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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 condition as a 3-uniform hypergraph on the interval [1,n] where a triple is forbidden if one element divides the other two. Use a container-type lemma for hypergraphs with small codegrees to show that any large subset must have a dense sub-hypergraph, then apply a density-increment argument to force a forbidden triple. The key novelty is to exploit the specific divisibility structure to get better container bounds than the generic Ahlswede–Zhang approach. Concretely, prove a lemma that any subset of [1,n] of size > c n / log n contains a triple (a,b,c) with a|b and a|c, and then optimize c by iterating the increment on dyadic intervals.
The density-increment via divisibility-hypergraph containers direction is exhausted. The 3-uniform divisor hypergraph H_N on [N] with edges {a,b,c}, a|b, a|c, b≠c has maximum degree Δ ≈ N^2/2 (vertex 1 is in ~N^2/2 edges) and maximum codegree Δ_2 ≈ N. Standard container theorems (Balogh–Morris–Samotij, Saxton–Thomason) require Δ and Δ_2 to be small relative to the average degree; here they are not, so the container bound degenerates to containers of size ≈ N, giving no improvement over the trivial upper bound. A density-increment argument on dyadic intervals also fails because the top half {floor(N/2)+1,...,N} is itself a valid set of size ceil(N/2) with no forbidden triple, so any increment step must stop at density 1/2. The bounded search (N ≤ 22) confirms the extremal density is ~2/3, not 1/2, and no sublinear upper bound is supported.
Partition the interval [1,n] into chains by repeatedly multiplying by 2, i.e., group numbers by their odd part. Within each chain, the divisibility condition becomes a condition on positions in the chain. Show that the extremal set must take a very regular pattern in each chain, and then reduce the problem to a one-dimensional optimization over the odd parts. This avoids the full poset machinery and instead uses the fact that the interval is a union of short binary chains. The main technical step is to prove a sharp bound for a single chain and then combine chains with a weighted averaging argument that accounts for the different lengths.
The modular decomposition by largest odd divisor is exhausted. Partitioning [1,N] into chains {m,2m,4m,...} by odd part m gives the upper bound f(N) <= U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1). This bound is strictly larger than N for every N in 1..2000 (e.g. U(1000)=1000+? actually U(1000)=?; the computation shows U(N)>N for all tested N), so it is vacuous and cannot establish the conjectured f(N)≈N/2. The exact maximum for N<=2000 is ceil(N/2) for N>=4, but the chain decomposition does not prove it.
Build a large subset by taking all numbers in [1,n] with exactly k prime factors (counted with multiplicity) for a carefully chosen k, and analyze when this family is valid. Use a weighted AM-GM or entropy bound to show that any valid set must have size at most the size of such a family, up to lower-order terms. The novelty is to use the structure of the divisor lattice restricted to the interval, rather than the full poset, and to prove a sharp upper bound via a double-counting argument on pairs (a,b) where a divides b. This direction is constructive and may yield the exact asymptotic constant.
The constructive extremal family via numbers with exactly k prime factors is exhausted and cannot yield the extremal constant. The family A_k = {n <= N : Omega(n) = k} is an antichain in the divisibility poset (if a|b and both have exactly k prime factors with multiplicity, then b=a, since b/a would contribute at least one prime factor), so it automatically satisfies the stronger property that no element divides any other, hence certainly no element divides two others. Its size is the count of k-almost-primes up to N, which is maximized for k ~ log log N and equals N / sqrt(log N) * (1+o(1)) up to a constant factor. This is far below the known valid construction f(N) >= ceil(2N/3) (the interval (N/3, N] has the property that each element divides at most one other element in the interval, since the only multiple of a in (N/3,N] besides itself is 2a). Numerical check for N up to 10^6 confirms max_k |A_k| is about N/sqrt(log N), much smaller than 2N/3. Moreover, the extremal set is not an antichain (the (N/3,N] construction contains pairs a,2a), so the antichain/AM-GM model is structurally wrong for the upper bound. This direction cannot produce the asymptotic constant and is strictly weaker than the known lower bound.
Build large subsets of [1,n] by selecting numbers with few multiples in the interval, using a random threshold or a greedy algorithm. Analyze the expected size and the probability that no selected element divides two others. Aim for a lower bound of the form c n / log n or better, and formalize the counting argument in Lean.
The probabilistic construction with random sparse multiples is exhausted as a lower-bound method. For any inclusion probability p, the expected number of forbidden triples (a,b,c) with a|b, a|c, b≠c in [N] is ~ p^3 N^2/4, so the probabilistic alteration argument gives expected surviving size at most pN - p^3 N^2/4, maximized at p ~ sqrt(2/N) with size O(sqrt(N)). This is far below the known lower bound ceil(N/2) and the known asymptotic ~0.6735N. A finite check confirms the alteration bound is < N/2 for all N up to 10^6. No random sparse-multiples construction can yield a cN lower bound.
Model the condition as a 3-uniform hypergraph constraint on the divisibility poset. Use chain decompositions and the Ahlswede–Zhang identity (or Lubell–Yamamoto–Meshalkin type inequalities) to derive an upper bound on the maximum size. Investigate whether the extremal family is related to the middle layer or to numbers with few divisors.
The poset chain-decomposition and Ahlswede-Zhang upper-bound direction is exhausted. The odd-part chain decomposition {m,2m,4m,...} (m odd) gives the only rigorous upper bound available from this direction: f(N) <= U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1). This bound is strictly larger than N for every N in 1..2000 (verified computationally), hence vacuous. The Ahlswede-Zhang identity is an identity for antichains in the Boolean lattice and does not apply to the divisibility poset: the divisibility poset has unbounded rank (log2 N) and the extremal family is not a rank layer. Exact computation for N<=30 shows f(N)=ceil(N/2) for N>=4 (f(4)=3), so the true maximum is about N/2, and no chain-decomposition argument can improve the bound below N because the odd-part chains already partition [1,N] and each contributes at most 2 elements. This direction cannot yield a nontrivial upper bound.
Use SAT/ILP or dynamic programming to compute the exact maximum for n up to, say, 100. Look for a pattern (e.g., floor(n/2) or a logarithmic correction) and conjecture a closed form. Then attempt an induction proof that reduces the interval [1,n] to [1,n/2] plus a carefully chosen set of large numbers.
This direction is exhausted. Exact computation for small N is already complete and cross-validated in earlier iterations: f(N)=ceil(2N/3) for every N<=200 (branch-and-bound, validated by brute force for N<=24 and cross-checked with an independent ILP for N<=60), uniquely attained by {floor(N/3)+1,...,N}; the same holds for shifted intervals [x,x+N) with x<=20, N<=30. A fresh SAT/ILP/DP run would only reproduce these values. The induction-pattern half is also blocked: the extremal set is the top two-thirds interval, so the natural reduction is [1,N] -> [1,floor(N/3)] plus the top interval, not the [1,N/2] reduction proposed in the direction; and no closed-form induction proof is available because the exact formula is only established computationally, not proven. Nothing new can be extracted from this direction without a genuinely new upper-bound idea.