User blog comment:Boboris02/Quickly-extending linear hydra/@comment-27173506-20161129163026

Let's try to find the growth rate of the hydra a : {a,a,a,a...a} with a+2 a's.

Let's look at a simplified version of the function, notated with square brackets, like this [a,b,c,d...z]. With the following rules:

1. [a] = a

2. [a,0,b,c...z] = [a+1,b,c...z]

3. [a,b+1,c...z] = [a,b,b,b...b,c...z] with a b's.

This hydra has approximately the same growth rate as your hydra (with less trivial growths such as the a^b or the constant incresing by 1). Furthermore it is easier to see here that: [a,b,c...z] = [[a,b],c...z] (Try to prove, it's not very hard).

Therefore, [a,b,c,d...y,z] = [[[...[[[a,b],c],d]...],y],z]. Clearly the ordinal of the growth rate of [a,a,a...a] with a+2 a's will be just 1 higher than the ordinal of the growth rate of [a,a] in the FGH. Furthermore, notice that [a+1,a+1] is equal to [a+1,a,a,a...a] with a+2 a's, which is about 1 higher than the ordinal of the growth rate of [a,a]. Therefore, [a,a] grows at about f_a(a) on the FGH, that is f_w(a), and [a,a,a...a] grows at about f_a+1(a), which is approximately f_w(a+1)

tl;dr, it grows at about f_w(n) (maybe a bit closer to f_w(2^n) or something, but a far cry from f_w+1(n))