What is the number of representations of a real number?

Question

Let $f:\omega\to\mathbb N$ be a function such that $\sum_{n=0}^\infty\frac{f(n)}{2^n}<\infty$.

We identify each natural number $n\in\mathbb N$ with the set $\{0,\dots,n-1\}$.

Then the map $$\sigma_f:\prod_{n=0}f(n)\to \mathbb R,\;\;\sigma_f:(x_n)_{n\in\omega}\to\sum_{n=0}^\infty\frac{x_n}{2^n},$$ is well-defined.

For a real number $y$ the elements of the set $\sigma_f^{-1}(y)\subset \prod_{n=0}^\infty f(n)$ are called $f$-representations of $y$.

It is well-known that for the constant function $f\equiv 2$, each real number $y$ has at most two $f$-representations.

Question 1. How many $f$-representations has a real number $y$ for a constant function $f\equiv c\in\mathbb N$?

Question 2. How many $f$-representations has a real number $y$ for the function $f(n)=n$, $n\in\omega$. Is the number of $f$-representations of $y$ at most countable?

Answer 1

Consider the case where $f(n)=4$ for all $n$. If $0\leq a\leq 1$ then there is usually a unique sequence $x\in\{0,1\}^\omega$ with $\sigma_f(x)=a$. Now for any subset $S\subseteq\mathbb{N}$ define $y^S$ by $(y^S_{2i},y^S_{2i+1})=(x_{2i},x_{2i+1})$ if $i\not\in S$, and $(0,x_{2i+1}+2x_{2i})$ if $i\in S$. This gives uncountably many different sequences $y^S$ with $\sigma_f(y^S)=a$. It seems like this should be typical.