We will prove, that every postive rational number is in sequence. $q_{2k}=\frac{n_{2k}}{n_{2k-1}}=\frac{n_k+n_{k-1}}{n_{k-1}}=1+\frac{n_k}{n_{k-1}}=1+q_k>1$ $q_{2k+1}=\frac{n_{2k+1}}{n_{2k}}=\frac{n_k}{n_k+n_{k-1}}=\frac{1}{1+\frac{n_{k-1}}{n_k}}=\frac{1}{1+\frac{1}{q_k}}<1$ Let $q=\frac{a}{b}$ is rational, that is not in sequence and $a+b$ is minimal. If $a>b$ then $\frac{a-b}{b}=q_k$ is in sequence,then $q_{2k}=1+q_k=\frac{a}{b}=q$ - contradiction If $a<b$ then $\frac{a}{b-a}=q_k$ is in sequence, then $q_{2k+1}=\frac{1}{1+\frac{b-a}{a}}=\frac{a}{b}=q$ - contradiction For proving uniqueness we can use same method. Let $k$ - minimal such number, that exists $m>k$ and $q_k=q_m$ Then obvious, that $m,k$ have same parity, so $q_{[k/2]}=q_{[m/2]}$ so $k$ can not be minimal