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It appears that the
<math>
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I Don't speak English.
Hello, I will demonstrate a new notation that I created.
Basic:I will give an example:
As we saw in the resolution, this gave us a very large number, but it can get stronger:
You can use all the letters of the alphabet (It follows the same logic)But if you think it's over, you're wrong:
I did some calculations and realized that:Now, imagine if we used this:
That's it, could you tell me the rate of growth?
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The Final Part. Idk if there would be any blog posts after this one, but i am done with this kind of activity, made from creating large numbers. Today we are going to create not an average number. Today, it is time for a transfinite ordinal number. So. First transfinite ordinal is Omega (ω, if you want). It is larger than any natural number and is the first transfinite one. Watch Vsauce's "How to count past infinity" video, if you want to know how numbers after infinite ω are making some sense.
So, if you watched that video at least once, you know that numbers like "ω + 1" or "Gω" have sense. Now we have to use some of the following transfinite stuff:
ε0 = ω^^ω (Epsilon Naught / EpsilonZero)
ε1 = ε0^^ω (Epsilon One)
So, ω^^ω is Epsilon Naught…
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I am reposting, because few people have seen (at least I think, because there was only one person comment), if you want me to delete this, there is no problem, so as not to be just a copy of my other post, I'll put some other information on this.
Note: In the penultimate example, I used any symbol to represent an ordinal which is an infinite "tower" of the last example, in this case an infinite tower of Gamma 0.
To demonstrate how powerful this is, if we use 2f!, we have a number greater than the G64!
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I summarise historical background of BMS (Bashicu Matrix System) for readers who want to study it. I appreciate any corrections with precise sources.
BMS is a system to generate a large number, which was created by Bashicu to define several large numbers, e.g. Bashicu matrix number. The main article of BMS in googology wiki is here.
BMS has many variants created by Bashicu and other googologists, and they are also called BMS. Mainly, googologists focus on the second version of BMS called BM2.
The original "definitions" of Bashicu matrix numbers are written by Bashicu here. He states that they are written in BASIC language, which is one of a progamming laguage, but they contain many compileerrors. Namely, they are never welldefined.
Bashicu a…
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I am back with some numbers. So, for first we need to introduce a function:
A[n] = 9(n of ^s here)9. So, for example, A[9] = 9^^^^^^^^^9. Also, we need to note that in this system, x^^y is tetration (x^x^... (y times)), x^^^y is pentation (x^^x^^... (y times)) and so on. So, A[9] = 9^^^^^^^^^9. This is... A big number. But we need to go further. A1[n] = A[n](n ^s here)A[n]. Example: A1[9] = A[9]^^^^^^^^^A[9].
So, global notation is: Ax[n] = A[x  1](n ^s here)A[x  1].
Then, we will use this notation to increase the ammount of A's.
AA[n] = A[A[n]  1](n ^s here)A[A[n]  1]
AA[n] = U2[n]. AAA[n] = U3[n]. etc.
UU2[n] = A...(U2 A's in total)...A[n]
Then, the final notation.
Гx[n] = U... (x U's in total) ....U[n]
So, my new number is Г(10^^^100)[9]. I …
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As people are having trouble reading the source code of Bashicu Matrix calculator, which is the definition of BM itself, I wrote a document to guide for the reading the source code.
https://kyodaisuu.github.io/basmat/definition.html
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So i know all ordinals from \(\omega\) to \(\varphi(\omega, 0)\), but I don't know anything beyond that. Can anybody give a list of ordinals and their definitions beyond \(\varphi(\omega, 0)\)? For example, what is ϑ()?
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Here is a new notation I am thinking about. I've already reached w^w in FGH, which was my initial goal; now my goal is to hit epsilonzero. When I saw PlantStar's Notation and how easy it was to create your own notation, I got the inspiration to create my own. However, the first time I tried, I utterly and completely fucking failed. I extended upon Graham's function, which has growth w+1, but I did a bunch of complicated stuff and only got to w+9 :( For my second attempt, I will try to make my own version of BEAF, based off of the factorial function. Right now, I have not developed very much and my notation is not very strong yet, but it will be! I promise.
Here is the basics:
 {a} = a!!!...!!! with a factorial signs. Pretty basic for 1entry…

Thanks to Ecl1psed276 for the sequence reduction method and for helping me test and fix the rules! (and also for cleaning up the rules)
Copied from the googology Discord:
Let x be the number in the square brackets. If the sequence is empty, output x. End. If the first number of the last element is 0, delete the element and replace x with x^2. End. Otherwise:
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Sequence reduction definition, for row k: 1  Ignore all rows other than row k. 2  Let the rightmost element be x. 3  Move to the rightmost nondiscarded element to your left. If you are at the leftmost element, stop this subprocess. 4  Let the current element be y. 5.1  If y < x, set x to be y. 5.2  If y >= x, label the current element as "discarded". 6  Go back to step 3.
Le…

Does anybody here knows the expansion rules for BM2?
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On hiatus. For now, check this instead.
NOTE: This proof uses this definition for BM2. There is a very good chance that it's right, and in fact ecl1psed276 has tested the rules even on quad sequences and hasn't found any sequences that give different results than the BMS calculator, but it's not proven yet, so we cannot be 100% sure.
First, let's see how (0,0)(1,1) evaluates. As there are already only two nondiscarded elements, the bad root is (0,0) and the bad part is (0,0). Delta is then (10,0) = (1,0), C_{0} is (1,1), S_{0} is (0,0), and each S_{n} adds 1 to the first term and keeps the second term the same (it adds 0 to it), giving us S_{n} = (n,0). The resulting sequence is therefore (0,0)(1,0)(2,0)(3,0)...(x^{2},0).
Collorary 1.1: If the last term of ea…
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I've really only made functions/notations that have a growth rate of only \(\omega ^ {3} + 1\), so I really, really hope that this goes higher than that.
RandomFunction(n) = The largest number that any F(n) defined within the rules stated below can return. Rules: You have to define the largest F(x) such that it can only use the lvl function (defined on my first blog post), recursion, you could also define other functions and use them in F(x), you can use inverses to lvl(x, y, z) like subtraction, logarithms, roots, etc., you can only have one parameter of F(x), and only whole numbers are allowed. For example, define F(x) = x. This would be 3 symbols because of the F(), the =, and the x. But why not the x inside of F? Because, when you use …
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Since we've lost the Croutonillion, I feel we should have a Kroutonillion.
Start at \(X = \{\underbrace{9, 9, \ldots, 9}_{999999999999999999999999999999999999999999}\}\)
1. \(f_{\varphi^{X}(X, X, X, X, X)}(X)\)
Write down in the comments the next step (for example, 2. 2X, 3. {X, X, X, X, X})
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As I implemented ordinal conversion of pair sequence to BM calculator, next step is to go further. To implement it we have to find a rule to convert BM to some known ordinal expression.
As it was easy to convert pair sequence to Buchholz psi function, I want to know if the same approach can be applied to trio sequence to some extent. Although (0,0,0)(1,1,1) is normally written as ψ(Ω_ω), it would be easier to use the notation of ψ0(ψω(0)) = p0(pw(0)) to find a rule to convert. Once we can find a rule, we can implement it in the BM calculator.
 (0,0,0)(1,1,1) = p0(pw(0))
 (0,0,0)(1,1,1)(0,0,0) = p0(pw(0))+1
 (0,0,0)(1,1,1)(0,0,0)(1,1,1) = p0(pw(0)) + p0(pw(0))
 (0,0,0)(1,1,1)(1,0,0) = p0(pw(0))w = p0(pw(0)+p0(0))
 (0,0,0)(1,1,1)(1,0,0)(2,1,1) = w^w^(…

The (n+1)separator in DAN cannot be embedded into the nth system of Taranovsky's ordinal notation. In Taranovsky's ordinal notation, consider this kind of expressions: let \(f(x)=C(C(C(C(x,\beta_t),\beta_{t1})\cdots,\beta_2),\beta_1)\), and \(\alpha>f(\alpha)\), then \(f(f(\cdots f(f(\alpha))))\) (with n+1 f's) is not standard in the nth system.
Suppose that the DAN rules (especially the case of the (n+1)ple comma) apply on Cexpressions, with \(\alpha\) to be an (n+1)separator term and \(f(\alpha)\) to be an nseparator term. Consider the least and simplest situation: if we need adding rule from nseparator and below, we will get \(f^m(\alpha)\) as an (nm+1)separator term, and finally \(f^{n+1}(\alpha)\) as a "0separator" term. Fo…
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This blog post is basically a compilation of all of the BM2 analysis that the community has done. It will not be very detailed, I will leave that to other people. I'm mainly intending this to be a sort of reference point for people if they want to know how big a BMS expression is. (All these comparisons will be done using UNOCF for its simplicity.)
BM2 Expression Ordinal
empty matrix \(0\)
(0) \(1\)
(0)(0) \(2\)
(0)(0)(0) \(3\)
(0)(1) \(\omega\)
(0)(1)(0) \(\omega+1\)
(0)(1)(0)(1) \(\omega2\)
(0)(1)(0)(1)(0)(1) \(\omega3\)
(0)(1)(1) \(\omega^2\)
(0)(1)(1)(0) \(\omega^2+1\)
(0)(1)(1)(0)(1) \(\omega^2+w\)
(0)(1)(1)(0)(1)(1) \(\omega^22\)
(0)(1)(1)(1) \(\omega^3\)
(0)(1)(2) \(\omega^\omega\)
(0)(1)(2)(0) \(\omega^\omega+1\)
(0)(1)(2)(0)(1) \(\omega^\omega+\om…
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I do not speak English.
Hello, I wanted to thank everyone who helped me in my last post and thanks to your help I created a factorial with ordinal!I used one photo, because I do not know how to put symbols like omega.
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the penultimate example is an ordinal representing an infinite "tower" of "Г" and you can continue to infinitely increase your amount of factorials. 
As I implemented a function to determine if Bashicu Matrix is standard in BM calculator, now I am trying to implement ordinal conversion.
Koteitan made a Pair sequence to Buchholz's Ψ function in four lines Javascript, and I read the code and translated. Actually I think koteitan's code only works for the standard form of pair sequence. In the BM calculator, standard form is checked and therefore I can just exclude nonstandard form.
The code converts the BM to omega and epsilon expression when ordinal is below zeta_0. It would be convenient because people are used to this kind of ordinal expression. Right now I just wrote a test code and paste the result here to verify before actually implementing it.
Update: I made some ordinal calculation …
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This is an OCF I made that has four entries. It has the property Θ(ω^{x},0,0,0) = ψ(Ω^{Ωx}).
Define:
 cof(0) = 0
 cof(ω) = ω
 cof(Ω) = Ω
 cof(S(x)) = 0
 cof(x+y) = cof(y)
 maxcof(S) = max{xx=cof(y)∧y∈S}
 maxcof({a,b,c,d}) = 0: cof(Θ(a,b,c,d)) = ω
 maxcof({a,b,c,d}) > 0, maxcof({c,d}) < Ω: cof(Θ(a,b,c,d)) = ω
 maxcof({a,b,c,d}) > 0, maxcof({c,d}) = Ω: cof(Θ(a,b,c,d)) = Ω
 ω[n] = n
 Ω[n] = n
 x+0 = x
 x+S(y) = S(x+y)
 cof(y) > 0: (x+y)[n] = x+(y[n])
 Θ(0,0,0,0) = ω
 Θ(0,0,0,x+1)[0] = 0
 Θ(0,0,0,x+1)[n+1] = Θ(0,0,0,x+1)[n]+Θ(0,0,0,x)
 cof(x) > 0: Θ(0,0,0,x)[n] = Θ(0,0,0,x[n])
 f = Θ(a,b,c): f(x) = Θ(a,b,c,x)
 ∃x∀y>x(f(y) > g(y)): f < g
 lim(F)(0)[0] = 0
 lim(F)(x+1)[0] = lim(F)(x)+1
 f ∈ F, ¬∃g∈F(g 0: enum(F,x) = min{ff∈F∧∀yenum(F,y))}
 cof(x) = 0, cof(F) = 0, f = max{gg∈F}: lim(F)(x)[n+1] = g…

I don't speak English.
Hello, I'm here to talk about something that could be added on the site.
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Wikia Googology could add a new area in the community section, where would be a place that would create articles of our numbers, where we could use our notations and notations of other people. Obviously, this would not interfere with official numbers, or search, as they would be the last of the search suggestions, and if any number is large enough, an official page could be added for that number.
What do you think? This would not interfere with anyone (because it would be in another "part"), and it would be cool for others to share their idea and perhaps win their place in an official article (as well as helping us reach the goal of 1 googol articl…

\(A(n) = n + 1\) which growth rate is \(f_0(n)\)
\(A(a, b) = \underbrace{A(A(\ldots A(b) \ldots))}_{a}\) which growth rate is \(f_{0}^{a}(b)\) or \(f_{1}(b)\)
\(A(a, b, c) = \underbrace{A(A(A(\ldots A(a, b)}_c \ldots, \ldots A(a, b) \ldots), A(\ldots A(a, b) \ldots, \ldots A(a, b) \ldots)), A(A(\ldots A(a, b) \ldots, \ldots A(a, b) \ldots), A(\ldots A(a, b) \ldots, \ldots A(a, b) \ldots)))\) Not really sure about the growth rate.
\(A(a, b, B, c) = \underbrace{A(A(A(\ldots A(a, b, B) \ldots)}_c, A(\ldots A(a, b, B) \ldots), B, A(\ldots A(a, b, B) \ldots)), A(A(\ldots A(a, b, B) \ldots), A(\ldots A(a, b, B) \ldots), B, A(\ldots A(a, b, B) \ldots)), B, A(A(\ldots A(a, b, B) \ldots), A(\ldots A(a, b, B) \ldots), B, A(\ldots A(a, b, B) \ldots)))\) …
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I construct a completely welldefined recursive ordinal notation, which presents a "prooftheoretic analogue" of the PTO of \(\textrm{ZFC}\) together with a canonical choice of a system of fundamental sequences, and define a computable large number.
I expect that it presents very large recursive ordinals, but have not been satisfied yet. I update it in order to make it stronger and to reduce the saladability.
Throughout this blog post, I work in \(\textrm{ZFC}\) set theory. I note that a similar argument is valid for any effectively axiomised theory containing arithmetic.
I will define a map \(\textrm{RWO}\) sending an \((n,d) \in \mathbb{N} \times \mathbb{N}\) to (the Goedel number of) a formula \(\textrm{RWO}(n,d)\) in an internal set theor…
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(I Don't speak English)
In my last post, I demonstrated a very powerful factorial, but weaker than HAN and said that I would try to create a more powerful factorial than HAN and I believe I did!
Let's start by demonstrating how it works (Your basic idea):
[m; n] ^! = n ^ ... ^ n
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with n ^ ... ^ n arrows
with n ^ ... ^ n arrows
.
. (repeated by m times)
.
with n ^ ... ^ n arrows
with n ^ ... ^ n arrows
.
. (repeats by (m1) times)
.
...
with n ^ ... ^ n arrows (repeats 1 time)
with (n1) ^ ... ^ n arrows
with (n1) ^ ... ^ n arrow…

(I Don't speak English)
I was seeing the factorials and decided to create a more powerful factorial of all ... at least of which I know.
From what I saw on the site, the most powerful factorial is the superfactorial (correct me if I'm wrong) which is defined by:
(n!)
n$ = (n!)After thinking a little, I created the Arrow Factorial and it is defined like this:
n^! = n ^ ... ^ n
with (n1) ^ ... ^ (n1) arrows
with (n2) ^ ... ^ (n2) arrows
...
with 2 ^ ... ^ 2 arrows
with 1 arrowExample:
3^! = 3 ^ ... ^ 3
with 2 ^ ... ^ 2 arrows
with 1 arrow,that is,
3 ^ ... ^ 3 = 3^^^^3 > 3$
with 2 ^ 2 arrowsTo have a …
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\(F(0) = \omega\)
\(F(n) = lvl(F(n  1) + 1, F(n  1) + 1, F(n  1) + 1)\) where lvl is defined in my first blog post.
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Take any function F from ordinals to ordinals. Then, we define:
 ψ_{0}^{F}(0)[0] = 0
 ψ_{β+1}^{F}(0) = min{x"x is regular"∧∀y ψ_{β+1}^{F}(0): ψ_{β}^{F}(α) = ψ_{β}^{F}(ψ_{β+1}^{F}(α))
The cofinalities of some ordinals are not taken into account and unspecified, but these are either (1) supposed to already be known by the reader or (2) able to be deduced from the ruleset. (NOTE: There are no extended FSes, like ω^{2}[ω+1] or Ω[Ω_{2}] (although something like Ω_{2}[Ω] is fine, as the FS of an ordinal has the same length as its cofinality))
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Sorry for grammatical errors. I do not speak English.
Well, today I will show a very simple and easy to understand but very strong notation (it may not compare with the larger ones, but it is very strong).
It uses as base the notation of knuth, the notation with arrows.
How it works:[m; n] x = n ^^ ... ^^ n
with n ^^ ... ^^ n of arrows
with n ^^ ... ^^ n of arrows
... (It is repeated for "m" times)
with n ^^ ... ^^ n of arrows
with "x" arrowsExample: [2; 3] 4 = 3 ^^ ... ^^ 3
with 3 ^^^^ 3 arrowsAs you can see this was repeated "m" times (2) and in the end there are "x" arrows (4).
We can still do this:
(I will use abc to get easier)[a; b; c] x = [a; b]; …
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First, I prove that (0,0,0)(1,1,1)(2,1,1)(1,1,1) is standard...
 (0,0,0,0)(1,1,1,1)
 = (0,0,0)(1,1,1)(2,2,2)(3,3,3)(4,4,4)...
 > (0,0,0)(1,1,1)(2,2,2)
 = (0,0,0)(1,1,1)(2,2,1)(3,3,1)(4,4,1)...
 > (0,0,0)(1,1,1)(2,2,1)
 = (0,0,0)(1,1,1)(2,2,0)(3,3,1)(4,4,0)(5,5,1)...
 > (0,0,0)(1,1,1)(2,2,0)
 = (0,0,0)(1,1,1)(2,1,1)(3,1,1)(4,1,1)...
 > (0,0,0)(1,1,1)(2,1,1)(3,1,1)
 = (0,0,0)(1,1,1)(2,1,1)(3,1,0)(4,2,1)(5,1,1)(6,2,0)(7,3,1)(8,1,1)...
 > (0,0,0)(1,1,1)(2,1,1)(3,1,0)
 = (0,0,0)(1,1,1)(2,1,1)(3,0,0)(4,1,1)(5,1,1)(6,0,0)(7,1,1)(8,1,1)...
 > (0,0,0)(1,1,1)(2,1,1)(3,0,0)
 = (0,0,0)(1,1,1)(2,1,1)(2,1,1)(2,1,1)...
 > (0,0,0)(1,1,1)(2,1,1)(2,1,1)
 = (0,0,0)(1,1,1)(2,1,1)(2,1,0)(3,2,1)(4,1,1)(4,2,0)(5,3,1)(6,1,1)...
 > (0,0,0)(1,1,1)(2,1,1)(2,1,0)
 = (0,0,0)(1,1,1)(2,1,1)(2,0,0)(3,1,1)(…

Uma duvida: se eu criar um número maior do que um incomputável, o meu número se torna automaticamente incomputável?
É possível criar um número computável maior do que um incomputável? Porque?
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(Sorry for grammar mistakes, I do not speak English) Hi! I created a new notation and wanted to share with you:
It works as follows:(n)
X = X ^ X
n;Where X is the basis and "n" defines the number of arrows, for example:
2 = 2 ^^^ 2
3;However, we can make it even more interesting if we add another "n":
(n²)
(n¹) (X ^ X) (2 ^^^^ 2)
X = ((X ^ X) ^ X) Example: 2 = ((2^^^2) ^ 2)
n¹; n²; …
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Hey all, I'd like to give an explicit example of a standard BM2 expression that never terminates. This expression is (0,0,0)(1,1,1)(2,2,1)(3,1,0)(2,2,1).
First, let's show how (0,0,0)(1,1,1)(2,2,1)(3,1,0)(2,2,1) is in fact standard:
We start from (0,0,0,0)(1,1,1,1). Then, we have:
(0,0,0,0)(1,1,1,1)
(0,0,0)(1,1,1)(2,2,2)
(0,0,0)(1,1,1)(2,2,1)(3,3,1)
(0,0,0)(1,1,1)(2,2,1)(3,3,0)
(0,0,0)(1,1,1)(2,2,1)(3,2,0)
(0,0,0)(1,1,1)(2,2,1)(3,1,1)
(0,0,0)(1,1,1)(2,2,1)(3,1,0)(4,2,1)
(0,0,0)(1,1,1)(2,2,1)(3,1,0)(4,2,0)
(0,0,0)(1,1,1)(2,2,1)(3,1,0)(4,1,0)
(0,0,0)(1,1,1)(2,2,1)(3,1,0)(4,0,0)
(0,0,0)(1,1,1)(2,2,1)(3,1,0)(3,0,0)
(0,0,0)(1,1,1)(2,2,1)(3,1,0)(2,2,1)So (0,0,0)(1,1,1)(2,2,1)(3,1,0)(2,2,1) is standard. Now, I will show how this expression never terminates.…
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Hi again ! Today I will make a little walkthrough of Chained Array Notation. I'm doing this because over the past weeks there have been a slew of changes to the notation all over the place, and it's just been a mess in general.
Expressions in CAN are of the form n[array]. n is called the base, and it's a number that is used nearly everywhere, so it's very important. The most basic expression in CAN is a[0]. This is equal to \(a^a\).
n n[0] 11 24 327 4256 53,125 646,656 7823,543 816,777,216 9387,420,489 1010,000,000,000This is equal to \(g_{\omega^\omega}\) in the SGH and very roughly corresponds to \(f_2\) in the FGH. Then, n[1] is n[0] repeated the base times. \(n[1] = (\cdots((n\underbrace{[0])[0])\cdots[0])}_\textrm{the base}\). 3[1] alo…
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Hey there, I'm back again. This isn't actually the second revision at all, but i wanted to sum up all of my changes in a new blog.
If you thought CAN looked a lot like HAN, well you are going to be even more correct, since i'm changing rule 4 :
4) \(a[0,\ldots,0,[\cdots[c+1]\cdots],\#] = a[0,\ldots,[0,\ldots,0,[\cdots[0]\cdots],\#],[\cdots[c]\cdots],\#]\)
This subtle change makes NCAN converge towards \(\varphi(\omega,0)\) much faster.
There is only one thing changed with HDCAN : when you nest an array, you look outwards for the innermost set of square brackets.
Meaning that \([0\{0,1\}1] = [0\{[0\{0,0\}1],0\},1]\)
Okay, a few changes ahead here :
First, separator arrays don't work the same anymore :
 They use curly brackets
 \(\{n@\} = \underbrace{@…
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I introduce a very easy technique to define an ordinal notation associated to a proper class of ordinal numbers. This blog post is just for beginners of googology who knows usual mathematics, and hence deals with a countable ordinals smaller than SVO.
Throughout this blog post, I work in ZFC set theory. Although the notion of a class is not a term in ZFC set theory, I use the traditional covention on a class in ZFC set theory.
I denote by \(\textrm{ON}\) the proper class of ordinal numbers. Let \(X \subset \textrm{ON}\) be a proper class. I will construct an ordinal notation associated to \(X\) and a class \(F(X) \subset X\).
For a class \(Y \subset \textrm{ON}\), I denote by \(\textrm{Lim}(Y) \subset Y\) the class of limit points of \(Y\), a…
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If you're interested or want to add pages to this wiki.
Module III: Lightspeed Slash NotationModule IV: Lightspeed Slash Notation  Googolisms
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This allows more than one chained arrow between each argument.
Same as chained arrow notation, but it allows multiple arrows between each argument. Here are a few examples:
 \(a \rightarrow\rightarrow b = \underbrace{a \rightarrow a \rightarrow a \ldots a \rightarrow a \rightarrow a}_b\)
 \(a \rightarrow\rightarrow b \rightarrow c = \underbrace{a \rightarrow a \rightarrow a \ldots a \rightarrow a \rightarrow a}_b \rightarrow c\)
 \(a \rightarrow\rightarrow b \rightarrow\rightarrow c =
\underbrace{a \rightarrow a \rightarrow a \ldots a \rightarrow a
\rightarrow a}_b \rightarrow\rightarrow c =\)
\( \underbrace{\underbrace{a \rightarrow a \rightarrow a \ldots a \rightarrow a \rightarrow a}_b \rightarrow \underbrace{a \rightarrow a \rightarrow a \ldot…
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Note: This tale is not completely original. The original story is entitled "Jugemu," a classical Japanese joke (rakugo). I just made it go wrong with googology.
Once upon a time, there was an old couple who have just got a son. It's the seventh day of the son, so they have to give him a name. Since they can't come up with a good name, the grandpa went to his friend to get an advice for the name. The friend agreed to give him some idea.
"OK. Now, what kind of boy do you want him to grow up to?"
"I just want him to live a long life."
"A 'long' life?"
Unfortunately, the friend was a googologist!
"Now, let's start with a googol. It is one followed by 100 zeroes."
"Googol Smith. This is a good name. I'll keep it."
"How about googolplex? It's one follow…
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Should we delete all numbers by Sbiis Saibian's numbers? Because in the "Explore" section, when I click on "Random Page", then it always takes me to an article about a number by Sbiis Saibian. Also, the numbers are just salad numbers so there isn't really much purpose to it.
I say we have a vote about deleting numbers by Sbiis Saibian.
I vote yes, and I'll update this vote as people comment.
Yes: 0
No: 5
OTHER OPTIONS:
Deedlit's first comment explains this: 1
Rpakr's first reply to Deedlit's first comment explains this option: 3
Kamafadelagato's first comment: 1
nnn6nnn's first comment: 1
MachineGunSuper's first comment: 1
If you have a vote or any comments, then please leave a comment below. Also, as I am not an admin, if "Yes" is the majority, then…
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This is a fast growing function inspiered by a recent numberphile video . Define \( \Pi(x) \) to be the minemum number \( N \) for which it is impossilbe to split all the numbers upto \( N \) in \( x \) groups without a group containing a pythagorian triplet.
\( \Pi(1)=5 \) because {1,2,3,4,5} contains the triplet \( 3²+4²=5² \)
The proof for \( \Pi(2)=7825 \) was outlined in the video.
It is unknown how fast this function grows. It's not even know if \( \Pi(3) \) is welldefined.
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Hi. I would like to know what the greatest WELLDEFINED ordinal notation used in googology up to now is.
There are many great ordinal notations, but some of them are not completely defined. (For them, we just have plans, directions, or dreams to define ones satisfying good desired excellent properties.)
So among ordinal notations which have already been completely defined, what is the greatest one? Is it Taranovsky's C?
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Hello, I do not speak English, and unfortunately there is no wiki for my language, so I will write in my language (I believe you will not have problems in understanding it, as it can be translated with google translator)
Para entender melhor, veja meu último post: https://googology.wikia.com/wiki/User_blog:Nedherman1/HDXT
Resolvi melhorar o meu fatorial e minha notação, por isso vou explicar como funciona:
YNotação: X + # + Z
Essa notação funciona da seguinte forma:
X será o número base que será elevado a ele mesmo Y! vezes.
Z será o número que elevará X, que é elevado a X Y! vezes, Y! vezes.
Y será o número que indicará quantas vezes X e Z serão elevados a eles mesmos Y! vezes e no final da conta Você ainda elevará o resu…
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Notation name: V.E.D.D.M.D.D.T.D.T.D.D.T.D.D.D.I.T.D.W.T.D.D.L. (Very Extended Dumb Dumbo Made Dumb Dumplings That Demanded The Dumb Doctors To Dump Dumb Dumplings In The Dumps Where Dumb Dumbo Lives) I made the name up
For simplicity's sake, shorten DDMDDTDTDDTDDDITDWTDDL to D\(D(a) = lvl(a, a, a)\) where the lvl function is defined at https://googology.wikia.com/wiki/User_blog:MilkyWay90/My_Level_Function so D(n) has a growth rate for \(f_{\omega}(n)\)
Note: # means the rest of the parameters
\(D(a, b, #, c, 0) = \underbrace{D(D(D(\ldots D(D(a, b, #, c), b, #, c) \ldots), b, #, c)}_{c},\underbrace{D(a, D(a, \ldots D(a, D(a, b, #, c), #, c)\)\ldots ,#, c), #, c)}_c1, #, \underbrace{D(a, b, #, D(\ldots D(a, b, #, D(a, b, #, c)) \ldots\)))}_{…
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In this page, I am going to analyze the fusible margin function, which is the difference between a number and the next fusible number after it. It is defined here .
The Main conjecture is listed in this paper. If it ever gets disproved, I am going to make a new section. I am going to display the negative base2 log in the array notation and FGH columms. The FGH column is empty until the number is no longer notated in decimal form.
Number Ordinal Margin Log (Array notation) FGH
0 1 0
1/2 2 1
3/4 3 2
7/8 4 3
15/16 5 4
31/32 6 5
63/64 7 6
\(12^n\) n+1 n
1 \(\omega\) 3
9/8 \(\omega+1\) 4
19/16 \(\omega+2\) 5
39/32 \(\omega+3\) 6
79/64 \(\omega+4\) 7
5/4 \(\omega 2\) 4
21/16 \(\omega 2 + 1\) 5
43/32 \(\omega 2 + 2\) 6
87/64 \(\omega 2 + 3\) 7
11/8 \(\omega 3\) 5
4…
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Thanks to Nish/Alemagno12 for some of the suggestions, especially for dimensional and above ^.^ Okay, this is my second attempt at Chained Array Notation, and this one should be much better.
So we're starting at LCAN, i.e these rules :
1) \(a[0] = a^a\)
2) \(a[\#,0] = a[\#]\)
3) \(a[b+1,\#] = a\underbrace{[b,\#]\cdots[b,\#]}_{a}\)
4) \(a[0,\ldots,0,c+1,\#] = a[a,\ldots,a,c,\#]\)
This reaches \(\omega^\omega\).
But then we're going to make changes regarding NCAN (Nested CAN) :
First off, let's change rule 4 :
4) \(a[0,\ldots,0,[\cdots[c+1]\cdots],\#] = a[0,\ldots,[0,\ldots,0,0,\#],[\cdots[c]\cdots],\#]\)
This acts like HAN, and this makes it much, MUCH more powerful.
But we first need to add our process :
If none of the rules apply, begin this process,…
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I know \(\zeta _ {0} = \varepsilon _ {\zeta _ {0}}\), but what about \(\zeta _ {1}\) or \(\zeta _ {2}\)?
Any help would be appreciated.
If you have any comments or/and the answer to my question please leave a comment!
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How do you use LaTeX in here?
\(\omega\) doesn't work
\(10\) doesn't work
\10 ^ {100} doesn't work.'
EDIT: realized it does work (most of it). No help required?
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How exactly do you use LaTeX? 
Hi, this is my new attempt at making a powerful notation, and this time I think I got it.
Stage 1 : Barebones (BBCAN) :
This has only one rule.
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This function is called the “extension function”. Within uncomputable numbers, users extend the theory to utterly explode the uncomputable number I.e Rayos FOST function extended to FOOT function to make BIG FOOT. All of these numbers use no more than one extension on the previous largest uncomputable number to make the new number much larger. But what about if more extensions were used on a particular uncomputable number?
Ext(0) = Rayos number or any other number that hasn’t been extended to make the number utterly larger
Ext(1) = BIG FOOT,Little Biggedon/Sasquatch and even utter oblivion as they had all one extension to make the number utterly larger I.e Oblivion was extended to make utter oblivion.
Ext(2) = The largest finite number with 2…
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This function is called the “extension function”. Within uncomputable numbers, users extend the theory to utterly explode the uncomputable number I.e Rayos FOST function extended to FOOT function to make BIG FOOT. All of these numbers use no more than one extension on the previous largest uncomputable number to make the new number much larger. But what about if more extensions were used on a particular uncomputable number?
Ext(0) = Rayos number or any other number that hasn’t been extended to make the number utterly larger Ext(1) = BIG FOOT,Little Biggedon/Sasquatch and even utter oblivion as they had all one extension to make the number utterly larger. Ext(2) = The largest finite number with 2 extensions. Ext(3)
Ext(Ext(2) = Ext(2) extensio…
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