Theorems demonstrate unbounded growth of f(n) in Erdős problem 684, suggesting a polylogarithmic upper limit.
For \(0≤ k≤ n\), write \({n}{k}=uv\) where the primes dividing \(u\) are at most \(k\) and the primes dividing \(v\) exceed \(k\), and let \(f(n)\) be the least \(k\) with \(u>n²\); Erdős problem 684 asks for bounds on \(f(n)\). We resolve the problem at the order level. By a short-multiplier construction \(nM=tLM-1\), where \(LM=lcm(1,…,M)\) and \(t\) is a multiplier of size \(exp(o(M))\) extracted from a Fourier sieve, we prove that for every fixed \(C>1\) there exist integers \(n\) with\[ f(n)>(C-o(1))log n,\]hence\[ n→∞f(n)/log n=∞.\]We thus refute the widely expected upper bound \(f(n)log n\) and place the order of \(f(n)\) strictly above \(log n\) infinitely often. A matching polylogarithmic upper bound \(f(n)(log n)²\) is known by Alexeev, Putterman, Sawhney, Sellke, and Valiant (arXiv:2603.29961). The reduction of the multiplier sieve to a dyadic fixed-\(Ω\) arithmetic-progression estimate, including a \(QM=M!/LM\) box parametrization, a local harmonic-height cap, and an exact-\(a\) product-shell extraction, is new. The required estimate uses Timofeev's mean-in-progressions framework together with a Burgess-based mod-\(p\) saving on the relevant prime band.
No takes yet. Share an insight, caveat, or question.
Ji Ho Bae (2026) studied this question.
Synapse has enriched 5 closely related papers on similar clinical questions. Consider them for comparative context: