Counting natural numbers in a set $\{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$, for every $n$

Question

Define a (set-valued) function $ n \to \{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$. Now let $f$ be a function that counts the number of natural numbers in the set $\{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$.

A question is: Is $f(n)=1$ for only a finite number of values of $n$?

Here, we study, for each natural number $n$, a set $\{\dfrac{n!}{k!(n-k)!k}:k=1,2,...n-1,n\}$ and seek to determine is there a finite number of natural numbers $n$ for which only the first term is a natural number.

I expect a finite number of natural numbers for which $f(n)=1$ because it seems to me that, although there will be much cancellations in calculation of a number $\dfrac{n!}{k!(n-k)!}$, there should be enough richness in a structure of $\dfrac{n!}{k!(n-k)!}$ that allows $\dfrac{n!}{k!(n-k)!k}$ to be a natural number for some $k \in \{1,2,...,n\} \setminus \{1\}$

This question (almost in the same form) was also asked here on MSE about five hours ago, and, I must admit that I was, kind of, in a dilemma should I post it on MO or MSE, but since there were no serious progress there, I am also posting it here.

There, Woett gave three constructive comments, but we still do not have a solution.

He also did computational check and wrote, that, if he did no mistakes, the only $n \leq 5 \cdot 10^5$ for which $f(n)=1$ are $1, 2, 3, 6, 7, 14, 15, 22, 23, 95$, which corresponds nicely with intuition of mine, that is, that there should be a finite number of such an $n$.

Steven Stadnicki gave OEIS link that gives values of $f(n)$ for $n=1,2,...,99$, here it is: https://oeis.org/A081372

Answer 1

First of all, let $p$ and $q$ be some prime numbers with $q/2<p<q$. Then $pq\mid\binom{n}{pq}$ if

$$n\equiv a \pmod {p^2}$$

and

$$n\equiv b \pmod {q^2}$$

with some $a,b>0$ and $a<p(q-p)$, $b<pq$. To prove this, apply Lucas's theorem on binomial coefficients and also note that $pq$ is a two-digit number $\overline{p0}$ in base $q$ and a three-digit number $\overline{1(q-p)0}$ in base $p$. Now lets assume $n$ is large enough.

It is a well-known fact that the interval $(x,x+x^{2/3})$ contains at least one prime for $x$ sufficiently large. Now let us choose $q$ to lie in the interval $(\sqrt{n}-n^{3/8},\sqrt{n}-n^{0.35})$ (which is clearly possible for $n$ large enough) and $p$ to lie in the interval $(q/2,\sqrt{n}/2)$.

Then we have the following:

$$q=\sqrt{n}-A$$

with $0<A<n^{3/8}$. Therefore,

$$q^2=n-2A\sqrt{n}+A^2,$$

so $0<n-q^2<2n^{7/8}=o(n)=o(pq)$, which is required. And we also have

$$p=\sqrt{n}/2-B$$

with $0<B<n^{0.35}$, therefore

$$0<n-4p^2=O(n^{0.85})=o(n)=o((q-p)p),$$

which is required. So, for $n$ large enough one can find some $k$ around $n/2$ that satisfies your condition, so $f(n)>1$ (and in fact even grows with $n$).