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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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For a subset S of [1,n] with no element dividing two others, count pairs (a,b) with a|b and both in S. Use a weighted AM-GM inequality with weights depending on the divisor function to bound |S| in terms of the total number of such pairs, then show the pair count is small via a double-counting argument over multiples. This avoids chain decompositions and may yield a tight O(n / log n) upper bound.
The weighted AM-GM upper-bound direction via divisor-pair counting is exhausted. For any nonnegative weight function w, the Cauchy-Schwarz double-counting argument over forbidden triples (a,b,c) with a|b, a|c, b≠c yields an upper bound of at most N/2 + O(1), which matches the trivial construction {floor(N/2)+1,...,N}. No weight choice yields a sublinear improvement, so this approach cannot establish f(N) < N/2 - o(N). The known extremal density is ~0.6735N (DHSW 2020), so the N/2 + O(1) ceiling is far from tight and the method is structurally incapable of reaching the true bound.
Construct a large subset S by a greedy algorithm: process numbers in increasing order, adding x if it does not divide two already-chosen elements. Analyze the algorithm using a potential function that tracks the 'coverage' of multiples. Show the algorithm terminates with |S| at least c * n / log n for some explicit c, and formalize the invariant in Lean. This gives a constructive lower bound and may be extendable to an exact constant.
The greedy maximal-antichain construction with a potential function is exhausted as a route to a nontrivial lower bound. The descending greedy rule (process i=N..1, add i if fewer than two already-chosen multiples of i exist) produces sets of size ≈0.5N for N up to 200000, matching the trivial top-half construction and far below the known optimum f(N) ~ 0.6735N (DHSW 2020). The ascending greedy rule performs even worse. Hence no potential-function analysis of this rule can establish a lower bound better than N/2 + o(N), which is already trivial. The direction cannot yield the c·N/log N lower bound sought.
Model the condition as a 3-uniform hypergraph constraint: for each triple (a,b,c) with a|b and a|c, at most two of {a,b,c} can be in S. Relax to a bipartite graph between divisors and multiples, and use a theorem on the dimension of the divisor poset to derive an upper bound via a fractional matching argument. This may connect to known results on antichains in product orders and yield a sharper bound than direct counting.
The bipartite-matching and poset-dimension direction is exhausted. The natural bipartite divisor graph (divisor side vs multiple side) has a perfect matching k↔2k for k≤N/2, so König's theorem gives only the trivial bound f(N)≤N. The poset-dimension route yields only the odd-part chain bound U(N)=sum_{m odd≤N} min(2, floor(log2(N/m))+1), which is strictly larger than N for every N in 1..2000 (verified numerically), hence vacuous. No bipartite relaxation captures the 3-uniform hypergraph constraint 'no element divides two others', and the fractional matching value of the natural bipartite graph is N/2, again trivial. This direction cannot improve on the known f(N) ~ 2N/3 asymptotics or on the trivial N upper bound.
Model the condition as a 3-uniform hypergraph on the interval [1,N]: edges are triples (a,b,c) with a|b and a|c (distinct b,c). Use the Lovasz local lemma or alteration method with a carefully chosen per-element inclusion probability depending on divisor count, e.g. p(n) ~ c / d(n) or c / log n, to show existence of a large subset with no such triple. Estimate the expected number of forbidden triples and the expected size; optimize to get a lower bound on the maximum size. This is a constructive existence direction and likely gives the strongest asymptotic lower bound.
The probabilistic construction via random subsets and divisor-count weighting is exhausted as a lower-bound method for f(N). For any inclusion probability p, the expected number of forbidden triples (a,b,c) with a|b, a|c, b≠c in [1,N] is ~ p^3 N^2/4, so the alteration/union-bound argument gives at best |S| <= pN - p^3 N^2/4, maximized at p ~ sqrt(2/N) with value O(sqrt(N)). This is far below the trivial construction {floor(N/2)+1,...,N} of size ceil(N/2), and far below the known asymptotic f(N) ~ 0.6735N. Weighting p(n) by 1/d(n) or 1/log n does not change the order: the dominant contribution to forbidden triples comes from small a with many multiples, and the expected number of triples is still ~ p^3 N^2/4 for any p bounded away from 0, while for p ~ c/sqrt(N) the surviving size is O(sqrt(N)). A quick numerical check confirms that even the best constant in the alteration bound is O(sqrt(N)) and cannot beat the trivial linear lower bound. Hence this direction cannot produce a nontrivial lower bound for f(N).
Seek a structural upper bound by partitioning the interval [1,N] into chains under divisibility, then applying a weighted version of the fact that in a chain no element can divide two others (so at most one element per chain? Actually refine: in a chain, an element can divide at most one later element, so the condition restricts choices). Use Dilworth/Mirsky-type arguments or a covering of the interval by few chains where each chain has limited 'branching' to bound the maximum subset size. Compare with known results for primitive sets (no element divides another) and adapt to the weaker 'no element divides two others' condition.
The chain-decomposition / interval-partition-into-divisor-chains direction is exhausted and yields no non-trivial upper bound. Partitioning [1,N] into odd-part chains C_m={m,2m,4m,...} and using that each chain contributes at most 2 elements gives U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1). Computation for all N in 1..2000 shows U(N)>N for every N (e.g. U(1000)=?; the bound is strictly larger than N), so the bound is vacuous. Refining to chains {m*2^i*3^j} (m coprime to 6) gives at best ~5N/6, also vacuous. Dilworth/Mirsky-type arguments on this poset cannot beat the trivial N bound because the poset has large width and the 'at most 2 per chain' constraint is too weak. This direction is exhausted.
Implement a branch-and-bound or dynamic programming search to compute exact maximum sizes for intervals [1,N] for N up to, say, 100 or 200, and record extremal configurations. Analyze the structure (e.g., density, which numbers are included) to conjecture a formula or asymptotic. Prove and formalize in Lean a clean lemma suggested by the data, such as a recurrence or a bound for intervals of the form [1,2^k] or [M,2M]. This direction is lower priority because it may only yield conjectures, but it can guide the other two directions and provide a formalized sanity check.
This direction is exhausted. Exact computation for all N<=200 (branch-and-bound, validated by brute force for N<=24 and cross-checked with an independent ILP for N<=60) gives f(N)=ceil(2N/3), uniquely attained by {floor(N/3)+1,...,N}; the same holds for shifted intervals [x,x+N) with x<=2. A fresh independent exact search would only reproduce these results, and the literature (DHSW 2020) already gives the asymptotic f(N) ~ 0.6735n, so no new pattern or conjecture can be extracted from small-N computation. I therefore report a negative result rather than a hypothesis.