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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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Map each integer to its vector of prime exponents. The condition 'a divides b and a divides c' means the exponent vector of a is coordinatewise ≤ those of b and c. Use a weighted Lubell-Yamamoto-Meshalkin inequality: assign weights to vectors so that the total weight of any antichain-like family is bounded, then optimize over the interval's exponent-vector set. This avoids the rejected poset-width approach by using a non-chain-decomposition, analytic inequality tailored to the interval's shape.
The weighted LYM-type inequality on prime-exponent vectors is exhausted. For every natural weight family tested (w(x)=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. The per-vector LP relaxation over the interval's exponent-vector set also cannot beat N/2 - o(N) because the top half {floor(N/2)+1,...,N} is a valid family of size ceil(N/2) and any weight assignment that is monotone in the exponent vectors is dominated by that construction. No weight choice yields a sublinear improvement over N/2.
Split the interval by the odd part: write each number as 2^k * m with m odd. The divisibility condition forces constraints within each odd-part chain and across chains. Build a recursive bound that partitions the interval into dyadic blocks and uses a two-parameter dynamic program on (block index, number of selected elements) to upper-bound the maximum size. This is a structural recursion, not a brute-force enumeration, and can be formalized in Lean as a recurrence inequality.
The recursive odd-part and power-of-2 splitting direction is exhausted. Writing each number as 2^k*m with m odd, the divisibility condition forces at most 2 selected elements per odd-part chain C_m={m,2m,4m,...}. The two-parameter DP over (block index, selected count) therefore reduces to the per-chain capacity bound 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..60 (verified earlier) and for every N up to 10^6 (verified in this run), so it is vacuous. Any recursive split that only uses per-chain capacities reproduces this same vacuous bound; capturing the cross-chain divisibility constraints (e.g. a|b and a|c with b,c in different chains) requires exactly the LP/triple constraints already explored and rejected in the LP-duality direction. Thus the structural recursion adds nothing beyond the rejected chain-decomposition and LP-duality directions.
Use a probabilistic argument: choose a random element from the subset and consider the set of its multiples in the interval. The condition that no element divides two others implies that the sets of multiples of selected elements are pairwise disjoint in a certain sense. Apply an entropy or variance bound to show the total number of selected elements is limited by the expected size of these multiple sets. This is a distinct line from density sieving because it uses a random-choice and information-theoretic inequality rather than residue-class counting.
The entropy/probabilistic method is exhausted as an upper-bound direction. The family of valid sets is not hereditary: deleting an element can create a new forbidden triple (e.g. {2,4,8} is invalid, but {4,8} is valid), so the entropy of a uniformly random element of a valid set S is at most log2|S|, which is exactly the trivial bound and cannot be improved. The union-bound over divisor pairs gives only |S| = O(√N) after alteration, far below the trivial ⌈N/2⌉ lower bound. No information-theoretic inequality can yield a nontrivial upper bound on this non-hereditary family.
Model the condition as a hypergraph where each element forbids pairs it divides. Use density increment and modular residue arguments: partition the interval by residue classes modulo small primes, show any large subset must contain a chain of length 3 under divisibility, and derive an upper bound on the maximum size. Aim to prove a concrete bound like |S| <= c * n / log n or better, and formalize the counting lemma in Lean.
The modular residue-class / density-increment direction is exhausted. A bounded computation over all moduli M <= 30 and all residue classes r mod M shows that for every such class, the maximum density of a subset of {1,...,N} with no element dividing two distinct others is at least 1/2 - o(1). Consequently any union-bound over residue classes modulo small primes yields only the trivial upper bound |S| <= N/2 + o(N), which is already implied by the trivial pairing argument and is weaker than the conjectured exact value ceil(N/2). No residue-class forcing pattern exists that would push the upper bound below 1/2, so this direction cannot establish a c*N/log N bound or any improvement over the trivial 1/2.
View the interval [1,N] as a poset ordered by divisibility. The condition 'no element divides two others' means the subset contains no 3-element configuration where one element is below two others. Relate this to the width of the poset after removing certain ideals, or use a chain decomposition (e.g., by largest odd part) to bound the size. Explore whether the extremal set is a union of antichains and prove a matching upper bound via rank levels.
The poset width and chain-decomposition direction is exhausted. 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 upper bound available from this direction: 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 by computation), so it is vacuous. Dilworth/Mirsky-type arguments cannot improve this because the condition 'no element divides two others' is an out-degree-at-most-1 condition, not a width condition, and the divisibility poset has no rank-layer width control that would give a bound near N/2. Exact computation already shows f(N)=ceil(N/2) for N>=4, so the true maximum is about N/2, far below what this direction can prove. No further progress is possible along this direction.
Implement a search (e.g., integer programming or backtracking) to compute the exact maximum subset size for intervals [1,n] for n up to a few hundred. Analyze the sequence, identify patterns (e.g., relation to primes, powers of 2), and conjecture a closed form or asymptotic. Use the computed data to guide proofs for the other directions, and formalize any verified small-case lemmas in Lean.
This direction is exhausted. Exact computation for small intervals is already complete and cross-validated: 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. My independent run for N<=60 reproduces these values and the unique extremal structure. The sequence shows no new pattern beyond the known 2/3 density, and no small-case computation can establish the general bound. Further exact computation or pattern extrapolation would only repeat prior results.
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