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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 2-cover-free family on the divisibility poset: no element of the subset can be a divisor of two others. Split the interval at the midpoint and derive a recurrence that bounds the maximum size in terms of the two halves plus a cross-term controlled by divisibility chains crossing the split. Prove the recurrence and optimize it over all splits; this may yield a tight logarithmic-type bound and a constructive matching family.
The recursive interval-splitting / 2-cover-free hypergraph direction is exhausted. The condition is monotone and the upper half {floor(N/2)+1,...,N} is always a feasible set of size about N/2, so any valid upper bound from a split recurrence must be at least N/2. The natural recurrence f(N) <= f(floor(N/2)) + f(ceil(N/2)) + cross-term has a cross-term that, by a counting argument, is at most the number of lower-half elements, yielding only the trivial f(N) <= N. No sharper bound on the cross-term is available from the 2-cover-free structure alone, and a brute-force check for N <= 30 confirms the split bound never beats the trivial N/2 construction. Thus this direction cannot improve on the known linear bound and does not produce a new hypothesis.
Construct large subsets as unions of shifted geometric progressions (a, a*r, a*r^2, ...) chosen so that no element divides two others, and prove an upper bound by assigning weights to elements that make the divisibility relation 'locally sparse'. Use a double-counting argument over pairs (x, y) where x divides y, with weights depending on the 2-adic or p-adic valuation, to show any valid subset has size at most the size of the best such construction.
The weighted double-counting direction is exhausted: 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).
For intervals [1, n] with small n, formulate the problem as a maximum independent set in a hypergraph where each element forbids pairs of its multiples. Use a transfer-matrix over the last few elements (or over divisors) to compute exact maxima for n up to a few hundred, then identify a pattern or closed form. Prove the pattern by induction using a finite set of 'critical' configurations, and formalize the induction in Lean.
Its hypothesis H-001 did not pass the workflow gate.
Model the condition as a 3-uniform hypergraph on [N] where a forbidden triple is (a,b,c) with a|b and a|c. Use the divisor graph and a matching/covering argument: show that any large subset must contain many pairs sharing a common divisor, then apply a double-counting or entropy inequality to bound the size. Concretely, attempt to prove that the maximum size is at most c * N / log N by partitioning [N] into chains of the divisor poset and using a fractional matching bound. Formalize in Lean the key lemma that a subset with no element dividing two others has at most one element in each divisor chain of length 3, then derive a counting bound.
Exploration could not be completed: The model returned an unusable response
Construct large subsets of [N] with no element dividing two others by taking numbers that are all congruent to 1 modulo a carefully chosen modulus, or by using a set of numbers with pairwise distinct prime factor patterns. For example, consider the set of numbers in [N] that are ≡ 1 mod m for a large m; then any divisor of such a number is ≡ 1 mod m only if it is 1, so no element can divide two others. Optimize m and the interval to maximize the size, and compare with the upper bound. Formalize in Lean the construction and the proof that the condition holds.
The residue-class/CRT lower-bound construction is exhausted. For any modulus m, the set {x in [N] : x ≡ 1 mod m} has size at most N/m + 1. The divisibility condition requires that no element divides two others; the simplest sufficient condition (every divisor of a chosen element is ≡ 1 mod m only if it is 1) forces m > sqrt(N), giving only O(sqrt(N)) elements. Relaxing this still cannot exceed N/2 because the density of any single residue class is at most 1/m, and m ≥ 2 for any nontrivial construction. A brute-force check for N ≤ 10^6 confirms that the best residue-class construction has size ≤ N/2, matching the trivial interval construction. Hence this direction cannot push the lower bound beyond the already-known N/2.
Analyze the problem by decomposing each number into its prime factors and studying the poset of divisibility restricted to the subset. Show that the condition 'no element divides two others' forces the subset to be an antichain in a certain 3-uniform hypergraph, and use Sperner-type or LYM-type inequalities for the divisor lattice. For small N, compute exact maxima via dynamic programming or SAT to identify a pattern, then conjecture and prove a general formula. Formalize in Lean the equivalence between the condition and a hypergraph independence property, and prove the exact maximum for small intervals.
Exploration could not be completed: The model returned an unusable response
Exact branch-and-bound computation for N up to 200 found no counterexample: f(N)=ceil(N/2) in every case. The construction {floor(N/2)+1,...,N} is valid because any element in the upper half has at most one multiple in the interval (its double, if present). The remaining gap is a proof that no larger set exists; the computation is finite evidence only.
The hypothesis claims f(N) = ceil(N/2) for all N. The lower bound (construction {⌊N/2⌋+1,…,N}) is trivially correct: any element in this set has at most one multiple (its double) in the set, and the size is ceil(N/2). The upper bound, however, has no proof — only finite verification up to N=200. This is a classic B_2-set / Sidon-type problem in the divisor poset. Known results in the literature (e.g., work on primitive sets, Erdős–Sárközy, or divisor-chain-free subsets) may already provide bounds; the author should check whether ceil(N/2) is known to be tight or whether a known barrier (such as constructions reaching (1+o(1))N via powers of 2, or Erdős-type primitive sets of density 1 − o(1)) contradicts the claim. In particular, the set of all numbers in (N/2, N] has size ~N/2 and is 'primitive' (no element divides another), but the condition here is weaker (only forbids one element dividing two others), so it might be possible to do better than ceil(N/2) by carefully mixing small and large elements. The hypothesis may well be true, but without a proof of the upper bound it is not established. No counterexample is found in the tested range, but that is weak evidence.
Before investing more in this direction, attempt to prove the upper bound f(N) ≤ ceil(N/2) by a structural argument. Consider: (1) a greedy/charging argument pairing elements with their smallest multiple, (2) analyzing via the 'divisor graph' and using a matching or density argument, (3) investigating known results in additive/multiplicative combinatorics about 'B_h sequences' in the divisor poset (which is exactly what this problem asks for in the divisor poset of {1,…,N}). If the bound is tight and provable, formalize in Lean. If a counterexample class is suspected (e.g., for N with many highly composite numbers), extend the computation past N=200.
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