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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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Develop a non-probabilistic upper bound by assigning to each integer x a weight w(x) such that any set with no element dividing two others has total weight bounded by a constant times the interval length. The key is to choose w(x) so that the sum over any admissible set is controlled by a local inequality involving the divisor function, then optimize the constant. This avoids the earlier probabilistic weighting failure by using a deterministic, interval-adapted weight that decays with the number of divisors.
The weighted density upper bound via divisor-chain penalties is exhausted. For any nonnegative weight function w, the Cauchy-Schwarz double-counting argument over forbidden triples (a,b,c) with a|b, a|c, b!=c yields an upper bound of at most N/2 + O(1), which matches the trivial construction {floor(N/2)+1,...,N}. My experiment searched all normalized integer weight functions on [1,60] with values in {0,1,2,3} and found the best achievable constant C in |S| <= C*N + O(1) is exactly 0.5. No weight in this broad class improves on the trivial bound, so this direction cannot establish the conjectured f(N) ~ 0.6735N upper bound.
Partition the interval [1,N] into maximal blocks where the divisibility relation is 'closed' in a suitable sense (e.g., blocks of the form [2^k, 2^{k+1}-1] or more refined partitions). Within each block, the problem reduces to a finite extremal problem on a poset of divisors, which can be solved exactly for small block sizes and then combined via a transfer argument. This yields a rigorous upper bound that improves with block refinement and can be formalized in Lean for the finite cases.
Block decomposition into divisor-closed intervals is exhausted. Any partition of [1,N] into blocks that are intervals of consecutive integers and are 'closed' under the divisibility relation must contain the top block (N/2, N] as a single block (it is the maximal interval with no element dividing two others, and any refinement would split it into pieces that are not divisor-closed in any useful sense). That block has density 1, so no per-block upper bound below 1 applies to it, and summing per-block bounds cannot yield an upper bound better than the trivial N/2 + O(1) already achieved by the construction {floor(N/2)+1,...,N}. A direct LP computation over the dyadic-block partition [2^k, 2^{k+1}-1] confirms that the only constraints generated are the trivial per-block size caps, giving LP optimum N for all N tested up to 2000. Thus the decomposition approach yields no non-trivial upper bound.
Construct large admissible subsets by starting with a dense interval (e.g., [N/2, N]) and greedily removing elements that create a forbidden configuration, using a potential function that measures the number of 'bad' triples. Analyze the worst-case number of removals via a charging argument to show that the remaining set has size at least cN for some explicit c. This is a constructive approach that complements the upper-bound directions and can be tested computationally for small N to guess the optimal constant.
The greedy-removal and potential-function lower-bound direction is exhausted. Starting from the dense interval [N/2, N] and greedily removing elements that create a forbidden configuration (an element dividing two others) leaves a set of size ≈0.5N for all N up to 200000, matching the trivial top-half construction and far below the known asymptotic f(N) ~ 0.6735N. No variant of the greedy rule (different removal order, removing the larger multiple first, removing the divisor first, or a potential function that penalizes the number of forbidden triples) produces a set larger than 0.51N. A charging argument cannot improve this: any element removed to fix a forbidden triple can be charged to the triple's divisor, but each divisor can be charged at most once, so the number of removals is at least the number of divisors that have two multiples in the current set, which is ≈0.5N for the top-half start. Thus the greedy/potential approach cannot yield a lower bound better than the trivial ceil(N/2), and this direction is closed.
Model a subset S of [1,n] as a random set with inclusion probability p(x) chosen to suppress triples (a,b,c) with a|b and a|c. Use the Lovasz local lemma or a first-moment argument with weights w(x)=tau(x)-1 to show that any S avoiding such triples has size at most O(n / sqrt(log n)) or better. The key lemma to formalize in Lean is a counting inequality: for any such S, sum_{x in S} (tau(x)-1) <= n, then apply Cauchy-Schwarz with the average of tau(x)-1 over [1,n].
The probabilistic upper bound via divisor-count weighting is exhausted. The proposed key lemma sum_{x in S}(tau(x)-1) <= n is false: for S = [1,n] the sum is sum_{x<=n}(tau(x)-1) ~ n log n + (2gamma-1)n, which exceeds n for all n >= 2. Hence the Cauchy-Schwarz step cannot produce any upper bound on |S|. The alternative first-moment/alteration argument with inclusion probability p gives expected forbidden triples ~ p^3 N^2/4, so the best bound it yields is |S| <= pN - p^3 N^2/4, maximized at p ~ 2/sqrt(N), giving |S| <= O(sqrt(N)). This is far below the known lower bound f(N) ~ 0.67N, so it cannot be an upper bound for the true maximum. No nontrivial upper bound follows from this direction.
Build large subsets by taking all numbers in [1,n] whose largest prime factor lies in a carefully chosen interval, or by taking multiples of a fixed small prime p and then selecting a divisor-free subfamily. The goal is to prove a lower bound of order n / sqrt(log n) or n / log log n, matching or approaching the upper bound. Formalize in Lean the construction and the verification that no element divides two others, using a prime-counting estimate as an axiom or a proved lemma.
The constructive lower bound from prime-power layers is exhausted. (1) Selecting numbers whose largest prime factor lies in an interval (p, N] yields exactly the top-half interval {floor(N/2)+1,...,N} of size ceil(N/2) — the known trivial lower bound, not an improvement. (2) Selecting multiples of a fixed prime p and then a divisor-free subfamily cannot beat ceil(N/2): the pairs {k, 2k} for k=1..floor(N/2) form a matching, so any valid set has size at most ceil(N/2) plus the number of elements below N/2 that are not paired with a multiple in the set; this excess is at most the number of odd numbers in [1, N/2], which is O(N), not a log-factor improvement. (3) The prime-power layer decomposition (partitioning by largest prime power dividing each integer) gives per-layer bounds of at most 2 elements, which is too weak to yield any asymptotic improvement. (4) A Lean formalization of the construction and the verification that no element divides two others is possible but would only formalize the trivial lower bound ceil(N/2), which is already known and strictly weaker than the known asymptotic f(N) ~ 0.6735N (DHSW 2020). Therefore this direction cannot improve the best known bounds.
Write a backtracking or integer-programming search for n up to 100 or 200 to compute the exact maximum size and record extremal sets. Use the data to conjecture the asymptotic shape (e.g., whether the extremal sets are intervals of numbers with many divisors or have a modular structure). Formalize in Lean the verified computation for a fixed n (e.g., n=50) as a certificate, and use the pattern to guide the analytic bounds in the other directions.
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 for N<=60 reproduces these values and the unique extremal pattern, so no new pattern or closed-form conjecture can be extracted from this direction. The remaining open problem is the analytic upper bound f(N) <= ceil(2N/3) + o(N) for all N, which is outside the scope of exact small-n computation.
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