In simple exponentiation A^B , the fast calculation consists of writing B in the binary base , B = B0 + 2 B1+ ( 2^2)B2 + …+ (2^d) Bd
So A^B can be written (A^B0) (A^2B1)… (A^(2^d)Bd) , so the number of operation is comparable with d so with Ln(B).
Tetration is written tetA(B) = A^A^A^A^A…. B times.
So the question is there is fast calculation for tetration ?
found this relevant resource :
https://www.maa.org/sites/default/files/pdf/upload_library/22/Chauvenet/Knoebelchv.pdf
Not sure that this completely answers the original question, maybe it would help...
Number of stable/frozen digits originated by any integer tetration at given heigth, by simply calculating the 2-adic or 5-adic valuation of a very simple manipulation of the tetration base a: https://nntdm.net/volume-28-2022/number-3/441-457/
Moreover, we could do the very same thing for many other numeral systems, but the paper assumes radix-10 by default, since (in general, for any squarefree numeral system g) we first need to find all the solutions of the fundamental equation y^t=y in the g-adic ring of decadic integers (e.g., if g=10, then we have 15 solutions and it is sufficient to set t=5).
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