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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 divisibility poset and seek an upper bound on the largest subset with no element dividing two others by inductively splitting the interval at n/2. The key idea: elements in (n/2,n] are mutually non-dividing, and any element x <= n/2 can divide at most one element in the upper half. Use a rank/level decomposition and a chain-antichain inequality (e.g., Mirsky/Dilworth-type) to derive a recurrence for the maximum size. Prove the recurrence in Lean and solve it to get a closed-form upper bound. This avoids the rejected divisor-count weighting by using structural induction rather than double counting.
The inductive interval-halving poset-decomposition direction is exhausted. Splitting [1,N] into (N/2,N] and [1,N/2] and applying a chain/antichain inequality gives only the trivial upper bound f(N) <= N/2 + O(1), which is strictly weaker than the known lower bound f(N) >= ceil(2N/3) (the interval {floor(N/3)+1,...,N} has no element dividing two others). The reason is structural: the forbidden configuration is a 3-chain (x,2x,4x), not an antichain, so Dilworth/Mirsky-type chain decompositions cannot capture the constraint. The natural recurrence from the halving decomposition is f(N) <= N/2 + f(N/2) + O(1), whose solution is f(N) <= N/2 + O(log N), still far above 2N/3. I verified numerically that this recurrence cannot be sharpened below 2N/3 for N <= 1000, and that the top-third interval is valid for N <= 2000. No new upper bound follows from this direction.
Reformulate the problem as a measure-theoretic extremal question: for a set A in [1,n] with no element dividing two others, consider the logarithmic density of A and of its multiples. Derive a continuous integral inequality (e.g., involving sum over a in A of 1/a and the measure of multiples) that must hold, then optimize over measurable sets to get an upper bound on |A|/n. This is a genuinely different line from divisor-count weighting because it uses a continuous relaxation and variational methods rather than discrete double counting. Formalize the key integral inequality in Lean and numerically solve the variational problem to conjecture the sharp constant.
The logarithmic-density continuous relaxation direction is exhausted. The continuous relaxation of the extremal problem is vacuous: it admits sets of density arbitrarily close to 1 (e.g., A = (εN, N] has log-density log(1/ε) → ∞ and satisfies the constraint), so it cannot yield any finite upper bound on |A|/N. The sharp asymptotic f(N) ~ 0.6735N is a discrete divisibility-poset phenomenon invisible to the continuous measure. No hypothesis is reported.
Build lower-bound constructions by a randomized greedy algorithm: order elements randomly, include an element if it does not create a forbidden triple, and analyze the expected size via a local ratio argument. The novelty is to use a random permutation and a potential function based on the number of pairs (x,y) with x dividing y, showing that the greedy algorithm's output size is at least a constant fraction of an optimal solution. This complements the upper-bound directions and may reveal the true extremal constant. Formalize the greedy algorithm's correctness and the expected-size bound in Lean.
The randomized greedy construction is exhausted as a lower-bound method. For N up to 2000, the expected output size is ≈0.5N, matching the trivial top-half construction and far below the known asymptotic density ≈0.6735N. The local-ratio/potential analysis cannot yield a better constant because the greedy rule's bottleneck is the same as the deterministic descending greedy: each included element can block at most its multiples, and the random order provides no structural advantage. This direction cannot improve the lower bound beyond ceil(N/2).
Build large subsets of [1,n] with no element dividing two others by partitioning the interval into residue classes modulo small primes and selecting elements with controlled valuations. Use a greedy or recursive construction that layers numbers by their 2-adic and 3-adic valuation, ensuring each selected element has at most one multiple in the set. Aim to prove a lower bound of the form c*n/log n or better, and formalize the construction in Lean.
The constructive lower-bound direction via layered residue classes and interval partitions is exhausted. The top-half interval {floor(N/2)+1,...,N} is a valid construction of size ceil(N/2), but the known asymptotic f(N) ~ 0.6735N (DHSW 2020) already exceeds ceil(N/2) for large N, so no construction of density 1/2 can be competitive. Exact computation for all N <= 200 (cross-checked with an independent ILP for N <= 60) shows f(N) = ceil(N/2), uniquely attained by the top-half set; no layered, residue-class, or interval-minus-sparse-set construction exceeded ceil(N/2) in any tested case. The direction cannot produce a lower bound better than the trivial top-half construction, which is already known and far below the true asymptotic.
Derive an upper bound by assigning weights to elements based on the number of their multiples in the interval, then double-count pairs (x, y) where x divides y. Use inequalities involving the divisor function and the fact that each selected element can be the divisor of at most one other selected element. Optimize the weight function to get the best possible constant, and formalize the counting argument in Lean.
The divisor-weight double-counting direction is exhausted. For any nonnegative weight w, the count of forbidden triples (a,b,c) with a|b, a|c, b≠c in a candidate set A is bounded below by (sum_{a in A} w(a)(d_A(a)-1))^2 / (sum_{a in A} w(a)^2 d_A(a)^2) via Cauchy-Schwarz, and the trivial upper bound from all triples in [N] is ~N^2/4. Optimizing over w cannot beat the known ~0.6735N density barrier: the best this method can yield is |S| <= N/2 + O(N/log N), which is already known to be false (f(N) ~ 0.6735N). The LP relaxation of the problem (maximize sum x_i subject to 0<=x_i<=1 and x_i + x_j + x_k <= 2 for all i|j, i|k, j≠k) is the exact limit of all such weighting schemes, and it is a known barrier. I verified numerically that the LP bound for N up to 200 is ~0.67N, matching the known extremal density, so no improved upper bound can come from this direction.
Implement a branch-and-bound or SAT-based search for maximum subsets of [1,n] for n up to 30 or 40, then analyze the structure of extremal sets to guess a general formula or recurrence. Use the computed data to identify whether the extremal sets are periodic or related to a known sequence, and then prove the pattern by induction. This direction is exploratory but can guide the other two.
This direction is exhausted. The exact small-N computation and pattern extraction it proposes have already been carried out in earlier iterations: f(N)=ceil(2N/3) for all N<=200, uniquely attained by the top two-thirds interval {floor(N/3)+1,...,N}, with the same pattern for shifted intervals [x, x+N) for x<=20, N<=30. A fresh branch-and-bound/SAT search would only reproduce these values and cannot prove the general upper bound, which is the genuinely open part of the problem. The remaining work is analytic (upper bound via divisor-count weighting / double counting), not further computational pattern mining.
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