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Lemma   Let $n$ be a positive integer and $p$ be a prime. The highest power of $p$ dividing $n!$ is $\displaystyle\sum_j\left\lfloor\frac{n}{p^j}\right\rfloor$ , where $\lfloor x\rfloor$ is the floor of $x$ .

Proof. Let $p^k$ divide $n!$ with $k$ as large as possible. Every $p$ th term of the sequence $1,\dots,n$ is divisible by $p$ , so contributes a factor to $p^k$ . There are $\displaystyle\left\lfloor\frac{n}{p}\right\rfloor$ such factors. Every $p^2$ th term contributes an extra factor above that, providing $\displaystyle\left\lfloor\frac{n}{p^2}\right\rfloor$ new factors. In general, the $p^j$ th terms contribute $\displaystyle\left\lfloor\frac{n}{p^j}\right\rfloor$ extra factors to $p^k$ . So the highest power of $p$ dividing $n!$ is $\displaystyle\sum_j\left\lfloor\frac{n}{p^j}\right\rfloor$ .

Proof. [Bertrand's conjecture] Let $n>2$ be a minimal counterexample to the claim. Thus there is no prime $p$ such that $n<p<2n$ .

In the sequence of primes$$ 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503,$$ each succeeding term is smaller than the double of its predecessor. This shows that $n\ge 2503$ .

The binomial expansion of $(1+1)^{2n}$ has $2n+1$ terms and has largest term $\binom{2n}{n}$ . Hence$$ \binom{2n}{n}\ge\frac{4^n}{2n+1}\ge\frac{4^n}{(2n)^2}.$$

For a prime $p$ define $r(p,n)$ to be the highest power of $p$ dividing $\binom{2n}{n}$ . To compute $r(p,n)$ , we apply the lemma to $(2n)!$ and $n!$ . We get that$$ r(p,n)=\sum_{j\ge 1}\left\lfloor\frac{2n}{p^j}\right\rfloor -2\sum_{j\ge 1}\left\lfloor\frac{n}{p^j}\right\rfloor =\sum_{j\ge 1}\left(\left\lfloor\frac{2n}{p^j}\right\rfloor -2\left\lfloor\frac{n}{p^j}\right\rfloor\right).$$ The terms of the sum are all 0 or 1 and vanish when $\displaystyle j>\left\lfloor\frac{\log(2n)}{\log p}\right\rfloor$ , so $\displaystyle r(p,n)\le\left\lfloor\frac{\log(2n)}{\log p}\right\rfloor$ , that is, $p^{r(p,n)}\le 2n$ .

Now $\displaystyle\binom{2n}{n}=\prod_p p^{r(p,n)}$ . By the inequality just proved, primes larger than $2n$ do not contribute to this product, and by assumption there are no primes between $n$ and $2n$ . So$$ \binom{2n}{n}=\prod_{\substack{1\le p \le n\\ p\text{\ prime}}}p^{r(p,n)}.$$

For $n>p>\frac{2n}{3}$ , $\frac{3}{2}>\frac{n}{p}>1$ and so for $p>\frac{2n}{3}>\sqrt{2n}$ we can apply the previous formula for $r(p,n)$ and find that it is zero. So for all $n>4$ , the contribution of the primes larger than $\frac{2n}{3}$ is zero.

If $p>\sqrt{2n}$ , all the terms for higher powers of $p$ vanish and $\displaystyle r(p,n)=\left\lfloor\frac{2n}{p}\right\rfloor-2\left\lfloor\frac{n}{p}\right\rfloor$ . Since $r(p,n)$ is at most 1, an upper bound for the contribution for the primes between $\sqrt{2n}$ and $\frac{2n}{3}$ is the product of all primes smaller than $\frac{2n}{3}$ . This product is $\exp\left(\vartheta(\frac{2n}{3})\right)$ , where $\vartheta(n)$ is the Chebyshev function$$ \vartheta(n)=\sum_{\substack{p\le n\\p\text{\ prime}}}\log p.$$ There are at most $\sqrt{2n}$ primes smaller than $\sqrt{2n}$ and by the inequality $p^{r(p,n)}\le 2n$ their product is less than $(2n)^{\sqrt{2n}}$ . Combining this information, we get the inequality$$ \frac{4^n}{(2n)^2}\le\binom{2n}{n}\le(2n)^{\sqrt{2n}}\exp\left({\textstyle\vartheta(\frac{2n}{3})}\right).$$ Taking logarithms and applying the upper bound of $n\log 4$ for $\vartheta(n)$ , we obtain the inequality $\frac{n}{3}\log 4\le(\sqrt{2n}+2)\log(2n)$ , which is false for sufficiently large $n$ , say $n=2^{11}$ . This shows that $n<2^{11}$ .

Since the conditions $n\ge 2503$ and $n<2^{11}$ are incompatible, there are no counterexamples to the claim.

proof of Bertrand's conjecture is owned by Chi Woo, Lieven Marchand, Michael Slone.