Recently there was a paper connecting the zeros of the GKW operator to the zeros of zeta, this is due to the fact that the Mellin transform of the Gauss map is a linear function of zeta . Transfer operator for the Gauss' continued fraction map. I. Structure of the eigenvalues and trace formulas and Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems

Why can the same relations not be used for the mapping associated with the Engel expansion formed by iterating the function

$\begin{array}{ll} g (x) & = x \left( \lfloor \frac{1}{x} \rfloor + 1 \right) - 1 \text{}\\ & = \sum_{n = 1}^{\infty} \left\{ \begin{array}{ll} x (n + 1) - 1 & \frac{1}{n + 1} < x \leqslant \frac{1}{n}\\ 0 & {otherwise} \end{array} \right. \end{array}$

whose Mellin transform is also a linear function of the Riemann $\zeta$ function ?

We have

$\begin{array}{ll} \int_0^{1} g (x) x^{s - 1} d x & = \sum_{n = 1}^{\infty} \int_{\frac{1}{n + 1}}^{\frac{1}{n}} (x (n + 1) - 1) x^{s - 1} x\\ & = \frac{\zeta (s + 1)}{s + 1} - \frac{1}{(s + 1) s}\\ & = \frac{\zeta (s + 1) s - 1}{(s + 1) s} \end{array}$

and the associated Transfer operator (I'll call it the Engel operator) analogous to the GKW operator is defined by \begin{equation} [\mathcal{L}_g f] (x) = \sum_{y : g (y) = x} \frac{f (y)}{\left| \frac{d}{d z} g (z) \right|_{z = y}} = \sum_{n = 1}^{\infty} \frac{f \left( \frac{x + 1}{n + 1} \right)}{n + 1} \end{equation} since $\frac{d}{d x} (x (n + 1) - 1) = n + 1$ and the inverse of the $n$-th component is $\frac{x + 1}{n + 1}$ which is found by solving $x (n + 1) - 1 = y$ for $x$.

The operator $[\mathcal{L}_g f] (x)$ applied to the function $x \rightarrow x^s$ results in

$[\mathcal{L}_g x \rightarrow x^s] (x^{}) = (x + 1)^s (\zeta (s + 1) - 1)$

The iteration function for the Gauss map is :

$h(x)=1/x-\lfloor 1/x \rfloor.$

where: $\lfloor 1/x \rfloor$ denotes the floor function

The Gaussâ€“Kuzminâ€“Wirsing $G$ acts on functions $f$ as

$[Gf](x) = \sum_{n=1}^\infty \frac {1}{(x+n)^2} f \left(\frac {1}{x+n}\right).$