aka huehuehue

  • I live in somewhere that is omega kilometers away from you
  • I was born on April 14
  • D57799

    According to Knuth Arrows Theorem by Sbiis Saibian,

    for all \(a≥2, b≥1, c≥1, k≥2: (a\uparrow^kb)\uparrow^kcn+3\)





    \(>a\uparrow^{k+1}(n+2)\times M(n-1)\)

    \(>a\uparrow^{k+1}(n+2)\times (n+2)\)







    for all \(n∈N:a\uparrow^k(M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)>M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}(n+1)\)





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  • D57799

    Proofs not included. Some of the following have formal proof, some don't.

    1. \((a\uparrow^nb)\uparrow^nc=a\uparrow^n(b\times c)\)

    The equation holds when n=0 or 1 (n=0 for multiplication)

    It's a simple property but it doesn't have a name.

    This is useful for extention to rational numbers.


    Also, \((a\uparrow^n{1/p})\uparrow^np=a\), so \(a\uparrow^n{1/p}=a\downarrow_R^np\)

    Therefore, \(a\uparrow^n(q/p)=(a\uparrow^nq)\downarrow_R^np\)

    (Notation explanation in 10.)

    2.\((a\uparrow\uparrow b)\uparrow\uparrow c1\)

    6.for all \(n \geq 1\), \(y=a\uparrow^n x\) passes (0,1), (-1,0), ..., (1-n,2-n), n points in total. All on the magic line of y=x+1

    Define that  \(K_n\) is the biggest solution smaller than zero t…

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  • D57799

    Originally I wanted to put the proof that uparrow notation is strictly increasing, but it turned into an extremely long proof. I cut it off and will probably put it in another post.

    For all a,b,c > 2

    When m = 1, \(a\uparrow(b\uparrow c) > (a\uparrow b)\uparrow c = a\uparrow(b\times c) \) holds.

    Assume that : \(a\uparrow^m(b\uparrow^mc)> (a\uparrow^mb)\uparrow^mc\) for \(m\geq1\)

    \((a\uparrow^{m+1} b)\uparrow^{m+1} c\)



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  • D57799

    Ranging FGH before omega

    October 7, 2014 by D57799

    The basic definitions:

    1. \(f_0(n)=n+1\)

    2. \(f_{\alpha+1}(n)=f_\alpha^n(n)\)

    FGH here is smaller than \(f_\omega(n)\), so the situation of when \(\alpha\) is a ordinal is ignored.

    FGH and intuitional upper & lower bounds
    lower FGH upper
    1 \(2n\) \(2n\) \(2\uparrow n\)
    2 \(2\uparrow n\) \(n2^n\) \(2\uparrow\uparrow n\)
    3 \(2\uparrow\uparrow n\) \(...\) \(2\uparrow\uparrow\uparrow n\)

    The most obvious lower bound of \(f_m(n)\) is \(2\uparrow^{m-1}n\), so let's prove it.  

    1. Take \(f_2(n)\) as our starting point, for \(n\geq1\) , \(f_2(n)=n2^n\geq 2\uparrow n\) , the inequality holds.

    2. Assume that \(f_m(n)\geq2\uparrow^{m-1}n\) for \(n\geq1\),



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  • D57799

    feeding FGH into FGH

    October 5, 2014 by D57799

    There may be some informal or invalid step in my proof. Just point it out. There's still a lot for me to learn.

    Feeding FGH into itself was considered before. I just want to find the limit of it and probably some usage of it.

    It looks like this:

    1. . It's theoretically the limit of FGH itself. But with Ordinal Collapsing Function that is more powerful and inaccessible ordinals, it is possible for FGH to go beyond its limit.

    --D57799 (talk) 06:24, October 5, 2014 (UTC)

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