User blog:Mh314159/Help me understand a natural number recursion

In my attempts to understand how recursions can generate large numbers with a few rules, I came across this from a different website:

Here's a somewhat boring entry: f(0,x)=x f(a,x)=f(g(a,a,x),x+1) g(4b+2,x,n)=b g(a,x,0)=0 g(4b+1,x,n+1)=2^g(x,x,n)*(4b+2) g(2^a*(4b+2),x,n+1)=2^g(a,a,n+1)*(4*g(2^a*(4b+2),x,n)+2) g(2^a*(4b+2)-1,x,n+1)=2^g(a,x,n+1)*(4*g(2^a*(4b+2)-1,x,n)+2)-1 Submit f(2^2^16,n) If you want to test this function, use something like f(8,n) since even f(2^2^16,1) is massive. Approximations of other functions: Successor: f(2,n) Weak Goodstein sequences/Ackermann function: f(512,n) Strong Goodstein sequences/Simple hydras/tree: f(4,n) TREE: f(2^(2^2046-1),n) Something stronger than TREE: f(2^2^16,n) Odd first arguments of f are not guaranteed to work correctly.

http://conwaylife.com/forums/viewtopic.php?f=12&t=2753&start=150

I understand why f(2,n) is the successor function, but I was unable to make progress understanding the recommended f(8,n). Also, it seems unusual to me, but perhaps not to those in the field, that f(4,n) is stronger than f(512,n). Can anyone walk through some of this and show me a little more about how this recursion works?