Problem Statement
Is it true that for every $1\leq k\leq n$ the largest prime divisor of $\binom{n}{k}$, say $P(\binom{n}{k})$, satisfies\[P\left(\binom{n}{k}\right)\geq \min(n-k+1, k^{1+c})\]for some constant $c>0$?
Categories:
Number Theory Primes Binomial Coefficients
Progress
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, whenever $k\leq n/2$,\[P\left(\binom{n}{k}\right)\gg k\log k.\]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, for $k\leq n/2$, in fact\[>e^{c\sqrt{k}}\]for some constant $c>0$.This is essentially equivalent to [961].
Source: erdosproblems.com/683 | Last verified: January 16, 2026