Tetration (literally "4th operator iteration") is iterated exponentiation, much like how exponents are iterated multiplying. For example, $2$^^$3$ is the same as $2^{2^2}$, which is $2^4$ or 16. This applies to any two positive real numbers, and possibly even all real numbers.
There are two inverses of tetration, iterated logarithms "super-logarithms" or "tetralogarithms" and iterated roots "super-roots". For instance, the value of $\sqrt{256}_s$ is $4$ because $4$^^$2 = 4^4 = 256$.
It is very difficult to calculate with or approximate either of these inverse operations to tetration. Super-logarithms use some complicated recursive algorithms (see this wiki article), but super-roots are even harder; the example I showed just turned out to be easy-to-find because $256$ is a perfect super-square (that is, $4^4$). I could only manage to find a formula for super square roots on Wikipedia using the Lambert W function. I originally did not know what this Lambert W function was, but soon after I found out it's just the inverse function of $xe^x$. This expression stated that $ssrt(x) = ln(x)/W(ln(x))$.
However, I cannot find a formula or even approximation for higher super-roots at all. I tried using linear approximation along with a Euclidean distance version of linear approximation on $^3\sqrt{2}_s$ and both were very innacurate. I could eventually calculate $^3\sqrt{2}_s$ to 42 digits as $1.4766843373578699470892355853738898365517$ using this high-precision calculator and binary guess-and-check methods, but I still can't seem to find a generic formula or approximation for these super-roots which have orders higher than $2$, so I would like to know if anyone knows of one.
Best Answer
$\def\srt{\operatorname{srt}}$ Although there are integral solutions, the super root $\sqrt[n]x_s,\srt_n(x)$ has an expansion using Lagrange reversion:
$$\sqrt[k]x_s=\srt_k(x)= 1+\sum_{n=1}^\infty\frac{\ln^n(x)}{n!}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0$$
Applying general Leibniz rule and $e^z$ Maclaurin expansion each $k-2$ times if $2<k\in\Bbb N$:
$$ \frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^\infty\sum_{n_2=0}^{n-1}\dots\sum_{n_{2k-5}=0}^\infty\sum_{n_{2k-4}=0}^{n-1-\sum_\limits{j=1}^{k-3}n_{2j}}\left.\frac{d^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}{dt^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}e^{(n_{2k-5}+1)t}\right|_0\prod_{j=1}^{k-2}\frac{n_{2j-3}^{n_{2j-1}}}{n_{2j-1}!}\left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_{2m}}\\n_{2j}\end{matrix}\right)\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0$$
where $n_{-1}=-n$. Next, use Kronecker delta and factorial power $m^{(n)}$ in $\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0=\delta_{n_{2j},n_{2j-1}}n_{2j-1}^{(n_{2j})}$. As hinted by Quantile Mechanics $(96)$ to $(97)$, remove the odd indexed sums substituting each $n_{2j}=n_{2j-1}$. We reindex $n_{2j}\to n_j$ and simplify:
$$\begin{align}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^{n-1}\dots \sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j} (n_{k-2}+1)^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-2}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)= \sum_{n_1=0}^{n-1}\dots \sum_{n_{k-1}=0}^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-1}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)\end{align}$$
where $n_0=n$. These links support the last equality. Therefore:
$$\bbox[1px, border: 2px solid red]{\sqrt[k]z_s=\srt_k(z)=1+\sum_{n=1}^\infty \sum_{n_1=0}^{n-1}\dots\sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j}\frac{(-1)^{n_1}\ln^n(z)}{\Gamma \left({n-\sum_\limits{j=1}^{k-2}n_m}\right)(n_{k-2}+1)^{1-n+\sum\limits_{j=1}^{k-2}n_j} n}\prod_{j=1}^{k-2}\frac {n_{j-1}^{n_j}}{n_j!}}$$
where all can be infinite sums or all, except the $n$ sum, can have an upper bound of $n$. Assume improper sums are empty. We find that:
$$\srt_1(z)=z$$ $$\srt_2(z)=1+\ln(z)+\sum_{n=2}^\infty\frac{\ln^n(z)}{n!}(1-n)^{n-1}$$ $$\srt_3(z)=1-\sum_{n=1}^\infty\sum_{k=1}^{n-2}\frac{(-1)^kk^{n-k+1}n^{k-2}\ln^n(z)}{(n-k)!k!}$$
shown here
$$\srt_4(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=1}^{n-k-1}\frac{(-1)^k n^{k-1}k^{m-1}\ln^n(z)}{(n-k-m)!k!m!m^{m+k-n-1}}$$
Shown here
$$\srt_5(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{j=1}^{n-k-m-1}\frac{(-1)^k k^m n^{k-1} m^{j-1}\ln^n(z)}{(n-j-k-m)!m!k!j!j^{j+m+k-n-1}}$$ Shown here
$$\srt_6(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-1-k}\sum_{j=0}^{n-1-k-m}\sum_{p=1}^{n-k-m-j}\frac{(-1)^k n^{k-1}k^m m^j j^{p-1}\ln^n(z)}{(n-1-k-m-j-p)!k!m!j!p!p^{p+j+m+k-n-1}}$$ shown here $$\vdots$$ $$\srt_\infty(z)=\sqrt[z]z$$
Compare both links for verification. The region of convergence is near $|z|=1$. Help is wanted with the region of convergence for the multiple series expansion and other representations of $\srt_k(z)$.