User blog:GamesFan2000/Exploding Factor-Multiple Function

The Intention
The Exploding Factor-Multiple Function, or EFMF, is a companion to the Factor-Multiple Function, this time with an aim of actually building a really fast-growing function. FMF is capped at omega+1 in the FGH, so we have quite a bit of room for improvement.

The Rules
We will define EFM(n) as our general expression. n will act as our base for the system. The system, as advertised, will use factors and multiples to build the function. The rules are as follows:

Rule 1: 0, fractions, decimals, negative integers, imaginary numbers and complex numbers are illegal. Natural numbers are the only valid base numbers.

Rule 2: EMF(n) will default to 1 if n=1. For all n greater than 1, start by finding every factor except for 1. Then, find every unique multiple up to and including the nth multiple of n (not counting n itself). Follow this up with finding all unique multiples up to and including the (n-1)th multiple of the nth multiple of n, then finding all unique multiples up to and including the (n-2)th multiple of the (n-1)th multiple of the nth multiple of n. Repeat the pattern up to and including the 1st multiple of the (n-(n-1))th multiple of…

Rule 3: With all of the numbers you’ve found, create an expression. This expression is built in order of smallest on the left to biggest on the right. The expression is to be based in Knuth’s up-arrow notation. The number of arrows is equal to the largest number in the expression. Solve this expression and call it a. Create an a-long expression of a’s, with an a-high arrow tower between each number. A tower in this context is defined as n^n^…nn, where the number of arrows on each level is determined by the expression on the previous, from top to bottom. In each level will be a a’s, and the highest level of each tower has a arrows between each a. Remember, there are essentially a branching paths between each section of each level. Call the answer of this a1. Using Rule 2 with a1 as the determinant, create another expression. Call this b1. Using the same parameters for the levelled expression, but with b1 as the determinant, create another expression. Call this a2. Using the logic shown here, continue the process until you’ve reached the (a1)th subscript of a.

Rule 4: Now that we’ve reached the (a1)th subscript of a, build a Conway chained arrow expression. It will be a_{a1} numbers long, with each number being equal to a_{a1}. Between each number is a_{a1} levels of right arrows, each level containing a_{a1} a_{a1}’s. The top level has a_{a1} arrows between each number. In context, it’s similar to what we did with the Knuth arrows, where whatever was on the level above your current level is how many arrows are on the current level. This expression will follow Peter Hurford’s rules. Call the answer of this expression c1. Using c1 as the determinant, create another expression with the same parameters. Call the answer of this new expression c2. Using this logic, continue the process until you’ve reached the (c1)th subscript of c. The number that corresponds to c_{c1} is the final answer of the entire expression.