Ackermann fixed points
#1
Edit: Posting while asleep Smile

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
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#2
the sum is real and the function is logical i think
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