[2015] 4th Zeration from base change pentation
#6
In case it was not clear , I think the zeration will be the one that satisfies 2[0]2[0] ... = x + 2 or this one a[0]b = ln(exp(a)+exp(b) ).

The ackermann recursion makes no sense in this context for pentation.

Numerical experiments seem hard at first because of the speed of pentation.

And I have 0 theory.

But maybe others have ideas.

regards

tommy1729
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RE: [2015] 4th Zeration from base change pentation - by tommy1729 - 03/29/2015, 05:47 PM

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