# Reasoning about least common multiples using ratios of factorials.

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

As I understand it, there is a well-known relationship between a factorial and the ratio of least common multiples (see my question here):

$$x! = \prod\limits_{i \ge 1}\text{lcm}\left(\frac{x}{i}\right)$$

For example:

$$6! = \text{lcm}(6)\text{lcm}(3)\text{lcm}(2) = (5\times4\times3)(3\times2)(2) = 720$$

Given:

$$\frac{(x^2+x)!}{(x^2)!} = \frac{\prod\limits_{i \ge 1}\text{lcm}((x^2+x)/i)}{\prod\limits_{j \ge 1}\text{lcm}((x^2)/j)}$$

Let $$p$$ be the highest prime less or equal to $$x^2+x$$.

Would it follow that there exists a combination of factorials such that:

$$\frac{(x^2+x)!\left\lfloor\frac{x^2}{2}\right\rfloor!\dots\left\lfloor\frac{x^2}{p}\right\rfloor!\left\lfloor\frac{x^2+x}{2\times3}\right\rfloor!\dots}{(x^2)!\left\lfloor\frac{x^2+x}{2}\right\rfloor!\dots\left\lfloor\frac{x^2+x}{p}\right\rfloor!\left\lfloor\frac{x^2}{2\times3}\right\rfloor!\dots} = \frac{\text{lcm}(x^2+x)}{\text{lcm}(x^2)}$$

The answers seems to me to be yes as long as all the combinations of primes $$\{2, \dots, p\}$$ are included in the correct way since this should cancel out the other $$\text{lcm}$$ values in the original equation.

Am I wrong?

Edit 1:

Here's an example with $$x=4$$:

$$\frac{20!\left(\frac{16}{2}\right)!\left\lfloor\frac{16}{3}\right\rfloor!\left\lfloor\frac{16}{5}\right\rfloor!\left\lfloor\frac{16}{7}\right\rfloor!\left\lfloor\frac{20}{6}\right\rfloor!\left(\frac{20}{10}\right)!}{16!\left(\frac{20}{2}\right)!\left\lfloor\frac{20}{3}\right\rfloor!\left(\frac{20}{5}\right)!\left\lfloor\frac{20}{7}\right\rfloor!\left\lfloor\frac{16}{6}\right\rfloor!\left\lfloor\frac{16}{10}\right\rfloor!}=\frac{\text{lcm}(4^2+4)}{\text{lcm}(4^2)} = 17\times19 = 323$$

Edit 2:

Mathlove pointed out a mistake in my original logic. I have changed the definition of $$p$$ to be the highest power less or equal to $$x^2 + x$$ (previously, it was $$p$$ to be the highest power less or equal to $$x$$ which was incorrect.

• For the last "equality", can you add an example for a small $x$, say $4$? It is difficult for me to understand what you want to say. – mathlove Oct 18 '18 at 4:52
• Great suggestion. $x=4$ works. I've added it as an example to the question. – Larry Freeman Oct 18 '18 at 5:24
• It looks you have $p=7$ for $x=4$, which seems to contradict "$p$ be the highest prime less or equal to $x$". – mathlove Oct 18 '18 at 5:52
• Good point. That was a mistake. I should say that p is the highest power equal or less than $x^2+x$. Otherwise, it won't cancel out when $i$ gets higher than $x$ in the above equation. – Larry Freeman Oct 18 '18 at 5:54
• "Let $p$ be the highest prime less or equal to $x^2+x$." Then, we have $p=19$ instead of $p=7$ for $x=4$, don't we? – mathlove Oct 18 '18 at 6:03