User blog:Luckyluxius/Improved version of Arrow Growing Hiearchy

\(Q_{0}(3,4) = 3 \uparrow 4 = 81\)

\(Q_{1}(3,4) = Q_{0}((Q_{0}(3,3)),(Q_{0}(4,4)))= (3 \uparrow 3) \uparrow (4 \uparrow 4) = 3^{3^{4^{4}}} \approx 10^{10^{121.821}}\)

\(Q_{m}(3,4) = Q_{m-1}(Q_{m-1}(3,3),Q_{m-1}(4,4))\) \text{for} m > 0 \)

Anything above \(\omega\), you add a third variable, but above \(\omega 2\), the third variable acts differently, and there is a 4th variable.

\(Q_{\omega}(n,m,o) = n \underbrace{\uparrow^{n \uparrow^{\cdots}^{n}}}_{o} m\)

\(Q_{\omega + 1}(n,m,o) = n \underbrace{\uparrow^{n \uparrow^{\cdots}^{n}}}_{o + 1} m\)

\(Q_{\omega + m}(n,m,o) = n \underbrace{\uparrow^{n \uparrow^{\cdots}^{n}}}_{o + m} m\)

\(Q_{\omega 2}(n,m,o,p) = Q_{\omega}(Q_{p-1}(n,n),(Q_{p-1}(m,m)) \text{for} p > 1\)

\(Q_{\omega \alpha}(n,m,o,p) = Q_{\omega \alpha -1}}(Q_{\alpha -1}(n,n),Q_{\alpha -1}(m,m))\)

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