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Let $x\#$ be the primorial for $x$.

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

It occurs to me that for $x \ge 4$, it is straight forward to find a lower bound of $\dfrac{(x^2+x)\#}{\left(\frac{x^2+x}{2}\right)\#}$.

I find it interesting that an upper bound can lead to a lower bound.

Is my reasoning correct?

Here's my argument:

(1) $\text{lcm}\left(\dfrac{x^2+x}{2}\right)\text{lcm}(x)\left(\dfrac{(x^2+x)\#}{\left(\frac{x^2+x}{2}\right)\#}\right) > \text{lcm}(x^2+x) $

  • if a prime $p \ge (x+1)$, then $p^2 > (x^2+x)$

  • if a prime $(x^2+x) \ge p^a > \frac{x^2+x}{2}$, then $p^{a+1} > (x^2+x)$ and $p^{a-1} \le \frac{x^2+x}{2}$

(2) $\text{lcm}(x^2+x) \ge {{x^2+x}\choose{\frac{x^2+x}{2}}}$

This follows from Legendre's Formula since:

  • Let $v_p(x)$ be the highest power of $p$ that divides $x$

  • $v_p({{x^2+x}\choose{\frac{x^2+x}{2}}}) = \sum\limits_{i \ge 1}\left\lfloor\dfrac{x^2+x}{p^i}\right\rfloor - 2\left\lfloor\dfrac{x^2+x}{2p^i}\right\rfloor$

  • It is well known that for each $i$, the difference is at most $1$ and that if $i > \log_p(x^2+x)$, then the difference is $0$.

(3) ${{x^2+x}\choose{\frac{x^2+x}{2}}} > \dfrac{2^{x^2+x}}{\frac{x^2+x}{2}}$

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

(4) Using Hanson's result that $\text{lcm}(x) < 3^x$, gives:

$$\left(\frac{(x^2+x)\#}{\left(\frac{x^2+x}{2}\right)\#}\right) > \left(\frac{4}{3}\right)^{(x^2+x)/2}\left(\frac{2}{(x^2+x)3^x}\right)$$

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