Problem #376: Are there infinitely many $n$ such that $\binom{2n}{n}$ is...

---

Progress

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.

The best result in this direction is due to Bloom and Croot [BlCr25], who proved that, if $p_1,p_2,p_3$ are sufficiently large primes, then there are infinitely many $n$ such that almost all of the base $p_i$ digits are $<p_i/2$. In other words, for all $\epsilon>0$, there are infinitely many $n$ such that $\binom{2n}{n}$ is coprime to $p_1p_2p_3$, except for a factor of size $\leq n^\epsilon$.

This is mentioned in problem B33 of Guy's collection [Gu04]. It is also discussed in an article of Pomerance [Po15c].

Graham offered \$1000 for a solution to this problem (as mentioned in [Gu04] and [BeHa98]).