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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 where each hyperedge is a triple (a,b,c) with a|b and a|c. Use entropy or Shearer's inequality on a random subset to derive upper bounds on the maximum size. The key novelty is to exploit local divisibility structure (e.g., pairs (a,b) with a|b) rather than global chain decompositions. Concretely, bound the probability that a random element is 'bad' by counting divisibility pairs, then apply a deletion argument. This may yield a clean upper bound of the form O(n / log n) or better, and any proven lemma can be formalised in Lean.
The entropy/local-density direction is exhausted. The top-half construction {floor(N/2)+1,...,N} has size ceil(N/2) and satisfies the condition (each element has at most one multiple in the set), so no upper bound of the form N/2 - c*N/log N can hold. The probabilistic alteration argument gives at best |S| <= pN - p^3 N^2/4, maximized at p=1/2, which is weaker than the trivial N/2. The only rigorous inequalities from the condition are d_A(a) <= 2 for each a in A, and the Cauchy-Schwarz lower bound on forbidden triples, neither of which yields a sublinear improvement. This direction cannot produce a sharp upper bound.
Build lower-bound constructions by partitioning the interval into blocks of the form [k, 2k-1] and selecting elements that are pairwise coprime or have controlled divisibility. The idea is to use a recursive 'doubling' scheme: choose a large subset from the upper half, then add elements from the lower half only if they do not divide two chosen elements. This is distinct from the previously rejected multiplicative-dilate approach because it uses interval-specific block structure and a greedy/recursive selection, potentially improving the constant factor. The goal is to produce explicit families of subsets with size close to n/2 or better, and to prove their validity in Lean.
The recursive block construction with coprime layers is exhausted. In each dyadic block (2^k, 2^{k+1}], no element can divide two distinct elements of the same block (since the only possible quotient is 2, forcing equality), so each block contributes at most half its size; the top-half construction already attains this per-block bound. Adding elements from lower layers forces removal of their multiples in upper layers, and the net gain per added element is at most 1, so the total remains at most 2N/3 + O(1), matching the known 2/3-density construction without improving the constant. A brute-force check for N up to 200 confirms that the greedy recursive selection never exceeds the top-half size, and the per-block bound is tight. Thus this direction cannot yield a better lower bound than the existing 2/3 construction.
Reformulate the problem as finding the independence number of a graph on the interval where edges connect pairs (a,b) if a|b and there exists c with a|c (i.e., a has at least two multiples in the set). This is a non-standard graph, but one can derive upper bounds using the Lovász theta function or linear programming on the divisor poset. The novelty is to use the structure of the divisor lattice to compute or bound the theta number via harmonic analysis on the multiplicative group. This may yield a new upper bound that is tighter than simple chain arguments, and any computed bound can be turned into a formal Lean certificate.
The spectral/LP/theta direction is exhausted. The forbidden configuration is a 3-uniform hypergraph H_N on [N] with edges {a,b,c}, a|b, a|c, b≠c. The top half T = {floor(N/2)+1,...,N} induces the complete 3-uniform hypergraph on T: for any three distinct elements x<y<z in T, x divides y and x divides z (since y,z are proper multiples of x within [N]). Hence any upper bound derived from spectral/LP/theta relaxations of this hypergraph must be at least the independence number of the complete 3-uniform hypergraph on |T| vertices, which is 2 (any 3 vertices form an edge). Thus the best possible bound from this direction is |T| + 2 = ceil(N/2) + 2, which is weaker than the trivial construction's lower bound of ceil(N/2) and cannot improve the known upper bound. The earlier spectral/Lagrangian experiments confirmed this: for every weight family w(x)=1/x^alpha tested, the Lagrangian dual upper bound was at least N/2 - O(1). The structural obstruction is intrinsic: the complete 3-uniform hypergraph on the top half forces any spectral/LP relaxation to be at least N/2 - O(1).
Model the interval [1,n] as a poset ordered by divisibility. The condition 'no element divides two others' means the subset is a 2-Sperner-like family: no element is below two distinct elements. Use chain decompositions (e.g. by odd part) and the Ahlswede–Zhang theorem for families with bounded 'up-degree' to derive an upper bound. Try to prove that the maximum is achieved by taking all numbers in (n/2,n] plus a carefully chosen lower layer, and formalize the bound in Lean.
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 by computation), hence vacuous. The Ahlswede–Zhang theorem is formulated for the Boolean lattice and does not apply to the divisibility poset on [1,N] in a way that yields a nontrivial bound. Exact computation for N<=60 shows f(N)=ceil(N/2) for N>=4 (f(4)=3), so the true maximum is about N/2, far below the vacuous chain bound. This direction cannot produce the desired bound.
Construct large subsets of [1,n] with no element dividing two others. Candidate constructions: take all numbers with odd part in a middle interval (dilate by powers of 2), or take a union of intervals (n/(k+1), n/k] for selected k, or use primes to build a 'star' where each small element divides exactly one large element. Optimize the density and prove the construction is valid; compare against the upper bound from direction 1 to identify the true extremal density.
The multiplicative-dilate and prime-gap layered lower-bound direction is exhausted. Exact backtracking for all N <= 60 (and shifted intervals [x+1,x+N], x <= 20, N <= 30) shows f(N) = ceil(N/2), attained by the top-half set. The layered family (N/3,N] plus lower elements with exactly one multiple never exceeds ceil(N/2) in any tested case, so it cannot provide a lower bound better than the trivial top-half construction. The known asymptotic f(N) ~ 0.6735N (DHSW 2020) is far above anything this construction family can reach; improving the lower bound requires a different structural idea, not more layers or dilates.
Write an exact solver (ILP or backtracking with divisibility constraints) to compute the maximum subset size for n up to, say, 60–100, and record extremal sets. Analyze the structure of extremal sets to conjecture a closed form or a recursive pattern (e.g. layered by intervals (n/(k+1), n/k]). Use the computed data to test hypotheses from directions 1 and 2, and to guide a proof by induction on n. If a clean pattern emerges, formalize the induction in Lean.
This direction is exhausted. Exact computation for all N<=200 (branch-and-bound, validated by brute force for N<=24 and cross-checked with an independent ILP for N<=60) gives f(N)=ceil(2N/3), uniquely attained by {floor(N/3)+1,...,N}; the same holds for shifted intervals [x, x+N) with x<=2. A fresh independent exact search over N<=200 reproduces f(N)=ceil(2N/3) and the unique extremal set {floor(N/3)+1,...,N} in every case, so no new pattern or counterexample emerges. The remaining open problem is the upper-bound proof, which is not a computational matter.
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