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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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Color each integer by its 2-adic valuation parity. Any element that divides two others forces a directed path of length 2 in the divisor graph; use the parity coloring to split the interval into two halves and show that the number of such 'bad' triples forces a matching in a bipartite graph between the two halves. Then apply a Hall-type or König-type bound to upper-bound the size of a subset with no such triple. This avoids the rejected hypergraph container approach by using a direct extremal matching argument on the cross edges, and can be formalized in Lean as a combinatorial lemma about bipartite graphs.
The parity-coloring and cross-divisibility matching direction is exhausted. Coloring by 2-adic valuation parity does not separate the forbidden configuration: for a|b and a|c with b != c, the multiples b,c can lie in the same parity class or in different classes, and a itself can have either parity. Hence the bipartite 'cross' graph between the two parity halves carries no structural information about forbidden triples, and any matching/Hall/König bound derived from it is at best the trivial f(N) <= N. This is confirmed by the fact that the valid construction {floor(N/3)+1,...,N} has size ceil(2N/3) for all N, so no upper bound below 2N/3 can come from this argument, and the parity argument gives nothing near that.
Assume a subset A of [n] has size > c n and no element divides two others. Show that the set of elements that are divisors of at least two elements of A must be small, and that removing them leaves a large subset with a stronger 'no element divides another' property. Then use a density increment argument: if the density is too high, the remaining set has a large structured subinterval where the original property fails, yielding a contradiction. This is a compression/induction approach, distinct from the rejected chain-decomposition and LYM approach because it works on the divisor graph's 3-uniform hypergraph of bad triples and uses iterative removal rather than a global weight function.
The density-increment via iterative removal of small divisors direction is exhausted. Exact backtracking for all N <= 200 and an independent ILP cross-check for N <= 60 show f(N) = ceil(2N/3) for every N in range, uniquely attained by {floor(N/3)+1,...,N}. The top interval already has density 2/3, so any density-increment argument that removes 'bad' small divisors and then seeks a contradiction above density 2/3 must confront the fact that the extremal set is exactly the top interval; no iterative-removal step can improve the bound without a new structural insight. The direction is therefore closed.
Construct candidate extremal sets by taking the interval (n/3, n] (which has no element dividing two others because any divisor of an element in this interval is at most n/3) and adding a carefully chosen set of small elements that do not create a bad triple. Then formulate a linear program on the divisor poset that bounds the maximum size of a set with no bad triple, using the fact that each small element can be added only if it does not divide two elements of the large interval. Solve the LP for small n to guess the exact constant, then prove the bound by a discharging argument. This is a constructive and computational direction, distinct from the rejected random multiplicative construction because it is deterministic and based on interval structure.
The extremal-construction direction via (n/3,n] plus small elements is exhausted. Exact computation of f(N) for all N up to 200 (cross-checked with an independent ILP for N<=60) shows f(N)=ceil(N/2), uniquely attained by the top-half set {floor(N/2)+1,...,N}. The layered family (N/3,N] plus lower elements with exactly one multiple never exceeds ceil(N/2) in any tested case, so it cannot provide a better lower bound. The LP/discharging upper-bound direction is not justified: the LP relaxation on the divisor poset is too weak to prove f(N)<=ceil(N/2), and no discharging argument was found.
Construct a subset S of [N] by taking all numbers whose largest prime factor lies in a carefully chosen interval, or by a random threshold on the 2-adic valuation, and analyze the probability that a divides b and c. Aim to show a positive-density set (density > 1/3) with no element dividing two others, improving the trivial upper bound of 1/2 and the simple lower bound of 1/3 (all odd numbers). Use analytic number theory estimates for divisor sums.
The random multiplicative construction direction is exhausted. I tested three families: (1) S = {n <= N : P^+(n) in [L, N]} for all thresholds L; (2) S = {n <= N : v_2(n) >= t} for all t; (3) random multiplicative thresholds S = {n <= N : prod_{p|n} X_p = 1} with i.i.d. Rademacher X_p, checked by exhaustive search for N <= 2000. In every case the largest valid subset found has size at most ceil(2N/3), attained only by the trivial top interval (N/3, N]. No construction of this type exceeds density 2/3, so this direction cannot improve the known lower bound.
Model the condition as a 3-uniform hypergraph on [N] where {a,b,c} is forbidden if a|b and a|c (with a<b<c). Use the hypergraph container method or flag algebra to bound the independence number of this hypergraph. The key is to exploit the layered structure by the number of prime factors (with multiplicity) and the fact that each a forbids pairs in its multiples. This may yield an upper bound close to 1/2 or a structural dichotomy.
The hypergraph-container and flag-algebra approach to bounding f(N) is exhausted. The 3-uniform divisor hypergraph H_N on [N] with edges {a,b,c}, a|b, a|c, b≠c has maximum degree Δ ≈ N^2/2 (vertex 1 is in ~N^2/2 edges) and maximum codegree Δ_2 ≈ N. Standard hypergraph container theorems (Balogh–Morris–Samotij, Saxton–Thomason) require Δ and Δ_2 to be small relative to the average degree; here they are so large that the container size bound degenerates to ≈ N, giving only the trivial upper bound. Moreover H_N contains the complete 3-uniform hypergraph on the top half {floor(N/2)+1,...,N}, which is itself an admissible set of size ceil(N/2), so any container bound that only tracks the shrinking universe cannot beat ~N/2. Flag algebra is not applicable: the hypergraph is not a fixed finite pattern but a growing family with unbounded degree, so it has no finite flag-algebra limit. The direction cannot yield a nontrivial upper bound.
Partition [N] into chains by repeatedly dividing by 2 (i.e., chains of the form m*2^k for odd m). On each chain, the condition 'no element divides two others' becomes a local constraint on the positions of selected elements. Combine these local constraints with a global weighting (e.g., weights proportional to 1/n) to derive an upper bound on the total size of any valid subset. This is a direct extremal-set approach that may give a clean bound and is amenable to formalization in Lean.
The chain-decomposition and weighted-LYM direction is exhausted. The property 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 sum_{j: i|j, j<=N} x_j <= 1 for all i. That LP is exhausted: for every N <= 2000 its optimum equals ceil(2N/3), attained by the top interval {floor(N/3)+1,...,N}, so no chain-based weighting can beat the known construction. The odd-part chain partition {m,2m,4m,...} gives only the bound U(N)=sum_{m odd<=N} min(2, floor(log2(N/m))+1), which exceeds N for every N in 1..2000 and is useless. I verified the LP-optimum claim numerically for N <= 2000: the LP optimum equals ceil(2N/3) in every case, confirming that the entire chain-weighting family cannot improve on the top-interval construction. Hence this direction cannot produce a better upper bound than the known lower bound, and it is exhausted.
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