User blog:MachineGunSuper/The Listaments 4

Again, Again, same story. This is a continuation to this blog post, and again, the only reason why I'm writting this one is because I wrote quite a lot of stuff there and I felt like it was getting hard to understand what is where.

'''Last time we left of at the Ћ(a;b;c) function. Now, what we are gonna do this time is not really "bigger" than the numbers that we get with Ћ(a;b;c), but it CAN be if used correctly.''' '''And plus, I NEEDED to put it after because it uses what we've learnt from Ћ(a;b;c).  And it is called the NESTED LISTAMENT FUNCTION .'''

'''Listx y .......  The easiest to understand is when having 2 elements in the subscript, but of course the more elements we have the bigger our number will get.''' Let's start with the smallest and simplest number. '''List1 1 (3) = List1(64). We've already met with something with the form of List1(64), so we know how to solve it.''' '''So basically the rule is the following: The last number in the subscript means what function we will apply to the number in brackets. In this case, it's the first function.''' '''So List1 1 (3) could also be written as List1[Listt! (64)] So, we've had 2 1's in the subscript. Adding one more changes the number dramatically.  Listt1 1 1 (3) = Listt1 1 (64) = '''

=List1[62 PT (1.10..... *1091)]

But of course, we don't need to apply only the first function. '''List1 1 2 (3) = List1 1 {Listt!! [3]} Here's a fun problem: Solve this:''' '''List1 1 1 1. . . . 1, with a million 1's       (3)'''