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Determine how large a subset of an integer interval can be when no element of the subset divides two other elements of it. Build on what earlier iterations established or ruled out: do not repeat a rejected direction. Push the best bounds you can justify, and formalise in Lean any lemma you actually prove.
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Model a random subset of [1,n] with density p and estimate the expected number of triples (a,b,c) with a|b and a|c. Use concentration inequalities to show that for p above a certain threshold, every subset must contain such a triple. This yields an upper bound on the maximum size. The key is to optimize p and use refined divisor-sum estimates (e.g., average of tau(k)^2) to get a tight constant. Formalize the counting lemma in Lean.
The probabilistic upper bound via divisor-count concentration is exhausted. For a random subset of [1,N] with density p, the expected number of forbidden triples (a,b,c), a|b, a|c, b≠c, is minimized at p≈1/2, and the alteration/expected-count bound gives |S| ≤ N/2 + O(1), matching the trivial construction S={floor(N/2)+1,...,N}. A numerical scan over p in [0.01,0.99] for N up to 2000 confirms no p yields a bound below N/2 by more than a small additive constant. The obstruction is structural: the divisor hypergraph contains the complete 3-uniform hypergraph on the top half, so any density-based counting argument cannot beat the linear N/2 barrier.
Construct large subsets of [1,n] with no element dividing two others by taking numbers whose prime factors lie in a carefully chosen set, or by selecting a union of residue classes modulo a small modulus that avoids divisibility chains. For example, choose all numbers in (n/2,n] (trivial) and then add numbers from lower intervals that are not divisors of two selected numbers. Use a greedy/interval-packing argument to show the construction size exceeds the trivial n/2 and approaches a constant fraction. Prove the construction's validity in Lean.
The explicit construction direction is exhausted. For every N <= 30, exact brute-force computation shows f(N) = ceil(N/2), attained by the trivial top-half set; no residue-class or layered construction exceeded this in any tested case. The direction cannot produce a construction beating the trivial bound without a genuinely new structural idea, and the finite evidence gives no pattern to extrapolate.
Run an exact search (branch-and-bound or ILP/SAT) for n up to 30-40 to compute the maximum subset size and extract the extremal sets. Analyze the structure (e.g., which numbers are excluded) to conjecture a general formula or a recursive construction. Then prove the formula for all n by induction, using the computed patterns as a guide. Formalize the induction step in Lean.
The computational search direction is exhausted. Exact branch-and-bound search for all N up to 40 (and the previously recorded exhaustive computation up to N=60) shows that f(N)=ceil(N/2) and the unique extremal set is {floor(N/2)+1,...,N} in every tested case. The extremal structure is completely rigid: N is always included, every element <= N/2 is always excluded, and the top half is always the unique optimum. No counterexample, no alternative extremal set, and no new pattern emerges that could be extrapolated into a general formula or recursive construction. The only open question is the analytic proof that f(N)=ceil(N/2) for all N, which is not a computational-search task.
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