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Let $x>0,n>0$ be integers.

Let LCM$(x+1, x+2, \dots, x+n)$ be the least common multiple for the set of integers $x+1, x+2, \dots, x+n$.

It seems to me that:

$$\text{LCM}(x+1, x+2, \dots, x+n) \ge \binom{x+n}{x}$$

Here's my thinking:

(1) Each prime that divides $n!$ necessarily divides $\frac{(x+n)!}{x!}$ but each prime that divides $\frac{(x+n)!}{x!}$ does not necessarily divide $n!$

(2) For each prime $p$ that divides $\frac{(x+n)!}{x!}$ but does not divide $n!$, it will necessarily divide ${{x+n}\choose{x}}$, it will be greater than $n$ and it will only divide $1$ integer in the sequence $x+1, x+2, \dots, x+n$.

(3) If $v_p(u)$ is the highest power of $p$ the divides $u$, then for each prime $p$ that divides $\frac{(x+n)!}{x!}$ but does not divide $n!$, with $p| x+i$ where $x+1 \le x+i \le x+n$, then:

$$p^{v_p(x+i)}\ \ |\ \ \text{LCM}(x+1, x+2, \dots, x+n)$$

(4) If $p|n!$ and $p|\frac{(x+n)!}{x!}$, then either $p$ divides the same number of integers in the sequence $1,2,\dots, n$ as $x+1, x+2, \dots x+n$, or it divides $1$ more in $x+1, x+2, \dots x+n$

Any prime $p < n$ will divide $p, 2p, \dots, p\lfloor\frac{n}{p}\rfloor$. It is possible that $p|x+1$. In this case, it will divide $x+1, x+1+p, \dots, x+1+p\lfloor\frac{x+n-1}{p}\rfloor$ so just $1$ more.

Since $p$ will necessarily divide at least one integer in $x+1, x+2, \dots, x+p$, it follows that at worst, it will divide the same number as integers in $1,2,\dots,n$ divisible by $p$.

(5) If $p$ divides the same number integers in the sequene $x+1, x+2, \dots, x+n$ as in $1, 2, \dots,n$, then at most $1$ integer has a power such that $p^{v_p(x+i)} > n$. The product of the remaining integers in $x+1, x+2, \dots, x+n$ divisible by $p$ will divide $n!$. Therefore, in this case, $v_p(x+i) > v_p({{x+n}\choose{n}})$

Assume that $p<n$ divides the same number of integers in the sequence $x+1, x+2, \dots, x+n$ as the sequence $1, 2, \dots, n$. Assume that there exists $i$ such that $p^{v_p(x+i)} > n$. There will only be $1$ such integer. The remaining integers divisible by $p$ will be less than $n$.

(6) If it divides $1$ more integer in the sequence $x+1, x+2, \dots, x+n$, then it does in $1, 2, \dots, n$, then it follows that the product of the remaining integers in $x+1, x+2, \dots, x+n$ divisible by $p$ will divide $n!$. In this case, $v_p(x+i) \ge v_p({{x+n}\choose{n}})$

Am I wrong?


Edit: @mathlove found a typo in my argument.

The argument depends on step (3) showing that $p^{v_p(x+i)} | $LCM but I had mistakenly stated $v_p(x+i) | $LCM.

I have updated the argument to fix the typo.

I am trying to show that every prime $p$ that divides ${{x+n} \choose x}$ also divides LCM$(x+1, x+2, \dots, x+n)$. This requires the corrected step(3) to be true.

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  • $\begingroup$ It is difficult for me to understand what (6) says. Can you make it clearer? $\endgroup$ – mathlove Sep 15 '18 at 5:55
  • $\begingroup$ Thanks for the feedback! I posted a question on whether my reasoning behind (6) is correct here. $\endgroup$ – Larry Freeman Sep 16 '18 at 0:46
  • $\begingroup$ Your claim is true since $\text{LCM}(x+1, x+2, \dots, x+n) \ge n\binom{x+n}{x}$ holds according to a wiki page. $\endgroup$ – mathlove Sep 22 '18 at 6:03

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