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.
 A: 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 
$$a_n=2\sqrt2\left(\dfrac{1+\left(\dfrac{4^{n-2}\sqrt2-1}{4^{n-2}\sqrt2+1}\right)^{2^{n-1}}}{1-\left(\dfrac{4^{n-2}\sqrt2-1}{4^{n-2}\sqrt2+1}\right)^{2^{n-1}}}\right).$$
A: Rewrite your expression to get $$a_{n+1}-{a_n}=a_n+\frac{1}{a_n}$$
The left hand is the usual difference sequence.  We are then motivated to look at the differential equation $$y'=y+\frac 1y$$  Which has the general solution $$y=\sqrt{\lambda e^{2x}-1}$$.  Of course, $e$ is the result of a continuous operation...it is clear that for your sequence $\frac{a_{n+1}}{a^n}\rightarrow 2$ so we expect that $$a_n\sim \sqrt{\lambda2^{2x}-1}$$  A little calculating shows that we do quite well with $\lambda \sim 1.29534632$  (I'm seeing a very close fit). 
A: Take the recurrence, and square:
$$
a^2_{n + 1} = 4 a^2_n + 4 + \frac{1}{a^2_n}
$$
We know that $a_n \to \infty$, so as a first approximation you have for $b_n = a^2_n$:
$$
b_{n + 1} = 4 b_n + 4
$$
For $b_0 = 1$ this one has solution:
$$
b_n = (n + 4) \cdot 4^{n - 1}
$$
So $a_n \sim 2^{n - 1} \sqrt{n}$. Replacing this in the recurrence would allow you to get tighter asymptotics.
A: The recursion formula is $A_{n+1}= 2A_n+ \frac{1}{A_n}$.  Assuming for the moment that this does converge to "A" then, taking the limit of both sides, as n goes to infinity, we have $A= 2A+ \frac{1}{A}$ or $A+ \frac{1}{A}= 0$.
Multiplying by A, $A^2+ 1= 0$.  Since the limit, if it exists, would have to be real, there is no limit.
