This question did not get anything from math.stackexchange. Therefore, and although probably very elementary, I pose it here. It is about an example in an 
[article](https://eudml.org/eudml/doc/183526) of Dekking and Mendez France:


For an integer $n$ let $s(n)$ be the number of ones in the binary expansion,
so that $s(2k)=s(k)$ and $s(2k+1)=s(2k)+1$, $\alpha \in (0,1/2)$ is fixed, $x = 2\pi i \alpha$ and $$ z\_n= \sum\_{k=0}^{n-1} \exp(xs(k)).$$
The authors claim that for all $n,m \in\mathbb N$ the following can be shown by induction:
$$ z\_n=z\_m \text{ and } z\_{n+1}=z\_{m+1} \Longrightarrow n=m.$$
I see this if $n,m$ have the same parity: One has 
$\exp(xs(n))=\exp(xs(m))$ and if $n,m$ are both odd this gives $\exp(xs(n-1))=\exp(xs(m-1))$ which rather
quickly implies $z\_{n-1}=z\_{m-1}$. 


If $n,m$ are both even one can use the identity
$$z\_{2n}=(1+e^x)z\_n$$ (each term $e^{xs(k)}$ for $k\in \lbrace n,\ldots,2n-1\rbrace$ corresponds to a term $e^{xs(j)}$ with $0\le j \le n-1$ such that
$s(k)=s(j)+1$) to obtain $z\_{n/2}=z\_{m/2}$ and $z\_{n/2+1}=z\_{m/2+1}$.


However, I do not see how to proceed if $n$ and $m$ have different parity.



Exponential sums and binary expansions
Always look for the counterexample would be a worthwhile motto it seems. 


For $\alpha = 1/3$ (so that $e^x$ is a complex cube root of unity), $n = 6$ and $m = 9$ provide one. This is because $s(6) = s(9)$ and $s(6), s(7), s(8)$ are all distinct modulo 3. 


For the record the first search for a counterexample was with $\alpha = 1/4$ and that failed (but without a proof -- certainly there are no examples less than $10^6$).


I suspect there may well be other counterexamples with $\alpha$ rational.


