Final Round

It is evident that all terms of the sequence are positive. Find $a_4$. Since $a_1=1$, $a_2=2$, $a_3=3$, we have $$a_4=\frac{a_3{a}_2+7}{a_1}=13.(1)$$ By condition, $$\begin{array}{rl}{a}_{n+1}& =\frac{a_n{a}_{n-1}+7}{a_{n-2}}=[7=a_n{a}_{n-3}-a_{n-1}{a}_{n-2}]=\\ & =\frac{a_n{a}_{n-1}+a_n{a}_{n-3}-a_{n-1}{a}_{n-2}}{a_{n-2}}=a_n\frac{a_{n-1}+a_{n-3}}{a_{n-2}}-a_{n-1},n\ge 4.\end{array}(2)$$ If we prove that all numbers $\frac{a_{n-1}+a_{n-3}}{a_{n-2}}$, $n\ge 4$, are integers, then we obtain that all numbers $a_n$, $n\ge 5$, are integers, since the numbers $a_1,a_2,a_3,a_4$ are integers. Let $b_n=(a_{n-1}+a_{n-3})/a_{n-2}$, $n\ge 4$. In particular $$b_4=\frac{a_3+a_1}{a_2}=\frac{3+1}{2}=2\text{and}{b}_5=\frac{a_4+a_2}{a_3}=\frac{13+2}{3}=5.(3)$$ Show that the sequence $(b_n)$, $n\ge 4$, is periodic with 2 as its period, i.e. $b_n=b_{n-2}$ for all $n\ge 6$. From definition of the sequences $(b_n)$ and $(a_n)$ it follows that $$\begin{array}{rl}{b}_n& =\frac{a_{n-1}+a_{n-3}}{a_{n-2}}=\frac{\frac{a_{n-2}{a}_{n-3}+7}{a_{n-4}}+a_{n-3}}{a_{n-2}}=\\ & =\frac{a_{n-2}{a}_{n-3}+7+a_{n-3}{a}_{n-4}}{a_{n-4}{a}_{n-2}}=[7=a_{n-2}{a}_{n-5}-a_{n-3}{a}_{n-4}]=\\ & =\frac{a_{n-2}{a}_{n-3}+(a_{n-2}{a}_{n-5}-a_{n-3}{a}_{n-4})+a_{n-3}{a}_{n-4}}{a_{n-4}{a}_{n-2}}=\\ & =\frac{a_{n-2}{a}_{n-3}+a_{n-2}{a}_{n-5}}{a_{n-4}{a}_{n-2}}=\frac{a_{n-3}+a_{n-5}}{a_{n-4}}=b_{n-2},n\ge 6.\end{array}$$ Therefore, taking into account (3), we have $b_{2k}=b_4=2$ and $b_{2k+1}=b_5=5$ for all $k=2,3,\dots$. Since (see (2)) $a_{n+1}=a_n{b}_n-a_{n-1}$ for all $n\ge 4$, and all numbers $b_n$, $n\ge 4$, and numbers $a_1,a_2,a_3,a_4$ are integers, it follows that all terms of the sequence $(a_n)$ are integers.