User:Ynought

My largest number so far is Yallun 7 "My website ""Array hierachy"My notation playground

My googolism playground

My : notation
I will try to make this my fastest growing recursive notation.Here it is.

My graph function "E"
I will try to make this my fastest growing graph function.Here it is.

Ordinal notation
Here it is.

A series of numbers "Yallun"
Here it is.

Monster notation(s)
Here it is.

Turing Machine stuff
Here it is.

Classes of numbers
\(\mathcal{O}(a,b)\) is the set of all numbers greater than \(a\) and less or equall to \(b\)

When i say \(\wp=\mathcal{O}(a,b)\) that means that \(\wp\) is the set of \(\mathcal{O}(a,b)\)

Alpha
\(\alpha=\mathcal{O}(0,1)\)

Beta
\(\beta=\mathcal{O}(1,10)\)

Gamma
\(\gamma=\mathcal{O}(10,100)\)

Delta
\(\delta=\mathcal{O}(100,10^{100})\)

Epsilon
\(\epsilon=\mathcal{O}(10^{100},^{100}10)\)

Zeta
\(\zeta=\mathcal{O}(^{100}10,G_{100})\) \(G\) is grahams function

Eta
\(\eta=\mathcal{O}(G_{100},f_{\omega^2}(100))\)

Theta
\(\theta=\mathcal{O}(f_{\omega^2}(100),f_{\omega^\omega}(100)\)

Iota
\(\iota=\mathcal{O}(f_{\omega^\omega}(100),f_{\varepsilon_0}(100))\)

Kappa
\(\kappa=\mathcal{O}(f_{\varepsilon_{0 }} (100),f_{\varepsilon_{100 }} (100))\)

Lambda
\(\lambda=\mathcal{O}(f_{\varepsilon_{100 }} (100)),f_{\zeta_0}(100)))\)

Mu
\(\mu=\mathcal{O}(f_{\zeta_{100 }} (100),f_{\eta_0}(100))\)

Nu
\(\nu=\mathcal{O}(f_{\eta_0}(100),f_{\varphi(\omega,0)}(100))\)

Xi
\(\xi=\mathcal{O}(f_{\varphi(\omega,0)}(100),f_{\text{LVO }} (100))\)

Omicron
\(\omicron=\mathcal{O}(f_{\text{LVO }} (100),f_{C(1)}(100))\)

Pi
\(\pi=\mathcal{O}(f_{C(1)}(100),f_{C(\omega)}(100))\)

Rho
\(\rho=\mathcal{O}(f_{C(\omega)}(100),f_{C(C(1))}(100))\)

Tau
\(\tau=\mathcal{O}(f_{C(C(1))}(100),f_{C(C(\omega))}(100))\)

Upsilon
\(\upsilon=\mathcal{O}(f_{C(C(\omega))}(100),f_{C^{100}(\omega)}(100))\)

Phi
\(\phi=\mathcal{O}(f_{C^{100}(\omega)},f_{C(\Omega)}(100))\)

Chi
\(\chi=\mathcal{O}(f_{C(\Omega)}(100),f_{C(\Omega_{100})}(100))\)

Psi
\(\psi=\mathcal{O}(f_{C(\Omega_{100})}(100),f_{\text{PTO}(\text{SMAH+})}(100))\)

Omega
\(\Omega=\mathcal{O}(f_{\text{PTO}(\text{SMAH+})}(100),n)\) where \(n\) is larger than the halting time of every turing machine with HUGE+ proof of halting with lenght of at most \(^{^{100}100}100\)

Digamma
\(\digamma=\mathcal{O}(n,\omega)\) the \(n\) from \(\Omega\)

Milestones:
1000.Edit:11.03.2019

...

TEST

\(\frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} }\frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}}}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} }\frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}}}}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} }\frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}}}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} }\frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }} \frac{\omega}{\omega\frac{\omega}{\omega}\frac{\omega}{\omega\frac{\omega}{\omega }}}}}}}} \)

\(\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+\frac{3+}{\Omega }}{\Omega}} {\Omega }} {\Omega }} {\Omega }} {\Omega }}{\Omega}} {\Omega }} {\Omega }} {\Omega }} {\Omega }}{\Omega}} {\Omega }} {\Omega }} {\Omega }} {\Omega }}{\Omega}} {\Omega }} {\Omega }} {\Omega}\)

\(2:(3,3,(1))=2:(3,3,(0)(0))=2:(3,3,(0)2)=f_2^2(2):(3,3,(0))(3,3,(0))\)

\(=f_2^2(2):(3,3,(0))(3,3,f_2^2(2))\) from there on it gets too big too write down i a reasonable amount of time and storage space

\(\begin{eqnarray*} & & 2:(0,2) \\ & & =2:((0,1)^{(0,1)},1) \\ & & =2:((0,1)^{((0,0)^{(0,0)},0)},1) \\ & & =2:((0,1)^{((0,0)^{(0)^{(0)}(0)^{(0) }} ,0)},1) \\ & & =2:((0,1)^{((0,0)^{(0)^{(0)}(0)^{2 }} ,0)},1) \\ & & =2:((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(2)))^{0})^{1 }} ,0)},1) \\ & & =2:((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(1)(1)))^{0})^{1 }} ,0)},1) \\ & & =2:((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(1)(0)(0)))^{0})^{1 }} ,0)},1) \\ & & =2:((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(1)(0)2))^{0})^{1 }} ,0)},1) \\ & & =f_2^2(2):((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(1)(0)))^{0})^{1 }} ,0)},1) \\ & & =f_2^2(2):((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(1)f_2^2(2)))^{0})^{1 }} ,0)},1) \\ & & =f_{f_2^2(2)}^{f_2^2(2)}(f_2^2(2)):((0,1)^{((0,0)^{(0)^{(0)}((0)^{1})^{1}(((0)^{0})^{0}(((0))(1)))^{0})^{1 }} ,0)},1) \end{eqnarray*}\)

\(\begin{eqnarray*} \gamma_{\beta}(\alpha) = \left\{ \begin{array}{ll} 1+\alpha & (\alpha < \beta < \varepsilon_0) \\ 1+\beta & (\beta < \varepsilon_0, \alpha \geq \beta) \\ \textrm{undefined} & (\beta \geq \varepsilon_0) \end{array} \right. \end{eqnarray*}\)

1    $ \langle ...,a,0,...\rangle[b,...] \rightarrow \langle ...,a-1,b,...\rangle[b,...] $ 2    $ \langle ...,a\rangle[b] \rightarrow \langle ...,a-1\rangle[(b,...,b)b] $ 3    $ \langle 0\rangle[...] \rightarrow [...] $ 4    $ (...,a,0,...)b \rightarrow (...a-1,b,...)b $ 5    $ [...,(...,a)b] \rightarrow [...,(...,a-1)b,...,(...,a-1)b] $ 6    $ (0)a \rightarrow a $ 7    $ [...,1] \rightarrow [...] $ 8    $ [...,1,a,...] \rightarrow [...,a,a-1,...] $ 9    $ [a,b,c,...,d] \rightarrow [[a,b-1,c,...,d],[a,b,c-1,...,d],...,d-1] $ 10    $ [a,b] \rightarrow [a+b] $ 11    $ [a] \rightarrow a $

A note on rule 9: in particular, $ \langle 1\rangle[(2)3,(4)5,6] \rightarrow \langle 1\rangle[(2)\langle 1\rangle[(2)3,(4)4,6],(4)\langle 1\rangle[(2)3,(4)5,5],5] $

We calculate the value of $ [(1)x] $:

$ [x,x] = x+x = f_1(x) $ $ [x,x,2] = [[x,x-1,2],[x,x,1],1] = [x,x-1,2] + x+x > x*x+x*(x-1)/2 > x^2 = f_2(x) $ $ [x,x,3] = [[x,x-1,3],[x,x,2],2] > [[x,x-1,3],x^2,2] > x^3 $ ... $ [x,x,x] > x^x = f_3(x) $ This can be extended, and we find that $ [(1)x] $ is about $ f_{2x-1}(x) $, roughly $ f_\omega(x) $.

We next calculate the value of $ [(2)x] $: $ [(1)x,x] = [(1)2x] = f_{4x-1}(2x) $ $ [(1)x,x,2] = [(1)[(1)x,x-1,2],[(1)x,x,1],1] = [(1)...[(1)x,1,2]...],f_{4x-1}(2x)] > f_{2x-1}^x(x) $ In fact, if $ g(y) = f_{2y-1}(y) $, then $ [(1)x,x,2] > g^x(x) $. This has power roughly $ f_{\omega2}(x) $

I conjecture, though I cannot prove it, that $ [(1)x,x,x] > g^{f_3(x)}(x) $, and $ [(1)x,x,...,x] > g^{f_{2x-1}(x)}(x) $. This has power roughly $ f_{\omega3}(x) $ This is still not close to $ [(2)x] $, which is $ [(1)x,(1)x,...,(1)x] $. We continue: $ [(1)x,(1)x] = [(1)x,x,x,...,x] = g^{f_{2x-1}(x)}(x) $ $ [(1)x,(1)x,2] = [(1)[(1)x,(1)x-1,2],g^{f_{2x-1 }} (x),1] $ We can see the pattern here, and predict that $ [(2)x] $ is roughly equivalent to $ f_{\omega^2}(x) $. If so, then further conjecture puts $ [(x)x] $ at $ f_{\omega^\omega}(x). $ $ [(1,0)x]=[(x)x] $, and $ [(1,1)x]=[(x+1)x] $ is only slightly larger. But when we get to $ [(1,2)x] $, we have $ [(1,1)x,(1,1)x,...,(1,1)x] $, meaning that it takes quite a few steps before rule 4 is applied more than once. I'm fairly certain that $ [(1,2)x] $ is at least $ f_{\epsilon_0}(x) $. I really am ill-equipped to judge. Further conjecture:

$ [(1,x)x] ~ f_{\epsilon_\omega} $ $ [(2,x)x] ~ f_{\epsilon_{\epsilon_\omega }} $ $ [(x,x)x] ~ f_{\sigma_0} $

\(f_{\sigma_0}(n)\)

\(\psi(\Omega^{\Omega}2) = \psi(\Omega^{\Omega} + \Omega^{\Omega})\) is the limit of the following sequence if I am correct: \begin{eqnarray*} & \psi(\Omega^{\Omega} + \Omega^{\omega}) & \\ & \psi \left( \Omega^{\Omega} + \Omega^{\omega^{\psi(\Omega^{\Omega} + \Omega^{\omega}) }} \right) & \\ & \psi \left( \Omega^{\Omega} + \Omega^{\omega^{\psi \left( \Omega^{\Omega} + \Omega^{\omega^{\psi(\Omega^{\Omega} + \Omega^{\omega}) }} \right)}} \right) & \\ & \vdots & \end{eqnarray*}\)

Let \(M\) denote the maximum of the left had side. The we have \begin{eqnarray*} f_a(n) & \sim & F_{2a}(n) \\ [a] & \sim & F_{2a}(a) = F_{2 \omega}(a) \\ [a,1] & \leq & F_{2 \omega + 1}(a) \\ [a,2] & \leq & F_{2 \omega + 2}^2(a) \\ [a,b] & \leq & F_{2 \omega + b}^2(a) \\ [a,b,1] & \leq & F_{2 \omega + \omega + 1}^2(M) \\ [a,b,2] & \leq & F_{2 \omega + \omega + 2}^2(M) \\ [a,b,c] & \leq & F_{2 \omega + \omega + c}^2(M) \\ [n,\ldots,n] & \leq & F_{\omega^2}^2(n) \\ [x][1] & \leq & F_{\omega^2}^4(M) \\ [x][2] & \leq & F_{\omega^2}^6(M) \\ [x][n] & \leq & F_{\omega^2}^{2n}(M) \\ [x][y] & \leq & F_{\omega^2 + 1}^2(F_{\omega^2}(M)) \\ [x][y][1] & \leq & F_{\omega^2 + 1}^2(F_{\omega^2}^4(M)), \end{eqnarray*} where \(F\) denotes the FGH. This is the limit of the current version, because [x][1][2] has an infinite loop.

?
\(a+b=c\) where \(a,b|\nexists(k|k\in\mathbb{N},\frac{a}{k}\in\mathbb{N}\land\frac{b}{k}\in\mathbb{N})\)

let \(\text{rad}(a,b,c)\) be solved by: \(p=\text{rad}(a,b,c)^k>c\)
 * 1) break a down into its prime factors
 * 2) if a prime appears multiple times then erase just so many of that prime that there is only one left(do as many times as necesary with all primes)
 * 3) multiply those together call it \(\text{rad}(a)\)
 * 4) break b down into its prime factors
 * 5) if a prime appears multiple times then erase just so many of that prime that there is only one left(do as many times as necesary with all primes)
 * 6) multiply those together call it \(\text{rad}(b)\)
 * 7) break c down into its prime factors
 * 8) if a prime appears multiple times then erase just so many of that prime that there is only one left(do as many times as necesary with all primes)
 * 9) multiply those together call it \(\text{rad}(c)\)
 * 10) then let \(\text{rad}(a)\times\text{rad}(b)\times\text{rad}(c)=\text{rad}(a,b,c)\)

the number of exceptions for \(p\) for \(k=n\) and \(a\) and \(b\) can be what they want,as long as they are following the rules above,is called \(p(n)\)

let \(d(n)=\text{max}(x|\text{yet to think of definition})\) (w.i.p.)

now let \(\ell\) be an formal theory and let \(q(n,\ell)=\text{max}(i|\forall_{i>j}p(f(j))>q(n-1,\ell))\) where \(f\) is an continuous function in \(\ell\) and \(f(j)>q(n-1,\ell)\) and if there is any fixed number in the definition of \(f\) it has to be smaller than \(10^{100}\)

except \(q(1,\ell)=\text{max}(k|\forall_{p(ji>,x=\text{the value of the first} p(f(n))<\omega)\) where \(f\) is an function provably recursive in \(\ell\)

MOD function
\(\begin{eqnarray*} \gamma_{\beta}(\alpha) = \left\{ \begin{array}{ll} \text{mod}(\text{mod}(a-b),b) & \text{if:}a-b)\geq 0 \\ b & \text{if:}(a-b)<0 \end{array} \right. \end{eqnarray*}\)