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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] according to the size of the largest prime factor (y‑smoothness). Within each class the divisor depth is bounded by O(log N / log y), limiting how many elements can share a common divisor. By choosing y as a slowly growing function of N, we can derive an upper bound of N / (log N)^{c} for some c>0, improving on trivial linear bounds. The approach will also explore whether a matching lower construction can be built from carefully chosen smooth numbers.
Its hypothesis H-003 did not pass the workflow gate.
Model the forbidden configuration (a divides b and a divides c) as a 3‑uniform hyperedge on vertices {a,b,c}. The problem becomes finding the largest independent set in this hypergraph. Apply known Turán‑type extremal results for 3‑uniform hypergraphs with bounded codegree (each pair appears in at most O(log N) edges) to obtain an upper bound of O(N / log N). Parallelly, adapt the dependent random choice technique to construct relatively large independent sets, giving improved lower bounds.
The hypergraph Turán method does not provide an upper bound better than O(N) for the maximum size of a subset of {1,…,N} with no element dividing two others; known constructions of size ⌈N/2⌉ show that any purported improvement would contradict these constructions. Hence this line of attack is exhausted for the goal of improving the asymptotic bound.
Implement an ILP formulation that enforces the divisor‑double‑free condition and solve it for N up to several thousand using modern solvers. Record the maximal cardinalities and the structure of optimal sets. Analyse the output to detect regularities (e.g., periodic gaps, preference for numbers with many distinct large prime factors) and formulate a conjectural recurrence or piecewise linear bound. Subsequent work will aim to prove the observed pattern analytically.
Exploration could not be completed: The model returned an unusable response
Treat the set {1,…,n} with the divisibility relation as a partially ordered set. Any admissible subset S has the property that each element belongs to at most one covering chain (an element can be a divisor of at most one larger element). Apply Dilworth’s theorem or Greene–Kleitman decomposition to bound the width of the poset, and translate this into an explicit upper bound on |S|, e.g. |S| ≤ n - ⌊log₂ n⌋. Formalise in Lean the definitions of divisor‑poset, chain, and prove the derived inequality.
Its hypothesis H-001 did not pass the workflow gate.
Select each integer in [1,n] independently with probability p, then delete any element that serves as a divisor of two retained numbers. Use the Lovász Local Lemma to show that for suitable p (≈1/2) the resulting set still has size Ω(n) while satisfying the no‑double‑divisor condition. Develop an algorithmic version (Moser‑Tardos) to produce an explicit set and prove its size bound. Formalise the random process and LLL application in Lean, yielding a lower‑bound lemma.
Its hypothesis H-002 did not pass the workflow gate.
Construct S by taking numbers in [1,n] whose smallest prime factor exceeds n^α for a chosen α (e.g., α=1/2). Such numbers can divide at most one other chosen number because any multiple would exceed n. Estimate the count of such ‘rough’ numbers via de Bruijn’s ψ‑function or Buchstab’s sieve, giving |S| = n - O(n/ log n). Refine the parameters to approach n - O(n/ (log n)^{c}) and compare with upper bounds. Formalise the rough‑number counting lemma and the resulting size estimate in Lean.
Exploration could not be completed: The model returned an unusable response
Consider the divisor poset ( {1,…,N}, | ). In any admissible set S each element can have at most one outgoing cover edge, so the induced subgraph is a disjoint union of chains. By Dilworth’s theorem the size of a minimum chain decomposition equals the width of the poset, i.e. the size of a largest antichain. The classic result of Erdős that the largest antichain in the divisor poset has size at most N / log₂ N gives a lower bound on the number of chains required. Since each chain contributes at most one element that is not a divisor of another element, at least ⌊log₂ N⌋ elements must be omitted to keep the out‑degree ≤1, yielding the stated inequality.
The hypothesis proposes a linear upper bound for the maximum cardinality of a subset S of an integer interval {1, ..., N} such that no element of S divides two distinct other elements of S. The argument relies on the divisor poset, Dilworth's theorem, and Erdős' result on the largest antichain. While the argument is mostly logically complete, there are missing assumptions and a need for formal proof to support some claims.
Revise and resubmit with additional formal proof and justification
A simple greedy algorithm that processes numbers from N down to 1 and adds a number whenever it does not cause any existing element to become a divisor of two retained numbers always produces a set of size at least ⌊N/2⌋+1 for N up to 200. This empirical evidence suggests a linear‑size construction may exist in general.
The hypothesis proposes a linear lower bound for divisor-double-free subsets, and the provided experiment code and report offer some empirical evidence to support this claim. However, the argument lacks formal proof and rigorous analysis of the greedy algorithm's performance for large N.
revise and resubmit with additional theoretical analysis
Partition [1,N] according to the size of the largest prime factor P⁺(n). Choose a smoothness threshold y = (log N)^α with α>1. Elements with P⁺(n)≤y are y‑smooth; any chain of divisibility among them has length at most O(log N / log y)=O(1/α). Hence each y‑smooth number can be the common divisor of at most O(1) pairs, limiting the number of y‑smooth elements that can appear in a divisor‑double‑free set to O(N / (log N)^{α}). The remaining numbers have a prime factor >y, so each such large prime can serve as a divisor for at most N / y numbers, again yielding a reduction by a factor of (log N)^{α}. Optimising α gives a concrete c>0 (e.g., α=2 yields c≈1). This heuristic suggests the claimed log‑power upper bound, while no construction is known to match it, indicating the bound may be non‑tight but improves on the linear bound.
The hypothesis proposes a log-power upper bound on the maximal cardinality of a subset with the property that no element divides two distinct other elements, using smooth-number decomposition. While the argument is partially convincing, it relies on heuristics and lacks explicit constructions and rigorous proofs, making it incomplete. The hypothesis runs into known barriers in number theory, particularly in establishing tight bounds on the number of smooth numbers and their divisibility properties.
revise and resubmit with additional rigor and explicit constructions
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