Since the fourth alternative definition of UNAN proposed by HaydenTheGoogologist2009 has some subtle problems, I will fix it again.
However, I merged this notation fix into the single notation called "Phenol notation".
The basic array notation has the following form:
a[c]b
where all variables (a, b, c) are non-negative integers.
Rules:
Base Rule: a[0]b = a×b
Prime Rule: a[c]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.
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. Whenever you see a {0}, you can replace it by a comma.
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 a separator, and n is a non-negative integer. Whenever you see a {0}, you can replace it by a comma.
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.
With all problems fixed up to the nested array notation, first, we could have some intended behavior for the fifth alternative definition based on the new processes right now.
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)
...