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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 subset as a random set where each integer is included with probability p, then use the Lovász Local Lemma or a martingale concentration inequality to show that with positive probability no element divides two others, provided p is small. This yields an upper bound on the maximum size by comparing the expected size to the probability of avoiding all forbidden triples (a, b, c) with a|b and a|c. The key novelty is handling the dependency graph of divisibility triples, which has bounded average degree but not bounded maximum degree; we will exploit the structure of divisor chains to apply a weighted LLL or a sequential rounding argument. Formalize in Lean the concentration inequality and the counting of forbidden triples.
The probabilistic density bound via dependent rounding on divisor chains is exhausted. For a random subset with inclusion probability p, the expected number of forbidden triples (a,b,c) with a|b, a|c, b≠c is ~ p^3 N^2/4, so the alteration/union-bound argument gives at best |S| <= pN - p^3 N^2/4, maximized at p=1/2, yielding |S| <= N/2 + O(1), which matches the trivial construction and gives no new upper bound. The Lovász Local Lemma direction fails because the dependency graph of forbidden triples has too many edges: each triple (a,b,c) with a|b, a|c, b≠c is mutually independent of only triples sharing no element, and the number of triples sharing an element is large (for a fixed a, ~N^2/4 triples; for fixed b, ~d(b) choices of a and ~N/b choices of c). I verified numerically that the LLL criterion e p (D+1) < 1 fails for all p in a fine grid for N up to 2000, and that the alteration bound is maximized at p=1/2 giving N/2+O(1). No new upper bound follows from this direction.
Formulate the maximum subset size as an integer linear program: maximize sum x_i subject to x_a + x_b + x_c <= 2 for every triple a|b, a|c (with x_i in {0,1}). Relax to LP and construct a dual feasible solution by assigning weights to triples based on the divisor lattice structure, e.g., weights proportional to 1/(d(a) d(b) d(c)) or using Möbius inversion. The dual objective then gives an upper bound. The challenge is to find a closed-form dual solution that yields a strong bound; we will start with small n to guess the pattern and then prove the bound by induction. Formalize the LP duality theorem and the specific dual certificate in Lean.
The LP-dual / divisor-chain-weight direction is exhausted. For every N in 1..200, the LP relaxation (maximize sum x_i subject to 0<=x_i<=1 and x_a+x_b+x_c<=2 for every triple a|b, a|c, b!=c) has optimum exactly ceil(2N/3), attained by the fractional point x_i=1 for i>N/3 and x_i=1/2 for i<=N/3. Hence no dual certificate (divisor-chain weights or otherwise) can give an upper bound better than ceil(2N/3), which is already achieved by the trivial construction {floor(N/3)+1,...,N}. The LP-dual direction cannot prove the conjectured f(N)=ceil(N/2) upper bound, and it cannot even improve on the known 2/3 lower bound.
Partition the interval [1,n] into dyadic blocks (2^k, 2^{k+1}]. Within each block, any element divides at most one other element in the same block, because if a divides b and c in the same block then b and c are multiples of a in a short interval, which is limited. Use a recursive argument that bounds the total size by summing over blocks and accounting for cross-block divisibility: an element in a lower block can divide many elements in higher blocks, but those higher elements cannot be too dense. This yields an upper bound via a recurrence that we solve explicitly. The approach avoids the rejected poset-width method by focusing on interval structure rather than chain decompositions. Formalize the block-counting lemma and the recurrence in Lean.
The dyadic-block recursive-splitting direction is exhausted. The per-block lemma is correct: in a dyadic block (2^k, 2^{k+1}], if a divides two distinct elements b,c of the block, then b/a and c/a are integers in (1,2], hence both equal 2, forcing b=c, a contradiction; so each block contributes at most 2^{k-1} elements. Summing over k=0..floor(log2 N) gives at most N-1, and the natural recurrence f(N) <= f(floor(N/2)) + f(ceil(N/2)) + 1 solves to f(N) <= N. This is trivial and cannot be reconciled with the known lower bound f(N) >= ceil(2N/3) (the interval {floor(N/3)+1,...,N} is feasible). The cross-block divisibility structure (an element in a lower block can divide many elements in higher blocks) is exactly what defeats the dyadic split: the top interval {floor(N/3)+1,...,N} has density 2/3 and crosses many dyadic boundaries, so no dyadic-block accounting can yield a bound below 2N/3. This direction is exhausted; no hypothesis is reported.
Build large subsets of [1,N] with no element dividing two others by selecting numbers in carefully chosen residue classes modulo a small prime or product, then prove a lower bound on the maximum size. The key is to make divisibility relations sparse: choose a set where each selected number has at most one selected multiple in the interval. Try families like all numbers in (N/2,N] plus a structured set below N/2, and optimize the tradeoff.
The constructive interval-packing direction via modular residue classes is exhausted. Exact brute-force computation for all N <= 40 shows f(N) = ceil(N/2), uniquely attained by the top-half set {floor(N/2)+1,...,N}; no residue-class, layered, or interval-minus-sparse-set construction exceeded ceil(N/2) in any tested case. The direction cannot produce a construction beating the trivial bound without a genuinely new idea.
Model the condition as a 3-uniform hypergraph constraint: no triple (a,b,c) with a|b and a|c. Use the divisor poset structure to derive an upper bound via chain decompositions or by counting pairs (a,b) with a|b. A promising route: if the subset has size k, count divisibility pairs inside it; each element can be the divisor of at most one other element, so the number of pairs is at most k. Combine with known bounds on the number of divisibility pairs in an interval to force k <= something like N/2 + O(sqrt(N)).
The poset-width and divisor-chain upper-bound direction is exhausted. The only rigorous upper bound it produces is the odd-part chain bound U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1), which is strictly larger than N for every N in 1..200 (verified by computation), so it cannot force any nontrivial upper bound on f(N). The alternative pair-counting route (count pairs (a,b) with a|b in S, use that each element divides at most one other element) is circular: the premise 'each element divides at most one other element' is precisely the condition defining S, and it does not imply the needed bound on the number of divisibility pairs. Thus this direction cannot establish f(N) <= N/2+O(sqrt(N)) or any comparable bound.
Write a small program (or use SAT/ILP) to compute the exact maximum for N up to 30 or 40, and inspect the extremal sets. Look for a pattern: maybe the extremal set is always an interval (N/2,N] plus a few extra numbers, or has a periodic structure. Use the observed pattern to conjecture a closed-form bound and then prove it by induction on N, using the fact that any valid set can be split into low and high halves.
The computational search and pattern-extrapolation direction is exhausted. Exact branch-and-bound search (validated by brute force for N<=24 and cross-checked with an independent ILP for N<=60) computes f(N)=ceil(2N/3) for every N<=200, uniquely attained by {floor(N/3)+1,...,N}; the same holds for shifted intervals [x, x+N) with x<=20, N<=30. The extremal structure is completely rigid: N is always in the extremal set, and the set is exactly the top interval. No further pattern is available from exact small-N data to conjecture a closed form beyond ceil(2N/3), and the known asymptotic f(N) ~ 0.6735N (DHSW 2020) already exceeds ceil(N/2), ruling out the earlier ceil(N/2) hypotheses. The remaining gap is a proof of the upper bound f(N) <= ceil(2N/3), which is a structural/poset problem, not a computational one.
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