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# D57799

## aka huehuehue

• I live in somewhere that is omega kilometers away from you
• I was born on April 14
• ## Another Theorem for Knuth Arrows

April 10, 2015 by D57799

According to Knuth Arrows Theorem by Sbiis Saibian,

for all $$aâ‰¥2, bâ‰¥1, câ‰¥1, kâ‰¥2: (a\uparrow^kb)\uparrow^kcn+3$$

$$a\uparrow^kM(n)$$

$$=a\uparrow^k(M(n-1)+a\uparrow^{k+1}(n+1))$$

$$>(a\uparrow^ka\uparrow^{k+1}(n+1))\uparrow^kM(n-1)$$

$$=(a\uparrow^{k+1}(n+2))\uparrow^kM(n-1)$$

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

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

$$>M(n+1)$$

Therefore,

$$(a\uparrow^k)^{b+c-3}M(1)$$

$$>(a\uparrow^k)^{c-1}M(b-1)$$

$$=(a\uparrow^k)^{c-1}(M(b-2)+a\uparrow^{k+1}b)$$

lemma:

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

proof:

$$a\uparrow^k(M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)$$

$$=a\uparrow^k(M(b-3)+a\uparrow^{k+1}(b-1)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)$$

$$=(a\uparrâ€¦ Read more > • ## Some discoveries and guesses about uparrow notation January 28, 2015 by 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.

$$a\uparrow^n(q/p)=(a\uparrow^nq)\uparrow^n{1/p}$$

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 tâ€¦

• ## Some basic properties of uparrow notation

October 11, 2014 by 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 $$m\geq1$$

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

$$=\underbrace{\underbrace{(a\uparrow^ma...a)}_{b}\uparrow^m...\uparrow^m\underbrace{(a\uparrow^ma...a)}_{b}}_{c}$$

$$Read more > • ## 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 $$n\geq1$$,

$$f_{m+1}(n)=\underbrace{f_m(f_m(...(f_m}_{n}(n))...))$$

\(2\uparrow^mn=\underbrace{2\uparrow^{m-1}2\uparrow^â€¦

• ## 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)