## Ultra oblivion

Ultra oblivion would be = to the largest finite number that can be uniquely defined using no more than an ultimate oblivion symbols in some K(Hyper oblivion) system in some K2(Hyper oblivion) 2-system in some K3(Hyper oblivion) 3-system in some K4(Hyper oblivion) 4-system in some... KHyper oblivion(Hyper oblivion) Hyper oblivion-system where the number Hyper oblivion can be represented with one symbol (byte).", where a Km(n) m-system is an arbitrary well-defined system of mathematics that can generate K(m-1)(n) (m-1)-systems and which can be uniquely described in at most n symbols and a K1(n) system is an arbitrary well-defined system of mathematics which can be uniquely described in n symbols.

## Cramo

Cramo would be = to Ultimate Croutonillion # Ultimate Croutonillion # Ultimate Croutonillion #Ultimate Croutonillion = X. Sam^{X}(Rayo^{X}(Sam(TREE^{X}(Garden^{X}(BB_{X}^{X}(BH^{X}(WW^{X}(FF^{X}(PP^{X}(Gerflo(Gerflo(Sam^{sam(X)}(G^{X}(WE(X,X@X!!!!!!!!) = Y

((Y ↑_{Y} ↑ Y) ↑(Y ↑ _{((Y ↑Y ↑ Y) ↑ Y) ↑ Y) = C}

Sam^{milton}(Gerflo^{Finity}(Rayo^{Ultimate croutonillion}(C^{XYFinityMiltonQuatterquinquagihtillion} ↑^{Milton} C + Hyper oblivion = D

((((LND(Rayo(Sam(Sam(D) # D # D # D) X {D,C,X,Y, Sam(Rayo(D) - Sam(Rayo(D) (not minus)} * D)))) @ D * (Yog sothoth + nyarlathotep) = Z

Expofacto(Z@Z#Z&Z%Z)$$$§§§§§§§§§§§§§§§§§§§§§§§§§§§§!!!!!!!! ↑_{Z} ↑ Z = B.

B^^^^^^^^^B!!!!!!! X Hyper oblivion = Q

Ban^{q}(RBAN^{Q}(Worm^{Q}(Pair^{Q}(Gar^{Q}(Fz^{Q}(box^{Q}(circle^{Q(}Green(End(Cg(Hydra(Bh(WW(FF(BB(PP(Se(G^{Q}(Tree^{Q}(TREE^{Q}(LND^{Q}rayo(Sam(Q ↑^{Q@Q} Q)))^{Q!Q!!Q$…}

## BEAF / BAN Multidimensional Array Problem

Hello!

Before getting into a heavy chunk of complainment, I just want to remind you that I really like these two notations. They're strong yet simple, and they can be used in like every large number video. I just want to point out to a few mistakes that these notations were made. Some of them do not even need to be fixed! Anyway, enjoy this blog post that gave me a rage attack while writing.

Lately I've been studying BAN (Bird's array notation) and I've found something very annoying. Let me explain.

So BEAF and BAN

Maybe THIS is the way Bird intended to make his notation, but I don't think it makes sense.

Same thing goes for BEAF, with the [x] being changed to (x-1), ab being changed to b & a, and stuff like that.

But do the problems end there? …

## My attempt at an improved notation

I'm fairly new to googology, and I've previously made my own notation. However, this time I want to focus more on making the notation less messy, and make the base rules themselves stronger rather than adding a bunch of extensions. All arrays are in the form of a(#)b, where # represents the rest of the array.

**Single entry:** a(c)b = a(c arrows)b = {a,b,c} in BEAF.

**Two entries:** a(0,1)b = a(a)b

a(c,1)b = a(a(c-1,1)b)b, so Graham's number is approximately 3(64,1)3.

a(c,2)b = a(a(c-1,2)b)b, the same thing as before. So in general a(c,d)b = a(a(c-1,d)b)b

When c reaches 0, a(0,d)b = a(a(d)b,d-1)b

So 3(2,2)3 = 3(3(1,2)3)3 = 3(3(3(0,2)3)3)3 = 3(3(3(3(2)3,1)3)3)3 ≈ {3,{3,3,2},1,2}

**Three entries and above:**

Zeros can only appear at the beginning of an array.

a(c,…

## Thoughts about Functional Domination

On this wiki, the notion of eventual domination is used to refer to a function's property to remain greater than another function after a certain constant. Rigorously, \(f\) eventually dominates \(g\) if there exists an \(N\) such that for all \(n\geq N\), we have \(f(n)>g(n)\). I have never been a huge fan of this definition since it implies the function \(\left(10\uparrow^{10}n\right)+1\) eventually dominates \(10\uparrow^{10}n\), and I believe this really loses sight on what we really mean about a function growing faster than another. I mean, clearly for these two functions (assuming continuity), their derivatives are the same, so they are growing at the same rate. Sbiis Saibian offered another suggestion using the notation \([>]\) to d…

## 75xsvbj4xS4pgUWgWg3ZB6tqHbsQz1UhpD9gycqmV4QfO64dnb

I don't know whether this sequence terminates or not, this sequence came to my mind when I was studying Godsteinn's theorem but still, maybe this sequence will continue to converge to infinity, maybe you can tell whether this sequence will end or not because I'm a bit no doubt about this, it's not for serious functions like Prss or Godsteinn sequence

actually the definition is quite short, chatgpt has made me the python code for this but still the original idea is my own, chatgpt said that maybe this sequence is unlikely to converge to infinity because the system is continuously cycling, maybe later I will add a new definition

(1) Take a number \(n\) in decimal

(2) Convert \(n)\ to binary number system

(3) Consider the binary representation as…

## I understand correctly BMS?

(0,0)(1,1)=(0)(1)(2)(3)…

(0,0)(1,1)(0,0)

(0,0)(1,1)(0,0)(1,0)

(0,0)(1,1)(0,0)(1,1)

(0,0)(1,1)(0,0)(1,1)(0,0)(1,1)

(0,0)(1,1)(1,0)

(0,0)(1,1)(1,0)(0,0)(1,1)(1,0)

(0,0)(1,1)(1,0)(1,0)

(0,0)(1,1)(1,0)(2,0)

(0,0)(1,1)(1,0)(2,0)(3,0)

(0,0)(1,1)(1,0)(2,1)

(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)

(0,0)(1,1)(1,1)

(0,0)(1,1)(1,1)(1,0)(2,1)(2,1)

(0,0)(1,1)(1,1)(1,1)

(0,0)(1,1)(2,0)

(0,0)(1,1)(2,0)(3,0)

(0,0)(1,1)(2,0)(3,1)

(0,0)(1,1)(2,0)(3,1)(3,1)

(0,0)(1,1)(2,0)(3,1)(4,0)

(0,0)(1,1)(2,0)(3,1)(4,0)(5,1)

(0,0)(1,1)(2,1)

(0,0)(1,1)(2,1)(2,1)

(0,0)(1,1)(2,1)(3,0)

(0,0)(1,1)(2,1)(3,0)(4,1)(5,2)

(0,0)(1,1)(2,1)(3,1)

(0,0)(1,1)(2,1)(3,1)(4,0)

(0,0)(1,1)(2,1)(3,1)(4,1)

(0,0)(1,1)(2,2)

(0,0)(1,1)(2,2)(1,0)

(0,0)(1,1)(2,2)(1,0)(2,1)(3,2)

(0,0)(1,1)(2,2)(1,1)

(0,0)(1,1)(2,2)(1,1)(1,1)

(0,0)(1,1)(2,2)…

## Ostrich function

The **Ostrich function** is a non-computable salad number created by me (Ilikepjns) in July 19, 2024.

The Ostrich function is denoted and defined as:

There is no possible computations.

## Unoffiical Biggest Finite Number

An *unofficial biggest finite number* (*UBFN*) is the largest, unofficial, defined finite number which is positioned as the closest number to infinity.

In order to make an *unofficial biggest finite number*, we must make the number as close to infinity as possible; however, infinity is not a number and unbounded meaning other numbers can go past it. Therefore, making the biggest finite number is not possible as if anything can be added by one. But let us make a custom number hierarchy that can prevent that paradox, but the question is: is it possible to make such a numerical hierarchy to compute the *unofficial biggest finite number*, and if so, what is the value of the *unofficial biggest finite number*?

The journey to create the *unofficial biggest finite …*

## Reverse mathematics and rank functions for directed graphs

Citation: Hirst, Jeffry L. "Reverse mathematics and rank functions for directed graphs." Archive for Mathematical Logic 39.8 (2000): 569-579.

A rank function for a directed graph $G$ assigns elements of a well ordering to the vertices of $G$ in a fashion that preserves the order induced by the edges. While topological sortings require a one-to-one matching of vertices and elements of the ordering, rank functions frequently must assign several vertices the same value. Theorems stating basic properties of rank functions vary significantly in logical strength. Using the techniques of reverse mathematics, we present results that require the subsystems $\mathbf{R C A}_{0}, \mathbf{A C A}_{0}, \mathbf{A T R}_{0}$, and $\boldsymbol{\Pi}_{1}^{1}-\m…

## new idea

I decided to apply one idea on the usual serial designation of the array.

I don't know how to describe it. You yourself will see everything.

(1)=1

(2)=2

(3)=3

(1,2)=ω

(2,2)=ω+1

((1,2),2)=ω*2

(((1,2),2),2)=ω*3

(1,3)=ω^2

((1,3),3)=ω^2*2

(1,4)=ω^3

(1,5)=ω^4

(1,(1,2))=ω^ω

(1,(2,2))=ω^ω+1

(1,(3,2))=ω^ω+2

(1,((1,2),2))=ω^ω*2

(1,(((1,2),2),2))=ω^ω*3

Here is the idea itself. Instead of (1,(1,3)) shift "and it turns out (1,((1,3),2)).

(1,((1,3),2))=ω^ω^2

(1,((1,4),2))=ω^ω^3

(1,((1,(1,2)),2))=ω^ω^ω

(1,((1,((1,(1,2)),2)),2))=ω^ω^ω^ω

(1,(1,3))=ψ(Ω)

((1,(1,3)),(1,3))=ψ(Ω)+ψ(Ω)

(1,(2,3))=ψ(Ω+1)

(1,((1,(1,3)),3))=ψ(Ω+ψ(Ω))

(1,(1,4))=ψ(Ω*2)

(1,(1,5))=ψ(Ω*3)

(1,(1,6))=ψ(Ω*4)

(1,(1,(1,2)))=ψ(Ω*ω)

(1,(1,(2,2)))=ψ(Ω*ω+1)

(1,(1,(3,2)))=ψ(Ω*ω+2)

(1,(1,((1,(1,(1,2))),2)))=ψ(Ω*ψ(Ω*ω))

(1,(1,((1,(…

## My attempt to beat Omega^Omega Function

I have previously tried (successfully!) to beat Graham's Function with a very simple and easy function and some people in the comments have encouraged me to expand this notation, so that's what I'm doing today.

First, I will recap the previous rules, but I'm going to change the way that I represented (the notation) it to make it easier to read.

Firsts Definitions:

a[1]b = a^b Ex: 2[1]3 = 2^3 = 8

a[n]b = a[n-1](a[n-1]···(a[n-1]a) with n times [n-1] Ex: 3[3]4 = 3[2](3[2](3[2](3[2]3))))

With this we can see that this notation is almost like the Knuth's Arrow one, but note that mine's grows a little faster, as we can see in: 2[2]2 = 23

More Definitions:

a[1,n]b = a[b,n-1]a Ex: 3[1,3]4 = 3[4,2]3

a[1,0]b = a[1]b

Take into account that the previous proper…

## Ordinals and Cardinals: Simplified

**Cardinals** are of how much of something there is. For instance “I have one banana” means there is a **Cardinality** of one, I.e. bananas.

**Ordinals** are of what place something came. For instance “I came second place in the race” means their **Ordinality** (or place)in the race was 2, or 2nd.

## Amr073's S notation

My S notation is based on a variant of the fast growing hierarchy. This variant works like this:

## some point parable imitates cantor normal form

I've actually been stuck in the ordinal ω^ω^ω for a long time and couldn't get out of it, why even though I had applied the Saibian method in inner argument notation but it still didn't work, reaching epsilon was my dream at that time one day I got enlightenment and finally I use a method that I call "the climbing method" this method appears in ordinal analysis, you might call this method CNF or something like that, the climbing method is quite easy to generalize, I will try to give an example, remember that the number at the end is not f_e0 but g_e0 in SGH, in the future I will just use the hardy hierarchy,

an infinite number of marbles are provided and a finite number of pots are also provided,

the exact number of pots is 6.02214076 × 10^{23…}

## rough prototype for nothing notation beyond e0 (1000th edit)

This is a prototype that still needs further analysis, this analysis comes from ordinal analysis and could be wrong depending on which fundamental sequence you use because it is different so sorry if there are mistakes because this is still premature, thank you

**Nothing notation sub notation:**

Basic nothing notation (BNN) up to (imitating multi argument veblen function, not sure, ill defined)

Legional nothing notation (LNN) up to **BEYOND** (It hasn't been created yet because I don't understand about OCF)

## My attempt to beat Graham's Function

This is my honest attempt to make a function that grows faster than Graham's function.

Let's define some notation before defining any kind of function.

First, I'll be using some small variation of the Knuth's up-arrow notation:

ax

Does anyone know how to check if this beats Graham? Or does anyone know a way to generalized even more?

## 8zXmsj1evC1x9IXsdjdH8s6XBc4VbYg7hTw6a70sYzGoTeOLr5

- 1 Air Conditioning Function
- 2 CB sequence
- 2.1 Definition
- 2.1.1 CB(n)

- 2.2 Analysis
- 2.2.1 Values For CB(n)
- 2.2.2 CB(84,000,000)

- 2.3 please make this better

- 2.1 Definition
- 3 Onagol
- 4 Passlorgulus
- 5 Wisslorgulus
- 6 Zentation
- 7 Chupplorgulus

The **Air Conditioning Function** or **AC Function**, denoted as **AC(n)** was a function that was created by **Franklin Lee** on July 13, 2024. It is a new function created just now.

**Layout**

The AC Function takes place in a hotel with an infinite number of rooms. A worker placed in the middle of the hotel needs to set the AC on to cool for as many rooms as possible without rooms becoming “hot”.

**Rules**

At the start of the game, the worker will set the AC to cool and go to another room.

The hotel worker can move one space up, down, left, or right, but not diagonally.

You may not …

## CB Sequence (taken from the page)

The CB Sequence is a fast-growing function, CB(n) arising from levels of exponentiation, including tetration, pentation, and so on.

- 1 Definition
- 1.1 CB(n)

- 2 Analysis
- 2.1 Values For CB(n)
- 2.2 CB(84,000,000)

Let's say that we have a line. For CB(1) to CB(2), we raise CB(1) to the power of 10. Making 10. Then For CB(3), you raise CB(2) to the power of itself CB(2) times. Then, you do the same for CB(4), but instead of raising it to the power of itself, you tetrate CB(2) to itself. Here, n equals the number inside the CB(n) function, and using up-arrow-notation, the amount of arrows you use is (n-1).

The first 3 values are CB(1) = 1, CB(2) = 10, and CB(3) = 10 tetrated to 10. This format of exponentiation getting more powerful, (explained at the top) goes…

## Notation thing

[a] by itself equals a,

[a,b] = a^b

[a,b,c] = a{c arrows}b

[a,b,c,d] = [a,b,[a,b,c],d-1], Trailing 1s are removed

So [2,3,2,3] = [2,3,[2,3,2],2] = [2,3,[2,3,[2,3,2]],1] = [2,3,[2,3,[2,3,2]]] = 2{2{2{2}3}3}3

Extended notation: [a,b/1] = [a,a,a, ... ,a,a,a] with b copies of a

[a,b,c,d, ... ,z/0] = [a,b,c,d, ... ,z]

[a,b,c, ... ,y,z/n] = [a,b, ... ,z,[a,b, ... ,z,[a,b, ... ,z,[...]]]] where there are [a,b,c, ... ,y,z/n-1] pairs of brackets.

So [a,b/2] = [a,a,a, ... ,a,a,a/1] with b copies of a = [a,a,a,...,[a,a,a,...,[a,a,a...,[...]]]] where there are [a,b/1] pairs of brackets

[3,3,3/2] = [3,3,3,[3,3,3,[3,3,3,[...]]]] with [3,3,3[3,3,3,[3,3,3,[...]]]] with [3,3,3] pairs of brackets

[a/n] is just [a,n/1]

Double slashes:

[a,b,c, ... ,y,z//n] = [a,b, ... ,z…

## Matrician systems

systems are on the following principle: the system n begins as a primitive sequence, only with a chech n "(0,n,n*2,n*3,…)". The system ends when the matrix becomes standard "(0,1,2,3,…)"

system 1 "ε0"

(0)=1

(0,0)=2

(0,0,0)=3

(0,1)=ω

(0,1,1)=ω^2

(0,1,2)=ω^ω

(0,1,2,2)=ω^ω^2

(0,1,2,3)=ω^ω^ω

(0,1,2,3,4)=ω^ω^ω^ω

…

system 2 "ψ(Ω_2)"?

(0)=1

(0,0)=2

(0,0,0)=3

(0,2)=ω

(0,2,0,2)=ω*2

(0,2,2)=ω^2

(0,2,2,2)=ω^3

(0,2,4)=ω^ω

(0,2,4,6)=ω^ω^ω

(0,2,4,6,8)=ω^ω^ω^ω

(0,1)=ψ(Ω)

(0,1,0)=ψ(Ω)+1

(0,1,0,2)=ψ(Ω)+ω

(0,1,0,2,4)=ψ(Ω)+ω^ω

(0,1,0,1)=ψ(Ω)+ψ(Ω)

(0,1,0,1,0,1)=ψ(Ω)+ψ(Ω)+ψ(Ω)

(0,1,3)=ψ(Ω+1)

(0,1,3,0,1,3)=ψ(Ω+1)+ψ(Ω+1)

(0,1,3,3)=ψ(Ω+2)

(0,1,3,5)=ψ(Ω+ω)

(0,1,3,5,7)=ψ(Ω+ω^ω)

(0,1,3,5,7,9)=ψ(Ω+ω^ω^ω)

(0,1,3,4)=ψ(Ω+ψ(Ω))

(0,1,3,4,3,4)=ψ(Ω+ψ(Ω)+ψ(Ω))

(0,1,3,4,6)=ψ(Ω+ψ(Ω+1))

(0,1,3,4,6,7)=ψ(Ω+ψ(Ω+ψ(Ω))) …

## concept

I have a strange idea for an inexperienced function (as it seems to me). We have a certain function F(x)=f(x), where "f" depends on the size "x". It turns out that the function is literally faster, the faster the function grows. Do you have ideas, how can this be realized, or how much such an idea is possible in principle?

## A small question about functions

Let's say we have the following function:

f(0)=0+2=2

f(1)=2+1=3

f(2)=3+1=4

f(3)=4+1=5

…

What is the growth rate of this function? In theory, a little more than n+1, but does not reach n+2.

## Linear booga notation analysis up to w^w

{{Infobox function|name=Linear booga notation|image=Booga(420)+1 or biqmal illustration.png|base=ordinal analysis|growthrate=(done)

**Wide-spread booga notation,** the limit is unknown because this notation is still in construction

## My notation (Ultra Factorial Notation) (UFN)

UFN is shown as this: U(n) and may be extended to U(a,b)

Simply, U(a) = a! (Factorial)

If we have U(a,2) then we have a!! not to be confused with the system a * (a-2) * (a-4) and so on. We mean (a!)! however

U(3,2) = 3!! or (3!)!. Since 3! is 6, we have 6! = 720

Therefore, U(3,2) is 720.

U(4,2) = 620 Sextillion ≈ \(f_2(73)\)

U(5,2) ≈ Gargoogol / 15 (6.689 x 10^{198}) ≈ \(f_2(651)\)

U(6,2) ≈ 2.601 x 10^{1749} ≈ \(f_2(5799)\)

U(7,2) ≈ 10^{16573} ≈ \(f_2(f_2(12))\)

U(8,2) ≈ 10^{168186} ≈ \(f_2(f_2(15))\)

so on so on.

If we have U(a,3) then it's equal to a!!! or ((a!)!)!

Even U(3,3) is = U(6,2) ≈ 2.601 x 10^{1749} ≈ \(f_2(5799)\)

We can say that: U(a,b) if a or b > 1:

U(a,b) = U(a!,b-1)

Let's call U(a,a) a¤ (Duriotion)

3% = U(3,3)

Then we have U(a,b,2)

U(a,b,2) = a¤¤¤¤¤.... (b ¤'…

## Just an interesting function

1)F0(a,b)=a+b

2)Fn+1(a,0)=Fn(a,a)

3)Fn(0,b)=Fn(b,b-1)

4Fn(a,b)=Fn(Fn(a-1,b),b-1)

My number-F10(10,10)

## Explaning the fast-growing hierarchy, but simplefied

We have a function where we have f_a (n)

f is a function

a is a level

n is an index.

The lowest level is 0.

\(f_0(n) = n + 1\)

\(f_1(n) = n2\)

## Adding to googol's

Onagol - 10^{100} + 1 Firstly named Addgol but sounded fake

Megol - 10^{100} + 2

Nofgol - 10^{100} + 3

Alagol - 10^{100} + 4

Mougol - 10^{100} + 5

Utagol - 10^{100} + 6

Sygol - 10^{100} + 7

Thifagol - 10^{100} + 8

Myfigol - 10^{100} + 9

Clogol - 10^{100} + 10

Magol - 10^{100} + 11

Hogol - 10^{100} + 12

Wegol - 10^{100} + 13

Pogol - 10^{100} + 14

Dirgol - 10^{100}+ 15

Hagol - 10^{100} + 16

Coigol - 10^{100} + 17

Nevabgol - 10^{100} + 18

Idegol - 10^{100} + 19

## My googol extensions

Cegol - 10^200. I do not remember why it's called Cegol. Called Gargoogol by other people

Degol - 10^300. In my country, Denmark the letter 3 is called 'tre' and evolved it to De

Stugol - 10^400

Falgol - 10^500

Nocgol - 10^600

Seggol - 10^700. Seven (syv in danish) got evolved to 'Seg'

Pomgol - 10^800. Evolved significantly after coining it. Was called 'Stigol'

Rolgol - 10^900.

Thougol - 10^1000. Name is very obvious

## strong 3-shiftedness

[WIP]

Hi everyone! ❀ In this blog-post, I aim to create a 3-shifted ordinal notation reaching the proof theoretic ordinal of \(\Pi_1\)-collection. Usually, 3-shiftedness only reaches \((^{+++})\)-stability, which is much weaker than \(\Pi_1\)-collection. My idea to strengthen 3-shiftedness is to make \(\psi(\alpha)\) regular when \(\text{dom}(\alpha) \ge \alpha^{\dagger}\) where \(\alpha^{\dagger}\) is the least ordinal \(\psi(x)\) with \(\text{dom}(x) = K\) after \(\alpha\), instead of only when \(\textrm{dom}(\alpha) = \psi(x)\) w/ \(\text{dom}(x) = K\). This should strengthen \(\psi(K)\) to be a stable ordinal instead of a \((^{++})\)-stable ordinal, as with usual 3-shiftedness.

Let \(S \subset R \subset T\) be sets of strings. Members of…

## analysis 3 systems

analysis part 1 "ψ(Ι)"

(((#)))=1

(((#)))(((#)))=2

(((#))(((#))))=ω

(((#))(((#))(((#)))))=ω^ω

(((#))((#)))=ψ(Ω)

(((#))((#))((#))=ψ(Ω*2)

(((#)(((#)))))=ψ(Ω*ω)

(((#)(((#)(((#)))))))=ψ(Ω*ψ(Ω*ω))

(((#)(((#)(((#)(((#))))))))=ψ(Ω*ψ(Ω*ψ(Ω*ω)))

(((#)((#))))=ψ(Ω^2)

(((#)((#))((#))))=ψ(Ω^3)

(((#)((#)(((#))))))=ψ(Ω^ω)

(((#)((#)(((#)))(((#))))))=ψ(Ω^ω^2)

(((#)((#)(((#))(((#)))))))=ψ(Ω^ω^ω)

(((#)((#)(((#))(((#))(((#))))))))=ψ(Ω^ω^ω^ω)

(((#)((#)(((#))((#)))))=ψ(Ω^ψ(Ω))

(((#)((#)(((#)(((#))))))=ψ(Ω^ψ(Ω*ω))

(((#)((#)(((#)(((#)(((#))))))))=ψ(Ω^ψ(Ω*ψ(Ω*ω)))

(((#)((#)(((#)((#))))))=ψ(Ω^ψ(Ω^2))

(((#)((#)(((#)((#))((#))))))=ψ(Ω^ψ(Ω^3))

(((#)((#)(((#)((#)(((#))))))))=ψ(Ω^ψ(Ω^ω))

(((#)((#)(((#)((#)(((#)((#)(((#))))))))))=ψ(Ω^ψ(Ω^ψ(Ω^ω)))

(((#)((#)((#)))))=ψ(Ω^Ω)

(((#)((#)((#)((#))))))=ψ…

## The function of investment systems

This idea came to me for a very long time, but only now I was able to implement it normally.

The essence is as follows: we have System n, a unit in a short-((.. (#)) ..)), in which n brackets. The limit of any system is considered ((.. ((## ... ##)) ..)). the number that turns out at the end is the answer

System 0 "ω"

=4

…

System 1 "εω"

(#)=1

(#)(#)=2

(#)(#)(#)=3

(#(#))=ω

(#(#))(#)=ω+1

(#(#))(#(#))=ω*2

(#(#))(#(#))(#(#))=ω*3

(#(#)(#))=ω^2

(#(#)(#)(#))=ω^3

(#(#(#))=ω^ω

(#(#(#)(#)))=ω^ω^2

(#(#(#(#))))=ω^ω^ω

(#(#(#(#(#)))))=ω^ω^ω^ω

(##)=ε0

(##)(#)=ε0+1

(##)(##)=ε0*2

(##)(##)(##)=ε0*3

(##(#))=ω^ε0+1

(##(#)(#))=ω^ε0+2

(##(#(#)))=ω^ε0+ω

(##(##))=ω^ε0*2

(##(##)(##))=ω^ε0*3

(##(##(#)))=ω^ω^ε0+1

(##(##(##(#))))=ω^ω^ω^ε0+1

(###)=ε1

(###(#))=ω^ε1+1

(###(###(#)))=ω^ω^ε1+1

(####)=ε2

(#…

## Pika kid10’s Array Notation long analysis (Pika kid10)

- 1 Pika_kid10’s Array Notation
- 2 Basic Array Notation
- 3 Nested Array Notation
- 4 Linear Array Notation

made by Pika_kid10 with help from Solarzone

Copy/Paste bank for PC:

ω

ε

ζ

η

φ

Γ

ψ

Ω

{0} = 1{0}{0} = 2

{0}{0}{0} = 3

{0}{0}{0}{0} = 4

{0}{0}{0}{0}{0} = 5

{0}{1} = ω

{0}{1}{0} = ω+1

{0}{1}{0}{0} = ω+2

{0}{1}{0}{0}{0} = ω+3

{0}{1}{0}{0}{0}{0} = ω+4

{0}{1}{0}{1} = ω*2

{0}{1}{0}{1}{0} = ω*2+1

{0}{1}{0}{1}{0}{0} = ω*2+2

{0}{1}{0}{1}{0}{0}{0} = ω*2+3

{0}{1}{0}{1}{0}{1} = ω*3

{0}{1}{0}{1}{0}{1}{0} = ω*3+1

{0}{1}{0}{1}{0}{1}{0}{0} = ω*3+2

{0}{1}{0}{1}{0}{1}{0}{1} = ω*4

{0}{1}{0}{1}{0}{1}{0}{1}{0}{1} = ω*5

{0}{1}{1} = ω^2

{0}{1}{1}{0} = ω^2+1

{0}{1}{1}{0}{0} = ω^2+2

{0}{1}{1}{0}{0}{0} = ω^2+3

{0}{1}{1}{0}{1} = ω^2+ω

{0}{1}{1}{0}{1}{0} = ω^2+ω+1

{0}{1}{1}{0}{1}{0}{1} = ω^2+ω*2

{0}{1}{1}{0}{1}{…

## Question Marks Notation discontinued (Polymations)

- 1 (DISCONTINUED UNTIL FULL ANALYSIS IS COMPLETED ON THE GOOGOLOGY WIKI PAGE FOR IT)
- 2 Question Mark Notation (v1)
- 2.1 Variables and Symbols (Under Construction)
- 2.1.1 Legend

- 2.2 Rules of notation
- 2.2.1 1. First Rule (Hyperoperator Rule)
- 2.2.2 2. Hyperoperator Rule 2
- 2.2.3 3. Hyperoperator Rule 3
- 2.2.4 4. BEAF Rule
- 2.2.5 5. Block Arrays Rule
- 2.2.6 6. Nested Commas in Block Arrays Rule
- 2.2.6.1 6.01. Nested Commas in Block Arrays 1b.

- 2.2.7 6.1. Nested Commas in Block Arrays Rule 2
- 2.2.8 7. Separator Rule

- 2.1 Variables and Symbols (Under Construction)

Before reading, if you can edit this page, then don’t. Also you can comment. Also please don’t suggest as I know what I’m doing. Also if you have a youtube channel then you are probably anonymous and should state what your username is here. Also drag the comment bar to …

## Cicada 3301, 2014 puzzle number list draft copy

Source: https://uncovering-cicada.fandom.com/wiki/What_Happened_Part_1_(2014)

- 1 The First Onion
- 1.1 The William Blake Collage
- 1.2 Finding the Private Key (or, Brute Forcing RSA)

- 2 The Second Onion
- 2.1 The Growing String

The hidden service featured , which is a collage of four paintings by William Blake.

Following the usual procedure for investigating images, it was noticed that this image contains a message, again, extractable by outguess. The following is its content, omitting the PGP header and signature:

e = 65537
n = **75579125746085351644267182920580212556413102071876330957950694457000592\**
10248050757270234679993673844203148013173091173786572116639

- -----BEGIN COMPRESSED RSA ENCRYPTED MESSAGE-----
Version: 1.99
Scheme: Crypt::RSA::ES::OAEP

eJwBswBM/zEwADE…

## Aarex's Superstrong Array Notation (Aarex Tiaokhiao)

- 1 Aarex's Superstrong Array Notation
- 2 Definition
- 3 Rules + Process
- 3.1 Comparison
- 3.2 Cases
- 3.3 Variants

- 4 Legacy Process (aSAN-1 to aSAN-3)
- 4.1 Relations (Legacy)

- 5 Code by AxiomaticSystem

Process

Aarex’s Superstrong Array Notation | Formal Definition (aSAN-3) | Aarex’s Googology

Version: aSAN-4 / aSBN-3a (A Step Till Four)

(Updated: 12/5/21 - Entrant Gateways)

[2/6/22: PROCESS HAS BEEN MINIMALIZED!]

WARNING: You are about to see a complex definition of aSAN! If this is too complicated for you, try reading the minimal version.

Arrays

- A = (a1, a2, a3, …, an) is an array with n entries, where all ax terms are called entries of A.
- Besides, an is the nth entry of A.

- Any entry can be either a positive integer or another array.
- If array A is an entry of ss(_), then A is …

## The name of the large number (moroo.com/uzokusou)

Source: https://www.moroo.com/uzokusou/misc/suumei/suumei1.html

I did a little research on the names of large numbers.

First, I had a dictionary of units on hand.