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<(2 \uparrow ^3 2 \uparrow ^3 4)-(3\uparrow^3 4)

=(2\uparrow^3 2\uparrow^2 2\uparrow^2 4)-(3\uparrow^3 3\uparrow^2 3\uparrow^2 3)



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

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