On Mathoverflow user leechlattice shows that $h(n) \ll n^{\frac 2 3}$.

1have lower_bound := sorry;2(lower_bound =O[Filter.atTop] fun n => ↑(Erdos160.erdos_160.h n)) ∧3  ∀ c > 0,4    ((fun n => Real.exp (c * Real.log ↑n ^ (1 / 12))) =O[Filter.atTop] fun n => ↑(Erdos160.erdos_160.h n)) →5      ∀ c > 0, (fun n => Real.exp (c * Real.log ↑n ^ (1 / 12))) =o[Filter.atTop] lower_bound