User blog comment:AJZajac/How fast does this sequence grow (problem in ordinal arithmetic)?/@comment-31607182-20171212204502

There's a problem with your second rule: Since the subscript of the function as well as the parameter decrease by one, we get: $$a(1) = f_{w+1}(0) = f_{w}^0(0) = 0 < a(0)$$

With the rule $$a(n+1) = f_{\omega+a(n)}(n+1)$$ this should work however.

The growth rate seems to be between $$f_{\omega \cdot 2}(n) $$

and

$$f_{\omega \cdot 2+1}(n)$$

With the rule $$a(n+1) = f_{\omega+a(n)}(a(n)) $$

the function is exactly $$ a(n) = f_{w2+1}(n)$$ (for n>0)