FANDOM


This is a sandbox.

$ 10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{{10\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{5.4\times10^{489}}10}}10}}10}10}10}10}10 $


$ 8\uparrow^{10} 9 \rightarrow 8\uparrow^{10} 9 \rightarrow 3 \rightarrow 4 $ .


$ \alpha\uparrow\uparrow\omega=\epsilon_{\alpha+1} $ と書かれるので、 この極限は

$ \vartheta(\epsilon_{\Omega+1}) $ となります。


$ J(a_1,a_2,\cdots ,a_{n-1},a_n)=D(a_{n [J(a_1,a_2,\cdots ,a_{n-1})]}) $ $ D(a_{n [1]}) $ $ \underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{\underbrace{10\cdots10}_{10000000000}}}}}}}}} $


Well, I don't understand it wery well, but if if "x←[1@(y+1)/y]→x" goes

$ f_{\vartheta(\Omega_{x})}(y) $ , the last suggestion of putting brackets to both sides of arrow, will go

$ f_{\vartheta(\Omega_{x})+1}(y) $ .


it would be in this page.


$ i\not\in\R $


$ a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{c}b $ $ a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{\underbrace{\vdots}_{a\underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{c}b}}b}b $