
0
unfinished and just an idea, nothing formal
define the function f for all ordinal inputs. \(f_\alpha(\beta) \) is defined where a,b are ordinals and a < (something), b< (something else)
\(f_0(\beta) = \beta+1 \)
\(f_{\alpha+1}(\beta) = f^\beta_\alpha(\beta) \) for all a, and superscript indicates functional recursion.
functional recursion for transfinite ordinals: let f be any function, then \( f^\alpha(\beta) = \sup(\{\gamma: \gamma = f^n(\beta)\}_{n \in \mathbb N}) \)
now f denotes transfinite FGH
\( f_{\alpha}(n) = f_{\alpha[n]}(n) \) if n in N
\( f_{\alpha}(\beta) = ??? \) if b not in N
examples:
\(f_0(\beta) = \beta+1 \)
\(f_1(\beta) = \beta\cdot 2\) proof: trivial
\(f_2(\beta) = \beta \cdot 2^\beta \)
\(f_\omega(\omega) = ???\) I don't know yet …
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I made "notationnesting base system (NNBS)". This is similar to shiftedness.
There are no definition(nondefined, undefined). I can't make them :(
0 0
ψ_0(0) 1
ψ_0(ψ_0(0)) ω
ψ_0(ψ_0(ψ_0(0),0)) ε_0
ψ_0(ψ_0(ψ_0(0),0)+ψ_0(ψ_0(0),0)) ψ(Ω×2) From here, I use EBOCF
ψ_0(ψ_0(ψ_0(0),ψ_0(0))) ψ(Ω×ω)
ψ_0(ψ_0(ψ_0(0),ψ_0(ψ_0(ψ_0(0),0)))) ψ(Ω×ψ(Ω))
ψ_0(ψ_0(ψ_0(0),ψ_0(ψ_0(0),0))) ψ(Ω^2)
ψ_0(ψ_0(ψ_0(0)+ψ_0(0),0)) ψ(Ω_2)
ψ_0(ψ_0(ψ_0(ψ_0(0)),0)) ψ(Ω_ω)
ψ_0(ψ_0(ψ_0(ψ_0(ψ_0(0),0)),0)) ψ(Ω_ψ(Ω))
ψ_0(ψ_0(ψ_0(ψ_0(0),0),0)) ψ(Ω_Ω)
ψ_0(ψ_0(ψ_0(0),0,0)) ψ(Ι) From here, I use UNOCF
ψ_0(ψ_0(ψ_0(0),0,ψ_0(0))) ψ(Ι+ψ_Ι(Ι)×ω)
ψ_0(ψ_0(ψ_0(0),0,ψ_0(ψ_0(0),0))) ψ(Ι+ψ_Ι(Ι)×Ω)
ψ_0(ψ_0(ψ_0(0),0,ψ_0(ψ_0(0),0,0))) ψ(Ι+ψ_Ι(Ι)^2)
ψ_0(ψ_0(ψ_0(0),0,ψ_0(ψ_0(0),0,ψ_0(0)))) ψ(Ι+ψ_Ι(Ι)^ω)
ψ_0(ψ_0(ψ_0(0),…
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I've recently created a number based on recursion in 300 characters in ruby. Here's the code:
# This character># and everything after it doesn't count as an character. It's a comment def a(m,n) if m == 1 return n**n # same as n^n elsif n == 1 return m**m else return a(m1,a(m,n1)) # You'll get an error here while running code telling you that a(m,n) is called way way too many times end end
def e(n) return a(n,n) end
def d(n) r = e(n) (e(n**n)).times do r = e(r) end return r end x = 999999999999999 (d(x)).times do x = d(x) end
p d(d(d(d(d(d(d(d(d(d(d(d(d(d(x)))))))))))))) # here the number is printed out
There are three functions in the code:
 a is the major function. Fed …
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�H!DeH A`
H#I.!M �@+IlH!A �� Ic�H#E.#An@ Do�@!i
#Ei�H*E �@ d�@ InH!o�@)La�@)Ii
H"i
#Ei�@!LdH @a
H!� ��!He� I �@(D
�@ Ao�"Ac
���Ee@(L ��(Ea�H*https://cryptii.com/pipes/baconcipher bitwise operator
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Before I continue with the next structures in my chevron notation (which I previously posted but which received no comments, sadly), I decided to look for something strong for the base function, but I did not want to simply adopt uparrows or Moser polygons or something else that someone already did. Here is my hierarchy starting from the successor function. I plan to use the base function ‹0›(x) to diagonalize this notation. I'm interested to hear what experts think of this extension in terms of growth and originality. Thank you.
f_{0}(x) = x+1For all functions F, F^{n}(x) = F(F^{n1}(x)) and F^{1}(x) = F(x)
Natural numbers separated by commas are strings of natural numbers. Commas are used in no other way. They do not indicate decimal places or group…
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From, Goodstein Sequence (https://googology.wikia.org/wiki/Goodstein_sequence),
I create new representation call "Extended Goodstein's baseb hereditary representation" or "xG's baseb hereditary representation".
xB[b,b_](n) means finding the xG's baseb hereditary representation of n and bumping the base from b to b_
xG Sequence isn't much different from normal Goodstein Sequence.
\begin{eqnarray*} xG_0(n) &=& n \\ xG_1(n) &=& xB[2,xG_0(n)](xG_0(n))  1 \\ xG_2(n) &=& xB[xG_0(n),xG_1(n)](xG_1(n))  1 \\ xG_3(n) &=& xB[xG_1(n),xG_2(n)](xG_2(n))  1 \\ &\vdots& \\ xG_{k+1}(n) &=& xB[xG_{k1}(n),xG_k(n)](xG_k(n))  1 \\ &\vdots& \\ \end{eqnarray*}
Until, it return to zero.
This is how xG's baseb hereditary representation work.
In case b = 2, if n …
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In my last post, I defined a function \(\text{ML}(n)\) and proved that \(\text{Rayo}(12145) > \text{ML}(65536)\). In this post, I will explore the function \(\text{ML}(n)\) itself.
Recall that I defined \(\text{Level}(\phi)\) for a formula \(\phi\) as follows:
1. \(\text{Level}(x \in y) = 0\)
2. \(\text{Level}(\neg \phi) = \text{Level}(\phi) + 1\)
3. \(\text{Level}( \exists x (\phi) ) = \text{Level}(\phi) + 1\)
4. If \(\text{Level}(\phi) = \text{Level}(\psi) = n\), then \(\text{Level}(\phi \wedge \psi) = n + 1\)
To make conjunctions of formulas with different levels, we can repeatedly take the conjunction of the lower level one with itself until it reaches the level of the higher one.
Since people do not generally like to see formulas like \( ( (…
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Okay, so this morning someone tried to add an invite link to the socalled "Googology Wiki Discord" to this wiki's main page.
I would like to make something clear. That Discord is in no way affiliated with this wiki or even the googology community.
the server is owned by Pi.jayk (aka DrCocktor), who is a banned troll (and may or may not be Edwin Shade) that tried to shut this wiki down once. Very few, if any, other regulars of this wiki are there.
There has been a (moreorless) official Discord for the googology community for almost four years now (from which Pi.jayk has been banned along with several of their sock puppets).
I decided to have a quick peak in there. Almost no googology was taking place, sarcastic comments about Plain'N'Simple …
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This is a formalization of an idea that occured to me a long time ago about how we can use the Rayo function to define a "smaller variant" of itself. By doing this, the Rayo function has an enormous "jump" that places it above most numbers we can define in theories weaker than standard set theory.
Specifically, the FOST formula I will define here defines an output of a function which I will denote as \(\text{ML}(n)\). \(\text{ML}\) is essentially a variant of a Rayo function defined so that FOST formulas are interpreted as being true in a certain set, rather than the entire universe. It also differs because the known lower bounds for \(\text{ML}(n)\) become large more quickly than the known lower bounds for Rayo (although the Rayo function …
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This is an English translation of my Japanese blog post submitted to a Japanese googology event focused on programming languages.
I implemented Weak Buchholz's OCF and Extended Weak Buchholz's OCF by the programming language \(\mathbb{Q}_p\), and explicitly gave ordinal notations associated to them except that I have not defined the notion of standard form because the definition of large functions does not use it. Weak Buchholz' OCF is a weak variant of Buchholz's OCF created by a Japanese Googology Wiki user Gaoji
Instead of the explanation of the code, I copy the automatic conversions into natural languages.
I define a set \(\mathbb{N}^{\omega^{\omega}}\) in the following recursive way:
 For any \(x \in \mathbb{N}\), \(x \in \mathbb{N}^{\ome…
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upcoming fandom events:
25 june: number contest
5 july: large game numbers (write numbers from spellen)
example:
ninja legends leaderboard hoogste statistieken 178 octillion
14 juli: fandom puzzel event
vind aanwijzingen in fandom wiki's en win een price: i will use 30 wiki's
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NUMMER WEDSTRIJD EVENEMENT
iedereen kan mijn pagina bewerken. iedereen zal een nummer schrijven. elk nummer is groter dan het vorige nummer, maar kleiner dan het volgende nummer
everyone can start now
https://googology.wikia.org/wiki/Googology_event link to the contest
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 En = 10^{n}
 Fn = {10,n,2}
 Gn = {10,n,3}
 Hn = {10,n,4}
 Jn = {10,10,n}
 Kn = {10,n,1,2}
 Ln = {10,n,2,2}
 Mn = {10,10,n,2}
 Nn = {10,10,10,n}
 Pn = {10,n+2 (1) 2}
 ln(n)
 log10(n) = ln(n)/ln(10)
 tetralog10(n) = log_10(log_10(log_10...log_10(log_10(n))...)) with n log_10's
 pentalog10(n) = tetralog_10(tetralog_10(tetralog_10...tetralog_10(tetralog_10(n))...)) with n tetralog_10's
 sup10(n) = F(log10(n))
 sqrt(n) = n^{1/2}
 cbrt(n) = n^{1/3}
 expofac(n) = n!1 = n^{n1......321}
 b pt n = 10^{10...10n} with b 10's
 sin(n)
 cos(n)
 tan(n)
 cot(n)
 sec(n)
 csc(n)

Diacosiaeikosievadienaeptakisekatommiriotetrakosiosarantodioeksakisekatommiriopentakosiopenintoctopentasekatommirioekatovpenintatriatetrakisekatommiriotriakosiotriantotriatriekatommiriodiakosioebbdomintodiodisekatommirioeksakosioeikositesseroekatimmirioenniakosiopenintoeksichilladoeikosievoarithmoi is 10^{221 442 558 153 333 280 624 956 021}.
It consist of very much greek words: Διακόσια είκοσι ένα επτακισεκατομμύριο τετρακόσια σαράντα δύο εξακισεκατομμύριο πεντακόσια πενήντα οκτώ πεντακισεκατομμύριο eκατόν πενήντα τρία τετρακισεκατομμύριο τριακόσια τριάντα τρία τρισεκατομμύρια διακόσια εβδομήντα δύο δισεκατομμύρια eξακόσια είκοσι τέσσερα εκατομμύρια εννιακόσια πενήντα έξι χιλιάδες είκοσι ένα αριθμοί (221 442 558 153 333 280 625 956 021).
THIS…
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It is 10^10^10...(repeated 1000^1000^1000^1000^1000x5 times total (includes the first 10)) ^10^10^100
Note the amount of "10^" is not 1000x1000x1000x1000x1000 because it would have the suffix "quintmilli", and one Millimilli is 1000x1000...(repeated 1000 times total)...x1000
This number can be expanded as every extra "Milli" equals another layer of the one thousands power. As in:
if we have six millis, the number becomes 1000^1000^1000^1000^1000^1000;
if we have seven millis, the number becomes 1000^1000^1000^1000^1000^1000^1000, and so on.
Googla means two "10^"s as one googol= 10^100, and we take the 100 and turn it into 10.
Googolplex means three "10^"s as one googolplex= 10^10^100, and we take the 100 and turn it into 10.
So, we add 2 an…
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Pentakosioeikosiduodiaekatommiriotetrakosiogdontoeksiekatommiriodiakosioennenitopentechilladeksakosiosarantopentearithm is 10^{522 486 295 945}.
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It consist of 16 greek words: πεντακόσια είκοσι δύο δισεκατομμύρια τετρακόσια ογδόντα έξι εκατομμύρια διακόσια εννενήντα πέντε χίλιάδες εξακόσια σαράντα πέντε αριθμοί (522 486 295 945 numbers). 
Ekatommirioenniakosioennenitaenneachilladenniakosioennenintoenneameden is 10^{1 999 999}. It consist of 9 greek words: εκατομμύριο (one million), εννιακόσια εννενήντα εννέα χίλιάδες (999 000), εννιακόσια ενενήντα εννέα (999), μηδενικά (zeroes)
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Ekplikikotromatikopentchildapithoeptatelaoct is 7.7 * 10^{5000}
The name consist of 5 greek words: Εκπληκτικός (arresting), τρομαχτικός (scaring), πέντε χιλιάδες (5000), αριθμοί (numbers), και επτά (and 7), τελεία (point), οκτώ (8).
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f(x) = x^(1/(1x))
For x>0, x≠1, f(x)^^2 = (f(x)^x)^^2) = g(x)
Ex. f(2) = 1/2, f(2)^2 = 1/4, (1/2)^^2 = (1/4)^^2
There are two equal values of g(x), one 0
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Megaleiodoktocontoctidksorosymvol is 10^{808}. The name of the number has 33 symbols! It consist of 5 greek words: μεγαλειώδης (grand), οχτακόσια (800), οκτώ (8), σύμβολο (symbol), ισχυρός (powerful).
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Does computable mean "Always Halts"? Does growth rate relate to Halting?
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(The OCF in question is this one.)
Take \(\psi_{\chi(M)}\). Up to \(\alpha = M\), like the OP pointed out, we get erratic behavior, and we get \(\psi_{\chi(M)}(M) =\) the first fixed point \(\alpha\mapsto\chi(\alpha)\). However, after \(\alpha = M\), we still get erratic behavior: the \(\chi(\alpha)\) function gets stuck from \(\alpha = \psi_{\chi(M)}(M)\) to \(\alpha = M\), since we don't get \(\chi(M)\) and thus \(\psi_{\chi(M)}(\alpha)\) for \(\alpha\geq M\) in \(C(\alpha,\beta)\), which means that we can only build up on lower values of \(\psi_{\chi(M)}(\alpha)\) using addition, \(\omega^{\gamma}\) and \(\Omega_{\gamma}\). This gets us that \(\psi_{\chi(M)}(M+\alpha) =\) the \(\alpha\)th fixed point \(\beta\mapsto\Omega_{\beta}\) after …
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Remember the Luxius Vandalism? He was just getting revenge for banning him. I guess he has anger issues, since he is nicer, especially on Googologyforeveryone Wiki. He has this array function that is basically a collab between him and GFE user POYO.
https://googologyforeveryone.fandom.com/wiki/LL%2BDEAF
This notation might actually be good.
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{x} = x^{2} {x,0} = {x}
{x,y} =
This notation might be at pentational level (as in the function's approximation is pentational)
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This is very large number. It is consist of 5 greek words: απίστευτος (incredible), μεγάλο (big), αφάνταστος (unimaginable), συγκεκριμένος (special), αριθμός (number).
It is 9 * 10^{101010...} Or 9 * 10^^58.
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The abbreviations below are powers of 1000
k
m
b
t
qu
qi
sex
sep
o
n
de
ud
du
td
qud
qid
sexd
sepd
od
nd
v
uv
dv
tv
quv
qiv
sexv
sepv
ov
nv
tg
utg
dtg
ttg
qutg
qitg
sextg
septg
otg
ntg
qug
uqug
dqug
tqug
ququg
qiqug
sexqug
sepqug
oqug
nqug
qig
uqig
dqig
tqig
quqig
qiqig
sexqig
sepqig
oqig
nqig
sexg
usexg
dsexg
tsexg
qusexg
qisexg
sexsexg
sepsexg
osexg
nsexg
sepg
usepg
dsepg
tsepg
qusepg
qisepg
sexsepg
sepsepg
osepg
nsepg
og
uog
dog
tog
quog
qiog
sexog
sepog
oog
nog
ng
ung
dng
tng
qung
qing
sexng
sepng
ong
nng
c
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What is greater: 100R{0,,, ...(100 commas)... ,,,1} in Hyp cos' R system or fC(С(C(Ω_3 2,0),0),0)(3) (i.e. tritar) with fixed FSs of Taranovsky's ordinal notation?
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I will begin this blog post by addressing an incident I had on another blog post on Sunday morning where I added the Mathematica code without posting a comment, and the guy just said I was rude. Next time, if there has been speculation about a precise method in the comments on my blog post and I clarify it in an edit to the post, I will post a comment, and in the future I will be more responsive to blog post comments in general. My apologies.
Anyway, I invented another new notation, which can encode any polynomial expression (and goes even further into hyperexponential territory), and then applies functional operators onto the function encoded in the expression.
I will begin by clarifying how I define functional hyperoperators. A functional…
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Calculator can be found here.
Starting Values:
 X=2.00000000055
 Y="P"
 Z=0.00040353607
 K=100000000000
date what number DJIA on
previous day ended with hundredth place of the
DJIA closing value (n) Z*n^2 X PEGG value (yes)
(1,0,0)10.0032220454
(1,0,1)_{3}(1,0,0)(7,2)_{2}(7,0)_{7}(6,5)_{8}(6,4)_{3}(6,3)(6,2)_{7}(6,1)_{4}(6,0)(5,5)_{4}(5,3)_{4}6.93273595623
(1,0,4)_{8}(1,0,2)_{4}(3,9)_{3}(3,8)_{8}(3,7)_{2}(3,6)(3,5)_{2}(3,3)_{5}(3,2)_{3}(3,1)(2,0)_{8}(1,1)_{2}6.90630367086
(1,1,1)_{2}(1,1,0)_{3}(1,0,8)(1,0,7)_{2}(1,0,6)_{6}(1,0,5)(1,0,3)_{3}(3,2)_{5}(2,3)_{8}(2,2)_{8}(2,1)_{5}(1,8)_{3}1.34328420574
(1,1,1)_{3}(1,1,0)(1,0,1)_{8}(7,5)(7,4)_{2}(7,2)_{6}(7,1)_{3}(6,6)_{3}(6,5)_{2}(6,4)(6,1)_{6}(6,0)_{2}2.03579880692
(1,2,2)_{2}(1,2,1)(1,2,0)_{6}(1,1,0)_{5}(1,0,4)_{8}(1,0,3)_{2}(1,0,2)_{7}(1,0,1)_{4}(1,0,0)_{7}(2,2)(2,1)_{3}(2,0)_{2}2.118947974…
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I am back after a long absence, with a new natural number recursion. This one I think is much simpler and cleaner looking than any of my previous attempts, and more powerful. Not being a mathematician, my rules are written in English supporting formulas and examples. I believe it is all welldefined and finite. Here are the first few rules; there are not many of them, and to this point I have used only chevrons, parentheses, numbers, commas, and superscripts. It starts with the standard FGH before switching to chevron notation and I have included quite a few examples that the experts out there probably don't need to see, but the novices might like to see them, and for me, a lot of the fun is in working out examples. There is an extension…
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I was just wondering how other competitors are more active. Maybe because of the "no sources" rule? maybe because of the Edwin shenanigans?
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So define ͳ(n) as f_{@}(n) in the fastgrowing hierarchy where @ is the biggest ordinal definable using Madore's ordinal trees using no more than n layers of branches, each branch having no more than n branches.
ͳ(x,y) = f_{@}(n) where @ is the biggest ordinal definable using Madore's ordinal trees using no more than x layers of branches, each branch having no more than y branches.
Here are my approximations.
Example:
(0,(0,(1,1,1),(0,(1,1,1,0,2),(0,1,1),0),0,1,1,(0,1,1),2))
(x) = x
(0(0,0)) = ω
(1(0,0)) = ω+1
(x(0,0)) = ω+x
(0,(0,1)) = ω2
(1,(0,1)) = ω2+1
(x,(0,1)) = ω2+x
(y,(0,x)) = ωx+y
(0,(0,(0,0))) = ω^{2}
(0,(0,(0,(0,0)))) = ω^{3}
(0,(1,0)) = ω^{ω}
Basically follows nesting rules in lefttoright order, bottomtotop, grouped by branch.
(0LxBy) = the biggest ordina…
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On my profile page, I listed the numbers with more than \(10^{10^6}\) digits whose first digits I have found, the largest of which is \(2^{2^{2^{32}}}\) which has more than \(9.341805\times10^{1292913985}\) digits. The calculation (which I did back in August 2015) took about 5 gigabytes of my computer's RAM (I used Mathematica), and a day and 6 hours of computation time (I used a desktop computer, not a laptop). The number begins with 315921269337233843004184822......... and ends ......717272631317364736.
Here is the Mathematica code, to address speculation in the comments on the precise method: nbrdgt = 100;
f[base_, exp_] :=
RealDigits[
10^FractionalPart[
N[ exp*Log10[ base], nbrdgt + Floor[ Log10[ exp]] + 2]], 10,
nbrdgt]; f[2, 2^2^32]Th…
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I am writing this in a new blog post, since this community tends to ignore blog posts older than a few days. Since I made my original Parenthesis Block Notation blog post, I have further analyzed the notation, including the growth rate and last digits of outputs.
I further examined the behavior of last digits of results of the 4argument function, and found that the last digits of (2, a, b, 2) will always be either 36 or 16, and if plotted, follow this pattern:
This is the same plot you will get if you shade the results of (3, a, b, 2) that end in 83 and the ones that end in 03.
I also found that the last digits of (2, 2, 2, n) (given below) appear to converge, much like tetration:(2, 2, 2, 2): ...566616377844323799083803775860736
(2, 2, 2, 3)…
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ERROR: This article has too many jokes.
Hi, did you watch Beyond Infinity Number Comparison by Reigarw Comparisons? It was a great video showing numbers of various sizes, starting from one to !!! ABSOLUTE INFINITY !!!
What's surprising is that they made up their own notation and well blended into other numbers! I cannot praise the video too much!!!
One slight defect is that they only show some expansion examples and did not show any of the definitions. However, the examples are ample enough so that we can easily guess the original definition.
So let's try it!
n and n' are natural numbers greater than 1, and m is a natural number greater than 0.
g_{1} = 3↑↑↑↑3
g_{n} = {3,3,g_{n1}} where {} is BEAF
g_{1,0} = g_{64}
g_{n,0} = g_{n1,g64}
g_{n,m} = g_{n1,gn,m1}
f_{Φ}(m) = g_{m+1,0}
f_{ΦΦ}(1) …
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We can encode a fundamental sequence for an ordinal α into an ordered tree in the follwing way:
 The vertexes are tuples (n_1, ..., n_m) such that α[n_1]...[n_m] is welldefined

https://www.youtube.com/watch?v=RJS3Z2DYEO4
This video is full of inaccuracies.
There is a comment by "Tsavorite Prince" pointing these out. You can like it so the creator sees it and fixes the video.
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Program language: python number and program name:
S2 = S[::] if not(S): return n if S[1] == 0: return S[:1] j = S[0] while type(j) == list and j != []: j = j[0]I forgot that this was needed, I'll just call the number TLAN&SLTANN, program name is LANdepartment is programming department because someone came up with this notation, I'm just implementing it. def expand(n,S):
e = S[0] if type(e) == list:
S2[0] = expand(n,S2[0]) return S2
elif e > 0:
S2[0] = 1
return S2
m = min([a for a in range(len(S)) if S[a] != 0])
if type(S[m]) == int: S2[m] = 1 for i in range(n): S2[m1] = S2[::]
elif type(S[m]) == list:
if S[m] …
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Nvm, this blog is closed.
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I have idea to create new hierarchy function with Droste effect.
https://en.wikipedia.org/wiki/Droste_effect
Let start with FastGrowing Hierarchy Function  style definition,
\begin{eqnarray*} D_0(n) &=& n+1 \\ D_{a+1}(n) &=& D_{a}^{\$}(n) = ( D_{a}(n)+1 ) \begin{cases} D_{a}^{\cdot^{\cdot^{\cdot^{D_{a}^{n}(n) }\cdot}\cdot}\cdot}(n) \end{cases} \\ \end{eqnarray*}
"\(\begin{cases} \\ \\ \end{cases}\)" mean, for example,
\begin{eqnarray*} ( 6 ) \begin{cases} D_{a}^{\cdot^{\cdot^{\cdot D_{a}^{n}(n) \cdot}\cdot}\cdot}(n) \end{cases} &=& D_{a}^{D_{a}^{D_{a}^{D_{a}^{ D_{a}^{D_{a}^{n}(n)}(n) }(n)}(n)}(n)}(n) \\ \end{eqnarray*}
Be careful,
\begin{eqnarray*} ( 6 ) \begin{cases} D_{a}^{\cdot^{\cdot^{\cdot D_{a}^{n}(n) \cdot}\cdot}\cdot}(n) \end{cases} &\n…
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In this post I will give some conjectures assuming arrayof operator were more formalized.
The limit of the trio sequences is about {L2,1}n,n. Why is it? By using analyses of BMS by Googleaarex and
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Now, this was simply just an idea based on Kapiluka 's. This may have contradiction with the normal FGH, but I will try.
Suppose \(f_0(x)\) means \(x+1\) for all ordinals not just if \(x\in\mathbb{N}\). If so, we can say that
\(f_0(\omega)\) = \(\omega+1\).
Then, if we can continue this to transfinite ordinals with \(f_1(x)\) which would be \(f^{n}_0(x)\) times or \(2x\). You are most likely asking how this works for transfinite numbers. Here is what I propose:
For \(f^{\beta}\)
Iff \(\beta\in\mathbb{N}\): nest \(f\beta\) times.
Iff \(\beta\) is a limit ordinal: find the growth rate when nesting the function \(f\).
Iff \(\beta\) is a successor ordinal and \(\beta\notin\mathbb{N}\): nest \(f\) until you reach a limit ordinal.
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If we follow the pat… 
Here's an analysis of the first system of Ordinal Tree Notation by FundamentalSeq. The linked blog post contains an analysis up to \(\varepsilon_0\), so I'm taking it from there.
Tree Ordinal
ABB000 \(\varepsilon_0\)
ABB00ABB000 \(\varepsilon_0 2\)
ABB00B00 \(\varepsilon_0\omega = \omega^{\varepsilon_0+1}\)
ABB00B0B00 \(\omega^{\varepsilon_0+2}\)
ABB00BA00 \(\omega^{\varepsilon_0+\omega}\)
ABB00BAB000 \(\omega^{\varepsilon_0+\omega^\omega}\)
ABB00BABB0000 \(\omega^{\varepsilon_0 2}\)
ABB00BB000 \(\varepsilon_1\)
ABB0A00 \(\varepsilon_\omega\)
ABB0A0BB000 \(\varepsilon_{\omega+1}\)
ABB0A0BB0A00 \(\varepsilon_{\omega 2}\)
ABB0AA00 \(\omega^{\omega^2}\)
ABB0B000 \(\zeta_0\)
ABB0B00BB0B000 \(\zeta_1\)
ABB0B0A00 \(\zeta_\omega\)
ABB0B0ABB0B0000 \(\zeta_{\zeta_0…
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h(n) = n +1
h(h(n)) = h(n)+1 = n+2
h(h(h(h......n h's....h(n)))..)) = n+n = 2n
g(n)a = h(h(h........a h's...h(h(n))..)) = n+a
g(n)n = 2n
g(g(n)n)n = 3n
g(g(g(n)n)n)n = 4n
g(g(g(g......g(g(n)n)n....)n)n with n g's = n^2
f(n)a = g(g(g.......a g's........g(n)n)n...)n)n = n*a
f(n)n = g(g(g(g......g(g(n)n)n....)n)n with n g's = n*n = n^2
f(f(n)n)n = n^3
e(n)n = f(f(f(f......n f's......f(f(n)n)n....n)n)n = n^n = n^^2
e(e(n)n)n = n^^3
d(n)n = e(e(e....n e's...e(n)n...n)n = n^^n = n^^^2
d(d(n)n)n = n^^^3
overall, for any letter, it's just the letter after it nested. i.e: c(n)n = the same nesting to d(d...d(n)n)..)n as the other ones
h(n) = {8}(n)n
{7}(n)n= g(n)n
{M}(n)n is just the mth letter of the alphabet with those two symbols after. The earlier the letter,…
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