v_{p}$ denote the number of times it divides $\binom{2N}{N}$. Hence $$\binom{2N}{N}=\prod_{p\leq2N}p^{v_{p}}.$$ Let $t=\log_{p}(2N)$ so that $\left\lfloor\frac{2N}{p^{j}}\right\rfloor=0$ when $j\geq t.$ By the lemma, we have $$v_{p}=\left(\left\lfloor\frac{2N}{p}\right\rfloor+\left\lfloor\frac{2N}{p^{2}}\right\rfloor+\cdots\right)-2\left(\left\lfloor\frac{N}{p}\right\rfloor+\left\lfloor\frac{N}{p^{2}}\right\rfloor+\cdots\right)=\sum_{i=1}^{t}\left\lfloor\frac{2N}{p^{i}}\right\rfloor-2\left\lfloor\frac{N}{p^{i}}\right\rfloor.$$ Since $\left\lfloor\frac{2N}{p^{i}}\right\rfloor-2\left\lfloor\frac{N}{p^{i}}\right\rfloor=0$ or $\left\lfloor\frac{2N}{p^{i}}\right\rfloor-2\left\lfloor\frac{N}{p^{i}}\right\rfloor=1$ for each $i$ we see that $v_{p}\leq t.$ Hence if $p^{r}$ divides $\binom{2N}{N}$ we must have $p^{r}\leq2N.$  Consequently $$\binom{2N}{N}=\prod_{p\leq2N}p^{v_{p}}\leq\prod_{p\leq2N}(2N)=(2N)^{\pi(2N)}.$$Applying the other lemma we get $\frac{4^{N}}{2N}\leq(2N)^{\pi(2N)}$and upon taking logarithms we find$$N\ln4-\ln(2N)\leq\pi(2N)\ln(2N)$$and hence $$\frac{2N\ln2}{\ln2N}\leq\pi(2N)+1.$$Since this holds for every even integer, and there is at most one prime between even integers we get
$$\frac{x\ln2}{\ln x}-2\leq\pi(x)\ \text{for all } \ x\geq2,$$ as desired.