Tetration is the next hyperoperation after more familiar addition, multiplication and exponentiation. It can be seen as a repeated exponentiation, similar to how exponentiation can be seen as a repeated multiplication, and multiplication — as a repeated addition. Tetration is denoted $^n a$, where $a$ is called the base and $n$ is called the height, and is defined for $n\in\mathbb N\cup\{-1,\,0\}$ by the recurrence $$ {^{-1} a} = 0, \quad {^{n+1} a} = a^{\left({^n a}\right)},\tag1$$ so that $${^0 a}=1, \quad {^1 a} = a, \quad {^2 a} = a^a, \quad {^3 a} = a^{a^a}, \, \dots \quad {^n a} = \underbrace{a^{a^{{.^{.^{.^a}}}}}}_{n\,\text{levels}},\tag2$$ where power towers are evaluated from top to bottom. To simplify matters, we restrict our attention to real bases in the interval $1<a<e^{1/e}$. Under this restriction, the sequence $\{^n a\}$ is strictly increasing and converges to a limit: $$^\infty a=\lim_{n\to\infty} {^n a} = \frac{W(-\ln a)}{-\ln a},\tag3$$ where $W(z)$ denotes the principal branch of the Lambert W-function. To simplify our following exposition it is convenient to introduce a new variable $q$ linked to $a$: $$q = \ln({^\infty a}) = {^\infty a} \cdot \ln a = -W(-\ln a).\tag4$$ The bounds on $a$ imply $0<q<1$. The value $q$ is closely related to the asymptotic growth of the tetration. In particular, it appears in remarkable limits: $$ \lim_{n\to\infty} \, \frac{{^{n+k} a}-{^n a}}{{^{n-k} a}-{^n a}} = -q^k, \quad \lim_{n\to\infty} \, \frac{{^{n+k} a}-{^\infty a}}{{^n a}-{^\infty a}} = q^k .\tag5$$ Definitions of addition, multiplication and exponentiation for fractional, negative and complex arguments are well-established. A natural extension of the tetration to fractional and complex heights is a long-standing question. It appears to be much less straightforward problem, mainly because exponentiation, unlike addition and multiplication, lacks both commutativity and associativity, and so tetration does not obey simple laws similar to $a^m \cdot a^n = a^{m+n}$ or $(a^m)^n = a^{m\,n}$.

Despite these obstacles, several proposals aiming to extend the definition of the tetration to fractional and complex heights have been put forth. Their common goal is to define a function that agrees with the discrete tetration for all positive integer heights, and satisfies the functional equation $$^{z+1} a = a^{\left({^z a}\right)}\tag6$$ for all $z$ in its domain. These conditions alone are not sufficient to determine a unique function, so additional natural restrictions (such as continuity, smoothness or analyticity) have been proposed, aiming to narrow down the set of candidate functions to a single and, in some sense, "the most natural" extension. The Wikipedia article on tetration is very cautious and indicates that, apparently, so far there is no consensus about which extension is "the one". Some of the proposals seem to lead to the same function, although apparently no rigorous proof of their equivalence has been published yet. The Citizendum article is written in a much bolder style, and refers to tetration as a unique well-defined function. But, overall, the study of tetration seems to be a quite isolated area of mathematics, with few connections to other topics.

A few months ago I made some interesting conjectures related to tetration that I hoped I would prove soon. It appears that there are some connections between tetration and the theory of $q$-analogs. Unfortunately, I have not made much progress in proving them since then (although I have learnt a lot about $q$-series), so I finally decided to post them here and ask for your help. First, let's recall some definitions.

The $q$-Pochhammer symbol: $$(p;\,q)_\infty = \prod_{k=0}^\infty(1-p\,q^k)\tag7$$ $$(p;\,q)_z = \frac{(p;\,q)_\infty}{(p\,q^z;\,q)_\infty}, \quad z\in\mathbb C\tag8$$ It easily follows that $$(p;\,q)_n = \prod_{k=0}^{n-1}(1-p\,q^k), \quad n\in\mathbb N\tag9$$

The $q$-binomial coefficients (also known as the Gaussian binomial coefficients): $${r \brack s}_q = \frac{(q;\,q)_r}{(q;\,q)_s \, (q;\,q)_{r-s}}.\tag{10}$$

From now on we assume that $a$ and, consequently, $q$ are fixed. Let us define an analytic function $t(z)$ by the following series:

$$ t(z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\binom {n-k} 2} {z \brack n}_q {n \brack k}_q ({^k a}).\tag{11} $$

This formula can be seen as a combination of the direct and reverse $q$-binomial transforms. It can be proved by induction that for $n\in\mathbb N, \, t(n)={^n a}$, so $t(n)$ satisfies the functional equation for tetration $(6)$ at least for positive integer arguments. Indeed, partial sums of $(11)$ can be seen as Lagrange interpolating polynomials in powers of $q^z$ that exactly reproduce tetration values for progressively larger initial segments of $\mathbb N$. Numeric evidence suggests that adding new points to the interpolating polynomial of this form does not result in erratic oscillations of its arcs between existing points (i.e. Runge's phenomenon does not occur), but rather makes them converge to a monotone function satisfying the functional equation for the tetration $(6)$. So we have:

Conjecture 1.The series $t(z)$ converges at least in the half-plane $\Re(z)>-2$ to an analytic function that satisfies the functional equation for tetration $t(z+1)=a^{t(z)}$ for all arguments in its domain of analyticity.

Conjecture 2.The function $t(z)$ is periodic with a purely imaginary period $\tau = 2\pi n i/\ln q$, and admits an analytic continuation to the half-plane $\Re(z) \le -2$ by repeated backward application of the functional equation $(6)$. There it has a countable set of isolated singularities at $z = m + \tau\,n, \,m,n\in\mathbb Z, \, m \le -2$, and a countable set of branch cuts that can be made along horizontal rays at $z = x + \tau\,n, \, n\in\mathbb Z, \, x\in\mathbb R, \, x < -2$.

Conjecture 3.For real $x>-2$, the derivative $t'(x)$ is a completely monotone function.

Conjecture 4.There is only one analytic function that vanishes at $z=-1$ and satisfies both the functional equation for the tetration $(6)$, and the complete monotonicity condition from the Conjecture 3.

10more comments