User blog comment:Planterobloon/How do you compare huge numbers to each other?/@comment-30754445-20170812190502

To compare two numbers, you need to write them in the same "standard" notation. The Fast-Growing Hierarcrhy is one way to do it. My own Lexicographic Letter Notation (an extension of ordinary decimal notation) is another.

At any rate, as Denis correctly stated, fω+1(3) eventually expands to:

fω(fω(6.89508×10121210694)

And there's no point to continue expanding this further, unless you want to actually write down the 121,210,695-digits of "6.89508...×10121210694)" in full.

Why? Well, remember that fω(n) roughly corresponds to having n-1 arrows in Knuth Up-Arrow notation. Also remember that the number of arrows in a number is much more important than anything else: a [n arrows] b would nearly always be much much bigger than a [n-1 arrows]c, regardless of the exact values of b and c.

IOW getting the precise arrow count should be our most priority when we're trying to estimate the szie of a number. And if the number of arrows is a 121,210,695-digit number... well, then writing as many digits of this number as we can takes priority over anything else.

Another way to look at it, is to look at this algorithm for estimating fω+1(3):

1. Set A = 6.89508×10121210694

2. Set B = 10 [A arrows] 10 (you could replace the 10's here with any smallish number greater than 2 and it won't make any significant difference)

3. C= 10 [B arrows] 10. The value of fω+1(3) is about C.

Not happy with this accuracy? Calculate more digits of A. There are 121,210,695 of them, and until you're done with that, there's no other way the make a better approximation.

Yet another way to look at it, is to examine how the number is written in Lexicographic Letter Notation:

fω+1(3) = K3-1-2-8-121210694-68950808...

This can be read as a summarized version of the algorithm given above:

1. K3 tells us that there are 3 steps in the algorithm (A,B and C).

2. K3-1 is the same thing, but also tells us that A is between 10 and 10↑↑10

3. K3-1-2 tells us that A is between 10↑↑2=1010 and 10↑↑3=1010 10

4. K3-1-2-8 tells us that A is between 1010 8 and 1010 9

5. K3-1-2-8-121210694 tells us that A is between 10121210694 and 10121210695

6. The next 121,210,695 digits are the actual digits of A

(and in case you're really really interested: the final digits of A can be computed fairly easily with modular arithmetics and they turn out to be ...2237413. After that, the lexicographical expansion continues with "-1-1-1-1-1-1..." for a very long stretch of 1's)