The following curious-looking fraction, with numerical value approximately $1.7302267782385217$, appears in this Reddit question:

$$1+\cfrac{2+\cfrac{4+\cfrac{8+\cdots}{9+\cdots}}{5+\cfrac{10+\cdots}{11+\cdots}}}{3+\cfrac{6+\cfrac{12+\cdots}{13+\cdots}}{7+\cfrac{14+\cdots}{15+\cdots}}}$$

This led me to consider the following functional equation; note that the fraction above is equal to $f(1)$.

**Proposition**: There is a unique meromorphic function $f$ on $\mathbb C\setminus[-1,0]$, satisfying
$$f(z)=z+\frac{f(2z)}{f(2z+1)}$$
for all $z\in\mathbb C\setminus[-1,0]$.

Proof sketch: Let $X$ be the function space of all bounded holomorphic functions on $|z|>10$, equipped with the sup norm. One checks that the map $\mathcal F$ defined by $$(\mathcal Fg)(z)=\frac{g(2z)+2z}{g(2z+1)+2z+1}$$ is a contraction on the ball of radius $2$ in $X$, so it has a unique fixed point $g_0$; then we can take $f(z)=g_0(z)+z$. Now if we define $$S_n=\{z\in\mathbb C\,:\,d(z,[-1,0])>10\cdot2^{-n}\},$$ then the functional equation gives the meromorphic extension of $f$ from $S_n$ to $S_{n+1}$. $\square$

I am interested in what we can say about $f$ in general, and the value $f(1)$ in particular. Here are a couple of my observations:

- Note that $f(1)\approx\sqrt3$ comes from the exceptionally good approximation $$f(z)=\sqrt{(z+1)^2-1}+O((z+1)^{-3}).$$
- Also, I suspect that near $z=0$ we have $f(z)\sim Az^\alpha$, where $\alpha=\frac{\log f(1)}{\log 2}$. This also controls the singularities of $f$ at the dyadic rationals in $[-1,0]$.

**Questions**:

- Is $f(1)$ irrational? transcendental? Of course, a closed-form expression for $f(1)$ would be great, but I'm not hopeful that there is one.
- How do we compute $f(1)$ to high accuracy, say 1000 digits?
- Have there been similar functional equations studied in the literature? The closest result I know of is the Woods−Robbin product identity, which can be written as $$2\times\cfrac{3\times\cfrac{5\times\cfrac{9\times\cdots}{10\times\cdots}}{6\times\cfrac{11\times\cdots}{12\times\cdots}}}{4\times\cfrac{7\times\cfrac{13\times\cdots}{14\times\cdots}}{8\times\cfrac{15\times\cdots}{16\times\cdots}}}=\sqrt2,$$ which comes from the nice closed-form solution $h(z)=\sqrt{z(z-1)}$ of the corresponding functional equation $h(z)=z\frac{h(2z-1)}{h(2z)}$. What happens when no closed-form is available?

`\backslash`

differs from`\setminus`

, thus: $$\begin{align} & \mathbb C\backslash[0,1] \\ {} \\ & \mathbb C\setminus[0,1] \end{align}$$ $\endgroup$