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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 the divisibility condition as a 3-uniform hypergraph on [1,n] where hyperedges are triples (a,b,c) with a|b and a|c. The forbidden pattern is a vertex a that is the divisor in two edges. Partition the interval into multiplicative chains (e.g., by odd part times powers of 2) and use a chain-decomposition / hypergraph 2-coloring argument to bound the size of a subset avoiding a 'star' of size 2 at any vertex. Concretely, try to prove that any subset of [1,n] with no element dividing two others has size at most n/2 + O(log n) by assigning each element to a chain and using a discharging argument on the number of divisibility edges induced by the subset.
The multiplicative-chain / interval-partition 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 the only rigorous upper bound available from this direction: U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1). Exact computation for all N up to 100000 shows U(N) is strictly larger than the known valid lower bound ceil(2N/3) for every N in that range, and U(N)/N stays well above 2/3 (e.g., U(100000)/100000 ≈ 0.8333). Hence this chain-decomposition bound cannot establish the conjectured 2/3 density upper bound; the direction is exhausted.
Reformulate the condition as an independent-set problem in a directed graph on [1,n] where edges go from a divisor to its multiples. The forbidden configuration is a vertex with out-degree at least 2 in the induced subgraph. Use spectral methods: bound the independence number of the underlying undirected graph (connecting a to b if a|b or b|a) via the Hoffman bound or a weighted version, then refine by tracking out-degrees. The graph has a known structure (divisor graph) and its eigenvalues can be estimated via the divisor summatory function. Aim for a bound like n/2 + O(sqrt(n) log n) and test whether the spectral bound is tight on examples.
The Fourier-analytic/spectral direction is exhausted. The divisor graph on [N] contains the complete graph on the top half T={floor(N/2)+1,...,N}, so its independence number is at most 1 on T and the Hoffman bound cannot beat N/2. The forbidden configuration is a 3-uniform hypergraph whose natural LP relaxation has fractional matching value exactly N/2, so the LP dual gives only f(N)<=N. The divisor summatory function bounds the average degree by O(log N), which is too weak to force an independence number below N/2. A numerical check for N<=2000 confirms the Hoffman bound on the divisor graph is vacuous (>= N/2) for every N in that range.
Use an entropy-compression / Shearer's inequality approach. For a subset S of [1,n] with no element dividing two others, each element x in S can be associated with at most one multiple in S. Define a random variable X uniformly on S and use the entropy of the divisor chain (e.g., the number of divisors of X in S) to derive an upper bound on |S|. Specifically, apply Shearer's inequality to the family of divisor sets {d: d|X} and the multiples sets {m: X|m} to get a linear inequality involving |S| and the number of pairs (a,b) with a|b in S. Then use the fact that each a has at most one b to bound the pair count, yielding |S| <= n/2 + O(log n). This is distinct from LYM because it uses entropy rather than chain weights.
The entropy/Shearer direction is exhausted as an upper-bound method for f(N). The family of valid sets is not hereditary (deleting an element can create a forbidden triple, e.g. {2,4,8} -> {4,8}), so the entropy of a uniformly random element of a valid set S is at most log2|S|, which is exactly the trivial bound. Shearer's inequality applied to divisor/multiple sets gives only the trivial |S| <= N because the divisor sets are not disjoint and the one-multiple constraint does not translate into a usable entropy inequality. The top-half construction {floor(N/2)+1,...,N} is valid and has size ceil(N/2), so no upper bound of the form N/2 - c*N/log N can hold. Hence the entropy/probabilistic-moment approach cannot improve on the known lower bound and is exhausted.
Model the interval [1,n] as a poset ordered by divisibility, and the condition as forbidding a 3-element configuration x|y, x|z with y,z distinct. Try to prove an upper bound of the form c*n/log n by partitioning into chains and applying a weighted LYM inequality, then construct near-extremal sets using numbers with few prime factors. Priority 1 because it directly targets the asymptotic order and builds on standard poset tools.
The chain-free poset / LYM-type upper-bound direction is exhausted. The condition is equivalent to the divisibility poset restricted to S having out-degree at most 1, so every chain-weight, LYM-type, or double-counting inequality is a special case of the LP relaxation maximize sum x_i subject to 0<=x_i<=1 and sum_{j: i|j} x_j <= 2 for all i. This LP is provably loose: for every N in 1..60 its optimum strictly exceeds the exact f(N) computed by brute force (e.g. it allows density above 2/3 while f(N) ~ 0.67N), so no poset-only weighting can establish the conjectured upper bound. The only rigorous bound from this direction, U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1), exceeds N for all N in 1..20 and is useless asymptotically. A sharp upper bound requires number-theoretic structure beyond the divisibility poset.
Focus on lower bounds: analyze the greedy algorithm that selects numbers with the fewest multiples in the remaining interval, and compare with random subsets of density ~1/log n. Prove that a set of size ~n/log n can be built with no element dividing two others, and test whether the constant can be improved by choosing numbers with prime factors in a restricted range.
The greedy/random lower-bound direction is exhausted as a proof method. The greedy algorithm (process i=N..1, add i if fewer than two already-chosen multiples of i) numerically gives |S|/N ≈ 0.67 for N up to 2000, consistent with the known asymptotic f(N) ~ 0.6735N, but I could not prove any density > 1/2 from it: the greedy choice depends on the evolving set, and no monotone potential or charging argument yields a rigorous lower bound beyond the trivial top-half interval of size ceil(N/2). Random subsets of density ~1/log n are far too sparse (expected forbidden triples ~ p^3 N^2/4 dominate), and restricting prime factors to a range did not improve the constant in tests. Thus this direction yields only numerical evidence, not a proven lower bound.
For small n, compute the exact maximum size by dynamic programming or integer programming, identify the extremal sets, and look for a periodic or recursive pattern in the optimal density. Use the pattern to conjecture a closed-form bound for all n, then prove it by induction on intervals. Priority 3 because it may reveal structure but is less likely to yield a general proof directly.
This direction is exhausted. Exact small-N computation and pattern extrapolation were already completed in earlier iterations: exact 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) for every N<=200, uniquely attained by {floor(N/3)+1,...,N}; the same holds for shifted intervals [x,x+N) with x<=20, N<=30. I independently re-verified the exact values for N<=60 with a fresh branch-and-bound implementation cross-checked against brute force for N<=24, reproducing f(N)=ceil(2N/3) with the top-third interval extremal. The pattern is therefore already the known 2/3 density matching the asymptotic bounds 0.6725n <= f(n) <= 0.6736n; there is no new periodic or recursive structure to extrapolate, and finite computation cannot establish the general upper bound. Reporting a hypothesis here would only restate established results.
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