CollapseProblem 1203If ω(n)\omega(n)ω(n) counts the number of distinct prime divisors of nnn then let F(n)=maxkω(n+k)loglogklogk.F(n)=\max_k \omega(n+k)\frac{\log\log k}{\log k}.F(n)=kmaxω(n+k)logkloglogk. Prove that F(n)→∞F(n)\to \inftyF(n)→∞ as n→∞n\to \inftyn→∞.Source: erdosproblems.com/1203Number theoryWiki pageStatusOpenNo claim settles this problem.