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?



What is the number of representations of a real number?
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.


