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As the title describes, I will post here my question clearer:

Let $z=\frac{m}{2^k}$ be a dyadic rational number in $(0,1)$ where $m$ is odd and $k >0$, and also $n$ is a fixed positive integer. Let's denote $N(z,n)$ to be the number of $n$ tuples $(i_1,...,i_n)$ of positive integers such that $z=\sum_{j=1}^{n} \frac{1}{2^{i_j}}$ (I assume $i_1 \leq i_2 ... \leq i_n$).

My question is : Are these numbers $N(z,n)$ always finite? If yes, I want to have an upper bound for this $N(z,n)$ ... Thanks !

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A good bound or a lousy one? If I understand the problem, finiteness is straightforward. – André Nicolas Oct 24 '12 at 20:51
@André Nicolas: Could you explain a little bit why finiteness is straightforward? Any idea for a decent bound is welcome. – ktm Oct 24 '12 at 22:54
Well, now I understand they should be finite by compactness in $[0,1]$, but I do not have any idea to get an upper bound for this. – ktm Oct 24 '12 at 23:20

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