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

    My version of NDAN

    February 11, 2018 by Bubby3

    I am going to shjow some milestone sequences in my proposed notation

    {1@2}  = {1`2}, {1{1,,3}2}, {1{1{1,,,3}2}2}, etc.

    {1@3} = {1`2@2}, {1{1,,3}2@2}, {1{1{1,,,3}2}2@2}, etc.

    {1@1@2} = {1@1`2}, {1@1{1,,3}2}, {1@1{1{1,,,3}2}2}, etc.

    {1{2@,2}2} = {1@2}, {1@1@2}, {1@1@1@2}, etc.

    {1{1,,2@,2}2} = {1{1,2@,2}2}, {1{1{1,2@,2}2@,2}2}, {1{1{1{1,2@,2}2@,2}2@,2}2}

    {1{1,,3@,2}2} = {1{1,2,,2@,2}2}, {1{1{1,2,,2@,2}2,,2@,2}2}, {1{1{1{1,2,,2@,2}2,,2@,2}2,,2@,2}2}

    {1{1@,3}2} = {1{1,,2@,2}2},  {1{1{1,,,3}2@,2}2}, {1{1{1{1,,,,3}2}2@,2}2}, etc.

    {1{1{2@,,2}2}2} = {1{1@,2}2}, {1{1@,1@,2}2}, {1{1@,1@,1@,2}2}, etc.

    {1@@2} = {1@2}, {1{1@,3}2}, {1{1{1@,3}2}2}, etc.

    {1@@@2} = {1@@2}, {1{1@@,3}2}, {1{1{1@@,,3}2}2}, etc.

    {1{1,2}@2} = {1@2}, {1@@2}, {1@@@2}, etc.

    {1{1,,2}@2} = {1{1,…

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  • Bubby3

    Taranovsky's ordinal notation (C(a,b)) might seem complex at the surface, but his 1st system is rather simple if you take an alternate way of explaining it. However, this function gets really complex and weird at 2nd system and higher, which I won't explain in this blog post. 

    Taranovsky's ordinal function is not your typical ordianl collapsing function. First of all, numbers (other than 0), operators, etc., are not defined in this system. Instead, you have to use this ordinal notation to define them with his notation. So when I am saying an expression is equlivant to an ordinal, I am saying that that the ordinal is shorthand for that expression for now and no, to save space.

    Another weird thing about it which seperates it from most other or…

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  • Bubby3

    Dropping Hydra notation

    December 19, 2017 by Bubby3

    First- and second-order brackets are like normal Bucholoz hydra notation, with n nests.

    However, third- and higher- order brackts diverge from it.

    My notation is () are first, [] are second, {} are third, and are fourth order brackets. 

    In my notation, all second or higher-order brackets must be directly or indirectly inside a first order bracket.

    My notation is a|b, where a is a number and b is a hydra, where a only increases when you remove a first-order bracket not inside any brackets.

    If you find an nth order bracket for n > 1

    1. Find the outermost n-1th-order bracket inside the n-th order bracket. If it is lower than an n-1th-order bracket, add an n-th order bracket directly inside it, B_(n-1) is the n-1st order seperator. 
    2. Set i to n
    3. Repeat th…
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  • 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…

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  • Bubby3

    FGH on ordinals

    November 5, 2017 by Bubby3

    Here I am going to do FGH on ordinals.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    \(f_3(f_3(\omega))\) =…

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