It is easy to see that $4\mid \binom{2n}{n}$ except when $n=2^k$, and hence it suffices to prove this when $n$ is a power of $2$.

Proved by Sárkzözy [Sa85] for all sufficiently large $n$, and by Granville and Ramaré [GrRa96] for all $n\geq 5$.

More generally, if $f(n)$ is the largest integer such that, for some prime $p$, we have $p^{f(n)}$ dividing $\binom{2n}{n}$, then $f(n)$ should tend to infinity with $n$. Can one even disprove that $f(n)\gg \log n$?

Sander [Sa92] proved that $f(n)\to \infty$, and later [Sa95] quantified this to $f(n) \gg (\log n)^{1/10-o(1)}$. Sander [Sa95] also proved that $f(n)\ll \log n$ for all $n$, and that $f(n) \gg \log n$ for almost all $n$. The proofs of the latter two facts are very easy using Kummer's theorem -- indeed, this immediately implies that any $p$ divides $\binom{2n}{n}$ with multiplicity $\ll \log_p n$, and that the multiplicity with which $2$ divides $\binom{2n}{n}$ is equal to the number of $1$s in the binary expansion of $n$, which is $\gg \log n$ for almost all $n$.

Erdős and Kolesnik [ErKo99] improved the lower bound for all $n$ to \[f(n) \gg (\log n)^{1/4-o(1)}.\]

It remains open whether $f(n) \gg \log n$ for all $n$.

Erdős, Graham, Ruzsa, and Straus [EGRS75] have shown that, for any two odd primes $p$ and $q$, there are infinitely many $n$ such that $\binom{2n}{n}$ is coprime to $pq$.

This is equivalent (via Kummer's theorem) to whether there are infinitely many $n$ which have only digits $0,1$ in base $3$, digits $0,1,2$ in base $5$, and digits $0,1,2,3$ in base $7$.

The sequence of such $n$ is A030979 in the OEIS.

A question of Erdős, Graham, Ruzsa, and Straus [EGRS75], who proved that if $f(n)$ is the sum in question then \[\lim_{x\to \infty}\frac{1}{x}\sum_{n\leq x}f(n) = \sum_{k=2}^\infty \frac{\log k}{2^k}=\gamma_0\] and \[\lim_{x\to \infty}\frac{1}{x}\sum_{n\leq x}f(n)^2 = \gamma_0^2,\] so that for almost all integers $f(m)=\gamma_0+o(1)$. They also prove that, for all large $n$, \[f(n) \leq c\log\log n\] for some constant $c<1$. (It is trivial from Mertens estimates that $f(n)\leq (1+o(1))\log\log n$.)

A positive answer would imply that \[\sum_{p\leq n}1_{p\mid \binom{2n}{n}}\frac{1}{p}=(1-o(1))\log\log n,\] and Erdős, Graham, Ruzsa, and Straus say there is 'no doubt' this latter claim is true.

Erdős and Graham state they can prove that, for $k$ fixed and large, the density of $n$ such that $\binom{n}{k}$ is squarefree is $o_k(1)$. They can also prove that there are infinitely many $n$ such that $\binom{n}{k}$ is not squarefree for $1\leq k<n$, and expect that the density of such $n$ is positive.

If $s(n)$ denotes the largest integer such that $\binom{n}{k}$ is divisible by $p^{s(n)}$ for some prime $p$ for at least one $1\leq k<n$ then it is easy to see that $s(n)\to \infty$ as $n\to \infty$ (and in fact that $s(n) \asymp \log n$).

A conjecture of Erdős and Selfridge. Proved by Ecklund [Ec69], who made the stronger conjecture that whenever $n>k^2$ the binomial coefficient $\binom{n}{k}$ is divisible by a prime $p<n/k$. They have proved the weaker inequality $p\ll n/k^c$ for some constant $c>0$.

Erdős and Graham write that 'a proof that this cannot happen infinitely often for $\binom{n}{2}$ seems hopeless; probably this can never happen for $\binom{n}{k}$ if $3\leq k\leq n-3$.'

Weisenberg has provided four easy examples that show Erdős and Graham were too optimistic here: \[\binom{7}{3}=5\cdot 7,\] \[\binom{10}{4}= 2\cdot 3\cdot 5\cdot 7,\] \[\binom{14}{4} = 7\cdot 11\cdot 13,\] and \[\binom{15}{6}=5\cdot 7\cdot 11\cdot 13.\]

Erdős once conjectured that $\binom{n}{k}$ must always have a divisor in $(n-k,n]$, but this was disproved by Schinzel and Erdős [Sc58].

Erdős and Graham write that $n+1$ always divides $\binom{2n}{n}$ (indeed $\frac{1}{n+1}\binom{2n}{n}$ is the $n$th Catalan number), but it is quite rare that $n$ divides $\binom{2n}{n}$.

Pomerance [Po14] has shown that for any $k\geq 0$ there are infinitely many $n$ such that $n-k\mid\binom{2n}{n}$, although the set of such $n$ has upper density $<1/3$. Pomerance also shows that the set of $n$ such that \[\prod_{1\leq i\leq k}(n+i)\mid \binom{2n}{n}\] has density $1$.

The smallest $n$ for each $k$ are listed as A375077 on the OEIS.

A theorem of Sylvester and Schur (see [Er34]) states that $P(\binom{n}{k})>k$ if $k\leq n/2$. Erdős [Er55d] proved that there exists some $c>0$ such that \[P(\binom{n}{k})>\min(n-k+1, ck\log k).\]

Erdős [Er79d] writes it 'seems certain' that this holds for every $c>0$, with only a finite number of exceptions (depending on $c$). Standard heuristics on prime gaps suggest that the largest prime divisor of $\binom{n}{k}$ is in fact \[> \min(n-k+1, e^{c\sqrt{k}})\] for some constant $c>0$.

This is essentially equivalent to [961].

A classical theorem of Mahler states that for any $\epsilon>0$ and integers $k$ and $l$ then, writing \[(n+1)\cdots (n+k) = ab\] where the only primes dividing $a$ are $\leq l$ and the only primes dividing $b$ are $>l$, we have $a < n^{1+\epsilon}$ for all sufficiently large (depending on $\epsilon,k,l$) $n$.

Mahler's theorem implies $f(n)\to \infty$ as $n\to \infty$, but is ineffective, and so gives no bounds on the growth of $f(n)$.

One can similarly ask for estimates on the smallest integer $f(n,k)$ such that if $m$ is the factor of $\binom{n}{k}$ containing all primes $\leq f(n,k)$ then $m > n^2$.

It is trivial that the number of prime factors is \[>\frac{\log \binom{n}{k}}{\log n},\] and this inequality becomes (asymptotic) equality if $k>n^{1-o(1)}$.

A problem of Erdős and Szekeres, who observed that \[\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \geq \frac{\binom{n}{i}}{\binom{j}{i}}\geq 2^i\] (in particular the greatest common divisor is always $>1$). This inequality is sharp for $i=1$, $j=p$, and $n=2p$.

This was resolved by Bergman [Be11], who proved that for any $2\leq i<j\leq n/2$ \[\textrm{gcd}\left( \binom{n}{i},\binom{n}{j}\right) \gg n^{1/2}\frac{2^i}{i^{3/2}},\] where the implied constant is absolute.

A problem of Erdős and Szekeres. A theorem of Sylvester and Schur says that for any $1\leq i\leq n/2$ there exists some prime $p>i$ which divides $\binom{n}{i}$.

Erdős and Szekeres further conjectured that $p\geq i$ can be improved to $p>i$ except in a few special cases. In particular this fails when $i=2$ and $n$ being some particular powers of $2$. They also found some counterexamples when $i=3$, but only one counterexample when $i\geq 4$: \[\textrm{gcd}\left(\binom{28}{5},\binom{28}{14}\right)=2^3\cdot 3^3\cdot 5.\]

A problem of Erdős and Szekeres. It is easy to see that $f(n)\leq n/P(n)$ for composite $n$, since if $j=p^k$ where $p^k\mid n$ and $p^{k+1}\nmid n$ then $\textrm{gcd}\left(n,\binom{n}{j}\right)=n/p^k$. This implies \[f(n) \leq (1+o(1))\frac{n}{\log n}.\]

It is known that $f(n)=n/P(n)$ when $n$ is the product of two primes. Another example is $n=30$.

For the second problem, it is easy to see that for any $n$ we have $f(n)\geq p(n)$, where $p(n)$ is the smallest prime dividing $n$, and hence there are infinitely many $n$ (those $=p^2)$ such that $f(n)\geq n^{1/2}$.

A problem of Erdős, Graham, Ruzsa, and Straus [EGRS75], who believed there is 'no doubt' that the answer is yes.

For example $(87,88)$ and $(607,608)$. Those $n$ such that there exists some suitable $m>n$ are listed as A129515 in the OEIS. In every known example $m=n+1$.

Erdős [Er96b] credits this to himself and Gordon 'many years ago', but it is more commonly known as Singmaster's conjecture. For $t=3$ one could take $a=120$, and for $t=4$ one could take $a=3003$. There are no known examples for $t\geq 5$.

Both Erdős and Singmaster believed the answer to this question is no, and in fact that there exists an absolute upper bound on the number of solutions.

Matomäki, Radziwill, Shao, Tao, and Teräväinen [MRSTT22] have proved that there are always at most two solutions if we restrict $k$ to \[k\geq \exp((\log n)^{2/3+\epsilon}),\] assuming $a$ is sufficiently large depending on $\epsilon>0$.