# Recurrence relation $a_{n+1}a_{n-1} = 1 + a_n$ [closed]

Consider the recurrence relation: $a_{n+1}a_{n-1} = 1 + a_n$ with initial values $a_1=x$ and $a_2=y$.

Is this an example of a homogeneous equation or just a linear one? In any case does anyone have an idea how to solve this?

## closed as off-topic by 3SAT, Davide Giraudo, Morgan Rodgers, Harish Chandra Rajpoot, Jack's wasted lifeFeb 9 '16 at 17:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 3SAT, Davide Giraudo, Morgan Rodgers, Harish Chandra Rajpoot, Jack's wasted life
If this question can be reworded to fit the rules in the help center, please edit the question.

• Should your expression read $a_{n+1}a_{n-1}=1+a_n$? – lulu Feb 9 '16 at 13:43
• Or $(a_{n+1})(a_n-1)=1+a_n$? – barak manos Feb 9 '16 at 13:44
• Also, format the formulas. Look at meta.math.stackexchange.com/questions/5020/… to see how. Then people will not have to guess whether an+1 means $an + 1$ or $a_n+1$ or $a_{n+1}$. – David K Feb 9 '16 at 13:45

This equation is neither homogeneous nor linear.

Hint This particular recurrence relation is a reasonably well-known example of a periodic recurrence, that is, for some $p > 0$, we have $$a_{n + p} = a_n$$ for all $n$. With this is mind, compute $a_1, a_2, a_3, \ldots$; eventually one will find terms $a_{1 + p} = x, a_{2 + p} = y$. (Often one sees this example instead written, equivalently, in terms of the iterates $f^n$ of the map $$f : (x, y) \mapsto \left(y, \frac{1 + y}{x}\right) .)$$

With this in hand, one can readily write down the general formula for $a_n$ by giving the values for each residue class mod $p$:

$$a_n = \left\{ \begin{array}{cc} x, & n \equiv 1 \pmod p \\ y, & n \equiv 2 \pmod p \\ \displaystyle{\frac{1 + y}{x}}, & n \equiv 3 \pmod p \\ \vdots & \vdots \end{array} \right. .$$