TITLE: If a power of 2 divides a number, under what conditions does it divide a binomial coefficient involving the number that it divides?
QUESTION [12 upvotes]: We have had many questions here about the divisibility of $\binom{n}{k}$, many of them dealing with divisibility by powers of primes, or expressions involving the $\textrm{gcd}(n,k)$ (I originally gave several more examples but took TheSimpliFire's advice to shorten the list, and many other examples can still be found by looking at this question's edit history):

How can we prove that a binomial coefficient (n choose k) is divisible by the ratio of n and the gcd(n,m)?
Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$

The topic has also lead to some interesting research papers recently:

In 2014: Some divisibility properties of binomial and $q$-binomial coefficients.
In 2017: Divisibility of binomial coefficients and generation of alternating groups.
In 2019: On the divisibility of binomial coefficients.

There's also many associated theorems:

Kummer's theorem
Lucas's theorem
Granville's theorem
Désarménien's theorem

However I've come across a related problem which is expressed completely in the title, and is remarkably not covered by the above extensive body of literature. In MathJax, if $s>0$ is an integer (let's also make it the largest one for which $2^s$ divides $n$), under what conditions of $k$ do we have the following:
$$
2^s \mid n \implies 2^s \left\vert \binom{n}{k} \right. . \tag{1}
$$
Since $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ and $2^s \mid n$ we are left with determining when $\frac{q}{k} \binom{n-1}{k-1}$ is an integer ($q$ being the result when $n$ is divided by $2^s$).
The implication works for a remarkable number of cases, but not always (for example if $\binom{n}{k}$ is odd).

REPLY [4 votes]: With the very helpful tip from user BillyJoe, we can say that for any prime $p$,
$$\tag{1}
\binom{n}{k} \equiv  \left\langle {n_{s-1}\cdots n_0 \atop k_{s-1}\cdots k_0} \right\rangle \prod_{j=1}^{r-s+1}\left\langle {n_{j+s-1}\cdots n_j \atop k_{j+s-1}\cdots k_j} \right\rangle\left\langle {n_{j+s-2}\cdots n_j \atop k_{j+s-2}\cdots _j} \right\rangle^{-1} \left( \textrm{mod} ~ p^s \right),
$$
where the indices 0 and $r$ represent the terms corresponding to the lowest and highest powers (respectively) in the $p$-adic expansions of $n$ and $k$, and the angular brackets correspond to a slight modification (described in more detail here) of the ordinary binomial coefficient symbols.
We then have that:
$$
\left\langle {n_{s-1}\cdots n_0 \atop k_{s-1}\cdots k_0} \right\rangle \prod_{j=1}^{r-s+1}\left\langle {n_{j+s-1}\cdots n_j \atop k_{j+s-1}\cdots k_j} \right\rangle\left\langle {n_{j+s-2}\cdots n_j \atop k_{j+s-2}\cdots _j} \right\rangle^{-1} \equiv 0 \left( \textrm{mod} ~ p^s \right),\tag{2}
$$
if and only if $p^s \mid \binom{n}{k}$.
We therefore have a general condition for when $p^s$ divides $\binom{n}{k}$, whether or not $p^s \mid n$.
For the special case where $p=2$ and $2^s \mid n$, which was the premise of the original question, we may be able to get a cleaner expression than what I have above in Eq. 2.