recurrence relation, all terms of the sequence positive Let $a_1=a$, $a_2=\frac{1}{a}-a$, $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ for $n=2,3,4,...$. 
Find all $a$ such that $(a_n)$ is a sequence of positive reals.

My attempt was to look at $a_3=\frac{3a^2-1}{a-a^3}$, $a_4=\frac{8a^3-4a}{3a^4-4a^2+1}$ and a few more, $a_1>0$ gives $a>0$, $a_2>0$ gives $a\in(0,1)$, $a_3>0$ gives $a\in(\frac{1}{\sqrt{3}},1)$, but this probably doesn't give important information and further terms are nasty.

Some observations...Posting it as answer since it is too long as comment. With $a_1 = a, a_2 = \frac{1}{a} - a$ and $a_n(a_{n+1} + a_n + a_{n-1}) = n$, we get
$$S = a_2a_3 + (a_3 + a_4)^2+(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 - 
(a_4^2 + a_5^2 + ...a_{n-1}^2) + a_na_{n+1}$$, where $S = \frac{n(n+1)}{2} - 3$.
So, $(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 = S - (a_2a_3 + a_na_{n+1}) + (a_4^2 + a_5^2 + ...a_{n-1}^2)$. Since LHS is $> 0$,
$S - (a_2a_3 + a_na_{n+1})$ is either $ > 0$ or It must be $S - (a_2a_3 + a_na_{n+1}) < (a_4^2 + a_5^2 + ...a_{n-1}^2)$. The lowest $n$ where this condition is applies is $n=5$ and the $a$ value we calculate must be applicable for $n>5$ I guess...
One more thing to try is to do alternate summation to cancel out products, like 
$a_3^2 + a_3a_2 + a_3a_4 - a_4a_5 -a_4^2 -a_3a_4 + a_5a_6 + a_5^2 +a_4a_5 -a_6a_7 - a_6^2 - a_5a_6 ... = 3 - 4 + 5 -6 + 7 = \sum_{1}^{n}{n}{(-1)}^{n-1}$ and see whether we can get anywhere simpler...
