I've been working on a project for a wiki that I'm a member of. It is the Sequence of the Day for September 2. You can see my progress at https://oeis.org/wiki/Template:Sequence_of_the_Day_for_September_2 .

In short I ask, when you consider tetrations with rational heights, and compare them to exponentials with rational powers, what is the equivalent form of tetration to the exponential $n^{1/n}$?

I was wondering if anyone here knew the answer and would share it with me. If so, I would like to rephrase my project.

  • $\begingroup$ Powers have the property that $(a^b)^c = a^{bc}$, so a rational power like $1/b$ has to have the effect of an inverse. Unfortunately I don't think tetrations have a similar property, so you might not be able to draw a meaningful parallel. $\endgroup$ – trutheality Jun 25 '11 at 3:16
  • 3
    $\begingroup$ Possible duplicate of How to evaluate fractional tetrations? $\endgroup$ – samerivertwice May 11 '17 at 12:36

For this answer, I use the tetration notation ${^b} a = a^{a^{...^a}}$ with height $b$.

You could just look at the sequence ${^{1/n}}n$ (assuming you have a definition of tetration for noninteger heights), but I imagine this is not what you are looking for because fractional exponentiation and fractional tetration are not good analogues to eachother. Many other answers have addressed this problem: e.g.: this, this, and this.

Perhaps a more interesting analogue is to look at the fact that $n^{1/n} = \sqrt[n]{n}$, so the sequence $a_n = n^{1/n}$ represents the unique positive number such that $(a_n)^n = n$. One could similarly define a sequence $b_n$ with tetration-inverse, so $b_n$ is the unique positive solution to the equation ${^n}(b_n) = n$. Fortunately for us, this sequence doesn't depend on how to compute tetration for non-integer heights so we don't have to do anything complicated here.

Is this sequence $b_n$ interesting? Probably not very much, but we can show one interesting fact: $$ \lim_{n\rightarrow\infty} b_n = e^{\frac{1}e} $$ Proof: Let $c_1 = \limsup b_n$ and $c_2 = \liminf b_n$. Suppose $c_1 > e^{\frac{1}e}$. Then there exists $\epsilon > 0$ such that for $n$ sufficiently large, $b_n > e^{\frac1e} + \epsilon$. This would imply $$ n = {^n}(b_n) > {^n}(e^{\frac1e} + \epsilon) $$ for all $n$ sufficiently large. This is impossible however, since for $\alpha > e^{\frac1e}$, ${^n} \alpha$ grows faster than $n$. Thus we can conclude $c_1 \le e^{\frac1e}$. Similarly, suppose $c_2 < e^{\frac1e}$. Then we would have for $n$ sufficiently large $b_n < e^{\frac1e}$, which would in turn yield $$ n = {^n} (b_n) < {^n}(e^{\frac1e}) < e $$ for all $n$ sufficiently large. (It is fairly straightforward to show that ${^n}(e^{\frac1e})$ is increasing and converges to $e$). This is a contradiction, and hence $c_2 \ge e^{\frac1e}$. Since $c_2 \le c_1$, we conclude that $c_1 = c_2 = e^{\frac1e}$. Therefore, $\lim b_n$ converges to $e^{\frac1e}$.


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