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  • Ubersketch

    In my last post I mentioned something called mensors. This is a concept I came up with after reading Bowers' Hypernomial page and writing tensors like so: [a,b[2]d,e[3]f,g[2]h,i]  Mensors are basically tensors except they use tensor dimensions and mensor dimensions. For example, a mensor could be like [a,b[c,d]e,f] I don't know what this would be used for but its something I'd like to mension.

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  • Ubersketch

    So this is pretty exciting, since I've officially found semi-formal rules for my function I explained a few days ago. Here are the rules. 1. f(a){[1]b} = f(f(... where the function is applied b times to a. 

    2. f(a){[b]c} = f(a){[b-1]a[b-1]a...} where there are c as.

    3. f(a){[b]c[d]e...} is evaluated like so: f(a){[b]c} = f(a) in f(a){[d]e} = f(a) in... until it is fully evaluated.

    f(a) can be any unary function, but to be self-contained, a+ = the successor of a.  For example: 2+{[1]3[1]2} = f(2){[1]2} = 8


    I have plans to getting around to mensor function inputs and mensor seperators/limit ordinal seperators.

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  • Ynought

    C function

    September 19, 2018 by Ynought

    imagine a map that can be coloured with n colors, given a set of rules that give the smallest non infinite amount of maps coloured that doesn’t violate any of the rules

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  • Ubersketch

    So the other day I posted this to reddit.
    https://www.reddit.com/r/googology/comments/9g5a1j/something_i_made/
    I haven't well defined it, and I haven't even made any remotely formal rules. Here goes nothing. Only the first entry can be followed with a +.
    {a} {a+} = a+1
    {a+[1]b} = {a+}+}...} with b pairs of braces. {a+[1]b[1]c} = {a+[1]b} c times (i.e. {a+[1]b[1]c} uses {a+[1]b} as the input for a c times)
    {a+[2]b} = {a+[1]a[1]a...} with a [1]s.  @a = {a+[2]b} {a+[3]b} = {@...a} with b @
    This goes on for a long (uotc) while. I'd love it if you guys pointed out and flaws, undefinedness, and left ideas for extensions.

    I suspect this function has the level of \({\varepsilon_0}\)

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  • Syst3ms

    I'll be straightforward : I don't know how to actually formalize stuff, so I won't bother saying this is formal, since it is extremely far from it. I'm just dumping this here so that it doesn't get lost in some Discord channel somewhere. This is definitely better than "transfinite FSes", however.


    \(\Omega_0 = 1 \\
    \pi_n(0,0) = \Omega_n \\
    n \ge 1: L(\Omega_n,\alpha) = \alpha \\
    m+1 > n: \pi_n(\Omega_{m+1},\alpha) = \pi_n(\pi_m(\Omega_{m+1},0),\alpha) \\
    \text{cof}(\alpha) = \Omega_{n+1}: L(\pi_n(\alpha,0),0) = 0 \\
    \text{cof}(\alpha) = \Omega_{n+1}: L(\pi_n(\alpha,\beta+1),0) = \pi_n(\alpha,\beta)+1 \\
    \text{cof}(\alpha) = \Omega_{n+1}: L(\pi_n(\alpha,\beta),\gamma+1) = \pi_n(L(\alpha,L(\pi_n(\alpha,\beta),\gamma)),0) \\
    \beta \in \text{Lim}: L(\p…







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  • Alemagno12

    Inspired by this blog post (in the japanese GWiki).

    When I say that f(x) = a0,a1,a2,...,an,{b0,b1,b2,...,bm} for some function f (like f(x) = A(2,x) mod 3), I mean that f(k) = ak for k ≤ n and that after that the values of f(k) loop through the bps.


    By definition A(0,x) = x+1, and it can be easily checked that A(1,x) = x+2, so:

    • A(0,y) mod 2z = {1,2,3,4,...,2z-2,2z-1,0}, and
    • A(1,y) mod 2z = {2,3,4,5,...,2z-2,2z-1,0,1}.

    The extremal case would be z = 1 - since A(0,y) is always odd and A(1,y) is always even, A(0,y) mod 2 = {1,0} and A(1,y) mod 2 = {0,1}.

    Next, we have that A(2,x) = 2x+3 - for x = 0, A(2,0) = A(1,1) = 1+2 = 3, and if A(2,n) = 2n+3, then A(2,n+1) = A(1,A(2,n)) = A(2,n)+2 = 2n+3+2 = 2(n+1)+3.  So:

    • A(2,y) mod 2z = {3,5,7,9,...,2z-3,2z-1,1}. …

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  • HitThePin

    So... I'm new here...

    September 15, 2018 by HitThePin

    And I tried to help by starting an article for the number 103. But I don't know how to put the big grid with the other numbers between 100 and 199 in it. Can someone more experienced here please tell me how? Thank.

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  • MilkyWay90

    How do I fix this?

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  • Hyp cos

    I compared OCFs with my array notation, but stopped at a weakly compact cardinal because I didn't understand how weak compactness works. Could someone help me about these questions:

    1. Weakly compact cardinals have many equivalent definitions, but what's a (set theoretical) definition of "weakly compact property" over a series of ordinals? This may be similar to "weakly inaccessible property" (to be a limit of a series of ordinals) and "weakly Mahlo property" (the set of a series of ordinals in that ordinal is stationary).
    2. If weak inaccessibility corresponds with recursive inaccessibility as a "recursive analogue", weak Mahloness corresponds with recursive Mahloness, and \(\Pi_n^1\)-indescribable cardinals correspond with \(\Pi_{n+2}\)-reflecti…
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  • ChromaticiT

    Infinite-order Xi Function

    September 10, 2018 by ChromaticiT

    The ITTM Busy beaver is to the Busy beaver as the Infinite-order Xi function is to the Xi function.

    We define a combinatory language with countably infinite symbols such that there is a bijection between these symbols and the natural numbers (including 0).

    0x := \(\iota\) x as in Chris Barker's iota combinator. Nxyzw := Where N is a natural number denoting a symbol, \(\Omega\)xyz (as in the oracle combinator) if x is well founded, w otherwise.

    Now that we have our language, let's just reuse the definition for the Xi function.

    If we start with a string of n symbols and we beta-reduce it, the largest possible finite output is called \(\Xi_{\infty}\)(n)

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  • Grand fuzili

    intervalux

    September 6, 2018 by Grand fuzili

    pense assim:

    g Meameamealokkapoowa oompa = J

    JJ = J elevado a quantidade de J

    J2 = JJ

    J3 = JJJ

    INTEVALUX = Jg Meameamealokkapoowa oompa

    grande numero

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  • DWither

    Doubt about busy beaver

    September 4, 2018 by DWither

    The busy beaver function is uncomputable because it's impossible to make a program that will evaluate BB(n) in a finite time, right? And this is because the computer will never know if it's going 1to halt until it halts. Well, I'm pretty sure that someone has to have asked this before, but why can't you just check for loops in the behavior of the turing machine to know if it's going to keep going forever?

    I thought of a step by step method to calculate BB(n) that definitely seems computable to me. I don't know enough about any programming language to directly show code though, so sorry about that. Tell me what I'm missing in all of this that makes my step by step method useless. 



    Step 1:

    Take all the possible combinations of states you can ha…



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  • DontDrinkH20

    This is a series of blog posts I will make to try to generalize the understanding of H-boogol-boogol to the general public of this wiki. It's quite a mind boggling concept, but the kind of person on this website is likely the kind of person who can figure it out if they just read a couple different detailed descriptions. So just believe in yourself, and feel free to get out a notebook and start taking notes.

    In the last blog post, Model Theory 101, we looked at systems of mathematics, and formalized the idea of one. We also formalized the idea of "this universe obeys the system of mathematics...." This idea is known as model theory. If you haven't read that yet, go read it now!


    In the last blog post, you may remember that I kept putting (fi…


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  • MilkyWay90

    Math number

    September 2, 2018 by MilkyWay90

    Let's create a "math language" called N.

    In N, you can: Add (F(x, y) = +(x, y))

    Subtract (F(x, y) = -(x, y))

    Use natural numbers (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...)

    Recurse (F(x, y) = F(x, -(y, 1)), F(x, 0) = +(x, x))

    Variables (a0, a1, ..., b0, b1, ..., z0, z1, ..., aa0, aa1, ...)

    Other functions (A(x) = x, F(x) = A(x + x))

    Equality (equal sign, but no greater than or less than or not equal)

    To define a function, do Function(v0, v1, v2, ..., vn) = (expression)

    Note: Addition and Substraction can only hold 2 arguments. Use +(2, +(2, 2)) instead of +(2, 2, 2)

    Now let's get into symbols!

    F(x) = +(x, x) has 6 symbols because F(x) is 2 symbols (F, and x), = is one symbol, +(x, x) is 3 symbols (+, x, and x)

    So Functions, natural nu…

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  • Nedherman1

    Arrow Destruction (AD)

    September 2, 2018 by Nedherman1

    I don't speak english.



    Hello, today I showed my new notation that I call Arrow Destruction!
    (I was inspired by Eners' notation)

    Basic:






















    we can put more than two entries:































    if we want, we can put more entries, which will follow this same logic

    We can make this notation more powerful in this way:
















    we can also put more than two entries that also follow the same logic



    However, it does not end here: ​​​​










    I explained in two ways and I showed an example, if I did not understand, see [a]^n is different from [a]^[n], since in the first example "n" is not between [ ], which means that we will use the rule shown above to solve, in the second example, "n" is between [ ], that is, first we will solve using the basic rule and, after we know the result of [n], we wil…

















































































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  • DontDrinkH20

    This is a series of blog posts I will make to try to generalize the understanding of H-boogol-boogol to the general public of this wiki. It's quite a mind boggling concept, but the kind of person on this website is likely the kind of person who can figure it out if they just read a couple different detailed descriptions. So just believe in yourself, and feel free to get out a notebook and start taking notes.


    If you know enough about mathematics folklore, you would know that Godel's incompleteness theorem shows that there are some things which just cannot be proven by a working system of mathematics; one that doesn't imply 0 = 1.

    In model theory, a theory \(T\) is that system of mathematics, aided by a language and a formal logic (but those ar…


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  • Gleive

    I originally planned to release this on September 1st, but meh.

    While going through this notation, you might notice that despite being finished in 2018 (bruh), it doesn't really grow as spectacularly fast or introduces some amazing new stuff. This is just a normal googologist's normal work, and I just want to say that I've tried many thing in googology and this is just a milestone I want to keep.

    Anyways, enough semi-emotional stuff, let's get into the notation.

    I mentioned in my previous post that this notation was refined and revisited several times. The original version was an extension on down-arrow notation (Maybe because BEAF is based on up-arrows?), the second version used my own functions, and the third version meddled with polygonal …

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  • DontDrinkH20

    This time I'm going to be defining some really big ZFC-definable countable ordinals (some even provably exist). These are based on the principle of making ordinals which are not definable within another very "wide" structure, like a transitive model of ZFC or \(V_\kappa\) for large enough \(\kappa\).


    Let \(\varphi\) be a formula and let \(x\) be a set. \(\varphi:x\) if and only if \(\varphi(A)\) if and only if \(A\in x\) for all \(A\).

    Then, \(\theta_\alpha\) is defined as the smallest ordinal \(\beta\) such that there is no formula \(\varphi\) such that \(\varphi^{V_\alpha}:\beta\). That is, the smallest ordinal which is not definable within \(V_\alpha\). Given a transitive structure \(\mathcal{M}=(M;\in)\), \(\theta(\mathcal{M})\) is the a…


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  • Bubby3

    BM2 doesn't terminate.

    August 28, 2018 by Bubby3

    There is an infinite descending chain at (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,2,0,0)(5,1,1,1)(6,2,1,1)(7,3,1,1). Here is to show it is standard. The result is here. It is actually the last example on the calculator

    • (0,0,0,0,0)(1,1,1,1,1)
    • = (0,0,0,0)(1,1,1,1)(2,2,2,2)(3,3,3,3)...
    • > (0,0,0,0)(1,1,1,1)(2,2,2,2)
    • = (0,0,0,0)(1,1,1,1)(2,2,2,1)(3,3,3,1)...
    • > (0,0,0,0)(1,1,1,1)(2,2,2,1)...
    • = (0,0,0,0)(1,1,1,1)(2,2,2,0)(3,3,3,1)...
    • > (0,0,0,0)(1,1,1,1)(2,2,2,0)
    • = (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,4,1,1)(5,5,1,1)...
    • > (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,4,1,1)
    • > (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,4,1,0)
    • > (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,4,0,0)
    • > (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,3,1,1)
    • > (0,0,0,0)(1,1,1,1)(2,2,1,1)(3,3,1,1)(4,3,1,0)
    • > (…
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  • PsiCubed2

    There's a guy here who seems to be fond of spamming my personal space here with nonsense. He does this both by spamming the comment sections of my blog posts, and by creating fake subpages to my user page.

    The first kind of abuse was already taken care of, by disabling the comments on all my blog posts.

    As for the second kind - I'm taking care of it with this post, by announcing the following statement:

    I never EVER create subpages to my username. Never did, and never will. So if you see such a page, you can be 100% certain that it is a fake. If you're an admin, please delete it. If you're a regular user, please ignore it because it most certainly isn't my work.

    Thank you.

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  • Scorcher007

    Analysis DAN up to Z2

    August 28, 2018 by Scorcher007

    My analysis is based on the following assumption:
    1) DAN is limit of П1n-CA0
    {n,n(1,,1,2)2} ~ П11-CA0
    {n,n(1,,,1,2)2} ~ П12-CA0
    {n,n(1,,,,1,2)2} ~ П13-CA0
    e.t.c.
    2) Hypcos hypothetical analysis, that:
    {n,n(1(1(1(1...(1,,2,,)...2,,)2,,)2,,)2)2} = {n,n(1(1',,2,,)2)2} ~ KP+Пn

    The analysis was conducted on the basis of pattern in the stable ordinals, which are similar to smaller ordinals
    Stable ordinals notation:
    S[a+n] = La1La+n
    S[a+S[a+1]] = La1La+Lb≺Lb+1
    S[Ωa+1] = La1Lωa+1CK = (+)-stable (zoo 2.8)
    S[Ia+1] = La1LIa+1 = inaccessibly-stable (zoo 2.11)
    S[Ma+1] = La1LMa+1 = Mahlo-stable (zoo 2.12)
    S[a2+1] = La1Lb1Lb+1 = doubly (a+1)-stable = (zoo 2.13)
    S[aω+1] = ω-ly (a+1)-stable = nonprojectible (zoo 2.15)
    S[aα+1] = α-ly (a+1)-stable
    S[I-aα+1] = 1st inacces…


















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  • GdawgGamerTV

    Help

    August 25, 2018 by GdawgGamerTV

    I did coin the numbers in my blog posts but somehow they are getting turned into pages and it says I did it. Will somebody fix this?

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  • GdawgGamerTV

    Heres my attempt at beating Jonathan Bowers' Utter Oblivian.

    I call it Oblivous Plex

    Utter OblivianUtter Oblivian * Graham's Number * (Bisuperiorgraham * Bisuperiorgraham * BIG FOOTGoogolplexianth * Oblivian * Oblivian * Oblivian * Googolplexian * 100Bisuperiorgraham * Bisuperiorgraham * Utter Oblivian

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  • GdawgGamerTV

    My old coined numbers, a small Superior series, I will be adding onto it. This may contain very VERY large numbers, such as Bisuperiorgraham and Bisuperior-Rayo

    Superior-quattordecillion =  QuattordecillionQuattordecillion 

    Superior-quindecillion = QuindecillionQuindecillion

    Superior-sexdecillion = SexdecilionSexdecillion

    Superior-septendecillion = SeptendecillionSeptendecillion

    Superior-octodecillion = OctodecillionOctodecillion

    Superior-googol = GoogolGoogolGoogol

    Superior-googolplex = GoogolplexGoogolplexGoogolplex

    Superior-rayo = Rayo's NumberRayo's Number

    Bisuperior-million = MillionMillionMillion

    Bisuperior-billion = BillionBillionBillion

    Bisuperior-trillion = TrillionTrillionTrillion

    Bisuperior-quadrillion = QuadrillionQuadrillionQuadrillion

    Bis…

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  • GdawgGamerTV

    Yearillion, as 10100,000,005

    Gamma-raytillillion as 10150,762

    Yihuh as 2,121,887,952

    Yiplex as 999,057,912

    Yifufi as 987,638,996

    Yinin as 819,276,947

    Yidid as 813,996,888

    Yikin as 804,887,450

    Yimen as 801,086,337

    Yiunu as 67,956,111

    Yicuc as 67,894

    Yidud as 3,195

    Yipup as 894

    Yimum as 847

    Yibase as 5.92342345666645584884585847645

    Yiutu as 1.4

    Short Yi-bucket as 0.0000000000000000000000000000007

    Long Yi-bucket as 0.00000000000767

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  • MachineGunSuper

    Hello. I have made many ordinal functions that reached the Omega Fixed Point, but what's next? How do you get past the Omega Fixed Point?

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  • MachineGunSuper

    Supertasks

    August 25, 2018 by MachineGunSuper

    I've got a few things to cover in this blog bost:

    I already know what they mean, but in case you don't, it means doing an infinite amount of stuff in a finite amount of time.

    A great example is with a cake:

    Cut a cake in half, then wait 30 seconds before cutting the half in half. Then wait 15 seconds and cut again. and if you keep cutting into half and each time wait half as much, you would in a finite amount of time cut and infinite amount of slices and place them in a finite space.

    My favourite examples are these:

    If we built a machine that could display the digits of pi like this: Start with the 3, then in 30 seconds the 1 then in 15 the 4 and so on, we would display all the digits of pi. What would be on the screen now?

    Second, if on a monit…

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  • Scorcher007

    Help those who understand the hierarchy of ordinals to the level of Z2. Have I built the list of ordinals correctly?
    (zoooo #) means number of ordinals in Madore's Zoo of ordinals

    countable ordinal ~ equiv. cardinal ~ PTO
    1• ω1CK = 1st admissible = П10-comprehension = (zoooo 2.1) ~ ω1 = 1st uncountable = 1st 1-regular ~ OCF→PTO ACA0
    2• ω2CK = 2nd admissible ~ ω2 = 2nd uncountable = 2nd 1-regular ~ OCF→KPω
    3• ωωCK = ω-th admissible = П11-comprehension = (zoooo 2.2) ~ ωω = ω-th uncountable = ω-th 1-regular ~ OCF→PTO П11-CA0
    4• I ~ recursively inaccessible ~ weakly inaccessible = (zoooo 2.3) ~ OCF→PTO KPi
    5• M ~ recursively Mahlo ~ weakly Mahlo = (zoooo 2.5) ~ OCF→PTO KPM
    6• K ~ П3-reflection = 2-admissible = (zoooo 2.6) ~ П11-indescribable = 2-regular = weakl…








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  • GdawgGamerTV

    Coining a new number

    August 25, 2018 by GdawgGamerTV

    I hereby state that I shall coin these three new numbers. King's dozen as 23, King's gross as 223 and Kingstillion as 1010001000050000450

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  • GdawgGamerTV

    hi

    August 24, 2018 by GdawgGamerTV

    hello, everyone, I am Sven and I like numbers k  I also claimed some new numbers, wilkillion, welkillion, welkillillion and more, you can see them here https://sites.google.com/site/numberwikipedia/the-wilkinson-array

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  • DatoXx8

    my array notation

    August 24, 2018 by DatoXx8

    {n}=n^[n]n         n^[n]n=n^^^^...n arrows…^^n

    {n,m}= {n,m-1} with {n,m-1} sets of { } brackets around n



    {n,m,o} =}}}}}

    {n,m,o,p}= }...{n,m,o,p-1}...}}}



    n_0= n^[n]n!!!!!.....n^[n]n!!!!!.....n^[n]n!!!!!..............

            itterating n^[n]n!!!!...n !`s…!  =

    n_1 is ((((((n_0)_0)...n_0 times…._0)))



    n_2 is ((((((n_1)_1)...n_1 times…._1)))

    ….

    a_b_c= a_(a_(a_…a_b_c-1 times…(a_b))))))....)



    a_b_c_d= a_b_(a_b_(a_b_(......a_b_c_d-1 times….a_b_c_d-1)))).....)

    ….

    =n[m]= n_n_n_n…….n[m-1] times…._n

    n[m{a}]=n[n[n[n[n[.....n[m{a-1}] times……n[m{a-1]]]]....]]]

    n[m{a\b}]=    n[m{a^[n[m{a]]b}]=(step 1)

    step 2

    step 1 but a replaced with step 1  2 times   

     step 3

    step 2 but a replaced with step 2 3 times  

    ….till step b

    n[m{a\b#c}] =

    n[m{a\b}] but b replaced with n[m{a\b#c-1}]  …









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  • BlauesWasser

    Just wondering... Because, you know... I'm a "noob" so.. Yeah, idk, just wondering..

    Let's take my recently created function, and find the growth rate. (Another user said that the limit to this function is F Omega+1(n) in the FGH) 

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  • IAmNotATRex

    I have decided to write an entire blog post of examples using my FGH extension to (hopefully) give people a greater understanding of it. To understand this blog post, you need to know the definition of my extension, which can be found here: https://googology.wikia.com/wiki/User_blog:IAmNotATRex/Array-Based_Extension_to_the_FGH. I advise that you keep two tabs open, one with this page and one with the definition of my extension, in case you ever get confused.


    I'll start off with a simple example: \(f[0, 0](2)\).

    If you also understand my old extension, know that this is equal to \(F1_{0}(2)\) using my old extension. However, that isn't very important, considering the fact than my new extension is significantly more powerful than my old extens…


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  • BlauesWasser

    So... I have created  new function... Tell me if it's ill-defined..  Hopefully it's clear (I'm technically still a beginner, so.. yeah)

    {a,b} = \(a^b\)

    {a,b,c} = \(a↑^c b\)

    {a,b,c,d} = \(a↑^{(c^d)}b\) or {a,b,{c,d}}

    {a,b,c,d,e} = \(a↑^{(c↑^ed)}b\) or {a,b,{c,d,e}}

    etc, etc... There can be more entries, just use the other letters in the alphabet; it's just the same logic..

    Here are some examples:

    {3,3} = \(3^3\)

    {3,3,3} = \(3↑^3 3\)

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  • IAmNotATRex

    After creating my own extension to the FGH, I have decided to create a more powerful extension using arrays. My previous extension can be found here: https://googology.wikia.com/wiki/User_blog:IAmNotATRex/My_Extension_to_the_Fast_Growing_Hierarchy.


    For this blog post, the notation for arrays I will use is \([a_{1}, a_{2},\ldots, a_{k-1}, a_{k}]\), where \(a\) is the name of the arrays and \(k\) is its length. For the sake of clarity, \([\langle a\rangle]\) will be equal to the array \(a\) instead of \(a\) nested inside a one-element array.


    I will first define a less powerful function, \(f[\langle a\rangle](n)\), and then a more powerful function, \(F_{a}[\langle b\rangle](n)\). If I made a mistake in my logic or a typo anywhere, please let m…



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  • Alemagno12

    Quick poll

    August 23, 2018 by Alemagno12

    survey

    How exactly do each of you guys come up with your results for a growth rate of a notation? (Not how you prove it, i.e if you either do analysis tables or normal justifications, but rather how you come up with it)

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  • Gleive

    Yes, we all know that the initials are GUN.

    I've been working on this notation for quite some time. Ever since I picked up googology this is the most refined work I've done. Most others are (Not naive, but not that significant either) extensions on existing notations or other projects that ended up being salad notations or not getting past linear beaf level.

    This will probably be finished a few days later. Will upload the contents to both my website and to here.

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  • BlauesWasser

    I still feel like a beginner, I don't believe I know enough googological stuff.. For example, I don't understand fgh.. What are some links to help get started? (again)

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  • MilkyWay90

    I am ready to start learning the easiest OCF!

    I have learned about finite parameter multi-variable Phi function (phi(1, 1, 0) = the limit of gamma_n), and the definition of an uncountable ordinal (an ordinal where there is no function f(n) where for every element in the ordinal (y), there will be a value f(x) = y where x is a natural number)



    So I am ready to start learning (the easiest) OCFs!



    So can somebody give a link/definition in the comments to whichever OCF is easiest to learn in their opinion?

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  • IAmNotATRex

    I don't know much about googology or ordinal numbers, but I thought it would be fun to try to create my own extension to the fast growing hierarchy. I don't know if this has been done before or how this compares to other notations or functions, so I'd appreciate it if someone could compare this to other forms of notation or other functions.

    I don't have much experience formatting with LaTeX, so please forgive me if something doesn't look good.


    I haven't really thought of a good name or notation, so I'll just use temporary notation for this blog post and refer to this as "my extension."

    My new extension uses the following format: \(Fn_a(b)\), where \(F\) is the name of the function.

    For \(n=0\), \(F\) simply reduces to the fast growing hierarch…


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  • Arasena

    The longest time, after which everything repeats, is Poincaré recurrence time for the Universe, which is around \(10^{10^{10^{10^{2.08}}}}\) (approximately a googolplexian) Planck times. Thus, it will have exactly the same state as for now. Is this time the largest number?

    If the Universe is infinite, does Poincaré recurrence time exist? Or if sub-atomic particles can be divided infinitely?

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  • Koteitan

    This is the table of the rule sets of (almost) all sub versions of Bashicu Matrix invented in 2015-2018.

    The aim of this entry is the comparison of the rule sets of them. The all sub versions can be categorized according to the difference of the three micro rules; Bad root searching rules, Bad part ascending rules and Ascension modification rules. In this article, the all sub versions are re-written in the mathematical definition with the same format. So that the comparison become easier. Please use this for your proof of the termination or analysis.

    Here is the Japanese version of this article.


    Rule sets
    Authors
    Dates Definitions Bad root searching Bad part ascending Ascension modification discussions


    BM1
    Bashicu
    21,August'15

    • BASIC pseudo code

    Left met…







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  • DontDrinkH20

    WARNING!!! You should be well-equipped with model theoretic terminology and computability theoretic terminology before you work with this paper, as well as some basic set theory. Feel free to grab a notebook and take some notes as you go along. You might need them; this one is a real doozy.


    There was originally going to be another subclass of Type C called the representable googolisms. These would be a weakening of the expressible googolisms where "metamathematical" was replaced by a less expressive type of language. However, I then realized that there was no known concise definition of "a natural number can be expressed by a language" so I decided to make one.

    It turns out that it doesn't require just a language, but rather a structure (whic…


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  • Rpakr

    In this blog post, I will define BM1 pair sequences formally without using things like ... which are ambiguous, and defines BM1 number.


    In this section I will define some functions, sets, etc. I will be using in the definition. When I use quotation marks, it means the string inside them should be treated as strings and not as numbers or expressions.


    \(\mathbb{N}_0\) is the set of all nonnegative integer.

    \(\mathbb{N}_0=\mathbb{N}\cup\{0\}\)


    When A and B are strings, \(con(A,B)\) is defined to be the string that is the concatenation of A and B.


    Define \(Pair(a,b)\) to be the string “(a,b)” when \(a\) and \(b\) are nonnegative integers.

    \(Pair(a,b)=P\Leftrightarrow (a\in\mathbb{N}_0∧b\in\mathbb{N}_0∧P=con(“(“,con(a,con(”,”,con(b,”)”)))))\)


    \(Pair(A…







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  • DatoXx8

    First post

    August 14, 2018 by DatoXx8

    I got a few functions that grow fairly fast but sadly i only got  them on google docs so.... here are the links

    smallest: 

    https://docs.google.com/document/d/1PQ08ZFbEu8g1qsqm4u8HCJfJp7PPhZ3aKrZsoa9EmgU/edit?usp=sharing

    medium:

    https://docs.google.com/document/d/1E18XQOEI-HKNFuEKsDrNIo7yZU9MvuwqTjcspl0XkfI/edit?usp=sharing

    giant:

    https://docs.google.com/document/d/14y_YlzB_i2I3njz5DrGOLBFwpy8HXPQPwmOIEdkW8sk/edit?usp=sharing 

    but they aren`t well defined

    and they have tons of opertunety`s for salad numbers    

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  • DontDrinkH20

    In my previous blog post, I rattled on for several paragraphs about my "Type A, B, and C" classifications of googological specimen. In this post, I will go into detail on the expanded full hierarchy I have since completed. The motivation behind these classifications is the fact that there are obvious differences in the ways certain googologies are defined, and usually those differences can show up in their size as well.


    The set of imaginable googolisms is the smallest type of googolism. These specimen can be easily imagined as quantities in the brain. For example, the amount of letters in a novel is imaginable. These are usually not considered true googolisms. Here are some examples of imaginable numbers:

    • \(6\)
    • \(100\)
    • \(1532\)
    • \(2167\)
    • Kaprekar's…


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  • Eners49

    New notation idea?

    August 14, 2018 by Eners49

    Here is the basics:

    • {a} = a!!!...!!! with a nested factorial signs. Pretty basic for 1-entry arrays.

    That's already a really powerful function. Just a single factorial sign is powerful by the layman's standards, but if we have a bunch of them, that's going to be amazing!

    • {0} = 0
    • {1} = 1! = 1
    • {2} = 2!! = 2! = 2
    • {3} = 3!!! = 6!! = 720! = 10^1,746.42
    • {4} = 4!!!! = 24!!! = 10^10^10^24
    • In general, {a} is about 10^^a

    With one-entry arrrays, we already have created a strong function, but we're only in between f2(n) and f3(n) in the fast-growing hierarchy, so we need to keep going. Let's see what two-entry arrays look like:

    • {a, b} = }...}}}, b-1} where there are {a, b-1} brackets. We've created a pretty strong recursion so far.
    • {a, 1} = {a}. Just like BEAF, …
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  • Gleive

    Okay so this just made my damn day.

    Warning : The comment in this image may be offensive for some people.

    Click here to go back if you don't want to look.

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  • P進大好きbot

    Hi. I am studying the strength and the termination of pair sequence system, and I need your help!

    Could you tell me the pair sequences of standard form corresponding to the following ordinals?

    1. \(\psi_0(\psi_1(\psi_2(\psi_3(0))))\)
    2. \(\psi_0(\psi_1(\psi_2(\psi_3(\psi_4(0))))\)
    3. \(\psi_0(\psi_3(0))\)
    4. \(\psi_0(\psi_4(0))\)
    5. \(\psi_0(\psi_3(\psi_2(0)))\)
    6. \(\psi_0(\psi_3(\psi_2(\psi_3(\psi_2(0)))))\)
    7. \(\psi_0(\psi_3(\psi_3(0)))\)
    8. \(\psi_0(\psi_3(0)+\psi_2(\psi_3(0)))\)
    9. \(\psi_0(\psi_3(0)+\psi_3(0))\)

    Here, \(\psi_n\) denotes Buchholz's OCF. (I know that I listed up the same ordinal more than one time.) Since they are smaller than \(\psi_0(\Omega_{\omega})\), they are strongly believed to be uniquely presentable by pair sequences of standard form. Please tell…

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  • DontDrinkH20

    There are really three types of googolisms. There are those which are very very large but have many equivalent definitions, like TREE(3) and Graham's number, and there are others which are very very large but are defined in such a way that they don't have many equivalent definitions, such as Rayo's number, BIG FOOT, and the current largest valid googolism, Little Bigeddon.


    The first type of googolisms mentioned above is what I like to call Type A, the second Type C. Type A objects are abundant on the wiki, by far the most often-made googolism. There are really only a few Type C googolisms (about 4 well-known ones). Type A has numbers like googolplex and Graham's number as well as functions like BEAF and FGH. Type A objects generally contain def…



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