User blog comment:Mh314159/Help me understand a natural number recursion/@comment-39541634-20191016114937

The general idea of such functions is simple: They re-order the natural numbers into a more complex ordering, and uses this new order to create your hierarchy of functions.

Here is a much simple example, which I'll walk you through:

(i) a(2,x) = 3^x

(ii) a(k,1) = 3

(iii) a(n+2,x+1)  = a(n,a(n+2,x))

(iv) a(1,x) = a(2x,3) (for x>1)

The first three rules give a simple recursion: a(2,x)=3^x, a(4,x)=3^^x , a(6,x)=3^^^x and so on. In general: a(2n,x)=3[n arrows]x.

So far there's nothing particularly clever here. The clever bit is the 4th rule, which basically starts a whole new hierarchy with n=1. It gives us:

a(1,x) = a(2x,3) = 3[n arrows]x ~ fω(x)

And from here, rule (iii) allows us to build more levels of recursion on this:

a(3,x) = a(1,a(1,a(1,...))) ~ fω+1(x)

a(5,x) = a(3,a(3,a(3,...))) ~ fω+2(x)

a(7,x) = a(5,a(5,a(5,...))) ~ fω+3(x)

And so on.

So basically, this function is comparable to ω*2 in the FGH. This is achieved by re-ordering the natural numbers for their usual order:

1,2,3,4,... (which corresponds to ω)

to:

2,4,6,8, ..., 1,3,5,7 , ... (which corresponds to ω*2)

Notice how in this example a(1,x) is stronger than a(3,x) even though 1 is less than 3. The reason for this is that 1 'appears later in the re-ordered list''. '''

Of-course the example you've given is much more complicated, but the idea is the same: The function basically reorders the natural numbers in some way that corresponds to some huge ordinal.

This is why f(4) is bigger than f(512). If you actually bother to write down the re-ordering in question, you'll find that 4 comes after 512. Much later, in fact.

As a side note: The fact that the numbers order is mixed like that, is crucial for the method to work. If we always had f(x,n) > f(y,n) for x>y, that means that we are ordering the numbers in their usual way:

1,2,3,4,...

And therefore our function will never get any stronger than fω(n).

Fun fact:

In principle, there is no limit to how high you can go with this method. It is possible to re-order the natural numbers so they represent any recursive ordinal you wish.

(the catch is that doing this is exactly as difficult as working with the ordinals directly. The two approaches are basically equivalent, even though they look superficially different)