# An alternative proof for Bertrand's Postulate when $n \ge 36$

I've been thinking about how factorials compare to primorials. I have not seen this argument before. If I made a mistake, please call it out. :-)

Let $x+1, x+2, \dots, x+n$ be a sequence of consecutive integers.

Let $r_{x,1},r_{x,2},\dots,r_{x,n}$ have the following properties:

• $r_{x,i} | (x+i)$
• $r_{x,1}, r_{x,2}, \dots, r_{x,n}$ are all relatively prime to each other.
• $\prod\limits_{i \le n}r_{x,i}$ is the least common multiple for $x+1, x+2, \dots, x+n$

Example:

• Let $x=35, n=5$ so that we have $36, 37, 38, 39, 40$
• $r_{35,1}=9, r_{35,2} = 37, r_{35,3} = 19, r_{35,4}=13, r_{35,5}=40$
• Each $r_i$ is relatively prime to each other and the product of all equals the least commmon multiple.

Let $r(x,n) = \prod\limits_{i \le n}r_{x,i}$

Claim 1: For $x \ge 36$, $\pi(x) \le \left\lfloor\frac{x}{3}\right\rfloor$

Except for $2,3$ all primes are $6x+1$ or $6x+5$.

If $x \ge 36$:

$\pi(x) \le \pi(36) + \left\lfloor\frac{x+1-36}{6}\right\rfloor + \left\lfloor\frac{x+5-36}{6}\right\rfloor \le \left\lfloor\frac{66}{6}+\frac{x-35}{6} + \frac{x-31}{6}\right\rfloor = \left\lfloor\frac{x}{3}\right\rfloor$

Claim 2: For $n \ge 36$, assuming no primes between $n$ and $2n$, there are at least $\left\lfloor\frac{2n}{3}\right\rfloor$ instances of distinct $i$ where $r_{n,i}=1$.

Note: These instances could occur anywhere between $n$ and $2n$.

Since $r_{n,1}, r_{n,2}, \dots, r_{n,n}$ are relatively prime to each other and there are less than or equal to $\left\lfloor\frac{n}{3}\right\rfloor$ primes, the rest must be $1$ and since each prime is uniquely found in one and only one $r_{n,i}$

Claim 3: For $n \ge 36$, $\dfrac{(2n)!}{n![r(n,n)]} \ge \dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!}$ and $r(n,n) \le \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$

Both of these follow directly from Claim 2:

• $r(n,n) < \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$ since there at most $\left\lfloor\frac{n}{3}\right\rfloor$ instances where $r_{n,i}\ne1$ and the product of any such instances is less than or equal to $\dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$

• $\dfrac{(2n)!}{n![r(n,n)]} > \dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!}$ since $\dfrac{(2n)!}{n![r(n,n)]} = \prod\limits_{i \le n \text{ and } r_{n,i}=1} (n+i)$ and there are at least $\left\lfloor\frac{2n}{3}\right\rfloor$ instances where $r_{n,i}=1$ and the product of any $\left\lfloor\frac{2n}{3}\right\rfloor$ instances will be greater or equal to $\dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!}$.

Claim 4: $\dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!} > \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$

• $\left\lfloor\frac{2n}{3}\right\rfloor \ge 2\left\lfloor\frac{n}{3}\right\rfloor$
• The product of any two instances of $(n+i)(n+j)$ will be greater than $(n+1)(n+2) = n^2+3n + 2 > 2n$ for $n \ge 1$.
• It follows that the $\left\lfloor\frac{2n}{3}\right\rfloor$ can be ordered into $\left\lfloor\frac{n}{3}\right\rfloor$ pairs of $2$ instances.

Claim 5: For $n \ge 36$, it is impossible that there are no primes between $n$ and $2n$.

Since $\dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!} > \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$, applying Claim 2 and Claim 3, it follows that:

$\dfrac{1}{2}\dfrac{(2n)!}{n!} < \dfrac{(2n)!}{n![r(n,n)]}$

But this is impossible since $r(n,n) \ge n\# > 2$

Edit 1: I found some minor problems with Claim 2 and Claim 3.

While Claim 1 establishes that $\pi(x) \le \left\lfloor\frac{x}{3}\right\rfloor$, these primes could be at any point between $n+1$ and $2n$. It is not correct to assume that $r_{n,i}=1$ for $i > \left\lfloor\frac{n}{3}\right\rfloor$

I have modified argument to remove this assumption.

Edit 2: It looks like my definition of $r_i$ does not work as I expected.

For example for the sequence $36, 37, 38, 39, 40$, by my previous definition: $r_1=36, r_2=37, r_3 = 19, r_4=13, r_5=10$. But then, $\gcd(r_1,r_5)=2$.

I have updated the definition which I believe fixes the issue.

Edit 3: Removed a reference to primorial which I had missed and updated the last sentence.

It is not essential to the argument but it is incorrect. Thanks to @mathlove for noticing it.

Edit 4: Updated claim 3 based on comment by @JonMark Perry.

Edit 5: Claim 5 is not valid. See answer or comments for detail.

• Why are the $r_{x,i}$ integers? – amakelov May 2 '17 at 22:37
• Yeah, choose $x=i=p$ for some prime $p>2$ – amakelov May 2 '17 at 22:42
• You might want to edit the last sentence before "Edit 1". – mathlove May 5 '17 at 5:56
• I think your arguments in Claim 1, 2, 3, 4 are fine, but I can't understand how you get $\dfrac{1}{2}\dfrac{(2n)!}{n!} < \dfrac{(2n)!}{n![r(n,n)]}$ from Claim 2, 3 and 4. Can you please elaborate? – didgogns May 5 '17 at 12:55
• your proof of claim 3 proves $\le$, but the statement claims $\lt$ – JMP May 6 '17 at 2:04

Claim $5$ claims too much. We only have:
$$\frac{(2n)!}{n![r(n,n)]} \ge \frac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!} \ge \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!} \ge r(n,n)$$