The board / combinatorics / #21
Erdős Problem #21
PROVED LEANdatabase updated 2026-08-24
The statement as posed is true, and a proof exists. A solution has also been checked by machine, in Lean.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
An Erdős–Lovász family of order `n`: an intersecting family of `n`-sets such that every set of size at most `n - 1` is disjoint from some member.
`f n` is the minimal size of an Erdős–Lovász family of order `n`.
Let \(f(n)\) be minimal such that there is an intersecting family \(\mathcal{F}\) of sets of size \(n\) (so \(A\cap B\neq\emptyset\) for all \(A,B\in \mathcal{F}\)) with \(\lvert \mathcal{F}\rvert=f(n)\) such that any set \(S\) with \(\lvert S\rvert \leq n-1\) is disjoint from at least one \(A\in \mathcal{F}\). Is it true that \[f(n) \ll n?\] Conjectured by Erdős and Lovász [ErLo75], who proved that \[\frac{8}{3}n-3\leq f(n) \ll n^{3/2}\log n\] for all \(n\). The upper bound was improved by Kahn [Ka92b] to \(f(n) \ll n\log n\). (The upper bound constructions in both cases are formed by taking a random set of lines from a projective plane of order \(n-1\), assuming \(n-1\) is a prime power.) This problem was solved by Kahn [Ka94] who proved the upper bound \(f(n) \ll n\). The Erdős-Lovász lower bound of \(\frac{8}{3}n-O(1)\) has not been improved, and it has been speculated (see e.g. [Ka94]) that the correct answer is \(3n+O(1)\). It is trivial that \(f(1)=1\) and \(f(2)=3\). The values \(f(3)=6\) and \(f(4)=9\) were established by Tripathi [Tr14]. Barát and Wanless [BaWa21] proved that \(f(5)=13\), and that \(13\leq f(6)\leq 18\).
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved · 3 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/21.Prize$500
Topicscombinatorics, intersecting family
Statement in LeanFormal Conjectures
Solution in Leanyes
OEISA391599
Original pageerdosproblems.com/21
Attempts
submit one →If you take a run at this one — with your own AI subscription or without — the guide says how, and the submission form gets you an honest read of what you found.