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So upon reading about the Fast growing hierarchy, I came up with this simple thing that avoids the use of ordinals:

? and d are either zero or involve @ symbols but don't involve # symbols.

F a = a + 1

F a # .... # b # d = F a # .... # b

F a # b # .... # c = F[a] a # b - 1 # .... # c

F[1] a # .... # b = F a # .... # b

F[k] a # .... # b = F[k-1] (F a # .... # b ) # .... # b

a # ? # .... # ? # 0 # b # .... # c = F a # ? # .... # ? # a # b - 1 # .... # c

F a # ? # .... # ? # d # b # .... # c = F a # ? # .... # ? # O(d) # .... # O(d) # a # 1@0 # b-1 # .... # c, with a amount of # O(d)'s and d involves at least one @ symbol.

Where we have the O function:

O(e) = e - 1

O(0@0) = a

O(d@0) = O(d)@a, where a is the first number in the FAIL.

O(d@e) = d@e-1

The function has a nice simple comparison to the fast growing hierarchy:

F a # b # c # ... # e = f[(ω^k)e + .... + ωc + ωb](a) where f[]() is the fast growing hierarchy.

Anyways, as an example expansion:

F 2 # 3 # 5@3 # 2

F (F 2 # 2 # 5@3 # 2) # 2 # 5@3 # 2

F (F (F (F 2 # 0 # 5@3 # 2) # 0 # 5@3 # 2) # 1 # 5@3 # 2) # 2 # 5@3 # 2

and then

F 2 # 0 # 5@3 # 2

= F 2 # 0 # 5@2 # 5@2 # 2 # 5@3 # 1

F 2 # 0 # 5@2 # 5@1 # 5@1 # 2 # 5@2 # 1 # 5@3 # 1

etc.  and we can go up to things like

F 5 # 5@5@5@5@5 # 5

which is decently large.

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