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  • Antimony Star

    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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  • Kanrokoti

    I made "notation-nesting base system (NNBS)". This is similar to shiftedness.

    There are no definition(non-defined, 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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  • Golapulusplex10
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  • Yabuszko

    Bignumber.rb

    July 3, 2020 by Yabuszko

    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(m-1,a(m,n-1)) # 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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  • Kaplikua

    �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/bacon-cipher bitwise operator

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  • Mh314159

    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 up-arrows 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.



    f0(x) = x+1

    For all functions F, Fn(x) = F(Fn-1(x)) and F1(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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  • Usedtosignin

    From, Goodstein Sequence (https://googology.wikia.org/wiki/Goodstein_sequence),

    I create new representation call "Extended Goodstein's base-b hereditary representation" or "xG's base-b hereditary representation".

    xB[b,b_](n) means finding the xG's base-b 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_{k-1}(n),xG_k(n)](xG_k(n)) - 1 \\ &\vdots& \\ \end{eqnarray*}

    Until, it return to zero.

    This is how xG's base-b hereditary representation work.

    In case b = 2, if n …

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  • Gonz0TheGreatt

    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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  • Username5243

    Clarification

    June 30, 2020 by Username5243

    Okay, so this morning someone tried to add an invite link to the so-called "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 (more-or-less) 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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  • Gonz0TheGreatt

    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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  • P進大好きbot

    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:

    1. For any \(x \in \mathbb{N}\), \(x \in \mathbb{N}^{\ome…


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  • Robloxpizzachef

    Ja of nee

    let me know in the comments

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  • Robloxpizzachef

    upcoming events

    June 24, 2020 by Robloxpizzachef

    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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  • Robloxpizzachef

    googology event

    June 24, 2020 by Robloxpizzachef

    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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  • Micrintel
    • En = 10n
    • 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) = n1/2
    • cbrt(n) = n1/3
    • expofac(n) = n!1 = nn-1......321
    • b pt n = 1010...10n with b 10's
    • sin(n)
    • cos(n)
    • tan(n)
    • cot(n)
    • sec(n)
    • csc(n)
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  • Micrintel

    W0

    June 24, 2020 by Micrintel
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  • Superlirarder

    Diacosiaeikosievadienaeptakisekatommiriotetrakosiosarantodioeksakisekatommiriopentakosiopenintoctopentasekatommirioekatovpenintatriatetrakisekatommiriotriakosiotriantotriatriekatommiriodiakosioebbdomintodiodisekatommirioeksakosioeikositesseroekatimmirioenniakosiopenintoeksichilladoeikosievoarithmoi is 10221 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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  • Mumuji

    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 milli-s, the number becomes 1000^1000^1000^1000^1000^1000;

    if we have seven milli-s, 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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  • Superlirarder

    Pentakosioeikosiduodiaekatommiriotetrakosiogdontoeksiekatommiriodiakosioennenitopentechilladeksakosiosarantopentearithm is 10522 486 295 945.


    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).

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  • Superlirarder

    Ekatommirioenniakosioennenitaenneachilladenniakosioennenintoenneameden is 101 999 999. 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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  • Superlirarder

    Ekplikikotromatikopentchildapithoeptatelaoct is 7.7 * 105000

    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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  • Tetrapyronia

    f(x) = x^(1/(1-x))

    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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  • Superlirarder

    Megaleiodoktocontoctidksorosymvol is 10808. 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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  • Golapulusplex10

    Does computable mean "Always Halts"? Does growth rate relate to Halting?

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  • Alemagno12

    (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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  • Kaplikua

    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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  • Kaplikua

    {x} = x2 {x,0} = {x}

    {x,y} =

    This notation might be at pentational level (as in the function's approximation is pentational)

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  • Superlirarder

    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 * 10101010... Or 9 * 10^^58.

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  • Keon175

    The abbreviations below are powers of 1000

    qu 

    qi 

    sex 

    sep 

    de 

    ud

    du 

    td 

    qud 

    qid 

    sexd 

    sepd 

    od 

    nd 

    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

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  • Tetramur

    What is greater?

    June 15, 2020 by Tetramur

    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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  • Allam948736

    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 hyper-exponential 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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  • Golapulusplex10
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  • Naruyoko

    PEGG detailed log/P

    June 14, 2020 by Naruyoko

    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)|46.93273595623
    (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)|26.90630367086
    (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)|31.34328420574
    (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)|22.03579880692
    (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)|22.118947974…











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  • Mh314159

    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 well-defined 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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  • Kaplikua

    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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  • Kaplikua

    Madore Tree FGH

    June 11, 2020 by Kaplikua

    So define ͳ(n) as f@(n) in the fast-growing 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))) = ω2

    (0,(0,(0,(0,0)))) = ω3

    (0,(1,0)) = ωω

    Basically follows nesting rules in left-to-right order, bottom-to-top, grouped by branch.

    (0LxBy) = the biggest ordina…


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  • Allam948736

    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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  • Allam948736

    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 4-argument 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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  • Tetramur

    [n] = 48^n

    [m, n] = [m-1, 48^n]

    [0, #] = [#]

    [1,1,1] is ill-defined.

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  • Nayuta Ito

    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.

    g1 = 3↑↑↑↑3

    gn = {3,3,gn-1} where {} is BEAF

    g1,0 = g64

    gn,0 = gn-1,g64

    gn,m = gn-1,gn,m-1

    fΦ(m) = gm+1,0

    fΦΦ(1) …

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  • Fejfo

    Tree building game

    June 7, 2020 by Fejfo

    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 well-defined
    (ie α[n_1]...[n_i] is a limit ordinal forall i Read more >
  • Ubersketch

    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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  • Antimony Star

    Program language: python number and program name: I forgot that this was needed, I'll just call the number TLAN&SLTANN, program name is LAN department is programming department because someone came up with this notation, I'm just implementing it. def expand(n,S):

        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]
        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[m-1] = S2[::]
        elif type(S[m]) == list:
            if S[m] …








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  • Windows NT 6.0
    • Note: trillion
    1. 31415926535897 = 10π trillion
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  • Tetramur

    Nvm, this blog is closed.

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  • Usedtosignin

    I have idea to create new hierarchy function with Droste effect.

    https://en.wikipedia.org/wiki/Droste_effect

    Let start with Fast-Growing 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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  • Tetramur

    BEAF and BMS

    June 3, 2020 by Tetramur

    In this post I will give some conjectures assuming array-of 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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  • Awesomeadndy

    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.


    If we follow the pat…

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  • Janek37

    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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  • VoidSansXD

    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 m-th letter of the alphabet with those two symbols after.  The earlier the letter,…

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