Notations
This page contains info about the different notations that define large numbers.
Some of which are used by real mathematicians and googologists, and some of them are quite niche and less familiar.
Notations
This page contains info about the different notations that define large numbers.
Some of which are used by real mathematicians and googologists, and some of them are quite niche and less familiar.
Scientific Notation
Tier 1 | Founder: Multiple People...
This should be the well-known notation for most people. Written either as aⁿ or a x bⁿ, this represents large numbers that are most likely encountered here and there in the real world. The n is simply repeated multiplication; for example, 3⁶ is 3x3x3x3x3x3, which is 729. Most calculators represent scientific notation with an E or e, which is then written as aEb or aeb.
While scientific notation doesn't allow stacking the exponent, normal exponents CAN stack, like 10^10^24 (in formulas the caret is omitted and the numbers are elevated), which is evaluated from right to left.
However, this will eventually become too inconvenient to write down with a lot of exponents: 10^10^10^10^10^10^10^10^10^10^10^10^10^10
That's when the next notation is introduced.
Arrow + Operator Notations
Tier 2 | Founders: Donald Knuth, Jonathan Bowers
This is where multiple larger operations beyond exponents are represented. Written as a↑ᶜb, the c represents how many arrows are between a and b, which represents hyperoperations such as tetration (2 arrows), pentation (3 arrows), hexation (4 arrows), etc. Exponentiation can be represented with one arrow, like 10↑10.
Each new hyperoperation is the repetition of the previous hyperoperation, such that 10↑ᶜ5 = 10↑ᶜ⁻¹10↑ᶜ⁻¹10↑ᶜ⁻¹10↑ᶜ⁻¹10. Some examples are:
10↑↑5 = 10↑10↑10↑10↑10
10↑↑↑5 = 10↑↑10↑↑10↑↑10↑↑10
10↑↑↑↑5 = 10↑↑↑10↑↑↑10↑↑↑10↑↑↑10
10↑↑↑↑↑5 = 10↑↑↑↑10↑↑↑↑10↑↑↑↑10↑↑↑↑10
etc.
Just like with the previous notation, these hyperoperations are evaluated from right to left. If arrows are not available, carets should be used instead, so 10↑↑↑10 is the same as 10^^^10.
You CAN stack these like the exponents, such that 10↑↑10 is c in 10↑ᶜ10, which does require formulas or text boxes that support layered superscripts. However, this too will eventually become inconvenient to write down, which brings us to the next section here.
We will introduce expansion into the equation. The last number above this text can also be written as 10{10↑↑10}10 or 10{10{2}10}10. What expansion does is similar to previous hyperoperators but with extra copies of a.
10{{1}}2 = 10{10}10
10{{1}}3 = 10{10{10}10}10
10{{1}}4 = 10{10{10{10}10}10}10
etc.
This is the binary function a{c}ᵈb, where d denotes how many curly brackets there are. In which case, 10{{1}}10 is 10{1}²10.
Repeated expansion is called multiexpansion: 10{{2}}4 = 10{{1}}10{{1}}10{{1}}10
Repeated multiexpansion is called powerexpansion: 10{{3}}4 = 10{{2}}10{{2}}10{{2}}10
Repeated powerexpansion is called expandotetration: 10{{4}}4 = 10{{3}}10{{3}}10{{3}}10
And then you keep doing this until the three-bracket tier, which is called an explosion. (is this supposed to be a pun? whatever.)
10{{{1}}}2 = 10{{10}}10
10{{{1}}}3 = 10{{10{{10}}10}}10
10{{{1}}}4 = 10{{10{{10{{10}}10}}10}}10
etc.
We continue this pattern for as long as feasible into multiexplosion, powerexplosion, explodotetration, etc., into detonation (4 brackets), pentonation (5 brackets), hexonation (6 brackets), and beyond.
Once again, these can be stacked like the exponents, such that 10{{2}}10 is d in 10{10}ᵈ10. But once again, this stops being usable after a certain point, which is when the next notation is introduced.
Chained Arrow Notation
Tier 3 | Founder: J. H. Conway and R. K. Guy
TO BE ADDED
Array Notation
Tier 3 | Founder: Jonathan Bowers
There are different iterations of the name of this notation, but this is the one I went with for the purpose of this page. This is written as {a,b,c,d,...} and this can represent many, many more hyperoperations than what the other notations can.
To understand this better, let's slow down with examples from previous notations compared to this notation.
{a,b} = aᵇ
{a,b,2} = a↑↑b
{a,{b,c,2},2} = a↑↑b↑↑c
{a,b,3} = a↑↑↑b
{a,b,c} = a↑ᶜb = a{c}b
{a,b,{c,d,e}} = a{c{e}d}b
{a,b,1,2} = a{{1}}b
{a,{b,c,1,2},1,2} = a{{1}}b{{1}}c
{a,b,c,2} = a{{c}}b
{a,b,{c,d,e,2},2} = a{{c{{e}}d}}b
{a,b,c,3} = a{{{c}}}b
{a,b,c,d} = a{c}ᵈb
{a,b,c,{d,e,f,g}} = a{c}ⁿb, n = d{f}ᵍe
Within these examples...
{10,3,2}, 10↑↑3, is {10,{10,10}} or 10↑10↑10
{10,3,3}, 10↑↑↑3, is {10,{10,10,2},2} or 10↑↑10↑↑10
{10,3,1,2}, 10{{1}}3, is {10,10,{10,10,10}} or 10{10{10}10}10
{10,3,2,2}, 10{{2}}3, is {10,{10,10,1,2},1,2} or 10{{1}}10{{1}}10
{10,3,1,3}, 10{{{1}}}3, is {10,10,{10,10,10,2},2} or 10{{10{{10}}10}}10
So the next hyperoperation is megotion (coined by Aarex) and that is {a,b,1,1,2}.
{10,3,1,1,2} is {10,10,10,{10,10,10,10}}
{10,4,1,1,2} is {10,10,10,{10,10,10,{10,10,10,10}}}
{10,5,1,1,2} is {10,10,10,{10,10,10,{10,10,10,{10,10,10,10}}}}
etc.
Basically, {a,b,1,1,2} is {a,a,a,{a,(b-1),1,1,2}}. If you are familiar with recursion, then the next hyperoperations are straightforward: Multimegotion: {a,b,2,1,2}, Powermegotion: {a,b,3,1,2}, Megotetration: {a,b,4,1,2}, Megoexpansion: {a,b,1,2,2}, Multimegoexpansion: {a,b,2,2,2}, Megoexplosion: {a,b,1,3,2}, Megodetonation: {a,b,1,4,2}, and now Gigotation, which is {a,b,1,1,3}.
The examples for each of these are:
{10,3,2,1,2} = {10,{10,10,1,1,2},1,1,2}
{10,3,3,1,2} = {10,{10,10,2,1,2},2,1,2}
{10,3,4,1,2} = {10,{10,10,3,1,2},3,1,2}
{10,3,1,2,2} = {10,10,{10,10,10,1,2},1,2}
{10,3,2,2,2} = {10,{10,10,1,2,2},1,2,2}
{10,3,1,3,2} = {10,10,{10,10,10,2,2},2,2}
{10,3,1,4,2} = {10,10,{10,10,10,3,2},3,2}
{10,3,1,1,3} = {10,10,10,{10,10,10,10,2},2}
We can keep going to Terotion: {a,b,1,1,4}, Petotion: {a,b,1,1,5}, Hatotion: {a,b,1,1,6}, Hepotion: {a,b,1,1,7}, and much more past that.
We'll eventually reach powiaination, which is {a,b,1,1,1,2}.
{10,3,1,1,1,2} is {10,10,10,10,{10,10,10,10,10}}
{10,3,1,1,1,3} is {10,10,10,10,{10,10,10,10,10,2},2}
etc.
You can probably guess what {a,b,1,1,1,1,2} corresponds to, as with {a,b,1,1,1,1,1,2} and beyond. The hyperoperation names get quite complicated beyond powiaination, though.
However, once we get to the stage of inconveniently typing many 10s in the array, such as {10,10,10,10,10,10,10,10,10,10,10,10,10}, there are two different notations within this setup that go on for a long time: Bowers' Exploding Array Function and Bird's Array Notation.
Bird's Array Notation
Tier 4 | Founder: Chris Bird
Now, you could call the previous notation "Array Notation" this as well; however, Bower's Exploding Array Function (BEAF) also shares the same pre-dimensional arrays, so to avoid confusion, I called the pre-dimensional arrays a different name.
So, while hyperoperation names are getting complicated, there are a few that are still worth noting. The one that counts the 10s in the array is called aperiogation or iteration, which is {a,b [2] 2}.
{10,5 [2] 2} = {10,10,10,10,10}
{10,7 [2] 2} = {10,10,10,10,10,10,10}
{10,10 [2] 2} = {10,10,10,10,10,10,10,10,10,10}
Now you repeat all the familiar recursions from the previous tab, BUT MAKE SURE the "[2] 2" part is present on ALL ARRAYS. For instance, {10,3,2 [2] 2} is {10,{10,10 [2] 2} [2] 2} NOT {10,{10,10} [2] 2}.
By the way, {10,10 [2] 2} is also equivalent to {10,2,2 [2] 2} and {10,2 [2] 3}; the first is because, from earlier, {10,10} is {10,2,2}, like how 10↑10 is 10↑↑2; the second is because in {a,b [2] c}, the b in each c increment counts how many copies of a are on the left-hand side of "[2]", so, for example, {10,5 [2] 3} is {10,10,10,10,10 [2] 2}, and the same is true for {10,5 [2] c} being {10,10,10,10,10 [2] c-1}.
The next hyperoperation is called expoiteration, which is {a,b [2] 1,2}.
{10,2 [2] 1,2} = {10,10 [2] 10}
{10,3 [2] 1,2} = {10,10 [2] {10,10 [2] 10}}
{10,4 [2] 1,2} = {10,10 [2] {10,10 [2] {10,10 [2] 10}}}
This is how the extra entries after "[2]" are formed, such that {10,3 [2] 1,e} is {10,10 [2] {10,10 [2] 10,e-1},e-1}.
{10,3 [2] 1,3} = {10,10 [2] {10,10 [2] 10,2},2}
{10,3 [2] 1,1,2} = {10,10 [2] 10,{10,10 [2] 10,10}}
The penultimate hyperoperation name to mention here is called trioteration, which is {a,b [2] 1 [2] 2}. At this point, I will start highlighting what is being nested in these upcoming numbers.
{10,5 [2] 1 [2] 2} = {10,10,10,10,10 [2] 10,10,10,10,10}
{10,5 [2] 1 [2] 3} = {10,10,10,10,10 [2] 10,10,10,10,10 [2] 2}
It's very important that the rows that are highlighted have the same number of 10s.
{10,3 [2] 1 [2] 1,2} = {10,10 [2] 10 [2] {10,10 [2] 10 [2] 10}}
{10,5 [2] 1 [2] 1 [2] 2} = {10,10,10,10,10 [2] 10,10,10,10,10 [2] 10,10,10,10,10}
The last name I can give is called trixxation, which is {a,b [3] 2}. There are no more notable names after that.
{10,5 [3] 2} = {10,10 [2] 10 [2] 10 [2] 10 [2] 10}
Except for the first row, there are 'b' number of 10s with "[2]" between each one.
{10,5 [3] n} = {10,10 [2] 10 [2] 10 [2] 10 [2] 10 [3] n-1}
2D sections from earlier are on the left-hand side of "[3]" since there is a hypothetical "1" on the right-hand side, even though it does not appear visually.
{10,5 [n] 2} = {10,10 [n-1] 10 [n-1] 10 [n-1] 10 [n-1] 10}
Extending to multiple dimensions is straightforward once you understand the recursion pattern. Just keep in mind that...
{10,5 [2] 2 [d] e} = {10,10,10,10,10 [d] e}
{10,5 [c] 2 [d] e} = {10,10 [c-1] 10 [c-1] 10 [c-1] 10 [c-1] 10 [d] e}
(where c is 3 or more and is at most 1 less than d.)
{10,5 [c] 1 [2] 2} = {10,10,10,10,10 [c] 10,10,10,10,10}
{10,5 [c] 1 [d] 2} = {10,10 [c] 10 [d-1] 10 [d-1] 10 [d-1] 10 [d-1] 10}
For multiple entries in the dimension brackets...
{10,b [1,2] 2} = {10,10 [b] 2}
{10,b [1,2] c+1} = {10,10 [b] 2 [1,2] c}
{10,3 [1,2] 1,2} = {10,10 [1,2] {10,10 [1,2] 10}}
{10,3 [1,2] 1 [2] 2}
[WIP]
Strong Array Notation
Founder: Hyp cos
Strong Array Notation has many similarities with Bower's array notations and Bird's Array Notation; however, its scaling in larger arrays is different from those aforementioned notations. There are currently 7 composed parts to this that are named separately. For the purpose of this page, I will group them into Strong Array Notation, but I will still mention the sub-notation names as we go through this.
LINEAR ARRAY NOTATION (LAN) | Tiers 1-3
The function we're looking at is s(a,b,c,...). The first three entries behave exactly like the previous functions, so just as a brief run-through...
s(a) = {a} = a = s(a,1)
s(a,b) = {a,b} = aᵇ = s(a,b,1)
s(a,b,2) = {a,b,2} = a↑↑b
s(a,b,c) = {a,b,c} = a↑ᶜb
The interesting part here is the fact that you can put a number on the left of the function, like 4s(10,6), and it will act like scientific notation so this equals 4 x 10⁶.
However, things are different with four entries and beyond. Instead of s(10,5,1,2) being equal to s(10,10,s(10,10,s(10,10,s(10,10,10)))), it is actually equal to s(10,10,5). Which means s(a,b,1,2) is s(a,a,b).
Recursion with four entries and beyond is a little different, too.
s(a,1,c,2) ≠ a
s(a,2,2,2) = s(a,s(a,a),1,2)
s(a,3,2,2) = s(a,s(a,s(a,a),1,2),1,2)
s(a,4,2,2) = s(a,s(a,s(a,s(a,a),1,2),1,2),1,2)
The general gist is that
s(a,3,c,#) = s(a,s(a,s(a,a),c-1,#),c-1,#)
Where # is anything after the third entry.
And also
s(a,a,b,c) = s(a,b,1,c+1)
so s(10,10,100,2) = s(10,100,1,3)
and s(10,10,100,3) = s(10,100,1,4)
and so on.
As far as five entries are concerned, it is very similar to how four entries are reached.
s(a,b,1,1,2) = s(a,a,a,b)
Yep. No need for a crazy amount of space just to get an extra entry. This does alter the previous examples for the third entry, though.
s(a,2,2,1,2) = s(a,s(a,a,a),1,1,2)
s(a,3,2,1,2) = s(a,s(a,s(a,a,a),1,1,2),1,1,2)
s(a,4,2,1,2) = s(a,s(a,s(a,s(a,a,a),1,1,2),1,1,2),1,1,2)
But otherwise, incrementing the fourth entry and beyond is the same.
s(a,a,b,c,d) = s(a,b,1,c+1,d)
s(a,a,a,b,c) = s(a,b,1,1,c+1)
And reaching the sixth entry is identical to getting the fourth and fifth entries!
s(a,a,a,a,b) = s(a,b,1,1,1,2)
The previous examples with the third entry would now be
s(a,2,2,1,1,2) = s(a,s(a,a,a,a),1,1,1,2)
s(a,3,2,1,1,2) = s(a,s(a,s(a,a,a,a),1,1,1,2),1,1,1,2)
s(a,4,2,1,1,2) = s(a,s(a,s(a,s(a,a,a,a),1,1,1,2),1,1,1,2),1,1,1,2)
And then incrementing the fourth entry and beyond is identical to before!
s(a,a,b,c,d,e) = s(a,b,1,c+1,d,e)
s(a,a,a,b,c,d) = s(a,b,1,1,c+1,d)
s(a,a,a,a,b,c) = s(a,b,1,1,1,c+1)
Reaching future entries is also very identical to getting the previous entries.
7 entries: s(a,a,a,a,a,b) = s(a,b,1,1,1,1,2)
8 entries: s(a,a,a,a,a,a,b) = s(a,b,1,1,1,1,1,2)
9 entries: s(a,a,a,a,a,a,a,b) = s(a,b,1,1,1,1,1,1,2)
10 entries: s(a,a,a,a,a,a,a,a,b) = s(a,b,1,1,1,1,1,1,1,2)
Once we get to the stage of having too many ones, we introduce the next stage of this notation series, which is...
EXTENDED ARRAY NOTATION (exAN) | Tier 4
The nesting structure we're working with is s(a,b{c}d}, which looks similar to {a,b [c] d} from Bird's Array Notation, but how the former works is very different. There must always be two entries at the start, though b = 1 doesn't always work the same way as we go through this.
Instead of b in s(a,b{2}2) counting how many copies of a are in the array, it counts how many ones (minus one) are between b and 2 with commas. Essentially, it is s(10,15{2}2) = s(10,15,1,1,...(14 ones)...,1,1,2) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2).
All of the examples for the third entry work like they did before, but the number in the brackets and the number on the right must be the same.
s(a,2,2{2}2) = s(a,a{2}2)
s(a,3,2{2}2) = s(a,s(a,a{2}2){2}2)
s(a,4,2{2}2) = s(a,s(a,s(a,a{2}2){2}2){2}2)
s(a,4,c{2}2) = s(a,s(a,s(a,a,c-1{2}2),c-1{2}2),c-1{2}2)
And getting the fourth entry and beyond is still about the same.
s(a,b,1,2{2}2) = s(a,a,b{2}2)
s(a,b,1,d{2}2) = s(a,a,b,d-1{2}2)
s(a,b,1,1,2{2}2) = s(a,a,a,b{2}2)
s(a,b,1,1,1,2{2}2) = s(a,a,a,a,b{2}2)
Do note that if there are ones at the end on the left-hand side of {2}, they are removed.
s(a,b,...,x,1{2}z) = s(a,b,...,x{2}z)
Incrementing the number on the right is about the same as earlier.
s(10,15{2}3) = s(10,15,1,1,...(14 ones)...,1,1,2{2}2) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2{2}2)
s(10,15{2}4) = s(10,15,1,1,...(14 ones)...,1,1,2{2}3) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2{2}3)
s(10,15{2}z) = s(10,15,1,1,...(14 ones)...,1,1,2{2}z-1) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2{2}z-1)
As for multiple entries after {2}, it's similar (again) to how you got the fourth entry and beyond.
s(a,b{2}1,2) = s(a,a{2}b)
And then you recursively do the same thing to increment the entries on the right.
s(a,3,c{2}1,2) = s(a,s(a,a,c-1{2}1,2),c-1{2}1,2)
s(a,b,1,2{2}1,2) = s(a,a,b{2}1,2)
s(a,b,1,d{2}1,2) = s(a,a,b,d-1{2}1,2)
s(a,b,1,1,2{2}1,2) = s(a,a,a,b{2}1,2)
s(10,15{2}2,2) = s(10,15,1,1,...(14 ones)...,1,1,2{2}1,2) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2{2}1,2)
s(10,15{2}3,2) = s(10,15,1,1,...(14 ones)...,1,1,2{2}2,2) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2{2}2,2)
s(10,15{2}z,2) = s(10,15,1,1,...(14 ones)...,1,1,2{2}z-1,2) or s(10,15,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2{2}z-1,2)
s(a,b{2}1,3) = s(a,a{2}b,2)
s(a,b{2}1,4) = s(a,a{2}b,3)
s(a,b{2}1,1,2) = s(a,a{2}a,b)
s(a,b{2}1,1,1,2) = s(a,a{2}a,a,b)
We then reach multiple {2}'s. The second {2} is similar to the first but without subtracting one of the ones. Unlike Bird's Array Notation we only do this to the second row.
s(10,15{2}1{2}2) = s(10,15{2}1,1,...(15 ones)...,1,1,2)
And then...
s(10,15{2}z{2}2) = s(10,15,1,1,...(14 ones)...,1,1,2{2}z-1{2}2)
s(a,b{2}1,2{2}2) = s(a,a{2}b{2}2)
s(10,15{2}1{2}3) = s(10,15{2}1,1,...(15 ones)...,1,1,2{2}2)
s(10,15{2}1{2}z) = s(10,15{2}1,1,...(15 ones)...,1,1,2{2}z-1)
Reaching the third {2} is the same as the second but one row to the right.
s(10,15{2}1{2}1{2}2) = s(10,15{2}1{2}1,1,...(15 ones)...,1,1,2)
(the middle rows are just kept as a 1.)
The fourth {2} isn't any different.
s(10,15{2}1{2}1{2}1{2}2) = s(10,15{2}1{2}1{2}1,1,...(15 ones)...,1,1,2)
As with the fifth {2}.
s(10,15{2}1{2}1{2}1{2}1{2}2) = s(10,15{2}1{2}1{2}1{2}1,1,...(15 ones)...,1,1,2)
And we keep this up until we get to the first {3}. The {3} counts how many {2}'s there are, with ones filling the blanks.
s(10,5{3}2) = s(10,5{2}1{2}1{2}1{2}1{2}2)
s(10,7{3}2) = s(10,7{2}1{2}1{2}1{2}1{2}1{2}1{2}2)
s(10,10{3}2) = s(10,10{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2)
(oh, and don't forget the second entry of the first row changes too.)
Lower levels from earlier occur on the left-hand side of higher levels. For this instance, {2} is on the left-hand side of {3}.
s(10,15{2}2{3}2) = s(10,15,1,1,...(14 ones)...,1,1,2{3}2)
s(a,b{2}1,2{3}2) = s(a,a{2}b{3}2)
s(10,15{2}1{2}2{3}2) = s(10,15{2}1,1,...(15 ones)...,1,1,2{3}2)
Incrementing the right-hand number of {3} is identical to how we got {3}.
s(10,5{3}3) = s(10,5{2}1{2}1{2}1{2}1{2}2{3}2)
s(10,7{3}3) = s(10,7{2}1{2}1{2}1{2}1{2}1{2}1{2}2{3}2)
s(10,10{3}3) = s(10,10{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2{3}2)
s(10,10{3}4) = s(10,10{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2{3}3)
s(10,10{3}z) = s(10,10{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2{3}z-1)
Now you have multiple entries on the right of {3}. What do we do now--
s(10,b{3}1,2) = s(10,10{3}b)
s(10,b{3}1,z) = s(10,10{3}b,z-1)
s(10,b{3}1,1,2) = s(10,10{3}10,b)
s(10,15{3}1{2}2) = s(10,15{3}1,1,...(15 ones)...,1,1,2)
s(10,15{3}1{2}1{2}2) = s(10,15{3}1{2}1,1,...(15 ones)...,1,1,2)
Okay, okay! You got me there. Now, what about multiple {3}'s?
s(10,5{3}1{3}2) = s(10,5{3}1{2}1{2}1{2}1{2}1{2}2)
s(10,7{3}1{3}2) = s(10,7{3}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2)
s(10,10{3}1{3}2) = s(10,10{3}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2)
s(10,10{3}2{3}2) = s(10,10{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2{3}1{3}2)
s(10,10{3}1{3}3) = s(10,10{3}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2{3}2)
s(10,15{3}1{3}1{2}2) = s(10,10{3}1{3}1,1,...(15 ones)...,1,1,2)
s(10,10{3}1{3}1{3}2) = s(10,10{3}1{3}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}1{2}2)
Now we get higher levels of the brackets (which is a separator by the way), starting with {4}.
s(10,5{4}2) = s(10,5{3}1{3}1{3}1{3}1{3}2)
s(10,5{4}3) = s(10,5{3}1{3}1{3}1{3}1{3}2{4}2)
s(10,b{4}1,2) = s(10,10{4}b)
s(10,15{4}1{2}2) = s(10,15{4}1,1,...(15 ones)...,1,1,2)
s(10,5{4}1{3}2) = s(10,5{4}1{2}1{2}1{2}1{2}1{2}2)
s(10,5{4}1{4}2) = s(10,5{4}1{3}1{3}1{3}1{3}1{3}2)
Now let's see how we get to {1,2}.
s(10,5{5}2) = s(10,5{4}1{4}1{4}1{4}1{4}2)
s(10,5{6}2) = s(10,5{5}1{5}1{5}1{5}1{5}2)
s(10,5{d}2) = s(10,5{d-1}1{d-1}1{d-1}1{d-1}1{d-1}2)
s(10,b{1,2}2) = s(10,10{b}2)
Oh, that's similar to Bird's Array Notation. What about {1,1,2}?
s(10,b{1,2}3) = s(10,10{b}2{1,2}2)
s(10,b{1,2}z) = s(10,10{b}2{1,2}z-1)
s(10,b{1,2}1,2) = s(10,10{1,2}b)
s(10,15{1,2}1{2}2) = s(10,15{1,2}1,1,...(15 ones)...,1,1,2)
s(10,5{1,2}1{3}2) = s(10,5{1,2}1{2}1{2}1{2}1{2}1{2}2)
s(10,b{1,2}1{1,2}2) = s(10,10{1,2}10{b}2)
s(10,b{1,2}1{1,2}1{1,2}2) = s(10,10{1,2}10{1,2}10{b}2)
s(10,5{2,2}2) = s(10,5{1,2}1{1,2}1{1,2}1{1,2}1{1,2}2)
s(10,5{d,2}2) = s(10,5{d-1,2}1{d-1,2}1{d-1,2}1{d-1,2}1{d-1,2}2)
s(10,b{1,3}2) = s(10,10{b,2}2)
s(10,b{1,1,2}2) = s(10,10{1,15}2)
Now we're in the prime position to get to {1{2}2}.
s(10,b{1,1,3}2) = s(10,10{1,15,2}2)
s(10,b{1,1,1,2}2) = s(10,10{1,1,15}2)
s(10,5{1{2}2}2) = s(10,5{1,1,1,1,1,2}2)
s(10,10{1{2}2}2) = s(10,10{1,1,1,1,1,1,1,1,1,1,2}2)
s(10,15{1{2}2}2) = s(10,15{1,1,...(15 ones)...,1,1,2}2)
To get higher nested levels like {1{3}2}...
s(10,15{1{2}3}2) = s(10,15{1,1,...(15 ones)...,1,1,2{2}2}2)
s(10,15{1{2}z}2) = s(10,15{1,1,...(15 ones)...,1,1,2{2}z-1}2)
s(10,15{1{2}1,2}2) = s(10,10{1{2}15}2)
s(10,15{1{2}1,1,2}2) = s(10,10{1{2}1,15}2)
s(10,15{1{2}1{2}2}2) = s(10,10{1{2}1,1,...(15 ones)...,1,1,2}2)
s(10,5{1{3}2}2) = s(10,5{1{2}1{2}1{2}1{2}1{2}2}2)
And then we find ourselves getting to {1{1{2}2}2}...
s(10,69{1{1,2}2}2) = s(10,10{1{69}2}2)
s(10,69{1{1,3}2}2) = s(10,10{1{69,2}2}2)
s(10,69{1{1,1,2}2}2) = s(10,10{1{1,69}2}2)
s(10,69{1{1{2}2}2}2) = s(10,15{1{1,1,...(69 ones)...,1,1,2}2}2)
We can keep this going with many nests until we land on the next stage of the notation, which is...
EXPANDING ARRAY NOTATION (EAN) | Tier 5
Welcome to the start of the more complicated separators. Don't worry, I have you covered! We'll be using grave accents here as separators.
Imagine {1`1} as the comma, {1`2} as the dot, {1`3} as {1·²}, and {1`ˢ} as {1·ˢ⁻¹}.
The first few values of b in s(a,b{c`d}e) are
s(10,2{1`2}2) = s(10,2{1,2}2)
s(10,3{1`2}2) = s(10,3{1{1,2}2}2)
s(10,4{1`2}2) = s(10,4{1{1{1,2}2}2}2)
s(10,5{1`2}2) = s(10,5{1{1{1{1,2}2}2}2}2)
s(10,b{1`2}2) = s(10,b{1{1{...{1{1,2}2}...}2}2}2) [b 2s]
Incrementing the right-hand side repeats this.
s(10,5{1`2}3) = s(10,5{1{1{1{1,2}2}2}2}2{1`2}2)
s(10,5{1`2}4) = s(10,5{1{1{1{1,2}2}2}2}2{1`2}3)
s(10,5{1`2}z) = s(10,5{1{1{1{1,2}2}2}2}2{1`2}z-1)
Incrementing numbers in the separators is similar.
s(10,5{1`2}1{1`2}2) = s(10,5{1`2}1{1{1{1{1,2}2}2}2}2)
s(10,5{2`2}2) = s(10,5{1`2}1{1`2}1{1`2}1{1`2}1{1`2}2)
s(10,15{1,2`2}2) = s(10,10{15`2}2)
s(10,15{1{2}2`2}2) = s(10,15{1,1,...(15 ones)...,1,1,2`2}2)
s(10,15{1{1,2}2`2}2) = s(10,10{1{15}2`2}2)
s(10,15{1{1{2}2}2`2}2) = s(10,15{1{1,1,...(15 ones)...,1,1,2}2`2}2)
VERY IMPORTANT:
s(10,5{1{1`2}2`2}2) = s(10,5{1{1{1{1{1,2}2}2}2}2`2}2)
(NOT s(10,5{1`3}2) YET)
The way you get to {1`3} is from
s(10,2{1`3}2) = s(10,2{1,2`2}2)
s(10,3{1`3}2) = s(10,3{1{1,2`2}2`2}2)
s(10,4{1`3}2) = s(10,4{1{1{1,2`2}2`2}2`2}2)
And incrementing the right-hand number of the grave is similar.
s(10,5{1{1`2}2`3}2) = s(10,5{1{1{1{1{1,2}2}2}2}2`3}2)
s(10,5{1{1`3}2`3}2) = s(10,5{1{1{1{1{1,2`2}2`2}2`2}2`2}2`3}2)
s(10,4{1`4}2) = s(10,4{1{1{1,2`3}2`3}2`3}2)
s(10,4{1`5}2) = s(10,4{1{1{1,2`4}2`4}2`4}2)
s(10,4{1`d}2) = s(10,4{1{1{1,2`d-1}2`d-1}2`d-1}2)
Then all the previous steps jump in.
s(10,15{1`1,2}2) = s(10,10{1`15}2)
s(10,15{1`1{2}2}2) = s(10,10{1`1,1,...(15 ones)...,1,1,2}2)
s(10,15{1`1{1,2}2}2) = s(10,10{1`1{15}2}2)
s(10,5{1`1{1`2}2}2) = s(10,5{1`1{1{1{1{1,2}2}2}2}2}2)
Now for multiple grave operators.
s(10,2{1`1`2}2) = s(10,2{1`1,2}2)
s(10,3{1`1`2}2) = s(10,3{1`1{1`1,2}2}2)
s(10,4{1`1`2}2) = s(10,4{1`1{1`1{1`1,2}2}2}2)
[WIP]
Bowers' Exploding Array Function
Founder: Jonathan Bowers
TO BE ADDED
Hyper-E Notation
Tier 1 | Founder: Sbiis Saibian
TO BE ADDED
Cascading-E Notation
Tier 2 | Founder: Sbiis Saibian
TO BE ADDED
Extending Cascading-E Notation
Tier 3 | Founder: Sbiis Saibian
TO BE ADDED
Numbers in Brackets
Founder: Guest
TO BE ADDED
Sources: Googology Wiki, smol twi's site, William Powell's N0TAI video, Hyperoperators (Aarex), Large Numbers Gradually Increasing (Douglas Shamlin Jr.), Strong Array Notation (D.S. Jr.), Steps Toward Infinity, SAN solver.
(Some sources were used to verify facts.)
[THIS IS A SIMPLIFICATION OF WHAT THE NOTATIONS DO.]