Nice problem! Hint:

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http://www.mathlinks.ro/Forum/viewtopic.php?search_id=611926086&t=294846 (Wed, 12, Aug, 2009 while this problem was in Ibero test 23, Sep, 2009, People, just try to do a better work).

Lousin Garckz wrote: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=611926086&t=294846 (Wed, 12, Aug, 2009 while this problem was in Ibero test 23, Sep, 2009, People, just try to do a better work). Unfortunately it seems that has become some kind of trend in this competition -more examples like this can be found in recent papers... In my opinion even if the jury tries to do its duty there might be some countries that are playing foul and do not propose completely original problems... However there are many people who are willing to improve the quality of the competition! You can still find original problems, like problem 3 that was proposed by the deputy leader of our country.

Do you have these examples you are talking about? You still don't know what really happened, maybe two people thought of the same thing (which I can guarantee happens sometimes) and you can't blame them as cheaters. Problems are supposed to be proposed way before the competition and maybe there was no time for them to check, so don't judge.

Exaktly, if \[ n=2^{k_0}+2^{k_0+k_1}+...+2^{k_0+k_1+...+k_m},\] then \[ a_k=k_0+\frac{1}{k_1+\frac{1}{k_2+\frac{1}{k_3+...}}}\]

Jutaro wrote: Consider the sequence $ \{a_n\}_{n\geq1}$ defined as follows: $ a_1 = 1$, $ a_{2k} = 1 + a_k$ and $ a_{2k + 1} = \frac {1}{a_{2k}}$ for every $ k\geq 1$. Prove that every positive rational number appears on the sequence $ \{a_n\}$ exactly once. It is easy to prove by induction that $a_{2^k}=k+1\cdots \cdots (1) $. In the same we can prove that if $a_m=\frac{a}{b} \iff a_{2^k\cdot m}=k+\frac{a}{b}$. For any $n \in \mathbb{N}$ we have $a_{2^{n-1}+1}=\frac{1}{n}$. So If we can prove that we can construct $\frac{a+1}{b}$ from any fraction $\frac{a}{b}$ we are done. We consider 3 cases: 1. $b|a+1$. In this case we can easily construct this from $(1)$. 2. $b<a+1, b\not| a+1$ 3. $b>a+1$ In case 2 we can write $a+1=bq+r$. So we have $\frac{a+1}{b}=q+\frac{r}{b}$, which yields that it is enough to prove case 3. For case 3, if we have $a_m=\frac{r}{b}$, we can say that $m$ is odd and $a_{m-1}=\frac{b}{r}$. Now using the same argument (Euclidean Algorithm) of case 2, and of case 1; we can conclude that we shall get some $a_i=r_j$, Now we prove the uniqueness of every term. If $a_i=a_j$ with $i>j$. We can can infer from the previous argument that $(a_i,a_j) \to (a_p,a_q)$ for some $p,q$ with $p\not=q$ (because we need the same number of steps for this transformation), and $a_p=a_q=n \in \mathbb{N}$. But from $(1)$ we get that $p=q$, which is a contradiction.

Jutaro wrote: Unfortunately it seems that has become some kind of trend in this competition -more examples like this can be found in recent papers... In my opinion even if the jury tries to do its duty there might be some countries that are playing foul and do not propose completely original problems... Anyways, the problem is folklore. I included it in a textbook I wrote in 1997 and even then I didn't know the source.