Does the set of products of $n$ consecutive positive integers have gcd $n!$, even after removing finitely many elements?

Question

Let $n$ be a positive integer greater than or equal to $2$, and let $S_n$ denote the set of all products of $n$ consecutive positive integers. For example, $S_2 = \{2,6,12,20,...\}$, $S_3 = \{6,24,60,120...\}$, etc. Now, I have noticed that the gcd (greatest common divisor) of $S_n$ is $n!$, or so I conjecture. However, now I am interested in a stronger conjecture. I conjecture that even after removing finitely many elements from $S_n$, the gcd of the resulting set is still $n!$. Is it true? Note, there are some infinite sets of positive integers whose gcd becomes different if you remove even a finite number of elements, like the set of positive integer powers of $2$: $\{2,4,8,16,...\}$. If you remove $2$ and $4$ from that set, the new gcd is $8$. I am interested in knowing if something similar does or does not happen for $S_n$.

Answer 1

The conjecture is true. An example of how it works is your set $S_2$. If we remove $2$ from the set the conjecture says the $\gcd$ of the remaining set is still $2$ and we can see that $\gcd(6,20)=2$ To prove it for $S_2$ we need to show that no prime except $2$ divides all but finitely many elements of $S_2$ and that $2^2$ does not divide all but finitely many elements of $S_2$. For any prime $p \gt 2$ the elements $(kp+1)(kp+2)$ for all natural $k$ do not have a factor $p$ and the elements $(4k+1)(4k+2)$ do not have a divisor $4$ so the $\gcd$ of the set after removing finitely many elements will still be $2$.

The same approach works for your other sets. It is not easy to write up in the general case, but for all primes $p$ that do not divide $n!$ you just take the elements that start with $kp+1$ and for all primes that do divide $n!$ you take multiples of the lowest power that does not divide $n!$ and add $1$ for the starting values.

Answer 2

As $(m+1)(m+2)...(m+n) = n! \binom{m+n}{n}$, the problem is equivalent to the fact that for any $m_0\in{\mathbb N}$, \begin{equation}1 = \gcd\left\{\binom{m+n}{n}, m\ge m_0\right\}\end{equation} Let $p$ be any prime number. By Kummer's theorem, the $p$-adic valuation of $\binom{m+n}{n}$ is equal to the number of carries in the addition of $m$ and $n$ written in base $p$.

For a given $n$, one can choose $m=p^k$ for $k$ large enough so that the number of carries is $0$ (this holds as soon as $n\lt p^k$), which proves that \begin{equation} n < p^k \quad\Longrightarrow\quad p \nmid \binom{p^k + n}{n}\end{equation}

Hence the conjecture is true.

Remark that Lucas's theorem gives the stronger result \begin{equation} n < p^k \quad\Longrightarrow\quad\binom{p^k + n}{n}\equiv 1 \pmod p\end{equation}