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).


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