FANDOM

Bubby3

aka Matthew

  • I live in Ohio, United States
  • I was born on February 17
  • My occupation is Too young to get a job
  • I am Male
  • Bubby3

    I was impatiently watiting for Sbiis Saibian to publish his extension of Extended-Cascading-E notation (and all the pages on his site that say that they are on hiatus). He even said that he was going to return to working on his site in 2017. Now, 2017 is almost over and he still hasn't updated his site since February (when he posted his 2017 Renewal Plan). This is a ncase of broken promises. He promised to work on his site more in 2017, but in 2017 he was no more active on his site than he was in 2016. Because of his inablity to keep promises, I predict he will never update his site again.

    Because of all of his inactivity, I plan on making my own fan extension to #xE^, which I will compare with his ad hoc extension. Here are some funermenta…

    Read more >
  • Bubby3

    FGH on ordinals

    November 5, 2017 by Bubby3

    Here I am going to do FGH on ordinals.

    \(f_0(\omega)\) = \(\omega + 1\)

    \(f_1(\omega)\) = \(\omega \cdot 2\)

    \(f_0^\omega(f_1(\omega))\) = \(\omega \cdot 3\)

    \(f_1(f_1(\omega))\) = \(\omega \cdot 4\)

    \(f_2(\omega)\) = \(\omega^2\)

    \(f_1(f_2(\omega))\) = \(\omega^2 \cdot 2\)

    \(f_1^\omega(f_2(\omega))\) = \(\omega^3\)

    \(f_1^{\omega \cdot 2}(f_2(\omega))\) = \(\omega^\4\)

    \(f_2(f_2(\omega))\) = \(\omega^\omega\)

    \(f_2(f_2(f_2(\omega)))\) = \(\omega^{\omega^\omega}\)

    \(f_3(\omega)\) = \(\varepsilon_0\)

    \(f_1^\omega(f_3(\omega))\) = \(\omega^{\varepsilon_0 + 1}\)

    \(f_2(f_3(\omega))\) = \(\omega^{\varepsilon_0 \cdot 2}\)

    \(f_2(f_2(f_3(\omega)))\) = \(\omega^{\omega^{\varepsilon_0 \cdot 2}}\)

    \(f_2^\omega(f_3(\omega))\) = \(\varepsilon_1\)

    \(f_3(f_3(\omega))\) =…

    Read more >
  • Bubby3

    BEAF ordinals, my analysis

    October 18, 2017 by Bubby3

    $ is a string of 1's (or can be empty)

    @ is a string of any entries (also can be empty

    Rules for funemental sequences:

    If the last entry is 1, then remove it

    the second entry is 1, {a,1,@}[n] = 1

    1. If the array has two entries, then it is Cantor Normal form
    2. If the second entry can be expressed in the form of n+X for any X, then {a,b+X,c,@}[n] = {{a,b,c,@},X,c,@}[n]
    3. If the second entry is a limit that can't be expressed in the form n+X, then {a,b,@}[n] = {a,b[n],@}
    4. If the third entry is a limit, then {a,b,c,@}[n] = {a,b,c[n],@}
    5. If the third entry is a 1 and the first non-1 entry after the second entry is a limit, then {a,b,1,$,c,@}[n] = {a,b,1,$,c[n],@}.
    6. Otherwise, consult the rules for normal BEAF.

    Seperators can have ordinals. Ordinals can also be in …


    Read more >
  • Bubby3

    Start with a givern collection of particles, and you can perform steps

    First-order step: Replace each atom with \(10^{80}\) atoms.

    n+1th order step, for \(n \geq 1\), count the number of atoms in the universe, call that number m, and repeat an n-1th order step m times

    Superstep. Count the number of atoms in the universe, call it n, and do an n-th order step on n atoms

    Bigger than G: Do a superstep on the observable universe (10^80 atoms) 64 times.

    Read more >
  • Bubby3

    In Strong Array Notation DAN, it has the idea of n-seperators. Tehy represent the 'level' of a seperator. For example, the grave accent and every seperator containing a double comma is a 1-seperator. 2-seperators are either double commas or seperators directly containing triple-commas. The level of a seperator is defined as follows. Clvl stands for comma level

    Clvl of a comma is 0 and the Clvl of a grave accent is 1 Clvl(,...,) with n commas is n for n > 1

    Clvl({a1A1a2...an-1An-1an}}, where a's are numbers and A's are separators = Max(Max(Clvl(A1),Clvl(A2)...Clvl(An-1),Clvl(An))-1,0). So the comma level of a separator is one less than the comma level of the separator with the highest comma level within that separator, and 0 if the the highes…

    Read more >

Ad blocker interference detected!


Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.