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.

$$z \uparrow^n \infty = a \implies z = a \uparrow^{n-1} (a \uparrow^{n-1} {-1})$$

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