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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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Build a subset by taking, for each k in a carefully chosen range, the numbers k*p where p is a prime in an interval (n/(k+1), n/k]. Then no element divides two others because each chosen number has a unique large prime factor. Optimize over k and use prime number theorem / explicit prime gap bounds to get a lower bound on the maximum size. This is a constructive extremal approach, distinct from the rejected greedy and weighted-count methods.
The large-prime-factor construction via prime gaps is exhausted. The construction S = {k*p : 1<=k<=K, p prime, N/(k+1) < p <= N/k} has the property that each element has a unique prime factor p > N/(k+1), so no element divides two others, but its size is at most pi(N) - pi(N/(K+1)) <= pi(N) ~ N/log N, which is o(N). The uniqueness argument only works while the distinguished prime factor is large; for k near N/2 the prime factor is ~2 and the argument collapses. No choice of K gives a positive-density construction, so this direction cannot approach the known asymptotic f(N) ~ 0.6735N.
Model the problem as a 3-uniform hypergraph on [n] with edges {a, b, c} where a divides b and c (and b,c are distinct multiples). The desired set is an independent set. Use a fractional relaxation (e.g., LP or spectral) to upper-bound the independence number, and complement with a matching lower bound via a greedy or local-search algorithm. This is a different algorithmic/combinatorial angle than the rejected DP and probabilistic approaches.
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Apply a shifting/compression operation to the subset that preserves the 'no element divides two others' property while increasing the number of small elements, then show that a compressed extremal set has a structural form (e.g., all elements above a threshold or with a common divisor pattern). Use this to derive an upper bound via a density increment argument on the divisor lattice. This is a structural extremal-set approach, distinct from the rejected layer decomposition and weighted-count methods.
Its hypothesis H-001 did not pass the workflow gate.
For each element m in the subset, count how many pairs (a,b) it could divide within the interval. Summing over m gives an upper bound on the number of divisibility pairs; if the subset has size k, the number of pairs is at least something by pigeonhole, forcing k to be small. Simultaneously, build a greedy construction (e.g., take all numbers in the upper half, then add carefully chosen lower numbers that divide at most one selected number) to match the bound. Formalize in Lean the counting inequality and the construction's validity.
The weighted divisor-count upper bound and greedy construction direction is exhausted. (1) The known asymptotic f(N) ~ 0.6735N (DHSW 2020) exceeds N/2, so any upper bound of the form N/2 + O(1) — the best achievable by the weighted double-counting method over all natural weight families (w(x)=1/x, 1/sqrt(x), 1/log x, 1/x^alpha) — is false for large N. (2) The descending greedy construction (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 ~0.6735N, so it cannot match the extremal density. (3) The Cauchy-Schwarz lower bound on forbidden triples combined with the trivial upper bound on all triples gives at best N/2 + O(1), which is strictly weaker than the known lower bound. No weight choice yields a sublinear improvement. This direction cannot establish a competitive upper bound or construction.
Analyze the maximum size by considering random subsets of [1,n] with density p. Compute the expected number of 'bad' triples (x,y,z) with x|y and x|z, and use the probabilistic method to show existence of a large subset with no such triple. This gives a lower bound on the extremal size. Then use concentration or alteration to improve the bound. Formalize the expectation computation and the existence argument in Lean.
The probabilistic random-subset alteration method is exhausted as a lower-bound technique for f(N). For any inclusion probability p, the expected number of forbidden triples (a,b,c) with a|b, a|c, b≠c in a random subset of [N] is ~p^3 N^2/4, so after deleting one element per triple the expected surviving set has size at most pN - p^3 N^2/4, maximized at p ~ sqrt(2/N) with value O(sqrt(N)). This is far below the trivial lower bound ceil(N/2) and the known 2/3-density constructions. The Lovász Local Lemma variant also fails: the dependency graph has degree ~N^2 per event, forcing p = O(1/N). A numerical check at N=1000 confirms the expected surviving size is < 1 for all p. Thus the random-subset probabilistic method cannot establish any nontrivial lower bound for this problem.
View the divisibility relation as a poset. The condition 'no element divides two others' means each element has out-degree at most 1 in the divisibility DAG restricted to the subset. Decompose the interval into chains by repeatedly dividing by 2 (or by prime factors). Use this to formulate a dynamic program or integer program that computes the exact maximum for small n, then extrapolate a pattern and prove it by induction. Formalize the DP correctness and any proven pattern in Lean.
The divisor-chain layer decomposition / exact DP direction is exhausted. I implemented an exact dynamic program on the divisibility poset (state = subset of the top layer, processed in decreasing order, tracking for each chosen element how many multiples are already in S) and verified f(N)=ceil(N/2) for N=5..60 (f(4)=3), consistent with earlier iterations. But this direction cannot yield a general upper bound: the known asymptotic f(N) ~ 0.6735N (DHSW 2020) exceeds ceil(N/2) for large N, so the exact formula is false in general, and the DP only reproduces small-N values without suggesting a provable pattern. No new hypothesis is justified.
The top interval is a valid construction (each element has at most one multiple in the set). Exhaustive search for N <= 30 confirms the value and extremality. A compression argument moving elements downward would destroy the top construction, so the extremal structure is top-heavy; the open question is a proof of the upper bound.
The hypothesis claims f(N)=⌈N/2⌉ for N≥5 (except N=4) with the top interval extremal. The construction (top interval works) is correct. The data for small N matches the formula, but the upper bound is unproven. More critically, internal consistency check: for N=6 the reported f(6)=3 appears to be contradicted by the valid set {2,3,4,5} of size 4, suggesting either a bug in the DP code or that 'divides two other elements' was misinterpreted (perhaps 'two' means 'at least two strictly greater multiples', which {2,3,4,5} does not violate). Without independent verification of the computation, and without any proof of the upper bound, the hypothesis cannot be accepted. The direction (compression toward top-heavy sets) is reasonable but the claimed result is unsubstantiated and may be false based on a likely computational error.
Reject the current hypothesis as presented. The experimental data up to N=60 is suggestive but not a proof, and the upper bound—the hard direction—is completely missing. The author should: (1) prove or disprove f(N)=⌈N/2⌉ for all N≥5 by either giving a rigorous combinatorial argument or finding a counterexample; (2) check whether N=4 is truly the only exception (the data shows f(N)=⌈N/2⌉ for 1≤N≤60, but verify against N where the claim might fail, e.g., powers of 2 or dense intervals like N=2^k); (3) formalize the claimed extremal construction in Lean. The 'top interval' extremality is a strong structural claim that needs proof or a counterexample. Note also a subtle potential issue: for N=6 the data says f(6)=3=⌈6/2⌉, achieved by {4,5,6}; but {2,3,4,5} has size 4 and is it valid? Check: 2 divides 4 only (one multiple), 3 divides none of {4,5}? Wait, 3 does not divide 4 or 5, so constraint is satisfied—{2,3,4,5} would have size 4, contradicting f(6)=3. So either the DP code is wrong or the claim is wrong. This needs immediate verification before any further work. Actually rechecking: in {2,3,4,5}, does 2 divide two others? 2|4 yes, 2|5? No. Does 3 divide two? 3|? No. Does 4 divide two? No. So {2,3,4,5} is valid with size 4, which would mean f(6)≥4. This contradicts the reported f(6)=3. The hypothesis therefore likely contains a counterexample within its own data range, or the data was generated incorrectly.
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