Find the closed-form expression for $c_n$ of this recursive sequence $$c_{n + 1} = 2\cdot|c_n| - \sqrt{c_n^2 + 16}$$ with $c_0 = 3$.
My question on math stackexchange was closed because lack of details. Let me clarify: I'm an amateur novelist, I don't know anything about Math. I'm not even English speaker. I accidentally derived this sequence when drawing an ellipse with geogebra. I only want to know if there is a closed-form expression for $c_n$ or not. Please don't be too harsh on me, bros. Normally I will not follow a Math problem that long. But I really found this sequence to be beautiful. I tried my best but can't find $c_n$. I'm hopeless. More than four weeks with no progress. Please don't close this.
How it was derived: ellipse $(E0)$ has $a = 5, b = 4, c = 3$. Ellipse $(E1)$ has the same $b$ but $c$ minus to $a - c$ so $c_1 = c_0 - (a_0 - c_0) = 2\cdot c_0 - a_0$. It is then generalized to $c_{n + 1} = 2\cdot c_n - a_n$ and $a_n = \sqrt{c_n^2 + b^2}$ then we have the above recurrence relationship. Note: because $c_n$ could change sign (direction) and we only talk about distant so it must take by absolute value as the equation on the top of my question.
If it just doesn't have any closed-form expression then I hope this sequence could help the research of chaotic theory. I don't know if I'm the first to discover it or it's already discovered but I don't know. I'm too tired and I quit. This problem is beyond my ability. Bye.
 A: As Gerhard Paseman noted, the existence of a closed-form expression here is unlikely. However, we can show that the sequence $(c_n)$ is bounded but not convergent, which explains your observations. 
Indeed, $c_{n+1}=f(c_n)$ for $n=0,1,\dots$, where $f(x):=2|x|-\sqrt{x^2+16}$. 
If $c_n\to c$ (as $n\to\infty$), then $f(c)=c$, so that $c=c_*:=-\sqrt2$. Also, $q:=|f'(c_*)|=|-5/3|=5/3>1$. So, $|c_{n+1}-c_*|=|f(c_n)-f(c_*)|\sim q|c_n-c_*|>|c_n-c_*|$ unless $c_n=c_*$ for all $n$, which is not the case, since $c_0=3\ne c_*$. So, the assumption $c_n\to c$ leads to a contradiction; that is, the sequence $(c_n)$ is not convergent. 
Let us now show that the sequence $(c_n)$ is bounded. First here, $f\ge f(0)=-4$. Next, the function $f$ is even, whence $f([-4,3])=f([-4,0])$. Also, $f$ is decreasing on $(\infty,0]$, whence $f([-4,0])=[f(0),f(-4)]\subset[-4,3]$. So, $f([-4,3])\subset[-4,3]$. Also, $c_0=3\in[-4,3]$. Thus, $c_n\in[-4,3]$ for all $n$. 

Here is a picture showing the graphs $\{(x,f(x))\colon x\in[-4,3]\}$ (yellow) and $\{(x,x)\colon x\in[-4,3]\}$ (blue), together with the "evolution" $(\overrightarrow{A_0A_1},\overrightarrow{A_1A_2},\dots,\overrightarrow{A_{19}A_{20}})$, where $(A_0,\dots,A_{20}):=\big((c_0,c_1),(c_1,c_1),(c_1,c_2),(c_2,c_2),\dots,(c_9,c_{10}),(c_{10},c_{10})\big)$: 

