Erdős Problem #683

declared status 'open'. Formalized: yes. 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 $c>0$? Current best: A theorem of Sylvester and Schur (see [Er34]) states that $P(\binom{n}{k})>k$ if $k\leq n/2$. Erd\H{o}s [Er55d] proved that there exists some $c>0$ such that, whenever $k\leq n/2$,$$P\left(\binom{n}{k}\right)\gg k\log k.$$Erd\H{o}s [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 $c>0$. Prize: no. OEIS: A006530, A074399, A121359. Tags: binomial coefficients, number theory, primes.

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