Let $X\_0,X\_1\in [0,1]$ and $b\_1,b\_2>0$ be integers. We are going to create a numeration system for vectors $(X\_0,X\_1)$, the base being the vector $(b\_1,b\_2)$, as follows.


Recursively define $X\_k=\{b\_2 X\_{k-1} + b\_1 X\_{k-2}\}$, for $k>1$. Here $\{\cdot\}$ represents the fractional part function and $X\_k\in [0,1]$. Clearly,
$$d\_k=b\_1 X\_k + b\_2 X\_{k+1} - X\_{k+2}$$
is an integer between $0$ and $b\_1+b\_2-1$. The sequence $d\_0, d\_1, d\_2,d\_3,\cdots$ represents, by definition, the digits of $(X\_0,X\_1)$ in base $(b\_1,b\_2)$. If $b\_1=0$ then the digits are just the standard digits of $X\_1$ in base $b\_2$.


**Questions:**


* Can two different vectors $(X\_0,X\_1)$ and $(X\_0',X\_1')$ have
the exact same digits in base $(b\_1,b\_2)$, assuming $b\_1,b\_2>0$?
* Can you reconstruct $(X\_0,X\_1)$ if you only know its digits in base
$(b\_1,b\_2)$?


My guess is that the answer to the first question is yes. So all that suffices is to provide an example. This would lead to a negative answer to my second question.


However, if the answer to the first question is negative, there would be the following interesting consequences. Let $b=b\_1+b\_2$. To each $(X\_0,X\_1)$ corresponds a unique number $f(X\_0,X\_1)\in[0,b]$ defined by its expansion in base $b$ as follows:


$$f(X\_0,X\_1)=\sum\_{k=0}^\infty \frac{d\_k}{b^{k}}.$$


The two consequences would be:


* Since for the immense majority of couples $(X\_0,X\_1)$ the
distribution of the digits $d\_k$ is NOT uniform on the
set $\{0,1,2,\cdots,b-1\}$ (see below why), the number $f(X\_0,X\_1)$ is not normal.
Since the set of non-normal numbers has zero Lebesgue measure, we
mapped $[0,1]^2$ onto a set of Lebesgue measure zero. The mapping
is bijective.
* We created an order on $[0,1]^2$. It is defined as follows:
$(X\_0,X\_1) < (X\_0',X\_1')$ if and only if $f(X\_0,X\_1) < f(X\_0',X\_1')$.


**Some useful results**


In order to prove or disprove my claims, I offer the following result. While at this stage I strongly believe that the formula below is correct, I did not technically prove it. This is just based on pattern recognition techniques and experimental math, yet I think the proof should be easy.


$$X\_k = \{A(k) X\_1\} \mbox{ with } A(0) =\frac{X\_0}{X\_1}, A(1) =1, \mbox{ and } A(k)= b\_2 A(k-1) + b\_1 A(k-2).$$


More about this in my former MO question, [here](https://mathoverflow.net/questions/372103/recursive-random-number-generator-based-on-irrational-numbers). In addition, as previously discussed, the digits of $(X\_0, X\_1)$ are almost surely NOT uniformly distributed over $\{0,1,\cdots b-1\}$, unlike classic digits of (say) $\log 2$ in base $b$. Just to give you an example (again based on strong empirical evidence but not a proof) this is the standard distribution of the digits in base $(b\_1=3, b\_2=3)$:


* digit $0$ appears with frequency $1/18$
* digit $1$ appears with frequency $3/18$
* digit $2$ appears with frequency $5/18$
* digit $3$ appears with frequency $5/18$
* digit $4$ appears with frequency $3/18$
* digit $5$ appears with frequency $1/18$


Essentially these are the frequencies you would observe in that base if you picked up $X\_0,X\_1$ randomly.



Hybrid numeration system on $[0,1]^2$
Here is a partial (negative answer) to your first question:


**Proposition 1:** Two different vectors $(X\_0,X\_1)$ and $(X\_0',X\_1')$ cannot have
the exact same digits $d\_0,d\_1,\dots$ in base $(b\_1,b\_2)$, assuming $b\_1,b\_2>0$ and $b\_1>b\_2+1$.


*Proof:* Suppose the contrary. Then for $k=0,1,\dots$ we have $X\_{k+2}=b\_1 X\_k+b\_2 X\_{k+1}-d\_k$, $X'\_{k+2}=b\_1 X'\_k+b\_2 X'\_{k+1}-d\_k$, and hence
$$Z\_{k+2}=b\_1 Z\_k+b\_2 Z\_{k+1},$$
where $Z\_k:=X'\_k-X\_k$. So, for some real $c\_+,c\_-$ and all $k=0,1,\dots$ we have
$$Z\_k=c\_+ u\_+^k+c\_- u\_-^k,$$
where
$$u\_+:=\frac{b\_2+\sqrt{b\_2^2+4b\_1}}2,\quad u\_-:=\frac{b\_2-\sqrt{b\_2^2+4b\_1}}2$$
are the roots $u$ of the equation $u^2=b\_1+b\_2 u$; see e.g. [linear difference equations with distinct characteristic roots](https://en.wikipedia.org/wiki/Linear_difference_equation#Solution_with_distinct_characteristic_roots).


Note that $u\_+>b\_2\ge1$ and also $u\_1>|u\_2|$. So, if $c\_+\ne0$, then $|Z\_k|\to\infty$ (as $k\to\infty$), which contradicts the conditions $Z\_k=X'\_k-X\_k$, $0\le X\_k<1$, $0\le X'\_k<1$. So, $c\_+=0$.


Now, for $b\_2>0$, the condition $b\_1>b\_2+1$ is equivalent to $|u\_-|>1$, whence $|Z\_k|=|c\_-|\,|u\_-|^k\to\infty$ if $c\_-\ne0$, which again contradicts the conditions $Z\_k=X'\_k-X\_k$, $0\le X\_k<1$, $0\le X'\_k<1$. So, $c\_-=0$, so that $Z\_k=0$ and $X'\_k=X\_k$ for all $k$. In particular, $(X\_0,X\_1)=(X\_0',X\_1')$. $\Box$


