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Let be maximal such that, for every , there exists some with such that for all .
Is it true that
Source: erdosproblems.com/771
An accepted solution exists. The statement is true.
Proved: a conjecture of Erdős and Graham. They observed the lower bound (for every , which one may take below , the multiples of the least prime not dividing , a prime below , avoid as a subset sum), and Alon and Freiman [AlFr88] proved the matching upper bound by exhibiting an , the least common multiple of the integers below with largest such that , whose avoiding sets have at most elements. The site labels the problem PROVED and credits the paper. Claim page: Alon and Freiman 1988 (accepted, refereed in Combinatorica).