For which pairs $k$ and $n$, $n\mid{{n-2} \choose {k}}$

Question

The question

This question that arose in a discussion with Ron Adin is quite simple:

For which pairs $k$ and $n$ does $n$ divide ${{n-2} \choose {k}}$?

Simple observations

1. It is easy to see that for every $k$ there are finitely many possible values of $n$. One way to see it is to note that if $n$ divides ${{n-2} \choose {k}}$ it also divides $(n-2)(n-3)\dotsb (n-k-1)$ and therefore also (if $n \ge k+2$) it must divide $(k+1)!$.

2. For $n=2k+2$, $\frac {1}{n} {{n-2} \choose {k}}=\frac {1}{2(k+1)}{{2k} \choose {k}}=\frac {1}{2}C_k$, where $C_k$ is the $k$-th Catalan number. So the question is when the $k$-th Catalan number is even. As $C_k={{2k+1} \choose {k}}-2 {{2k}\choose {k+1}}$, this is if and only if ${{2k+1} \choose {k}}$ is even and by Kummer's theorem this always happens, unless $k$ and $k+1$ have no common $1$ digits in base $2$, namely iff $k+1$ is a power of $2$. Maybe a similar analysis can be done in other cases as well, and perhaps a complete description is possible.

Experimenting

a little, it seems that given $n$ there are few values of $k$ such that $n \mid{{n-2} \choose {k}}$.

Motivation

These are the cases where vertex-regular $n$-vertex $k$-dimensional $\mathbb Q$-acyclic complexes with complete $(k-1)$-dimensional skeletons exist. We know (albeit by an indirect argument and not by an explicit construction) that whenever $d_1,d_2,\dotsc d_n$ are non-negative integers that sum up to ${{n-2} \choose {d}}$ then a $\mathbb Q$-acyclic complex with complete $(k-1)$-dimensional skeletons exists such that the degree of vertex $i$ (namely, the number of $k$-faces containing it) is $d_i + {{n-2} \choose {k-1}}$.

Related MO question: Seeking very regular $\mathbb Q$-acyclic complexes

Answer 1

I doubt there exists a simple description for the general solution, however it's possible to give a characterization in some cases.

For example, in the case of square-free odd $n$, Lucas' theorem implies that $(n,k)$ is a solution iff for every prime $p\mid n$, there exists a base-$p$ digit in $k$ that is greater than the corresponding base-$p$ digit of $n-2$. For the last digit, it means that $p\mid k+1$.

From here we can construct a particular series of solutions $(n,k)$ with semi-prime $n=pq$, assuming, say, that $p\mid (k+1)$, i.e. the last base-$p$ digit of $k$ is $p-1$, and $p\equiv 1\pmod{q}$, i.e. the second but last base-$q$ digit of $n-2$ is $0$. Taking primes $q$ and $p=qs+1$ and integer $k=pt-1$ for some integers $s>0$ and $t\in[0,q]$, where to satisfy the condition on the second but last base-$q$ digit of $k=pt-1=qst+t-1$, it's enough to have $q\nmid st$. That is, we need to have $q\nmid s$ and $1\leq t\leq q-1$.

Example. Take $q=23$ and $p=2\cdot 23+1=47$ giving $n=pq=1081$. Then any $k\in\{47t-1\ :\ t=1,2,\dotsc,22\}$ will do the job.