Ackermann fixed points Daniel Fellow Posts: 248 Threads: 80 Joined: Aug 2007 09/18/2022, 03:13 PM (This post was last modified: 09/19/2022, 03:26 AM by Daniel.) 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. $$z \uparrow^n \infty = a \implies z = a \uparrow^{n-1} (a \uparrow^{n-1} {-1})$$ Daniel « Next Oldest | Next Newest »

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