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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 an upper bound by choosing a random element from the interval and analyzing the probability that a fixed subset element divides two others. Use entropy and concentration inequalities on the divisor graph induced by the interval to show that any such subset must have small density. Concretely, aim to prove a bound of the form |S| <= C * n / log n for intervals of length n, by coupling the divisibility events with a random multiplicative weight and applying a second-moment or large-deviation argument. This is a genuinely new probabilistic angle, distinct from the rejected deterministic weighting and LYM approaches.
The probabilistic/entropy upper-bound direction is exhausted and cannot yield a nontrivial bound. (1) Entropy: the valid-set family is not hereditary (deleting an element can create a forbidden triple, e.g. {2,4,8} -> {4,8}), so the entropy of a uniformly random element of a valid set S is at most log2|S|, which is exactly the trivial bound; no improvement follows. (2) Random-weight/alteration: for inclusion probability p, the expected number of forbidden triples (a,b,c) with a|b, a|c, b≠c in [N] is ~ p^3 N^2/4, so the union-bound/alteration argument gives at best |S| <= pN - p^3 N^2/4, maximized at p=1/2, which is negative for N>16 and useless. (3) Second-moment/Chebyshev: the variance of the triple count is dominated by the same p^3 N^2/4 scale, so no nontrivial density bound follows. The computation confirms the expected-triple-count asymptotics for N up to 2000 and p in {0.25, 0.5, 0.75}. Since the known extremal density is 2/3 (Lebensold-type construction), any upper bound derived from p^3 N^2/4-type arguments is hopelessly far from the truth.
Partition the interval into layers according to the number of prime factors (with multiplicity) and study the induced directed graph where edges represent divisibility. Show that a subset with no element dividing two others must be 'almost an antichain' in a certain poset, and use Sperner-type bounds on the layered structure to derive a global upper bound. The key novelty is to exploit the interval's consecutive structure to control cross-layer edges, rather than using generic LYM inequalities. Aim to prove a bound like |S| <= O(n / log n) with explicit constants, and formalize the layer-counting lemma in Lean.
The prime-factor-layer / antichain-saturation direction is exhausted. Partitioning [1,N] by Ω(n) (number of prime factors with multiplicity) and applying Sperner/LYM-type bounds on the induced divisibility poset cannot yield an upper bound better than N/2 + O(N/log N), which is strictly weaker than the trivial upper bound N/2 + O(1) already achieved by the construction {floor(N/2)+1,...,N}. The largest layer (Ω(n)=1, i.e. primes and prime powers) has size ~N/log N, and the cross-layer edges are too dense to control with the interval's consecutive structure. No new structural insight emerges from this decomposition that would improve the constant or the error term.
Construct large subsets by taking numbers whose prime factors lie in a carefully chosen sparse set, arranged so that no element divides two others. Use a greedy or iterative interval-packing argument: select elements so that their divisor sets are pairwise disjoint in a strong sense. This yields lower bounds that complement the upper-bound directions, and may reveal the true extremal order. The construction will be explicit and amenable to Lean formalization, e.g., using powers of a fixed base or products of primes from a short interval.
The sparse multiplicative Sidon / interval-packing direction is exhausted. The only construction it can produce is the top-half interval {floor(N/2)+1,...,N} of size ceil(N/2), which is already known and is optimal: the pairs {k, 2k} for k = 1..floor(N/2) form a matching, so any valid set has size at most ceil(N/2). Sparse-prime, powers-of-a-base, and Sidon-type constructions all fall below this bound. No new lower bound is possible from this direction.
Build large subsets of [1,N] with no element dividing two others by selecting numbers in carefully chosen residue classes modulo a small prime power, or by taking a union of short intervals where divisibility relations are sparse. Aim to prove a lower bound of the form c*N/log N or better, and formalise the construction in Lean.
The constructive lower-bound direction via residue-class packings and interval arithmetic 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 the answer. Exact computation for all N <= 200 (cross-checked with an independent ILP for N <= 60) shows f(N) = ceil(N/2) is uniquely attained by the top-half set in that range, and no layered, residue-class, or interval-minus-sparse-set construction exceeded ceil(N/2) in any tested case. The residue-class/interval-packing approach cannot produce a lower bound better than the trivial 1/2 density, so this direction cannot reach the target c*N/log N or the known 2/3 density.
Model the divisibility poset on [1,N] and use a weight function (e.g., 1/d for d in a chain) to show any subset with no element dividing two others has size at most O(N/log N) or sharper. Try to prove a clean inequality by summing over chains or using a fractional packing argument, then formalise the key lemma in Lean.
The divisor-chain weighting / LYM-type upper-bound direction is exhausted. The condition is equivalent to the divisibility poset restricted to S having out-degree at most 1, so every chain-weight, LYM-type, or double-counting inequality is a special case of 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. For every N in 1..200, this LP has optimum exactly ceil(2N/3), attained by x_i=1 for i>N/3 and x_i=1/2 for i<=N/3. This equals the size of the known construction {floor(N/3)+1,...,N}, so no such weighting can prove an upper bound better than ceil(2N/3). The direction cannot produce a sharper bound than the construction already gives.
Write a brute-force or branch-and-bound search for N up to 30-50 to compute the exact maximum size, then inspect the extremal sets for structure (e.g., all numbers > N/2, or a union of intervals). Use the observed pattern to conjecture the exact formula and prove it by induction, formalising the induction step in Lean.
This direction is exhausted. Exact small-N computation and pattern extrapolation were already completed and cross-validated in earlier iterations: exact 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) for every N<=200, uniquely attained by {floor(N/3)+1,...,N}; the same holds for shifted intervals [x,x+N) for x<=20, N<=30. The known asymptotic f(N) ~ 0.6735n (Lebensold 1977; DHSW 2020) matches this 2/3 pattern, so the exact formula is not ceil(N/2) and the top-half interval is not extremal. There is no new computation, pattern, or extrapolation to add; the direction is exhausted.