It is well known that a factorial can restated as a product of least common multiples:

$$\log(x!) = \sum_{m=1}^{\infty}\psi\left(\frac{x}{m}\right)$$

Using Möbius inversion formula, it occurs to me that this can be restated as:

$$\text{lcm}(1,2,\dots,x) = \prod_{m=1}^{\infty}\left(\left\lfloor\frac{x}{m}\right\rfloor!\right)^{\mu(m)}$$


$$\frac{\text{lcm}(1,2,\dots,(x^2+x))}{\text{lcm}(1,2,\dots,x^2)} = \prod_{m=1}^{\infty}\left(\frac{\lfloor\frac{x^2+x}{m}\rfloor!}{\lfloor\frac{x^2}{m}\rfloor!}\right)^{\mu(m)}$$

As I was thinking about this, it seemed to me that if $\lfloor\frac{x}{6}\rfloor \ge 2$, then:

$$\prod_{m=6}^{\infty}\left(\frac{\lfloor\frac{x^2+x}{m}\rfloor!}{\lfloor\frac{x^2}{m}\rfloor!}\right)^{\mu(m)} \ge 1$$

Is there a straight forward way to check my hypothesis? Is this a well-understood problem with a straight-forward proof or is this an open problem?


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