explicit formula for recurrence relation $a_{n+1}=2a_n+\frac{1}{a_n}$ For $n\in\mathbb N$,
$$a_{n+1}=2a_n+\frac{1}{a_n},\quad a_1=1.
$$
Can any one give an explicit formula for all $a_n$? If such an explicit general formula doesn't exist, please explain it. I've tried to figure out the $n$-iterated function $f^{(n)}$ where $f(x)=2x+1/x$ or even $f(\tan(t))$. But in either cases, I failed.Since the recurrence isn't linear nor homogeneous,the generating function method doesn't apply here.

We can transform this equation in to $$\dfrac{b_{n+1}}{2}=b_n+\dfrac1{b_n}$$ by substituting $b_n=\sqrt2a_n.$
Note that $$b_{n+1}-4=\dfrac{2}{b_n}(b_n-1)^2$$ and $$b_{n+1}+4=\dfrac{2}{b_n}(b_n+1)^2$$ Now $$\dfrac{b_{n+1}/4-1}{b_{n+1}/4+1}=\left(\dfrac{b_n-1}{b_n+1}\right)^2$$ Continuing this processes we can obtain
$$\dfrac{b_{n+1}/4-1}{b_{n+1}/4+1}=\left(\dfrac{b_n-1}{b_n+1}\right)^2=\left(\dfrac{4b_{n-1}-1}{4b_{n-1}+1}\right)^{2^2}=\left(\dfrac{4^2b_{n-2}-1}{4^2b_{n-2}+1}\right)^{2^3}=\cdots=\left(\dfrac{4^{n-1}b_{1}-1}{4^{n-1}b_{1}+1}\right)^{2^n}$$ You can obtain $b_{n+1}$ and hence $a_n$ from here.
$$b_{n+1}=4\left(\dfrac{1+\left(\dfrac{4^{n-1}b_{1}-1}{4^{n-1}b_{1}+1}\right)^{2^n}}{1-\left(\dfrac{4^{n-1}b_{1}-1}{4^{n-1}b_{1}+1}\right)^{2^n}}\right).$$ Therefore 
$$a_n=2\sqrt2\left(\dfrac{1+\left(\dfrac{4^{n-2}\sqrt2-1}{4^{n-2}\sqrt2+1}\right)^{2^{n-1}}}{1-\left(\dfrac{4^{n-2}\sqrt2-1}{4^{n-2}\sqrt2+1}\right)^{2^{n-1}}}\right).$$
