- FINAL -
This is my dumbest notation created based on arrays, as being experimental and ill-defined. This notation is based off of the number 99.
This notation has 4 levels as follows:
Basic level
Linear level
Separator level
Death level
UPDATE October 19-20, 2025: I decided to create some examples and analyze the growth rath of some specific expressions.
Uses ? (question mark) as the notation format.
?<> = 99
?<p> = 99*p
?<#,1> = ?<#>
?<p,q> = ?<p,q-1>*99 for q > 1
Where # is a part of said array (a string of entries and commas, it can also be empty). If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied.
# is a part of said array (a string of entries and commas, it can also be empty).
?<> = 99
?<p> = 99*p
?<#,1> = ?<#>
?<p,q#> = ?<p,q-1#>*99 for q > 1
Otherwise, start the process below:
Scan the array:
?<p,1,q#> = ?<p,p,q-1#> for q > 1
?<p,1,1,...,1,1,q#> = ?<p,1,1,...,1,1,q-1#> for q > 1
If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied.
# is a part of said array (a string of entries, commas, and separators, it can also be empty).
?<> = 99
?<p> = 99*p
?<#,1> = ?<#>
?<p,q#> = ?<p,q-1#>*99 for q > 1
Otherwise, start the process below:
Scan the array:
Commas are the {1} separators
?<p,1,q#> = ?<p,p,q-1#> for q > 1
?<p,1,1,...,1,1,q#> = ?<p,1,1,...,1,p,q-1#> for q > 1
?<p{2}q#> = ?<p,1,1,...,1,1,2{2}q-1#> where there are p entries at the left of {2}, and q > 1
?<p{r}q#> = ?<p{r-1}1{r-1}1{r-1}...,1{r-1}1{r-1}2{r}q-1#> where there are p entries at the left of {r}, and q > 1, and r > 1
If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied.
Now, let's define another function.
f(n) = ?<99{n}99>
?<p*2> = f(?<p>)
?<p#*2> = f(?<p#>)
?<p*q> = f(?<p*q-1>) for q > 1
?<p**2> = ?<p*1*1*1*...*1*1*1*2> where there are p entries separated by *'s
?<p***2> = ?<p**1**1**1**...**1**1**1**2> where there are p entries separated by **'s
n! = ?<99****...****99> where there are n *'s (not to be confused with the factorial function)
?<p@2> = ?<p>!
?<p#@2> = ?<p#>!
?<p@q> = (?<p@q-1>)! for q > 1
If we keep defining functions like this, and then combine them all into one enormous function, we can create much more enormous numbers.
?<p@@2> = ?<p@1@1@...@1@1@1@2> where there are p entries separated by @'s
We can even do ?<p@@@@@@@...@@@@@@@q> with 99 @'s, and that is the limit of the notation.
?<> = 99
?<p> = ?<p,1> = 99 × p
?<p,2> = ?<p> × 99 = 99 × 99 × p = 9,801 × p
?<p,3> = ?<p,2> × 99 = 99 × 9,801 × p = 970,299 × p
?<p,4> = ?<p,3> × 99 = 99 × 970,299 × p = 96,059,601 × p
?<p,n> = ?<p,n-1> × 99 = 99^n × p
?<p,1,2> = ?<p,p> = 99^p × p
?<p,2,2> = ?<p,1,2> × 99 = (99^p × 99 × p) = 99^(p + 1) × p
?<p,3,2> = ?<p,2,2> × 99 = (99^(p + 1) × 99 × p) = 99^(p + 2) × p
?<p,n,2> = 99^(p + n − 1) × p
?<p,1,3> = ?<p,p,2> = 99^(2p − 1) × p
?<p,2,3> = ?<p,1,3> × 99 = 99^2p × p
?<p,p,3> = 99^(3p − 1) × p
?<p,p,4> = 99^(4p − 1) × p
?<p,p,n> = 99^(np − 1) × p
?<p,1,1,2> = ?<p,1,p> = 99^(p^2 − p − 1) × p
?<p,2,1,2> = 99^(p^2 − p) × p
?<p,1,2,2> = ?<p,p,1,2> = 99^(p^2 − 1) × p
?<p,1,3,2> = ?<p,p,2,2> = 99^(p^2 + p − 1) × p
?<p,1,1,3> = ?<p,1,p,2> = 99^(2p^2 − p − 1) × p
?<p,1,1,4> = ?<p,1,p,3> = 99^(3p^2 − p − 1) × p
?<p,1,1,1,2> = ?<p,1,1,p> = 99^(p^3 − p^2 − p − 1) × p
?<p,1,1,1,3> = ?<p,1,1,p,2> = 99^(2p^3 − p^2 − p − 1) × p
?<p,1,1,1,1,2> = ?<p,1,1,1,p> = 99^(p^4 − p^3 − p^2 − p − 1) × p
?<p,1,1,1,1,1,2> = ?<p,1,1,1,1,p> = 99^(p^5 − p^4 − p^3 − p^2 − p − 1) × p
?<p{2}2> = ?<p,1,1,1,...,1,1,1,2> with n−2 1's ≈ 99^p^(p − 2)
?<p,2{2}2> = ?<p{2}2> ×99 ≈ 99^(p^(p − 2) + 1)
?<p,1,2{2}2> = ?<p,p{2}2> ≈ 99^(p^(p − 2) + p)
?<p,1,1,2{2}2> = ?<p,p{2}2> ≈ 99^(p^(p − 2) + p^2)
?<p{2}3> = ?<p,1,1,1,...,1,1,1,2{2}2> with n−2 1's ≈ 99^(2p^(p − 2))
?<p{2}4> = ?<p,1,1,1,...,1,1,1,2{2}3> with n−2 1's ≈ 99^(3p^(p − 2))
?<p{2}1,2> = ?<p{2}p> ≈ 99^p^(p − 1)
?<p{2}2,2> = ?<p,1,1,1,...,1,1,1,2{2}1,2> with n−2 1's ≈ 99^(p^(p − 1) + p^(p − 2))
?<p{2}1,3> = ?<p{2}p,2> ≈ 99^(2p^(p − 1))
?<p{2}1,4> = ?<p{2}p,3> ≈ 99^(3p^(p − 1))
?<p{2}1,1,2> = ?<p{2}1,p> ≈ 99^(p^(p − 1) × (p − 1)) ≈ 99^p^p
?<p{2}1,1,3> = ?<p{2}1,p,2> ≈ 99^(p^(p − 1) × 2(p − 1)) ≈ 99^(2p^p)
?<p{2}1,1,1,2> = ?<p{2}1,1,p> ≈ 99^(p^p × 2(p − 1)) ≈ 99^p^(p + 1)
?<p{2}1,1,1,1,2> = ?<p{2}1,1,1,p> ≈ 99^(p^(p + 1) × 2(p − 1)) ≈ 99^p^(p + 2)
?<p{2}1{2}2> = ?<p{2}1,1,1,...,1,1,1,2> with n−1 1's ≈ 99^(p^(2p − 3) × 2(p − 1)) ≈ 99^p^(2p − 2)
?<p{2}1{2}3> = ?<p{2}1,1,1,...,1,1,1,2{2}2> with n−1 1's ≈ 99^(2p^(2p − 2))
?<p{2}1{2}1,2> = ?<p{2}1{2}p> ≈ 99^p^(2p − 1)
?<p{2}1{2}1,1,2> = ?<p{2}1{2}1,p> ≈ 99^p^2p
?<p{2}1{2}1{2}2> = ?<p{2}1{2}1,1,1,...,1,1,1,2> with n−1 1's next to the last {2} ≈ 99^p^(3p − 2)
?<p{2}1{2}1{2}1{2}2> = ?<p{2}1{2}1{2}1,1,1,...,1,1,1,2> with n−1 1's next to the last {2} ≈ 99^p^(4p − 2)
?<p{2}1{2}1{2}1{2}1{2}2> = ?<p{2}1{2}1{2}1{2}1,1,1,...,1,1,1,2> with n−1 1's next to the last {2} ≈ 99^p^(5p − 2)
?<p{3}2> = ?<p{2}1{2}...{2}1{2}2> with n−1 {2}'s ≈ 99^p^((p − 1)p − 2) ≈ 99^p^(p^2 − p − 2)
?<p{3}1,2> = ?<p{3}p> ≈ 99^p^(p^2 − p − 1)
?<p{3}1{2}2> = ?<p{3}1,1,1,...,1,1,1,2> with n−1 1's ≈ 99^p^(p^2 − 2)
?<p{3}1{3}2> = ?<p{3}1{2}1{2}...{2}1{2}2> with n−1 {2}'s ≈ 99^p^(2p^2 − p − 2)
?<p{3}1{3}1{3}2> = ?<p{3}1{3}1{2}1{2}...{2}1{2}2> with n−1 {2}'s ≈ 99^p^(3p^2 − p − 2)
?<p{4}2> = ?<p{3}1{3}...{3}1{3}2> with n−1 {3}'s ≈ 99^p^(p^3 − p^2 − p − 2)
?<p{4}1{4}2> = ?<p{4}1{3}1{3}...{3}1{3}2> with n−1 {3}'s ≈ 99^p^(2p^3 − p^2 − p − 2)
?<p{5}2> = ?<p{4}1{4}...{4}1{4}2> with n−1 {4}'s ≈ 99^p^(p^4 − p^3 − p^2 − p − 2)
?<p{6}2> = ?<p{5}1{5}...{5}1{5}2> with n−1 {5}'s ≈ 99^p^(p^5 − p^4 − p^3 − p^2 − p − 2)
f(1) = ?<99{1}99> = ?<99,99> = 99^99 × 99 = 99^100 ≈ 3.660323412732 × 10^199
f(2) = ?<99{2}99> ≈ 99^99^98
f(3) = ?<99{3}99> ≈ 99^99^9,701
f(4) = ?<99{4}99> ≈ 99^99^960,398
f(5) = ?<99{5}99> ≈ 99^99^95,079,401
f(6) = ?<99{6}99> ≈ 99^99^9,412,860,698
f(99) = ?<99{99}99> ≈ 99^99^99^98
?<p*2> = f(?<p>) = f(p × 99) = ?<99{99p}99> ≈ 99^99^99^(p^2 − p − 1)
?<p,2*2> = f(?<p,2>) ≈ 99^99^99^(p^3 − p^2 − p − 1)
?<p,1,2*2> = f(?<p,p>) ≈ 99^99^99^(p^(p + 1))
?<p{2}2*2> = f(?<p{2}2>) ≈ 99^99^99^99^p^(p − 2)
?<p{3}2*2> = f(?<p{3}2>) ≈ 99^99^99^99^p^(p^2 − p − 1)
?<p{n}2*2> = f(?<p{n}2>) ≈ 99^99^99^99^p^p^(p − 1)
?<p*3> = f(?<p*2>) ≈ 99^99^99^99^99^99^(p^2 − p − 1)
?<p*4> = f(?<p*3>) ≈ 99^99^99^99^99^99^99^99^99^(p^2 − p − 1)
?<p*5> = f(?<p*4>) ≈ 99^99^99^99^99^99^99^99^99^99^99^99^(p^2 − p − 1)
?<p*n> = f(?<p*n−1>) ≈ E[99](p^2 − p − 1)#(3n − 3) ≈ 99^^(3n − 2)
?<p*1,2> = ?<p*p> = f(?<p*p−1>) ≈ E[99](p^2 − p − 1)#(3p − 3) ≈ 99^^(3p − 2)
?<p*2,2> = f(?<p*1,2>) ≈ 99^^(3p + 1)
?<p*3,3> = f(?<p*1,2>) ≈ 99^^(3p + 4)
?<p*1,3> = ?<p*p,2> ≈ 99^^(6p − 2)
?<p*1,4> = ?<p*p,3> ≈ 99^^(9p − 2)
?<p*1,1,2> = ?<p*1,p> ≈ 99^^(3p(p − 1) − 2) ≈ 99^^(3p^2 − 3p − 2)
?<p*1,1,3> = ?<p*1,p,2> ≈ 99^^(6p(p − 1) − 2) ≈ 99^^(6p^2 − 6p − 2)
?<p*1,1,1,2> = ?<p*1,1,p> ≈ 99^^(3p^p − 3p^2 − 3p − 2)
?<p*1,1,1,1,2> = ?<p*1,1,p> ≈ 99^^(3p^4 − 3p^3 − 3p^2 − 3p − 2)
?<p*1{2}2> = ?<p*1,1,...,1,1,2> with p−1 1's ≈ 99^^(3p^(p − 1))
?<p*1{2}1,2> = ?<p*1{2}p> ≈ 99^^(3p^p)
?<p*1{2}1{2}2> = ?<p*1{2}1,1,...,1,1,2> with p−1 1's at the right of {2} ≈ 99^^(3p^(2p − 1))
?<p*1{3}2> = ?<p*1{2}1{2}...{2}1{2}2> with p−1 {2}'s ≈ 99^^(3p^(p^2 − p − 1))
?<p*1{4}2> = ?<p*1{3}1{3}...{3}1{3}2> with p−1 {3}'s ≈ 99^^(3p^(p^3 − p^2 − p − 1))
?<p*1*2> = ?<p*?<p*2>> = ?<p*f(?<p>)> ≈ 99^^99^99^99^(p^2 − p − 1)
?<p*2*2> = ?<f(p*1*2)*1*2> ≈ 99^^99^^99^99^99^(p^2 − p − 1)
?<p*3*2> = ?<f(p*2*2)*2*2> ≈ 99^^99^^99^^99^99^99^(p^2 − p − 1)
?<p*1,2*2> = ?<p*p*2> ≈ 99^^(p + 2)
?<p*2,2*2> = ?<f(p*1,2*2)*1,2*2> ≈ 99^^(p + 3)
?<p*1,3*2> = ?<p*p,2*2> ≈ 99^^2p
?<p*1,1,2*2> = ?<p*1,p*2> ≈ 99^^p^2
?<p*1{2}2*2> = ?<p*1,1,...,1,1,2*2> with p−1 commas ≈ 99^^p^p
?<p*1*3> = ?<p*?<p*1*2>*2> ≈ 99^^99^^99^99^99^(p^2 − p − 1)
?<p*1*4> = ?<p*?<p*1*3>*3> ≈ 99^^99^^99^^99^99^99^(p^2 − p − 1)
?<p*1*1,2> = ?<p*1*p> ≈ 99^^^(p + 1)
?<p*1*2,2> = ?<p*?<p*1*1,2>*1,2> ≈ 99^^^(p + 2)
?<p*1*1,3> = ?<p*1*p,2> ≈ 99^^^2p
?<p*1*1,1,2> = ?<p*1*p,2> ≈ 99^^^p^2
?<p*1*1{2}2> = ?<p*1*1,1,...,1,1,2> with p−1 commas ≈ 99^^^p^p
?<p*1*1*2> = ?<p*1*?<p*1*2>> ≈ 99^^^99^99^99^(p^2 − p − 1)
?<p*1*1*3> = ?<p*1*?<p*1*1*2>*2> ≈ 99^^^99^^^99^99^99^(p^2 − p − 1)
?<p*1*1*1*2> = ?<p*1*1*?<p*1*1*2>> ≈ 99^^^^99^99^99^(p^2 − p − 1)
?<p*1*1*1*1*2> = ?<p*1*1*1*?<p*1*1*1*2>> ≈ 99^^^^^99^99^99^(p^2 − p − 1)
?<p*1*1*1*1*1*2> = ?<p*1*1*1*1*?<p*1*1*1*1*2>> ≈ 99^^^^^^99^99^99^(p^2 − p − 1)
?<p**2> = ?<p*1*1*1*...*1*1*1*2> with p−2 1's ≈ 99{p−1}99^99^99^(p^2 − p − 1), where {} denotes the hyperoperator = {99, 99^99^99^(p^2 − p − 1), p − 1} in BEAF [FGH ordinal ω]
?<p,2**2> = ?<?<p,2>**2> ≈ {99, 99, 9801p} in BEAF [FGH ordinal ω]
?<p*2**2> = ?<?<p*2>**2> ≈ {99, 99, 99^99^99^(p^2 − p − 1)} in BEAF [FGH ordinal ω]
?<p*1*2**2> = ?<p*?<p*2>**2> ≈ {99, 99, 99^^99^99^99^(p^2 − p − 1)} in BEAF [FGH ordinal ω]
?<p**3> = ?<p*1*1*1*...*1*1*1*2**2> with p−2 1's ≈ {99, 99, {99, 99, 99^^99^99^99^(p^2 − p − 1)}} in BEAF [FGH ordinal ω]
?<p**4> = ?<p*1*1*1*...*1*1*1*2**3> with p−2 1's ≈ {99, 99, {99, 99, {99, 99, 99^^99^99^99^(p^2 − p − 1)}}} in BEAF [FGH ordinal ω]
?<p**1,2> = ?<p**p> ≈ {99, p, 1, 2} in BEAF [FGH ordinal ω + 1]
?<p**1*2> = ?<p**?<p**2>> ≈ {99, {99, 99^99^99^(p^2 − p − 1), p − 1}, 1, 2} in BEAF [FGH ordinal ω + 1]
?<p**1*1*2> = ?<p**1*?<p**1*2>> ≈ {99, {99, {99, 99^99^99^(p^2 − p − 1), p − 1}, 1, 2}, 2, 2} in BEAF [FGH ordinal ω + 2]
?<p**1**2> = ?<p**1*1*1*1*...*1*1*1*2> with p−2 1's at the right ≈ {99, 3, p, 2} in BEAF [FGH ordinal ω2]
?<p**1**1**2> = ?<p**1**1*1*1*1*...*1*1*1*2> with p−2 1's at the right ≈ {99, 3, p, 3} in BEAF [FGH ordinal ω3]
?<p***2> = ?<p**1**1**...**1**1**2> with p−2 1's ≈ {99, 3, p, p−1} in BEAF [FGH ordinal ω^2]
?<p****2> = ?<p***1***1***...***1***1***2> with p−2 1's ≈ {99, 3, p, p−1, p−1} in BEAF [FGH ordinal ω^3]
p! = ?<p*****...*****p> with p *'s ≈ {99, p + 1 (1) 2} in BEAF [FGH ordinal ω^ω]
?<p@2> = ?<p>! with p *'s ≈ {99, 99p (1) 2} in BEAF [FGH ordinal ω^ω]
?<p,2@2> = ?<p,2>! with p *'s ≈ {99, 9801p (1) 2} in BEAF [FGH ordinal ω^ω]
?<p*2@2> = ?<?<p,2>@2> with p *'s ≈ {99, 99^99^99^(p^2 − p − 1) (1) 2} in BEAF [FGH ordinal ω^ω]
?<p@3> = ?<p*****...*****p@2> with p *'s ≈ {99, p + 1 (1) 3} in BEAF [FGH ordinal ω^ω×2]
?<p@1,2> = ?<p@p> ≈ {99, p + 1 (1) p} in BEAF [FGH ordinal ω^(ω + 1)]
?<p@1*2> = ?<p@?<p@2>> ≈ {99, {99, 99p (1) 2} (1) p} in BEAF [FGH ordinal ω^(ω + 1)]
?<p@1@2> = ?<p@?<p@2>> ≈ {99, 99 (1) {99, 99p (1) 2}, 2} in BEAF [FGH ordinal ω^(ω + 1)×2]
?<p@@2> = ?<p@1@1@...@1@1@2> with p−2 1's ≈ {99, 99 (1) 99, p − 1} in BEAF [FGH ordinal ω^(ω + 2)]
?<p@@@2> = ?<p@@1@@1@@...@@1@@1@@2> with p−2 1's ≈ {99, 99 (1) 99, 99, p − 1} in BEAF [FGH ordinal ω^(ω + 3)]
?<p@@@@@...@@@@@2> with n @'s ≈ {99, n (1)(1) 2} in BEAF [FGH ordinal ω^ω2]
<99@@@@@@@...@@@@@@@99> with 99 @'s ≈ {99, 99 (1)(1) 2} in BEAF [FGH ordinal ω^ω2]
Conclusion: The intended limit of the jazzy array notation is the fast-growing hierarchy ordinal level ω^ω2 (BMS (0)(1)(2)(1)(2) and Y sequence (1,2,3,2,3)), which is comparable to Saibian's deutero-godgahlah (E100#^#*#^#) and Bowers' dubol ({10, 100 (1)(1) 2}). Actually, levels prior to the death level is mostly well-defined here.
?<3,1,2>
= ?<3,3,1>
= ?<3,3>
= ?<3,3> × 99
= ?<3,2> × 99^2
= ?<3,1> × 99^3
= ?<3> × 99^3
= 99 × 3 × 99^3
= 99^4 × 3
= 96,059,601 × 3
= 288,178,803