Austria2019

Answer. Yes.

Sequence $(a_n)$ starts with values $2$, $14$, $198$, $2786$, $39202$, $551614$. Sequence $(b_n)$ starts with values $2$, $14$, $82$, $478$, $2786$, $16238$, $94642$, $551614$. We therefore conjecture that $a_{2k+1}=b_{3k+1}$ holds for $k\ge 0$. Shifting the recurrence yields $$\begin{gathered} a_{n+2}-14a_{n+1}-a_n=0, \\ {a}_{n+1}-14a_n-a_{n-1}=0, \\ {a}_n-14a_{n-1}-a_{n-2}=0 \end{gathered}$$ for $n\ge 2$. Multiplying these recurrences by $1$, $14$ and $-1$, respectively, and taking the sum yields $a_{n+2}-198a_n+a_{n-2}=0$ and thus $$a_{n+2}=198a_n-a_{n-2}$$ for $n\ge 2$.

Shifting the recurrence of $(b_n)$ yields $$\begin{array}{rl}{b}_{n+3}-6b_{n+2}+b_{n+1}& =0,\\ b_{n+2}-6b_{n+1}+b_n& =0,\\ b_{n+1}-6b_n+b_{n-1}& =0,\\ b_n-6b_{n-1}+b_{n-2}& =0,\\ b_{n-1}-6b_{n-2}+b_{n-3}& =0\end{array}$$ for $n\ge 3$. Multiplying these recurrences by $1$, $6$, $35$, $6$ and $1$, respectively, and taking the sum yields $b_{n+3}-198b_n+b_{n-3}=0$ and thus $$b_{n+3}=198b_n-b_{n-3}$$ for $n\ge 3$. We see that the subsequences $(a_{2k+1})$ and $(b_{3k+1})$ have the same initial values $a_1=b_1=14$ and $a_3=b_4=2786$ and fulfil the same recurrence. This implies that $a_{2k+1}=b_{3k+1}$ for all $k\ge 0$. From the given recurrence, it is obvious that the sequence $(a_n)$ is strictly increasing. Thus we also get infinitely many values which occur in both sequences.