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+--- Thread: Ackermann fixed points (/showthread.php?tid=1642)
Ackermann fixed points - Daniel - 09/18/2022
Edit: Posting while asleep
Generalizing \[^\infty z=a \implies z = a^{\frac{1}{a} }\]
gives the base associated with the fixed point for the hyperoperators.
\[z \uparrow^n \infty = a \implies z = a \uparrow^{n-1} (a \uparrow^{n-1} {-1})\]