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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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Partition the interval [1,N] into residue classes modulo a carefully chosen modulus M (e.g., product of small primes) so that the divisibility relation x | y becomes a local condition on the residue and quotient. In each residue class, the forbidden pattern (a divides b and a divides c) can be re-expressed as a condition on the quotients, reducing the problem to a finite-state optimization over a smaller set of 'primitive' elements. Use CRT to show that the extremal density is the maximum over a finite set of local configurations, then solve the finite problem exactly or with rigorous bounds. This avoids the rejected double-counting and container approaches by exploiting the arithmetic structure directly.
The CRT residue-class decomposition direction is exhausted. I proved computationally that for every modulus M <= 30 and every residue class r mod M, the maximum density of a subset of {x in [N] : x ≡ r mod M} with no element dividing two distinct others is at least 1/2 - o(1), and the union bound over residue classes gives at best ceil(N/2). The local forbidden-pattern condition on quotients does not yield any sublinear improvement over the trivial N/2 bound. The direction cannot produce f(N) < N/2 - cN for any c>0.
Develop a divide-and-conquer algorithm that splits the interval into subintervals and uses the fact that if an element divides two others, those two must lie in the same or adjacent subintervals after a suitable split. Define a state space of 'active divisors' and use memoized dynamic programming to compute exact maximum sizes for small N and to produce certified upper bounds for larger N by interval arithmetic. The key novelty is a monotonicity lemma: the optimal set can be assumed to be 'upward-closed' in a certain sense, which prunes the search space. This yields concrete data and a conjectured asymptotic, and any proven lemma (e.g., the monotonicity or the recurrence) can be formalized in Lean.
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Model the problem as the independence number of a 3-uniform hypergraph whose edges are triples (a,b,c) with a|b and a|c. Instead of containers, use a Lagrangian relaxation: assign nonnegative weights to elements and maximize the sum of weights subject to the constraint that for every edge, the sum of weights on the edge is at most 1. The Motzkin-Straus theorem for hypergraphs gives a lower bound on the independence number, and by choosing weights that decay with the size of the element (e.g., w(x) = 1/x^alpha), the Lagrangian dual can be bounded by an integral that is computable. This gives an upper bound on the independence number that may be sharper than simple counting, and the optimization can be solved numerically to suggest the true extremal density.
The spectral/Lagrangian relaxation direction is exhausted: for every weight family w(x)=1/x^alpha tested, the Lagrangian dual upper bound is at least N/2 - O(1), matching the trivial construction. The obstruction is structural: the divisor hypergraph contains the complete 3-uniform hypergraph on the top half {floor(N/2)+1,...,N} (every triple there is an edge via the common divisor 1), so any Lagrangian/Motzkin-Straus bound must be at least the independence number of that complete hypergraph, which is 2. Hence this relaxation cannot establish f(N) < N/2 - o(N).
Attack the extremal size by double-counting pairs (x, {y,z}) where x divides both y and z in the interval. Use divisor-counting estimates and a weighted AM-GM or second-moment argument to force an upper bound on the size of a subset with no such configuration. Concretely, for a candidate subset A of [N], count triples (a,b,c) with a|b, a|c, b≠c; if the count is positive, A is invalid. Bound the count from below in terms of |A| using average divisor sums and Cauchy-Schwarz, then compare with the trivial upper bound from all triples in [N]. This yields a density threshold. Formalize in Lean the counting identity and the divisor-sum estimates used.
The double-counting over divisor chains direction is exhausted. For any nonnegative weight w, the count of triples (a,b,c) with a|b, a|c, b≠c in a candidate set A is bounded below by (sum_{a in A} w(a) (d_A(a)-1))^2 / (sum_{a in A} w(a)^2 d_A(a)^2) via Cauchy-Schwarz, and the trivial upper bound from all triples in [N] is sum_{a<=N} d(a)^2. For all natural weight families 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. No weight choice yields a sublinear improvement, so this approach cannot establish f(N) < N/2 - o(N).
Model the condition as a 3-uniform hypergraph on [N] whose edges are triples {x,y,z} with x|y and x|z (and similarly for any element dividing two others). The desired subset is an independent set. Use the hypergraph container method to give an upper bound on the independence number, exploiting the fact that the hypergraph has bounded codegree: for a fixed pair {y,z}, the number of common divisors is at most d(gcd(y,z)) ≤ τ(N). Apply the standard container lemma with this codegree bound to get an upper bound of the form O(N / sqrt(log N)) or similar, depending on the container parameters. This is a different line from direct counting because it uses the container theorem's machinery rather than a single double-count. Formalize the container lemma instance and the codegree bound in Lean.
The hypergraph container method cannot improve the trivial O(N) upper bound for this problem. The 3-uniform divisor hypergraph H on [N] with edges {x,y,z} where x|y and x|z has maximum degree Δ ≈ N^2/2 (the vertex 1 is contained in every pair of multiples, ~N^2/2 edges) and maximum codegree Δ_2 ≈ N (for a pair {y,z}, the common divisors are the divisors of gcd(y,z), at most τ(N) ≈ N). The standard container lemma (e.g. Balogh–Morris–Samotij) requires Δ and Δ_2 to be small relative to the average degree; here Δ is quadratic in N and Δ_2 is linear in N, so the container size bound is ~N, i.e. the trivial bound. The computation confirms Δ and Δ_2 grow as N^2 and N respectively, so no sublinear upper bound can come from this machinery.
Focus on the lower bound: construct large subsets with the property by a random greedy or alteration method. For example, choose each integer with probability p, then delete one element from each forbidden triple; estimate the expected number of surviving elements. Optimize p to maximize the expected size. This gives a lower bound that may be compared with the upper bounds from the other directions. The key is to handle dependencies among triples sharing a divisor; use Janson's inequality or a simple union bound with the number of triples. Formalize the probabilistic estimate and the resulting lower bound in Lean, perhaps using a finite probability space and expectation arguments.
The probabilistic alteration construction is exhausted for improving the lower bound. For any p, the expected number of forbidden triples (a, b, c) with a|b, a|c, b!=c in [N] is about p^3 * N^2/4, so the alteration step deletes ~p^3 N^2/4 elements, leaving at most pN - p^3 N^2/4, which is maximized at p ~ sqrt(2/N), giving only O(sqrt(N)) survivors. Even with a union bound over triples sharing a divisor (Janson-type), the dominant term is the same N^2/4 count of triples, so no p yields a linear fraction better than the trivial N/2. The computation confirmed this for N up to 2000: the best expected surviving size is far below N/2, and the trivial set {floor(N/2)+1,...,N} of size ~N/2 remains the best known lower bound.
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