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Problem 771
claims/: The 1 claim page of Problem 771, one per claimant's result; the problem's standing derives from them.
Statement. Let be maximal such that, for every , there exists some with such that for all .
Is it true that
Status. 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).
Source. erdosproblems.com/771, accessed 2026-09-04, and the cached snapshot of 2026-09-05 (refresh of 22:48 UTC): PROVED, header key [Er89], no last-edited line, an empty discussion thread and an empty proof-claim tab, no formalized statement, OEIS "Possible". Cite as: T. F. Bloom, Erdős Problem #771, https://www.erdosproblems.com/771.
References.
- [AlFr88] Alon, N. and Freiman, G., On sums of subsets of a set of integers. Combinatorica (1988), 297-306.
Formalization. None recorded.
Progress
Not yet compiled.
Known Results
Not yet compiled.
Linked library material
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