User blog:Simply Beautiful Art/Fast Array Iteration Leaping

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.

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

F a # ? # .... # ? # 1@0 # b #  .... # c = F a # ? #  ....  # ? # 0 #  ....  # 0 # a # 1@0 # b-1 #  .... # c, with a amount of '# 0's

F a # ? # .... # ? # d@0 # b # .... # c = F a # ? # .... # ? # O(d)@a # .... # O(d)@a # a # d@0 # b - 1 # .... # c, with a amount of '# O(d)@a's

F a # ? # .... # ? # d@e # b # .... # c =  F a # ? # .... # ? # d@e-1 # .... # d@e-1 # a # d@e # b - 1 # .... # c, with a amount of '# d@e-1's

Where we have the O function:

O(a) = a - 1

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.