1

Below I did a development of the recurrence starting in $k_0$ and $k_1$. (with $k_0, k_1 \neq 0$)

Note that the sequence is alternated and we get a constants values for each class of $n$:

$$a_n = \begin{cases} k_0, & \text{if $n$ is {0,6,12,...}} \\ k_1, & \text{if $n$ is {1,7,13,...}} \\ k_1/k_0, & \text{if $n$ is {2,8,14,...}} \\ 1/k_0, & \text{if $n$ is {3,9,15,...}} \\ 1/k_1, & \text{if $n$ is {4,10,16,...}} \\ k_0/k_1, & \text{if $n$ is {5,11,17,...}} \\ \end{cases}$$

That is, the sequence:

$$k_0, k_1, \frac{k_1}{k_0}, \frac{1}{k_0}, \frac{1}{k_1}, \frac{k_0}{k_1}, k_0, k_1, \frac{k_1}{k_0}, \frac{1}{k_0}, \frac{1}{k_1}, \frac{k_0}{k_1}, k_0, k_1, ...$$

My question is whether this method is correct and safe to resolve this type of relation?

Fractall
  • 103
  • Welcome to Mathematics Stack Exchange. Your work looks correct. Once you have determined that $a_6=a_0$ and $a_7=a_1$, since $a_n$ depends only on $a_{n-1}$ and $a_{n-2}$, it follows that $a_{6k+r}=a_r$ for all $k\in\mathbb N$ by mathematical induction – J. W. Tanner Feb 23 '20 at 15:10

0 Answers0