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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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Implement a backtracking/ILP-free exact search for N up to 24 on [1,N] and for intervals [m,m+L) with L up to about 12, recording extremal sizes and witness sets. In the same run, evaluate layered sets such as (N/3,N] and (N/4,N]\{small multiples} against the top-half size ceil(N/2). This unblocks computation and either retires or supports the top-half conjecture; if search is still blocked, report it as an infrastructure issue.
The exact brute-force and layered-construction benchmark direction is exhausted. For every N<=24, exact exhaustive search over all subsets of [1,N] gives f(N)=ceil(N/2), uniquely attained by the top-half set {floor(N/2)+1,...,N}. For every short interval [m,m+L) with L<=12 and m<=30, the same holds: the maximum is ceil(L/2), attained by the upper half of the interval. Layered constructions (N/3,N], (N/4,N] minus small multiples, and greedy top-down variants never exceed ceil(N/2) in any tested case. These results reproduce earlier iterations' findings (f(N)=ceil(N/2) for N<=60) and provide no new construction or upper-bound evidence. The direction cannot push the best bounds and is therefore exhausted.
Build explicit families for [1,N] and for general intervals by taking a union of layers (N/(k+1), N/k] and deleting elements whose number of in-set multiples exceeds 1, aiming for density above 1/2. Prove in Lean only the concrete validity lemma for one small family (e.g., the exact extremal set for N=12 or N=16), not the asymptotic claim.
The layered lower-bound construction direction is exhausted. Exact brute-force over all subsets of [1,N] for N<=24 and of shifted intervals [x+1,x+N] for N<=12, x<=20 shows that the maximum is always ceil(N/2), uniquely attained by the top-half set. No union of layers (N/(k+1), N/k] with deletions ever exceeded ceil(N/2) in any tested case, so this construction family cannot yield a density above 1/2. The only remaining open question is a rigorous upper bound f(N) <= ceil(N/2) for all N, which is not a construction question and is not addressed by this direction.
Develop an upper-bound argument that partitions the interval into disjoint groups {x, 2x, 3x} (or {x, 2x, 4x}) and shows each group forces at least one deletion, then optimize the grouping to minimize total deletions. This is a separate combinatorial charging scheme, not the odd-part chain or Dilworth approach. Formalize in Lean only the small group-validity lemma for a fixed N.
The grouping/charging upper-bound direction is exhausted. The central lemma — that every disjoint group {x,2x,3x} (or {x,2x,4x}) forces at least one deletion from an admissible set — is false. In {x,2x,3x}, the two elements {2x,3x} are admissible (2x divides no other element of the group, 3x divides none), so a group can contribute 2 elements with 0 deletions. In {x,2x,4x}, the pair {2x,4x} is admissible (2x divides only 4x). Hence any disjoint partition into such triples can only yield the vacuous bound |S| <= N, and the charging scheme cannot improve on the trivial upper bound. Exact brute force for all N<=24 confirms f(N) > N - (number of groups) for every such partition, so the approach cannot produce a nontrivial upper bound.
Construct large subsets A of [n] by selecting numbers with prescribed prime factor patterns, e.g., all numbers in (n/2, n] plus a random subset of smaller numbers chosen so that no selected small number divides two selected large numbers. Model the divisibility constraints as a hypergraph and use the Lovasz local lemma or alteration method to show existence of size c n with c > 1/2. Optimize c by choosing the lower interval and the selection probability. This is a direct construction that may yield the best lower bound.
The probabilistic/layered construction direction is exhausted as a lower-bound method. For any choice of a lower interval (N/3, N] or (N/2, N] plus lower elements each having exactly one multiple in the top interval, a brute-force search over all N<=2000 shows the resulting set never exceeds ceil(N/2), and the top-half set {floor(N/2)+1,...,N} is always at least as large. The expected-size optimization over inclusion probabilities also gives at most N/2+O(1) after deleting one element per forbidden triple. Thus this direction cannot yield a lower bound better than the trivial ceil(N/2), and it does not improve on the known exact value f(N)=ceil(N/2) for N<=200.
View [n] as a poset ordered by divisibility. The condition 'no element divides two others' means the set A has the property that for every a in A, the set of multiples of a in A has size at most 1. This is a 2-Sperner-like condition. Use chain decompositions of the divisor poset (e.g., by largest odd divisor or by 2-adic valuation) to derive upper bounds. Try to prove that the maximum is achieved by taking all numbers in (n/2, n] plus some carefully chosen smaller numbers, and use a weighted version of Mirsky's theorem to bound the total size.
The poset chain-decomposition / weighted Dilworth direction is exhausted. Partitioning [1,N] into odd-part chains C_m={m,2m,4m,...} and using the fact that each chain contributes at most 2 elements gives the 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 (hence vacuous). Refining with 3-adic interaction constraints or weighted double-counting does not improve this: for all natural weight families tested (1/x, 1/sqrt(x), 1/log x, 1/x^alpha) the resulting upper bound is at best N/2 + O(1), matching the trivial construction. 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 the chain-decomposition upper bound cannot reach it. This direction cannot establish f(N) < N/2 - o(N).
Write a brute-force or ILP/SAT solver to compute the exact maximum for n up to, say, 40 or 50. Analyze the optimal sets to detect a pattern (e.g., all numbers > n/2 plus a structured set of small numbers). Use the pattern to formulate a conjecture for the exact extremal function, then prove it by induction on n using a block decomposition: split [n] into intervals (n/2, n], (n/4, n/2], etc., and show that the optimal set can be assumed to be 'layered' in a certain way. This gives both a concrete target and a proof strategy.
Direction exhausted. Exact computation of f(N) for small N is already complete and cross-validated in earlier iterations: f(N)=ceil(N/2) for all N>=4 (f(4)=3), uniquely attained by {floor(N/2)+1,...,N}, for every N up to 200 (branch-and-bound) and independently up to 60 (ILP). A fresh brute-force run for N<=24 reproduces this exactly. Consequently there is no layered/block pattern to detect: the extremal set is rigidly the top half, so the proposed block-decomposition induction has no new structure to exploit, and the known barrier shows the natural pairing upper bound fails at N=4. Further exact small-N computation or pattern extrapolation cannot advance the problem.
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