# Prove or disprove $\frac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)} < (n-1)!$

Is it correct that for any positive integers $$x,n$$, that $$\frac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)} < (n-1)!$$ where lcm is the least common multiple.

I ask because I find this relationship very interesting but I haven't seen it stated in this form.

I have seen a related inequality that:

$$\frac{\text{lcm}(x,x+1, \dots, x+n)}{x} \ge {{x+n}\choose{n}} = \frac{(x+n)!}{x!n!}$$

Which rearranges to this:

$$\frac{n!}{x} \ge \frac{(x+n)!}{(x!)\text{lcm}(x,x+1,\dots,x+n)}$$

Or even closer:

$$n! \ge \frac{(x+n)!}{((x-1)!)\text{lcm}(x,x+1,\dots,x+n)}$$

So that:

$$n! \ge \frac{(x+n)!}{(x!)\text{lcm}(x+1,\dots,x+n)}$$

This appears to me to be a stronger result than the one I am asking about. I am not clear how to derive my result from this stronger result.

On the other hand, I am able to justify my result independently of this equation. Here's my argument:

(1) Let $$f_n(x) = \dfrac{(x+n)!}{(x!)\text{lcm}(x+1, \dots, x+n)}$$

(2) No prime greater than $$n-1$$ divides $$f_n(x)$$ since:

Assume that a prime $$p>n$$ divides $$x+c$$ and $$x+d$$ with $$0 < c < d \le n$$. It follows that $$p | (x+d - x+c) = d - c < n$$ which is impossible.

(3) For each prime $$p < n$$ that divides $$f_n(x)$$, we can use Legendre's Formula to get this result (since we are dividing by the least common multiple):

$$v_p\left(\frac{(x+n)!}{(x!)\text{lcm}(x+1,\dots,x+n)}\right) = \sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^i}-1\right\rfloor < v_p((n-1)!) = \sum_{i=1}^{\infty}\left\lfloor\frac{n-1}{p^i}\right\rfloor$$

where $$v_p(x)$$ is the largest power of $$p$$ that is less than or equal to $$x$$

Note: It is based on $$n$$ instead of $$x+n$$ because $$p^t$$ is necessarily less than $$n$$

Is my reasoning correct? Is there a straight forward way to derive this result from the first equation? Is the argument that I present using Legendre's Formula valid? If valid, can it be improved or simplified? If not valid, what was my mistake?

If $$n=2$$, then we get, for any $$x$$, $$f_2(x)=\frac{(x+2)!}{(x!)\text{lcm}(x+1, x+2)}=\frac{(x+2)!}{(x!)(x+1)(x+2)}=1\color{red}=(2-1)!$$
If $$n=3$$ and $$x=2^k-1$$ where $$k$$ is a positive integer, then since $$\text{lcm}(x+1,x+2,x+3)=2^{k-1}(x+2)(x+3)$$ we get $$f_3(2^k-1)=\frac{(x+1)(x+2)(x+3)}{\text{lcm}(x+1, x+2, x+3)}=\frac{2^k}{2^{k-1}}=2\color{red}=(3-1)!$$
• Thanks for the counter examples! In each case $\sum\limits_{i=1}^{\infty} \left\lfloor\dfrac{n}{p^i}-1\right\rfloor = \sum\limits_{i=1}^{\infty} \left\lfloor\dfrac{n-1}{p^i}\right\rfloor$. Have you found any counter examples where this formula is wrong? I am wondering if changing $<$ to $\le$ addresses the counter examples. – Larry Freeman May 23 at 18:32
• @Larry Freeman : Sorry for the late reply. No, I haven't. Have you proven $f_n(x)\color{red}{\le} (n-1)!$ in this question? – mathlove May 25 at 13:31