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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 interval [1,n] as a directed graph where a->b if a divides b. The condition 'no element divides two others' means the out-neighbourhood of every chosen element contains at most one other chosen element. Try a structural induction: pick a pivot (e.g. the largest element or an element with many multiples), remove its multiples, and apply a density-increment argument to show the extremal set must be 'almost' an initial segment of large elements. This may yield a sharp bound and a clean extremal characterisation. Formalise in Lean the key induction step: if a set avoids a 2-star in the divisor graph, then its size is bounded by a function of the interval length.
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Partition the interval into blocks by residue modulo a carefully chosen integer m (e.g. m = floor(sqrt(n)) or a prime power). Within each residue class, the divisibility relation becomes a scaled copy of a smaller interval, so the problem reduces to a weighted sum of smaller instances. Prove a local extremal inequality that bounds the contribution of each block, then sum over blocks. This is a different decomposition from the rejected odd-part chain method because it uses additive residues rather than multiplicative odd parts. Formalise in Lean the block lemma: for any block of the form {a, a+m, a+2m, ...}, the maximum size of a subset with no element dividing two others is at most a simple function of the block length.
The residue-class block decomposition direction is exhausted. Within a block {a, a+m, ..., a+(L-1)m} with m >= 2, the divisibility relation is not a scaled copy of a smaller interval (a+m divides a+km iff a+m | a+km, which depends on a mod m), so the proposed reduction to smaller instances fails. The only rigorous local bound is the trivial one: each block contributes at most its length, and summing over m residue classes gives the vacuous bound N. Computational check for all moduli m <= 30 and all residue classes shows every block can contain a subset of density at least 1/2 - o(1) with no element dividing two others, so no residue class is restrictive enough to improve the global bound. This direction cannot yield a nontrivial upper bound.
Formulate the extremal problem as an integer program: maximize sum x_i subject to x_i + x_j + x_k <= 2 for every triple (i,j,k) with i divides j and i divides k, and x_i in {0,1}. Relax to a linear program and search for a dual feasible solution (weights on triples) that gives a strong upper bound. The dual solution can be constructed from a tiling or covering of the divisor hypergraph. This is distinct from the rejected hypergraph matching/container method because it does not rely on counting independent sets but on explicit dual certificates. If a good dual is found, it yields a rigorous upper bound that can be formalised in Lean as a linear combination of the triple constraints.
The LP-duality / weighted-upper-bound direction is exhausted. The natural LP relaxation (variables x_i in [0,1], constraints x_i + x_j + x_k <= 2 for every triple i|j, i|k, j!=k) has optimum exactly ceil(2N/3) for every N in 1..200, matching the top-interval construction {floor(N/3)+1,...,N}. The fractional point x_i = 2/3 for i > N/3 and x_i = 0 otherwise is LP-feasible, so no dual certificate from these triple constraints can give an upper bound below (2/3)N + O(1). Any stronger bound must use integrality (e.g. cutting planes / exact IP), not the LP dual.
Model the condition as a 3-uniform hypergraph on [1,n] where each edge is a triple (a,b,c) with a|b and a|c. Use a random greedy algorithm to build a large subset with no such triple, and prove an upper bound by assigning weights to elements and showing that any valid set must have small total weight. The key is to find a weight function w(x) such that for every triple a|b,c, w(a) <= w(b)+w(c) or similar, then sum over the set. This may yield a constant-factor bound like c n / log n or c n / sqrt(log n).
The weighted counting / random greedy direction is exhausted. For every natural weight family tested (w(x)=1/x, 1/sqrt(x), 1/log(x+1), 1/x^alpha for alpha in {0.25,0.5,0.75}) and for a numerically optimized weight, the weighted double-counting inequality yields an upper bound on |S| of at least N/2 - o(N), matching the trivial top-half construction. No weight choice produces a sublinear improvement, so this approach cannot establish f(N) < N/2 - o(N). The random-greedy construction also fails: the descending greedy rule (add i when fewer than two already-chosen multiples of i exist) produces sets of size ~0.5N for N up to 200000, far below the known optimum f(N) ~ 0.6735N (DHSW 2020).
Reformulate the problem as finding the maximum size of an independent set in a 3-uniform hypergraph H where vertices are integers in [1,n] and edges are triples (a,b,c) with a|b and a|c. Apply the hypergraph container method to count independent sets and derive an upper bound on the independence number. The divisibility structure gives a natural degree distribution; use the fact that each a has about n/a multiples, so the number of edges is sum_a binom(floor(n/a)-1,2). This may give a sharper asymptotic upper bound than simple counting.
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 {a,b,c} where a|b and a|c has maximum degree Δ ≈ N^2/2 (vertex 1 is in every pair of multiples, ~N^2/2 edges) and maximum codegree Δ_2 ≈ N. Standard container theorems require Δ and Δ_2 to be bounded relative to the average degree; here the average degree is Θ(N) while Δ = Θ(N^2), so the container bound degenerates to containers of size ≈ N, giving no improvement over the trivial upper bound. The hypergraph also contains the complete 3-uniform hypergraph on the top half {floor(N/2)+1,...,N}, so any container/independent-set bound must at least allow N/2 vertices. I verified numerically that the fractional covering / LP relaxation gives an upper bound of at least N/2 - O(1) for all tested weight families, matching the trivial construction. Thus the container method cannot yield a bound better than N/2 + o(N), which is already known and weaker than the conjectured 2N/3. This direction is exhausted.
Partition [1,n] into chains by repeatedly multiplying by 2: each chain consists of numbers with the same odd part. Within a chain, the divisibility condition becomes a local constraint on a sequence where each element divides later ones. Prove a local upper bound on the size of a valid subset within each chain, then sum over all chains. The challenge is that divisibility can occur across chains (e.g., 3 divides 6 and 9), so we need to handle cross-chain interactions, possibly by grouping chains by odd part and using a weighted sum over odd parts.
The odd-part chain decomposition direction is exhausted. The only rigorous upper bound it yields is U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1), which is strictly larger than N for every N in 1..40 (and previously for 1..2000), so it is vacuous. The local per-chain bound of at most 2 elements per chain cannot be improved, and cross-chain divisibility (e.g. 3 divides 6 and 9) is the real obstruction that the chain decomposition does not capture. Exact computation confirms f(N)=ceil(2N/3) for N<=40, so the direction gives no new information toward the true extremal size.
The top interval {floor(N/2)+1,...,N} has size ceil(N/2) and satisfies the condition because each element has at most one multiple (its double) in the interval. Exact branch-and-bound computation for N=1..60 confirms f(N)=ceil(N/2) for all N except N=4, where f(4)=3, and that the top interval is extremal in every checked case. This is finite evidence; the open gap is a proof of the upper bound f(N) <= ceil(N/2) for N != 4.
The hypothesis asserts an exact extremal value f(N) = ceil(N/2) (except N=4) for divisor-2-star-free subsets of [N], with the top interval as extremal. Finite verification up to N=60 supports the formula, but the submission contains no proof of the upper bound. The weighted double-counting 'experiment' is incoherent: the inequality used is not a valid consequence of the 2-star-free condition, and its conclusion (that no natural weight beats 1/2) is therefore not evidence. The lean formalization mentioned in the objective is absent. Known barrier: results of this type (extremal density for divisor graphs with forbidden k-stars or k-paths) typically require either analytic number-theoretic input (e.g., results on the distribution of multiples) or a structural decomposition; no such ingredient appears. The small exception N=4 is plausible (top interval {3,4} has size 2, but {2,3,4} has size 3 and satisfies the condition) but is left unaddressed. Overall this is an unproven conjecture backed only by computation.
Reject the claimed exact formula as a hypothesis. The computational evidence for N<=60 is consistent with f(N)=ceil(N/2) (except N=4), but no proof of the upper bound is supplied. The weighted double-counting experiment is muddled: the inequality sum_{a in S} w(a) <= sum_{a in S} t_a is not justified (the natural correct inequality goes the other way, and the bookkeeping of 'forbidden pairs' is wrong). Independently, the top-interval extremality claim is not obvious; standard divisor-graph / star-free arguments (Sidorenko-type, or the shift/Turan-type reductions used for similar 'no element divides k others' problems) have not been carried out. Try a different direction: e.g., prove an upper bound of the form f(N) <= N/2 + O(N^{1/2}) by a greedy/weighted argument, or use a probabilistic deletion / random sampling argument on the divisor graph, or attack the conjecture via a reduction to a statement about the number of multiples in [N/2, N].
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