Solution April 10, 2007

Problem

Let a sequence be defined as follows: $a 1 = 3$ , $a 2 = 3$ , and for $n \ge 2$

$a_{n+1}a_{n-1}= a_{n}^{2}+2007.\qquad\qquad(1)$

Find the largest integer less than or equal to $Q := \frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}$ .

Solution

Take the difference of (1) for $n=t\,$ and $n=t+1\,$ .

$a_{t+1}a_{t-1} - a_{t+2}a_t = a_t^2 - a_{t+1}^2\,$

$a_{t+1}(a_{t-1}+a_{t+1}) = a_t(a_{t}+a_{t+2})\,$

$T:=\frac{a_{t-1}+a_{t+1}}{a_t} = \frac{a_{t}+a_{t+2}}{a_{t+1}}\,$

By induction we see that $T\,$ is independent of $t\,$ , so, perhaps surprisingly, $a_n\,$ is generated by the linear recurrence

$a_{n+1} = Ta_n - a_{n-1}\qquad\qquad(2)$ .

In our case, $a_3 = 672\,$ and $T = 225\,$ . The characteristic equation of (2), $t^2 = Tt - 1\,$ has solutions $t_{1,2} = \frac{225 \pm \sqrt{224}}2$ with $t_1t_2 = 1\,$ and $t_1+t_2 = T\,$ , and we get a closed form solution for $a_n\,$ , namely