User blog comment:MyNameIsKILLBOT/My Number/@comment-32783837-20171226021459

Ah! A new guy. Let's see what we have here.

First of all, the function f(n) = n^n^n^n^...^n^n^n^n with n "n"s can be represented in up-arrow notation as n↑↑n or n↑↑↑2. More specifically, the function itself you've described to get from i(n) to i(n+1) is called the megafuga- function.

However, the i(n) function itself appears to be:

i(0) = 2 ↑↑↑2 = 4

i(n) = i(n-1) ↑↑↑2

So, your number is equal to ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (  ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (  ( ( ( ( (4 ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2) ↑↑↑2)--that's 42 instances of " ↑↑↑2".

It's a decently large number for a beginner, comparable to f4(42) in FGH and (I think) exactly equal to gε 42 (4) in SGH given the sequence of  εn =  εn-1 ↑↑ ω. (Don't worry if you have no idea what I'm talking about.)

Basically, your function i(n) achieves pentational growth. That's ok.

However, if you're going for a very, VERY large number, instead of inputting a random number like 42 into your function, use something like i(10100) or better yet: i(i(i(i(...(i(i(i(n))))...))) where there are n "i"s, where n is probably something large. That achieves hexational growth, and of course you can go still beyond.