User blog comment:Deedlit11/The slow-growing hierarchy and other hierarchies/@comment-5150073-20130315163450/@comment-5529393-20130315185356

Yes. Of course H_{w^0}(n) = H_1(n) = H_0(n+1) = n+1 = F_0(n+1).

One can show that if a >= b, H_{w^a + w^b} (n) = H_{w^a} (H_{w^b} (n)). So H_{w^a * m} (n) = (H_{w^a})^m (n). For successor ordinals, H_{w^(a+1)} (n) = H_{w^(a+1) [n]} (n) = H_{w^a * n} (n) = (H_{w^a})^n (n) = (F_a)^n (n) (by induction) = F_{a+1} (n).

For a limit ordinal a, H_{w^a} (n) = H_{w^a [n]} (n) = H_{w^(a[n]) (n) = F_{a[n]} (n) (by induction) = F_a (n).

So H_{w^a} (n) = F_a (n) for a < epsilon_0.