Question mark notation

is a notation made by wiki user to express big numbers.Polymations. New notation. Retrieved 2024-01-14.Polymations. Question Mark Notation v2 up to dimensional qmn. Retrieved 2024-06-20. It is also made to fix the problem in BEAF that \(\{2,n,1,2\} = 4\) because \(2\{2\{...\}2\}2 = 4\). The creator wanted to make this notation so that an expression with \(\{2,2\}\) is 4 but an expression with \(\{2,3\}\) is way larger. The special trait of this notation is that if you simplify it all the way down, the final result will simplify down to \(a\{\text{very big number}\}b\). The creator will add more information to this notation, and it is currently under construction.
==Definition==
===Current version (ver.2)===
====Terminology====
Notation is written as \(a:[b]\), where a is a question mark structure, and b is the index. Think of the question mark structure as an ordinal. A question mark structure is a group of question mark blocks.
\(???[a]\) represents a question mark block. It is actually written like that in the notation. \(a \in \mathbb{N}\)
\(\#\) represents the rest of something.
\(\$\) represents an array.
\(\text{©}\) represents a separator or a rest of a separator.
====Rules (Basic):====
\(???[0]:[n] = n\)
\(???[a]:[n] = ???[a-1]:[n+1]\)
====Rules (Multi-Basic) ====
\(\#,???[1]:[a] = \#:[a]\)
\(???[1],\#,???[b]:[a] = ???[2],\#,???[b-1]:[a]\)
\(\#,???[0],???[0]:[a] = \#,???[1]:[a]\)
\(\#,???[0],???[b]:[a] = \#,???[\#,???[1],???[b-1]:[a]],???[b-1]:[a]\)
\(\#,???[b],???[c]:[a] = \#,???[\#,???[b-1],???[c+1]:[a]],???[c-1]:[a]\)
\(???[b],???[c]:[a] = ???[???[b+1],???[c-1]:[a]],???[c-1]:[a]\)
\(???[0]:[n] = n\)
\(???[a]:[n] = ???[a-1]:[n+1]\)
====Rules (Linear)====
Linear blocks have more power than one-entry blocks, therefore it is standard if all of them go before one-entry blocks.
\(\#,???[\#,1]:[a] = \#,???[\#]:[a]\)
\(\#,???[1,\#,b]:[a] = \#,???[2,\#,b-1]:[a]\)
\(\#,???[\#,0,0]:[a] = \#,???[\#,1]:[a]\)
\(\#,???[\#,0,b]:[a] = \#,???[\#,(\#,???[\#,1,b-1]:[a]),b-1]:[a]\)
\(\#,???[\#,b,c]:[a] = \#,???[\#,???[\#,???[b-1,c+1]:[a]],c-1]:[a]\)
\(\#,???[1,c]:[a] = \#,???[a,c-1],???[a,c-1],\cdots,???[a,c-1]:[a]\)
\(\#,???[b,c]:[a] = \#,???[b-1,c],???[b-1,c],\cdots,???[b-1,c]:[a]\)
\(\#,???[1]:[a] = \#:[a]\)
\(???[1],\#,???[b]:[a] = ???[2],\#,???[b-1]:[a]\)
\(\#,???[0],???[0]:[a] = \#,???[1]:[a]\)
\(\#,???[0],???[b]:[a] = \#,???[1],???[b-1]:[a]\)
\(\#,???[b],???[c]:[a] = \#,???[\#,???[b-1],???[c+1]:[a]],???[c-1]:[a]\)
\(???[b],???[c]:[a] = ???[b+1],???[c-1]:[a]\)
\(???[0]:[n] = n\)
\(???[a]:[n] = ???[a-1]:[n+1]\)
====Rules (Dimensional)====
\(\#,???[\$],???[b]:[a] = \#,???[\$],???[b-1]:[a+1]\)
\(\#,???[\#\text{©}b\text{©}c]:[a] = \#,???[\#\text{©}(\#,???[\#\text{©}b-1\text{©}c+1]:[a])\text{©}c-1]:[a]\)
\(\#,???[\$]:[a] = \#,???[e(\$,\#,???[e(\$,a)]:[a])]:[a]\)
\(\#,???[1,c]:[a] = \#,???[a,c-1],???[a,c-1],\cdots,???[a,c-1]:[a]\)
\(\#,???[b,c]:[a] = \#,???[b-1,c],???[b-1,c],\cdots,???[b-1,c]:[a]\)
\(\#,???[1]:[a] = \#:[a]\)
\(???[1],\#,???[b]:[a] = ???[2],\#,???[b-1]:[a]\)
\(\#,???[0],???[0]:[a] = \#,???[1]:[a]\)
\(\#,???[0],???[b]:[a] = \#,???[1],???[b-1]:[a]\)
\(\#,???[b],???[c]:[a] = \#,???[\#,???[b-1],???[c+1]:[a]],???[c-1]:[a]\)
\(???[b],???[c]:[a] = ???[b+1],???[c-1]:[a]\)
\(???[0]:[n] = n\)
\(???[a]:[n] = ???[a-1]:[n+1]\)
\(,\) is an abbreviation for \(\langle 1 \rangle_{1}\)
\(|\) is an abbreviation for \(\langle 1 \rangle_{2}\)
\(\omega\) is an abbreviation for \(\langle 1 \rangle_{3}\)
\(\text{♦}_{a}\) is an abbreviation for \(\langle 1 \rangle_{a}\)
\(\text{○}_{a}\) is an abbreviation for \(\langle 2 \rangle_{a}\)
\(\langle a \rangle\) is an abbreviation for \(\langle a \rangle_{1}\)
\(e(m,a)\) for non-zero integer \(m = m-1\)
\(e([\#\text{©}1],a) = [\#]\)
\(e([\#1\text{©}b],a) = [\#a f(\text{©},a) a f(\text{©},a)\cdots a f(\text{©},a) a \text{©}e(b,a)]\) for \(\text{©} \neq \langle 1 \rangle _1\)
\(e([1\text{©}\#\text{©}b],a) = [2\text{©}\#\text{©}b-1]\)
\(e([\#\text{©}0\text{©}0],a) = [\#\text{©}1]\)
\(e([\#\text{©}0\text{©}b],a) = [\#\text{©}a\text{©}b-1]\)
\(f(\text{©} \langle 1 \rangle _1, a) = \text{©}\)
\(f(\text{©} \langle \$ \rangle _b, a) = \text{©} \langle \$ \rangle _e(b,a) \langle \$ \rangle _e(b,a)… \langle \$ \rangle _e(b,a) \langle \$ \rangle _e(b,a)\)
\(f(\text{©} \langle \$ \rangle _1, a) = \text{©} \langle e(\$,a) \rangle _{a\text{©} \langle e(\$,a) \rangle _{a\text{©} \langle e(\$,a) \rangle _…{a\text{©} \langle e(\$,a) \rangle _{a\text{©} \langle e(\$,a) \rangle 2}2}\cdots 2}2}\)
===Old version===
====Ver.1 (Primary Question Mark Notation only)====
Variables
First version
First, we explain the first version of the variables: 
*\(a\) = base
*\(b\) = prime
*\(c\) = base block
*\(d\) = prime block
*\(P\) = question mark structure
* \(P_n\) in rule 4 = nth block in question mark structure
*\(P_n [o]\) in rule 4 is where o is the input of the nth block in a question mark structure. For example, if \(P_3\) is the third block, \(P_3 [6]\) is the block \(???[6]\).
*\(S\) is a valid expression with or without separators, such as \(1,2\) or \(1\text{♦}_3 2\).
*\(B\) is a valid expression with or without separators, either from the first or second bound.
Second version
Next, we explain the second version of the variables:Polymations. Question Marks Notation. Retrieved 2024-02-13.
*\(a\) = base, \(a \in \mathbb{Q}\)
*\(b\) = prime, \(b \in \mathbb{N}\)
*\(c\) = base block, \(c \in \mathbb{N}\)
*\(d\) = prime block, \(d \in \mathbb{N}\)
*\(P\) = question mark structure
*\(P_n\) in rule 4 = nth block in question mark structure
*\(P_n [o]\) in rule 4 is where o is the input of the nth block in a question mark structure. For example, if \(P_3\) is the third block, \(P_3 [6]\) is the block \(???[6]\).
*\(S\) is a valid expression with or without separators, such as \(1,2\) or \(1\text{♦}_3 2\).
*\(S[n]\) is the nth entry in a array expression. If \(S=1\text{♦}_3 2\), then \(S[2]=2\).
*\(S[n.5]\) is the nth separator in a array expression. If \(S=1\text{♦}_3 2\), then \(S[1.5]=\text{♦}_3\).
*\(B\) is a valid expression with or without separators, either from the first or second bound.
*\(\text{Upg}(f,\{e\},\{B\})\) is a function to be defined
*\(\text{⏀}\) is the WIP 3rd bound symbol
Primary Question Mark Notation
Rules of the notation
Here are the rules of this notation, what makes this notation so big.
1. Hyperoperator rule
Rule H1-1. \(\{a,b\}=a+b\)
Rule H1-2. \(???[c]\{a,b\}=a\uparrow^{c-1}b\) (Note:\(a\uparrow^{0}b = a \times b\))
Rule H2-1. \(???[1],???[2]\{a,b\}=???[a^b]\{a,b\}\)
Rule H2-2. \(???[c],???[2]\{a,b\}=???[???[c+1]\{a,b\}]\{a,b\}\)
Rule H3. \(???[c],???[d]\{a,b\}=???[???[c+1],???[d-1]\{a,b\}]\{a,b\},c>1,d>2\)
2. BEAF rule
Rule A1. If the last number is 1, 
\(P_1, P_2, \ldots P_{n-1}, P_n[1] = P_1, P_2, \ldots P_{n-2}, P_{n-1}\)
Rule A2. If the first number is 1, 
\(P_1[1], P_2, \ldots P_{n-1}, P_n[p] = P_1[2], P_2, \ldots P_{n-1}, P_n[p-1]\)
Rule A3. If the second to last entry is 0,
\(P_1, P_2, \ldots P_{n-1}[0], P_n[q] = P_1, P_2, \ldots P_{n-1}[P_1, P_2, \ldots P_{n-1}[1], P_n[q-1]], P_n[q-1]\)
Rule A4. Otherwise,
\(P_1, P_2, \ldots P_{n-1}[p], P_n[q] = P_1, P_2, \ldots P_{n-1}[P_1, P_2, \ldots P_{n-1}[p-1], P_n[q]], P_n[q-1]\)
3. Block Arrays Rule
Rule B1. \(\cdots,???[c,1]\{a,b\}=???[c]\{a,b\}\)
Rule B2. \(\cdots,???[c,d],???[f]\{a,b\}=???[???[c+1]\{a,b\},d],???[f-1]\{a,b\}\)
Rule B3. \(\cdots,???[c,d]\{a,b\}=\underbrace{???[c,d-1],???[c,d-1],???[c,d-1],...,???[c,d-1]}_{c}\{a,b\}\)
We can also apply the 3rd rule to this array and make structures like this:
\(\begin{align*}???[1,1,3]\{a,b\}&=???[1,???[1,1,2]\{a,b\},2]\{a,b\}\\&=???[1,???[1,???[1,1,1]\{a,b\},1]\{a,b\},2]\{a,b\}\\&=???[1,???[???[1]\{a,b\}]\{a,b\},2]\{a,b\}\\&=???[1,???[a^b]\{a,b\},2]\{a,b\}\end{align*}\)
4-1. Nested Commas in Block Arrays Rule
#\(???[c,,2]\{a,b\}=???[\underbrace{c,c,...,c,c}_{c}]\{a,b\}\)
#\(???[1,,2],???[2]\{a,b\}=???[a^b,,2]\{a,b\}\)
#\(???[c,,2],???[2]\{a,b\}=???[???[c+1]\{a,b\},,2]\{a,b\}\)
#\(???[c,,2],???[d]\{a,b\}=???[???[c+1]\{a,b\},,2],???[d-1]\{a,b\}\)
#\(???[c,,2],???[d,2]\{a,b\}=???[c,,2],\underbrace{???[d],???[d],???[d],...,???[d]}_{d}\{a,b\}\)
#\(???[c,,2],???[d,2],???[f,2]\{a,b\}=???[c,,2],???[d,2],\underbrace{???[f],???[f]...,???[f]}_{f}\{a,b\}\)
#\(???[c,,2],???[d,,2]\{a,b\}=???[c,,2],???[\underbrace{d,d,d,...,d}_{d}]\{a,b\}\)
#\(???[c,,2],???[c,,2]\{a,b\}=???[[c,,2],2]\{a,b\}\)
#\(???[[c,,2],n]\{a,b\}=\underbrace{???[[c,,2],n-1],???[[c,,2],n-1],...,???[[c,,2],n-1]}_{c}\{a,b\}\)
#\(???[[c,,2],1,2]\{a,b\}=\underbrace{???[[c,,2],c-1],???[[c,,2],c-1],...,???[[c,,2],c-1]}_{c}\{a,b\}\)
#\(???[[c,,2],1,1,2]\{a,b\}\) = refer to rule 3 or 5
#\(???[[c,,2],,2]\{a,b\}=???[[c,,2],\underbrace{c,c,...,c}_{c}]\{a,b\}\)
#\(???[[[c,,2],,2],,2]\{a,b\}=???[[[c,,2],,2],\underbrace{c,c,...,c}_{c}]\{a,b\}\)
#\(???\underbrace{[[[...,,2],,2],,2]}_{c}\{a,b\}=???[c,,3]\{a,b\}\)
4-2. Nested Commas in Block Arrays 1b. 
# \(???[c,,1,2]\{a,b\}=???[\underbrace{c,,???[c,,...}_{c}]\{a,b\}]\{a,b\}\)
#\(???[c,,n,2]\{a,b\}=???[???[\underbrace{...,,n-1,2]{a,b},,n-1,2}_{c}]{a,b}\)
#\(???[c,,1,3]\{a,b\}=???[\underbrace{c,???[c,...}_{c},2]\{a,b\},2]\{a,b\}\)
#\(???[c,,1,n]\{a,b\}=???[\underbrace{c,???[c,...}_{c},n-1]\{a,b\},n-1]\{a,b\}\)
#\(???[c,,1,1,2]\{a,b\}\) = refer to rule 3 or 5
#\(???[c,,1,,2]\{a,b\}=???[c,,\underbrace{c,c,c,...,c}_{c}]\{a,b\}\)
#\(???[c,,n,,2]\{a,b\}=???[???[\underbrace{...,,n-1,,2]\{a,b\},,n-1,,2}_{c}]\{a,b\}\)
#\(???[c,,1,,3]\{a,b\}=???[\underbrace{c,,???[...}_{c}]\{a,b\},,2]\{a,b\}\)
#\(???[c,,2,,3]\{a,b\}=???[???[\underbrace{...,,1,,3]\{a,b\},,1,,3}_{c}]\{a,b\}\)
#\(???[c,,1,,4]\{a,b\}=???[c,,???[\underbrace{...]\{a,b\},,3}_{c}]\{a,b\}\)
#\(???[c,,2,,4]\{a,b\}=???[???[\underbrace{...,,1,,4]\{a,b\},,1,,4}_{c}]\{a,b\}\)
#\(???[c,,1,,c]\{a,b\}=???[c,,???[\underbrace{...]\{a,b\},,c-1}_{c}]\{a,b\}\)
#\(???[c,,c,,c]\{a,b\}=???[???[\underbrace{...,,c-1,,c]\{a,b\},,c-1,,c}_{c}]\{a,b\}\)
#\(???[c,,c,,1,2]\{a,b\}\) = refer to these previous steps
#\(???[c,,c,,c,,c]\{a,b\} \approx ???[4,,,2]\{a,b\}\)
#\(???[\underbrace{c,,c,,c,,c,,c,,...,,c}_{c}]\{a,b\} \approx ???[c,,,2]\{a,b\}\)
4-3. Nested Commas in Block Arrays Rule 2 
#\(???[c,,,3]\{a,b\}\) = refer to rule 6.
#\(???[c,,,c]\{a,b\}\) = refer to rule 6.
#\(???[c,,,c,,,c]\{a,b\} \approx ???[3,,,,2]\{a,b\}\) (refer to rule 6.01)
#\(???[c,,,,2]\{a,b\} \approx ???[\underbrace{c,,,c,,,...,,,c}_{c}]\{a,b\}\)
#\(???[c,,,,,2]\{a,b\} \approx ???[\underbrace{c,,,,c,,,,...,,,,c}_{c}]\{a,b\}\)
#\(???[c,^f d]\{a,b\} \approx ???[\underbrace{c,^{f-1}c,^{f-1}...,^{f-1}c}_{c}]\{a,b\}\) (where \(,^{n}=\underbrace{,,,...,,,}_{n}\))
#\(???[c|2]\{a,b\}=???[c,^c c]\)
5. Separator Rule
#\(???[1|2],???[2]\{a,b\}=???[a^b|2]\{a,b\}\)
#\(???[c|2],???[2]\{a,b\}=???[???[c+1]\{a,b\}|2]\{a,b\}\)
#\(???[c|2],???[d]\{a,b\}=???[???[c+1]\{a,b\}|2],???[d-1]\{a,b\}\)
#\(???[c|2],???[d,f]\{a,b\}=???[???[c+1]\{a,b\}|2],???[d,f-1],???[d,f-1],...\{a,b\}\)
#\(???[c|2],???[d,,2]\{a,b\}=???[???[c+1]\{a,b\}|2],???[\underbrace{d,d,d,...,d}_{d}],...{a,b}\)
#\(???[c|2],???[d,,f]\{a,b\}\) = refer to rule 6.01
#\(???[c|2],???[d,,,2]\{a,b\} =???[???[c+1]{a,b}|2], ???[d,,d,,...(d times)]{a,b}\)
#\(???[c|2],???[d|2]\{a,b\}=???[???[c+1]\{a,b\}|2],???[d,^d d]\{a,b\}\)
#\(???[c|2],???[d|2],???[f|2]\{a,b\}=???[???[c+1]\{a,b\}|2],???[???[d+1]\{a,b\}|2],???[f,^f f]\{a,b\}\)
#\(???[c,2|2]\{a,b\}=\underbrace{???[c|2],???[c|2],...,???[c|2]}_{c}\{a,b\}\)
#\(???[c,n|2]\{a,b\}=\underbrace{???[c,n-1|2],???[c,n-1|2],...,???[c,n-1|2]}_{c}\{a,b\}\)
#\(???[c,c,c|2]\{a,b\}\) = refer to rule 3
#\(???[c,,c|2]\{a,b\}\) = refer to rule 6.01
#\(???[c,,,c|2]\{a,b\}\) = refer to rule 6b
#\(???[(c|2)|2]\{a,b\}=???[c,^c c|2]\{a,b\}\)
#\(???[\underbrace{((c|2)|2...|2)}_{c}|2]\{a,b\}=???[c|3]\{a,b\}\)
#\(???[\underbrace{((c|n-1)|n-1...|n-1)}_{c}|2]\{a,b\}=???[c|n]\{a,b\}\)
#\(???[c|1,2]\{a,b\}=???[\underbrace{c|(c|(c|...))}_{c}]\{a,b\}\)
#\(???[c|2,2]\{a,b\}=???[((c|\underbrace{1,2)|1,2)|...|1,2|1,2}_{c}]\{a,b\}\)
#\(???[c|n,2]\{a,b\}=???[((c|\underbrace{n-1,2)|n-1,2)|...|n-1,2|n-1,2}_{c}]\{a,b\}\)
#\(???[c|1,3]\{a,b\}=???[c|(\underbrace{???[c|(...}_{c}),2]),2]\{a,b\}\)
#\(???[c|2,3]\{a,b\}=???[\underbrace{((c|2)|1,3...|1,3)}_{c}|1,3]\{a,b\}\)
#\(???[c|1,d]\{a,b\}=???[c|\underbrace{(???[c|(...)}_{c},d-1]),d-1]\{a,b\}\)
#\(???[c|f,d]\{a,b\}=???[\underbrace{((c|2)|f-1,d...|f-1,d)}_{c}|f-1,d]\{a,b\}\)
#\(???[c|1,1,3]\{a,b\}\) = refer to rule 3 or 5
#\(???[c|c,,2]\{a,b\}\) = refer to rule 6.01
#\(???[c|c,,,c]\{a,b\}\) = refer to rule 6.1
#\(???[c|c,^c c]\{a,b\}=???[c|(c|c)]\{a,b\}\)
#\(???[c|(c|c,^c c)]\{a,b\}=???[c|(c|(c|c))]\{a,b\}\)
#\(???[c|\underbrace{(c|...)}_{c}]\{a,b\}=???[c||2]\{a,b\}\)
#\(???[\underbrace{c||(c||...c||(c}_{c}||2))]\{a,b\}=???[c||3]\{a,b\}\)
#\(???[\underbrace{c||(c||...c||(c}_{c}||n-1))]\{a,b\}=???[c||n]\{a,b\}\)
#\(???[\underbrace{c||(c||...c||(c}_{c}||c))]\{a,b\}=???[c||1,2]\{a,b\}\)
#\(???[c||1,3]\{a,b\}\) = refer to step \(???[c|1,3]\)
#\(???[c||1,1,3]\{a,b\}\) = refer to rule 3 or 5
#\(???[c|||c]\{a,b\}=???[\underbrace{c||(c||...)}_{c}]\{a,b\}\)
#\(???[c||||c]\{a,b\}=???[\underbrace{c|||(c|||...)}_{c}]\{a,b\}\)
#\(???[c|||||c]\{a,b\}=???[\underbrace{c||||(c||||...)}_{c}]\{a,b\} \approx ???[5\omega2]\{a,b\}\)
#\(???[c||||||||||c]\{a,b\}=???[\underbrace{c|||||||||(c|||||||||...)}_{c}]\{a,b\} \approx ???[10\omega2]\{a,b\}\)
#\(???[c\omega2]\{a,b\}=???[c|^c c]\{a,b\}\) (where \(|^{n}=\underbrace{|||...|||}_{n}\))
Now that we understand the rules for separators, we can define one rule for all separators:
#\(\omega = ♦_3\) (Omega is an abbreviation for \(♦_3\))
#\(a♦_n 2=a♦_{n-1}^{a}a\) (with applying earlier rules for other separators)
#For any separator \(\mu\), \(\mu^n=\underbrace{\mu\mu\mu...\mu\mu\mu}_{n}\).
#For example, \(???[3♦_4 2]\{a,b\}=???[3\omega\omega\omega3]\{a,b\}\)
This diamond is called a Super-separator, because it behaves like a separator, but breaks the limit of previous separators on their own. However, we need to expand more and more, so let’s add arrays to super-separators. It’s optional whether you want to write omega or \(♦_3\) for the 3rd separator.
#\(???[c♦_{1,2}2]\{a,b\}=???[c♦_{???[c♦_{???[c♦_{...}2]\{a,b\}}2]\{a,b\}}2]\{a,b\}\) with c "\(???[c♦\)"'s.
Comparisons with Fast-growing hierarchy
\(???[n]\{n,n\} \approx f_{\omega}(n)\)
\(???[1],???[2]\{n,n\} \approx f_{\omega}(n^{2})\)
\(???[n],???[n]\{n,n\} \approx f_{\omega+1}(n)\)
\(???[n],???[n],???[2]\{n,n\} \approx f_{\omega+2}(n)\)
\(???[n],???[n],???[3]\{n,n\} \approx f_{\omega+3}(n)\)
\(???[n],???[n],???[n]\{n,n\} \approx f_{\omega2}(n)\)
\(???[n],???[n],???[n],???[2]\{n,n\} \approx f_{\omega2+1}(n)\)
\(???[n],???[n],???[n],???[n]\{n,n\} \approx f_{\omega3}(n)\)
\(\underbrace{???[n],???[n],\cdots,???[n],???[n]}_{m}\{n,n\} \approx f_{\omega(m-1)}(n)\hspace{7mm}(m\leqq3)\)
\(???[n,2]\{n,n\} \approx f_{\omega^{2}}(n)\)
\(???[1,2],???[2]\{n,n\} \approx f_{\omega^{2}}(n^{2})\)
\(???[n,2],???[2]\{n,n\} \approx f_{\omega^{2}+1}(n)\)
\(???[n,2],???[n]\{n,n\} \approx f_{\omega^{2}+\omega}(n)\)
\(???[n,2],???[n],???[n]\{n,n\} \approx f_{\omega^{2}+\omega2}(n)\)
\(???[n,2],???[m,2]\{n,n\} \approx f_{\omega^{2}+\omega m}(n)\)
\(???[n,2],???[n,2]\{n,n\} \approx f_{\omega^{2}\times2}(n)\)
\(???[n,2],???[n,2],???[n,2]\{n,n\} \approx f_{\omega^{2}\times3}(n)\)
\(\underbrace{???[n,2],???[n,2],\cdots,???[n,2],???[n,2]}_{m}\{n,n\} \approx f_{\omega^{2}\times m}(n)\)
\(???[n,3]\{n,n\} \approx f_{\omega^{3}}(n)\)
\(???[n,4]\{n,n\} \approx f_{\omega^{4}}(n)\)
\(???[n,n]\{n,n\} \approx f_{\omega^{\omega}}(n)\)
In the Block Arrays Rule section, the calculation method for the case where there are three or more elements in square brackets cannot be analyzed because the "3rd rule" does not specify which item is the rule.
Secondary Question Mark Notation
==Sources==
Category:Functions
Category:Notations
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