# 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

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• 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

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