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.

I post it as a curiosity. Considering the difference equation
$$
\frac{n}{x_n^2} = 1 + \frac{x_{n+1}+x_{n-1}}{x_n}
$$
and considering 
$$
\Delta x_n = x_n-x_{n-1}
$$
we have
$$
\frac{n}{x_n^2} = 3+\frac{\Delta x_{n+1}-\Delta x_n}{x_n}
$$
which is a difference approximation for the DE
$$
\frac{t}{x^2(t)} = 3 + \frac{\ddot x(t)}{x(t)}
$$
now calling $x(1) = a = \frac{2 \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}$ and $x(2) = a-\frac 1a$ and integrating we have

In red the plot for $x_n = \sqrt{\frac n3 + \frac{1}{36n}+ \mathcal{O}(\frac{1}{n^3})}$ and in blue the DE solution. 
and now considering instead the initial conditions $x(2) = a-\frac 1a, x(3) = \frac{1-3a^2}{a^3-3}$ we obtain

This result points in the direction of good agreement between the DE and the recurrence. Now following with the initial conditions 
$$
x(3) = \frac{1-3a^2}{a^3-3}\\
x(4) = \frac{8 a^3-4 a}{3 a^4-4 a^2+1}
$$
we obtain

