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Preface: understand that this should all be modulo some fixed $m$; otherwise the numbers become so ridiculously large that the question makes no sense. That said, I'll leave $m$ off the notation to make it more clear.

Hyperexponentation can be defined recursively as follows:

$a\uparrow_n 0=1$

$a\uparrow_0 b=ab$

$a\uparrow_{n+1}b+1 = a\uparrow_n(a\uparrow_{n+1}b)$

For small $n$ you recover familiar operations: $a\uparrow_1 b=a^b$, and $a\uparrow_2 b = a^{a^{a^{\cdots a}}}$ (repeated $b$ times).

Computation is challenging. Since $a\uparrow_{n+1}b$ tends to be "quite large," computing $a\uparrow_{n+1}b+1$ is only plausible if there is an "efficient algorithm" for $\uparrow_n$.

There is such an algorithm for $a\uparrow_1b$ (exponentiation), which runs essentially in logarithmic time in $b$, which follows essentially from the fact that $a\uparrow_1 2b=(a\uparrow_1b)\uparrow_1 2$.

However, I don't see an obvious equivalent, even for $\uparrow_2$ -- while the iterative nature seems like it should work out, it doesn't seem true that $a\uparrow_2 (2b) = (a\uparrow_2 (b)) \uparrow_2 2$, or any variation of the right I can think of.

Is there a way to compute this more efficiently?

Note there may be a connection to the Ackermann function, where (at least according to Wikipedia), $A(m, n)=2\uparrow_{m-2}(n+3) - 3$.

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1 Answers1

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For modular arithmetic:

A modification of Euler's totient theorem can be given as

When $b>\log_2(p)$ we have

$$a^b\equiv a^{(b\bmod c)+c}\pmod p\tag{$\star$}$$

where $c=\varphi(p/\gcd(a^{\lfloor\log_2(p)\rfloor},p))$ and $\varphi$ is Euler's totient function.

which allows us to repeatedly push $\bmod$ into the exponent. This terminates when we reach either $p=1,\gcd(a^p,p)=p,$ or we've reached the end the power tower (or something we can directly compute).

After that, we can work our way down with modular exponentiation by squaring, as you have noted.

Try it online!

Example:

$818023620212985786400\uparrow\uparrow442883135576504326053\pmod{1158210822627198136417}$

$$a=818023620212985786400\\\begin{array}{c|c|c}p&\gcd(a^{\lfloor\log_2(p)\rfloor},p)&\varphi(p/\gcd)\\\hline1158210822627198136417&1&1158210822627198136416\\1158210822627198136416&32&24118439168417024424\\24118439168417024424&8&1676147829304314240\\1676147829304314240&640&1472026782156480\\1472026782156480&320&2787929510040\\2787929510040&40&46196895888\\46196895888&16&1437004800\\1437004800&25600&29160\\29160&40&486\\486&2&162\\162&2&54\\54&2&18\\18&2&6\\6&2&2\end{array}$$

Since $\gcd(a^1,2)=2$, we stop here. Now working back:

$$\begin{array}{c|c|c}p&(b\bmod c)+c&a^{(b\bmod c)+c}\bmod p\\\hline2&-&0\\6&2&1\\18&7&10\\54&28&10\\162&64&10\\486&172&172\\29160&658&18640\\1437004800&47800&260300800\\46196895888&1697305600&32572251856\\2787929510040&78769147744&910022074840\\1472026782156480&3697951584880&903930403260160\\1676147829304314240&2375957185416640&1215567107351528320\\24118439168417024424&2891714936655842560&14632068328029956416\\1158210822627198136416&38750507496446980840&373675920273457955680\\1158210822627198136417&1531886742900656092096&105094019098433015730\end{array}$$

Hence the remainder is $105094019098433015730$.

In the case of tetration, this directly gives us a method to compute it's last digits. In the case of higher hyperoperators, we can use the same strategy, working out a lower bound of the tetration height and then checking if the remainder is uniquely determined.

$(\star)$ This follows from a combination of the Chinese remainder theorem and Euler's totient theorem. $a^p$ is computed using modular exponentiation by repeated squaring as you have noted. The $\gcd$ is computed with the Euclidean algorithm. $\varphi$ is computed by iteratively searching for prime factors and be sped up using faster factorization algorithms.

For non-modular arithmetic:

There is no such algorithm possible since there are no nice relationships for $a\uparrow^nb$ for any $n\ge2$.

That problem aside, this isn't even reasonable to consider for any $a,b\ge2$ because the results quickly grow to a size that will make arithmetic on them slow, unless one chooses to go for a rough approximation.

This latter issue, however, can be circumvented if $0<a<e^{1/e}$ for tetration, where it converges. When $a>e^{-e}$ is in this range, then we may approximate it with

$$a\uparrow^2b\simeq q+[(a\uparrow^2n)-q]\ln(q)^{b-n}$$

for large $b$ and $q=a^q$. As $n\to\infty$ this approximation improves but relies on a large amount of precision. This can be avoided by using Aitken's delta squared acceleration method to make it converge faster. When $a<e^{-e}$, the same approach can be taken, but by considering solutions to $q=a^{a^q}$.

Of course, the same approach cannot be used for higher hyperoperations, since they require continuous lower hyperoperations and there is no agreed upon extension of tetration to non-integer heights.

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