We consider the two sequences $(a_n)_{n\ge 0}$ and $(b_n) _{n\ge 0}$ of integers, which are given by $a_0 = b_0 = 2$ and $a_1= b_1 = 14$ and for $n\ge 2$ they are defined as
$a_n = 14a_{n-1} + a_{n-2}$ ,
$b_n = 6b_{n-1}-b_{n-2}$.
Determine whether there are infinite numbers that occur in both sequences
The answer is negative.
First notice that the sequences are subsets of the positive integers.
Now notice that $a_n > 14a_{n-1} > 14^2a_{n-2} > \dots > 14^{n-1} a_1 = 14^n$
While $b_n < 6b_{n-1} < \dots < 6^{n-1}b_1 = 6^{n-1}.14$
But notice that $14^{n} > 6^{n-1}.14$, for every integer greater than $2$.
Meaning that $a_n > 14^n > 6^{n-1}.14 > b_n$. Meaning that we won't ever have that $a_n=b_n$, for some $n \geq 2$.
The answer is yes.
We will prove that for any $k \geq 0$, $a_{2k+1} = b_{3k+1}$ holds.
It is easy to prove by recurrence relations that if $\alpha = \sqrt{2} + 1$ and $\beta = \sqrt{2} - 1$, then $a_n = \alpha^{3n} + (-1)^n\beta^{3n}$ and $b_n = \alpha^{2n+1} - \beta^{2n+1}$.
Now it is clear that $a_{2k+1} = b_{3k+1}$ by trivial algebra.
Thus, there are infinitely many numbers occuring in both sequences.