# Using least common multiple to prove there exists a prime between $2x$ and $3x$

Let $$\text{lcm}(x)$$ be the least common multiple of $$\{1,2,3,\dots, x\}$$.

Hanson showed that $$\text{lcm}(x) < 3^x$$

I'm wondering if the following argument is valid for showing that there is always a prime $$p$$ such that $$2x < p < 3x$$ for $$x \ge 246$$.

Here is my argument:

(1) if prime $$p$$ satisfies $$2x < p < 3x$$, then $$p | {{3x}\choose{2x}}$$.

This follows from the fact that $${{3x}\choose{2x}}$$ is an integer and that $$p$$ will not divide out by $$2x!$$ or $$x!$$.

(2) $${{3x}\choose{2x}} > \dfrac{6^x}{x}$$ for $$x \ge 4$$

For $$x \ge 4$$, $${{2x}\choose{x}} \ge \dfrac{4^x}{x}$$ since $${8\choose4} = 70 > \dfrac{4^4}{4} = 64$$ and $${{2x}\choose{x}} = 2\left(\dfrac{2x-1}{x}\right){2(x-1)\choose{x-1}} > 2\left(\dfrac{2x-1}{x}\right)\left(\dfrac{4^{x-1}}{x-1}\right) > \dfrac{4^x}{x}$$ and $${{3x}\choose{2x}} > \left(\dfrac{3x}{2x}\right)^x {{2x}\choose{x}} > \left(\dfrac{3}{2}\right)^x\dfrac{4^x}{x}$$

(3) For any prime $$q$$ such that $$\frac{3x}{2} \le q \le 2x$$, it follows that $$2q > 3x$$ and $$q \nmid {{3x}\choose{2x}}$$ since it will be divided out by $$2x!$$.

(4) Assume that there is no prime greater than $$2x$$ that divides $${{3x}\choose{2x}}$$

(5) It follows that $${{3x}\choose{2x}} < \text{lcm}(\frac{3x}{2})\text{lcm}(\sqrt{3x})$$ since:

• From Legendre's Formula, it is well known that if $$v_p(x)$$ is the highest power of $$p$$ that divides $$x$$, then $$v_p({{3x}\choose{2x}}) = \sum\limits_{i \le \log_p(3n)} \left\lfloor\frac{3x}{p^i}\right\rfloor - \left\lfloor\frac{2x}{p^i}\right\rfloor - \left\lfloor\frac{x}{p^i}\right\rfloor$$ which equals $$1$$ or $$0$$ for each $$i$$ so that $$v_p({{3x}\choose{2x}}) \le \log_p(3x)$$

• If a prime $$p > \sqrt{3x}$$, then $$v_p({{3x}\choose{2x}}) = 1$$ and $$p | \text{lcm}(\frac{3x}{2})$$

• If a prime $$p \le \sqrt{3x}$$, then $$v_p({{3x}\choose{2x}}) \le \log_p(3x) \le v_p(\frac{3x}{2})+1$$

(6) So that: $$\dfrac{6^x}{x} < 3^{3x/2}3^{\sqrt{3x}}$$

(7) But this is not true for $$x \ge 246$$ since:

$$246\ln 6 - \ln 246 > 435.26 > 435.24 > \frac{3}{2}(246)\ln 3 + \sqrt{3\times246}\ln 3$$

and for $$x \ge 246$$, $$6 > (3^{3/2})(3^{\sqrt{3x+2} - \sqrt{3x}})\left(\dfrac{x+1}{x}\right)$$

(8) So we can reject step (4).