Let's extend the notation with separators, such as commas and brackets!
The linear array notation has the following form:
a[c, d, e, ..., n]b
where all variables are nonnegative integers, it can be any number of entries.
The extended array notation has the following form:
a[c X d X e X ... {X} n]b
where all variables are nonnegative integers, it can be any number of entries between separators and {X} can be any level of separators, denoted with {}. {} separators can be nested.
The number before the array (a) is the base, the number after it (b) is the iterator.
The 2-entry array notation has the following form:
a[c,d]b
where all variables (a, b, c, d) are non-negative integers.
Rules:
Base Rule: a[0,0]b = a[0]b = a×b
Tailing Rule: a[c,0]b = a[c]b
Prime Rule: a[c,d]1 = a, a[c,d]0 = 1 if c > 0, a[0,d]0 = 0
Recursion Rule: a[c,d]b = a[c-1,d](a[c,d](b-1)) if b > 1 and c > 0
Hyperoperation Rule: a[0,d]b = a[b,d-1]a
If there are two or more distinct rules to apply to a single expression, the lowest-numbered rule which is applicable and whose result is a valid expression will be applied.
The linear array notation has the following form:
a[c,d,e,...,n]b
where all variables (a, b, c, d, n) are non-negative integers. It can be any number of entries.
The number before the array is the base, the number after it is the iterator.
Rules:
Base Rule: a[0]b = a×b
Tailing Rule: a[#,0]b = a[#]b
Prime Rule: a[%]1 = a, a[c,%]0 = 1 if c > 0, a[0,%]0 = 0
Recursion Rule: a[c#]b = a[c-1#](a[c#](b-1)) if b > 1 and c > 0
If there are two or more distinct rules to apply to a single expression, the lowest-numbered rule which is applicable and whose result is a valid expression will be applied.
Otherwise, you start the process described below from the first number inside the brackets:
Case I: If the first few entries are 0, move to the next entry until the nonzero entry.
Case II: If the entry next to the zero entry is greater than 0, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base:
a[0,0,0,...,0,0,0,n,#]b = a[0,0,0,...,0,0,b,n-1,#]a if n > 0. The process ends.
The last process ends, check if rules apply.
The planar/dimensional array notation has the following form:
a[c X d X e X ... X n]b
where all variables (a, b, c, d, n) are non-negative integers, and the X's are separators.
Now, what is a separator? Separators are the items that come between the entries. In the previous parts there was only one type of separator, the comma. Now I am making a second separator type, the brackets. In that case, the separators are the comma and {n}, where n is a non-negative integer. Commas indicate {0} separators.
The number before the array is the base, the number after it is the iterator.
Rules:
Base Rule: a[0]b = a×b
Tailing Rule: a[# X 0]b = a[#]b (X is a separator)
Prime Rule: a[%]1 = a, a[c X %]0 = 0 if c > 0, a[0 X %]0 = 0 (X is a separator)
Recursion Rule: a[c%]b = a[c-1%](a[c%](b-1)) if b > 1 and c > 0
If there are two or more distinct rules to apply to a single expression, the lowest-numbered rule which is applicable and whose result is a valid expression will be applied.
Otherwise, you start the process described below from the first number inside the brackets:
Case I: If the first few entries are 0, move to the next entry until the nonzero entry.
Case II: If the entry next to the zero entry is greater than 0:
II1: If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base:
a[0 X 0 X 0 ... X 0 X 0 X 0,n,#]b = a[0 X 0 X 0 X ... X 0 X 0 X b,n-1,#]a if n > 0. The process ends.
II2: If there are brackets {n} before it, where n > 0, replace the "0{n}m" with "0{n-1}0{n-1}...{n-1}0{n-1}0{n-1}1{n}m-1" where there are b zeroes between {n-1} separators, and change the iterator (b) to the base (a). In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas. The process ends.
The last process ends, check if rules apply.
The hyperdimensional array notation has the following form:
a[c X d X e X ... X n]b
where all variables (a, b, c, d, n) are non-negative integers, and the X's are separators.
In this extension, the separators are the comma and {a, b, c, ..., z}, where all a, b, c, ..., z are non-negative integers. Commas are {0} separators.
The number before the array is the base, the number after it is the iterator.
Rules:
Base Rule: a[0]b = a×b
Tailing Rule: a[# X 0]b = a[#]b (X is a separator)
Prime Rule: a[%]1 = a, a[c X %]0 = 1 if c > 0, a[0 X %]0 = 0 (X is a separator)
Recursion Rule: a[c%]b = a[c-1%](a[c%](b-1)) if b > 1 and c > 0
If there are two or more distinct rules to apply to a single expression, the lowest-numbered rule which is applicable and whose result is a valid expression will be applied.
Otherwise, you start the process described below from the first number inside the brackets:
Case I: If the first few entries are 0, move to the next entry until the nonzero entry.
Case II: If the entry next to the zero entry is greater than 0:
II1: If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base:
a[0 X 0 X 0 ... X 0 X 0 X 0,n,#]b = a[0 X 0 X 0 X ... X 0 X 0 X b,n-1,#]a if n > 0. The process ends.
II2: If there are brackets {%} before it (% is not just 0):
II2.1: If there are brackets {n%} before it, where n > 0, replace the "0{n%}m" with "0{n-1%}0{n-1%}...{n-1%}0{n-1%}0{n-1%}1{n%}m-1" where there are b zeroes between {n-1%} separators, and change the iterator to the base. In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas. The process ends.
II2.2: If there are brackets {0%} before it (say, the first entry n = 0 in {n%}), move to the next entry until the nonzero entry, then do the same process analogous to the linear array notation: "If the entry next to the zero entry is greater than 0, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base." The process ends.
The last process ends, check if rules apply.
The nested array notation has the following form:
a[c X d X e X ... X n]b
where all variables (a, b, c, d, n) are non-negative integers, and the X's are separators.
In this extension, the separators are the comma and {m A n}, where x is an expression such that {x} is a separator, A is an inner separator, and n is a non-negative integer. Commas are {0} separators.
The number before the array is the base, the number after it is the iterator.
Rules:
Base Rule: a[0]b = a×b
Tailing Rule: a[# X 0]b = a[#]b (X is a separator)
Prime Rule: a[%]1 = a, a[c X %]0 = 1 if c > 0, a[0 X %]0 (X is a separator)
Recursion Rule: a[c%]b = a[c-1%](a[c%](b-1)) if b > 1 and c > 0
If there are two or more distinct rules to apply to a single expression, the lowest-numbered rule which is applicable and whose result is a valid expression will be applied.
Otherwise, you start the process described below from the first number inside the brackets:
Case I: If the first few entries are 0, move to the next entry until the nonzero entry.
Case II: If the entry next to the zero entry is greater than 0:
II1: If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base:
a[0 X 0 X 0 ... X 0 X 0 X 0,n,#]b = a[0 X 0 X 0 X ... X 0 X 0 X b,n-1,#]a if n > 0. The process ends.
II2: If there are brackets {%} before it (% is not just 0):
II2.1: If there are brackets {n%} before it, where n > 0, replace the "0{n%}m" with "0{n-1%}0{n-1%}...{n-1%}0{n-1%}0{n-1%}1{n%}m-1" where there are b zeroes between {n-1%} separators, and change the iterator to the base. In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas. The process ends.
II2.2: If there are brackets {0%} before it (say, the first entry n = 0 in {n%}), move to the next entry until the nonzero entry, then do the same cases analogous to the linear array notation: "If the entry next to the zero entry is greater than 0, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base." The process ends.
II2.3: If there are inner brackets nested inside outer brackets, do the same process analogous to the previous cases above. The process ends.
The last process ends, check if rules apply.
The level of this notation is quite easy, since it is based on function which can be provably recursive within Peano arithmetic, The fast-growing hierarchy here is given by Wainer hierarchy.
a[0,1]b has level ω - We can see that a[0,1]b = a[b]a in the basic notation (Extension 1).
a[1,1]b has level ω+1
a[2,1]b has level ω+2
a[3,1]b has level ω+3
a[0,2]b has level ω2
a[1,2]b has level ω2+1
a[2,2]b has level ω2+2
a[0,3]b has level ω3
a[1,3]b has level ω3+1
a[0,4]b has level ω4
a[0,5]b has level ω5
a[0,6]b has level ω6
In general, a[c,d]b is equal to {a,b,c,d+1} in BEAF, assuming c >= 1.
...
a[0,0,1]b has level ω^2
a[1,0,1]b has level ω^2+1
a[2,0,1]b has level ω^2+2
a[0,1,1]b has level ω^2+ω
a[1,1,1]b has level ω^2+ω+1
a[0,2,1]b has level ω^2+ω2
a[0,3,1]b has level ω^2+ω3
a[0,0,2]b has level ω^2*2
a[1,0,2]b has level ω^2*2+1
a[0,1,2]b has level ω^2*2+ω
a[0,0,3]b has level ω^2*3
a[0,0,4]b has level ω^2*4
a[0,0,0,1]b has level ω^3
a[1,0,0,1]b has level ω^3+1
a[0,1,0,1]b has level ω^3+ω
a[0,0,1,1]b has level ω^3+ω^2
a[0,0,0,2]b has level ω^3*2
a[0,0,0,3]b has level ω^3*3
a[0,0,0,0,1]b has level ω^4
a[0,0,0,0,2]b has level ω^4*2
a[0,0,0,0,0,1]b has level ω^5
a[0,0,0,0,0,0,1]b has level ω^6
In general, if the array has n 0's, the growth is ω^n. And the limit of this notation is ω^ω.
Moving on extended arrays...
a[0{1}1]b has level ω^ω. In particular, since a[0{1}1]b = {a,b+2 (1) 2} in BEAF, this level doesn't match exactly over {a,b (1) 2}-level.
a[1{1}1]b has level ω^ω+1
a[2{1}1]b has level ω^ω+2
a[0,1{1}1]b has level ω^ω+ω
a[1,1{1}1]b has level ω^ω+ω+1
a[0,2{1}1]b has level ω^ω+ω2
a[0,0,1{1}1]b has level ω^ω+ω^2
a[0,0,0,1{1}1]b has level ω^ω+ω^3
a[0{1}2]b has level ω^ω*2
a[1{1}2]b has level ω^ω*2+1
a[0,1{1}2]b has level ω^ω*2+ω
a[0{1}3]b has level ω^ω*3
a[0{1}4]b has level ω^ω*4
a[0{1}0,1]b has level ω^(ω+1)
a[1{1}0,1]b has level ω^(ω+1)+1
a[0,1{1}0,1]b has level ω^(ω+1)+ω
a[0{1}1,1]b has level ω^(ω+1)+ω^ω
a[0{1}2,1]b has level ω^(ω+1)+ω^ω*2
a[0{1}0,2]b has level ω^(ω+1)*2
a[0{1}1,2]b has level ω^(ω+1)*2+ω^ω
a[0{1}0,3]b has level ω^(ω+1)*3
a[0{1}0,0,1]b has level ω^(ω+2)
a[0{1}0,0,2]b has level ω^(ω+2)*2
a[0{1}0,0,0,1]b has level ω^(ω+3)
a[0{1}0{1}1]b has level ω^ω2
a[1{1}0{1}1]b has level ω^ω2+1
a[0{1}1{1}1]b has level ω^ω2+ω^ω
a[0{1}0{1}2]b has level ω^ω2*2
a[0{1}0{1}0,1]b has level ω^(ω2+1)
a[0{1}0{1}0,0,1]b has level ω^(ω2+2)
a[0{1}0{1}0{1}1]b has level ω^ω3
a[0{1}0{1}0{1}0{1}1]b has level ω^ω4
...
a[0{2}1]b has level ω^ω^2
a[1{2}1]b has level ω^ω^2+1
a[0,1{2}1]b has level ω^ω^2+ω
a[0{1}1{2}1]b has level ω^ω^2+ω^ω
a[0{1}0{1}1{2}1]b has level ω^ω^2+ω^ω2
a[0{2}2]b has level ω^ω^2*2
a[0{2}3]b has level ω^ω^2*3
a[0{2}0,1]b has level ω^(ω^2+1)
a[0{2}0,2]b has level ω^(ω^2+1)2
a[0{2}0,0,1]b has level ω^(ω^2+2)
a[0{2}0{1}1]b has level ω^(ω^2+ω)
a[0{2}0{1}0,1]b has level ω^(ω^2+ω+1)
a[0{2}0{1}0{1}1]b has level ω^(ω^2+ω2)
a[0{2}0{2}1]b has level ω^(ω^2*2)
a[0{2}0{2}2]b has level ω^(ω^2*2)*2
a[0{2}0{2}0,1]b has level ω^(ω^2*2+1)
a[0{2}0{2}0{1}1]b has level ω^(ω^2*2+ω)
a[0{2}0{2}0{2}1]b has level ω^(ω^2*3)
a[0{2}0{2}0{2}0{2}1]b has level ω^(ω^2*4)
a[0{3}1]b has level ω^ω^3
a[1{3}1]b has level ω^ω^3+1
a[0{3}2]b has level ω^ω^3*2
a[0{3}0,1]b has level ω^(ω^3+1)
a[0{3}0{1}1]b has level ω^(ω^3+ω)
a[0{3}0{2}1]b has level ω^(ω^3+ω^2)
a[0{3}0{3}1]b has level ω^(ω^3*2)
a[0{3}0{3}0{3}1]b has level ω^(ω^3*3)
a[0{4}1]b has level ω^ω^4
a[0{4}0{4}1]b has level ω^(ω^4*2)
a[0{5}1]b has level ω^ω^5
a[0{6}1]b has level ω^ω^6
Now, you reached the hyperdimensional arrays!
a[0{0,1}1]b has level ω^ω^ω
a[1{0,1}1]b has level ω^ω^ω+1
a[0,1{0,1}1]b has level ω^ω^ω+ω
a[0{1}1{0,1}1]b has level ω^ω^ω+ω^ω
a[0{2}1{0,1}1]b has level ω^ω^ω+ω^ω^2
a[0{0,1}2]b has level ω^ω^ω*2
a[0{0,1}3]b has level ω^ω^ω*3
a[0{0,1}0,1]b has level ω^(ω^ω+1)
a[0{0,1}0,0,1]b has level ω^(ω^ω+2)
a[0{0,1}0{1}1]b has level ω^(ω^ω+ω)
a[0{0,1}0{2}1]b has level ω^(ω^ω+ω^2)
a[0{0,1}0{0,1}1]b has level ω^(ω^ω*2)
a[0{0,1}0{0,1}0{0,1}1]b has level ω^(ω^ω*3)
a[0{1,1}1]b has level ω^ω^(ω+1)
a[0{1,1}2]b has level ω^ω^(ω+1)*2
a[0{1,1}0,1]b has level ω^(ω^(ω+1)+1)
a[0{1,1}0{0,1}1]b has level ω^(ω^(ω+1)+ω^ω)
a[0{1,1}0{1,1}1]b has level ω^(ω^(ω+1)*2)
a[0{2,1}1]b has level ω^ω^(ω+2)
a[0{3,1}1]b has level ω^ω^(ω+3)
a[0{0,2}1]b has level ω^ω^ω2
a[0{0,2}0,1]b has level ω^(ω^ω2+1)
a[0{0,2}0{0,2}1]b has level ω^(ω^ω2*2)
a[0{1,2}1]b has level ω^ω^(ω2+1)
a[0{2,2}1]b has level ω^ω^(ω2+2)
a[0{0,3}1]b has level ω^ω^ω3
a[0{0,4}1]b has level ω^ω^ω4
a[0{0,0,1}1]b has level ω^ω^ω^2
a[0{0,0,1}2]b has level ω^ω^ω^2*2
a[0{0,0,1}0,1]b has level ω^(ω^ω^2+1)
a[0{0,0,1}0{0,0,1}1]b has level ω^(ω^ω^2*2)
a[0{1,0,1}1]b has level ω^ω^(ω^2+1)
a[0{2,0,1}1]b has level ω^ω^(ω^2+2)
a[0{0,1,1}1]b has level ω^ω^(ω^2+ω)
a[0{0,2,1}1]b has level ω^ω^(ω^2+ω2)
a[0{0,0,2}1]b has level ω^ω^(ω^2*2)
a[0{0,0,3}1]b has level ω^ω^(ω^2*3)
a[0{0,0,0,1}1]b has level ω^ω^ω^3
a[0{0,0,0,2}1]b has level ω^ω^(ω^3*2)
a[0{0,0,0,0,1}1]b has level ω^ω^ω^4
a[0{0,0,0,0,0,1}1]b has level ω^ω^ω^5
a[0{0,0,0,0,0,0,1}1]b has level ω^ω^ω^6
Moving on nested arrays...
a[0{0{1}1}1]b has level ω^ω^ω^ω
a[1{0{1}1}1]b has level ω^ω^ω^ω+1
a[0,1{0{1}1}1]b has level ω^ω^ω^ω+ω
a[0{1}1{0{1}1}1]b has level ω^ω^ω^ω+ω^ω
a[0{0,1}1{0{1}1}1]b has level ω^ω^ω^ω+ω^ω^ω
a[0{0{1}1}2]b has level ω^ω^ω^ω*2
a[0{0{1}1}0,1]b has level ω^(ω^ω^ω+1)
a[0{0{1}1}0{1}1]b has level ω^(ω^ω^ω+ω)
a[0{0{1}1}0{0,1}1]b has level ω^(ω^ω^ω+ω^ω)
a[0{0{1}1}0{0{1}1}1]b has level ω^(ω^ω^ω*2)
a[0{1{1}1}1]b has level ω^ω^(ω^ω+1)
a[0{0,1{1}1}1]b has level ω^ω^(ω^ω+ω)
a[0{0{1}2}1]b has level ω^ω^(ω^ω*2)
a[0{0{1}0,1}1]b has level ω^ω^ω^(ω+1)
a[0{0{1}0{1}1}1]b has level ω^ω^ω^ω2
a[0{0{2}1}1]b has level ω^ω^ω^ω^2
a[0{0{3}1}1]b has level ω^ω^ω^ω^3
a[0{0{0,1}1}1]b has level ω^ω^ω^ω^ω
a[0{0{0,0,1}1}1]b has level ω^ω^ω^ω^ω^2
a[0{0{0{1}1}1}1]b has level ω^ω^ω^ω^ω^ω
a[0{0{0{0,1}1}1}1]b has level ω^ω^ω^ω^ω^ω^ω
a[0{0{0...{0{0{0,1}1}1}...1}1}1]b has level ω^^(2n+1)
Now, the limit of the alpha stage is ε0 with respect to the fundamental sequences based on Cantor normal form.
A complex example:
3[0{0{0,1}1}1]3
= 3[0{0{3}1}1]3 <= Case II1+II2.1: If there are brackets {0%} before it (say, the first entry n = 0 in {n%}), move to the next entry until the nonzero entry, then do the same cases analogous to the linear array notation: "If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base."
= 3[0{0{2}0{2}0{2}1}1]3 <= Case II2.1: If there are brackets {n%} before it, where n > 0, replace the "0{n%}m" with "0{n-1%}0{n-1%}...{n-1%}0{n-1%}0{n-1%}1{n%}m-1" where there are b zeroes between {n-1%} separators, and change the iterator to the base. In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas.
= 3[0{0{2}0{2}0{1}0{1}0{1}1}1]3 <= Case II2.1: If there are brackets {n%} before it, where n > 0, replace the "0{n%}m" with "0{n-1%}0{n-1%}...{n-1%}0{n-1%}0{n-1%}1{n%}m-1" where there are b zeroes between {n-1%} separators, and change the iterator to the base. In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas.
= 3[0{0{2}0{2}0{1}0{1}0,0,0,1}1]3 <= Case II2.1: If there are brackets {n%} before it, where n > 0, replace the "0{n%}m" with "0{n-1%}0{n-1%}...{n-1%}0{n-1%}0{n-1%}1{n%}m-1" where there are b zeroes between {n-1%} separators, and change the iterator to the base. In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas.
= 3[0{0{2}0{2}0{1}0{1}0,0,3}1]3 <= Case II1+II2.1: If there are brackets {0%} before it (say, the first entry n = 0 in {n%}), move to the next entry until the nonzero entry, then do the same cases analogous to the linear array notation: "If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base."
= 3[0{0{2}0{2}0{1}0{1}0,3,2}1]3 <= Case II1+II2.1: If there are brackets {0%} before it (say, the first entry n = 0 in {n%}), move to the next entry until the nonzero entry, then do the same cases analogous to the linear array notation: "If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base."
= 3[0{0{2}0{2}0{1}0{1}3,2,2}1]3 <= Case II1+II2.1: If there are brackets {0%} before it (say, the first entry n = 0 in {n%}), move to the next entry until the nonzero entry, then do the same cases analogous to the linear array notation: "If there are commas before it, decrease the nonzero entry by 1 and change the last zero entry placed before the first nonzero entry into the iterator, and change the iterator to the base."
= 3[0{0{2}0{2}0{1}0,0,0,1{1}2,2,2}1]3
= 3[0{0{2}0{2}0{1}0,0,3{1}2,2,2}1]3
= 3[0{0{2}0{2}0{1}0,3,2{1}2,2,2}1]3
= 3[0{0{2}0{2}0{1}3,2,2{1}2,2,2}1]3
= 3[0{0{2}0{2}0,0,0,1{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{2}0,0,3{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{2}0,3,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{2}3,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0{1}0{1}1{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0{1}0,0,0,1{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0{1}0,0,3{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0{1}0,3,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0{1}3,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0,0,0,1{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0,0,3{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}0,3,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0{1}3,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0,0,0,1{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0,0,3{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}0,3,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{2}3,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0{1}0{1}1{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0{1}0,0,0,1{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0{1}0,0,3{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0{1}0,3,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0{1}3,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0,0,0,1{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0,0,3{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}0,3,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0{1}3,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0,0,0,1{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0,0,3{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{0,3,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{3,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3 <= Case II2.1: If there are brackets {n%} before it, where n > 0, replace the "0{n%}m" with "0{n-1%}0{n-1%}...{n-1%}0{n-1%}0{n-1%}1{n%}m-1" where there are b zeroes between {n-1%} separators, and change the iterator to the base. In that case, if there are brackets {1} before it, replace the "0{1}m" with "0,0,0,...,0,0,0,1{1}m-1" where there are b zeroes between commas.
= 3[0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{1,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{1,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{1,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
= 3[0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{1,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{1,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{0,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{0,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}0{0,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2{2}2,2,2{1}2,2,2{1}2,2,2}1]3
...
as the FGH approximation by fundamental sequence:
FS0: f_{ε0[6]}(3) = f_{ω^ω^ω^ω^ω}(3)
FS1: f_{ω^ω^ω^ω^3}(3)
FS2: f_{ω^ω^ω^(ω^2*3)}(3)
FS3: f_{ω^ω^ω^(ω^2*2+ω3)}(3)
FS4: f_{ω^ω^ω^(ω^2*2+ω2+3)}(3)
...