Closed form for non-linear recurrence $a_n=\sqrt{a_{n-1}+6}$ with $a_1=6$ Does equation $a_n=\sqrt{a_{n-1}+6}$ with $a_1=6$ have a closed form? I've found no linearization method. Any suggestion or hint will be highly appreciated.
 A: There is no hope to find an explicit formula for $a_n$, but the asymptotics is clear.
One has $a_n=u(a_{n-1})$ where the function $u:x\mapsto\sqrt{x+6}$ has a unique fixed point $a=3$, hence $a_n-a=u(a_{n-1})-a=u(a_{n-1})-u(a)$ and one can suspect that $a_n\to a$. As a matter of fact, $u(a_n)-a=b_n\,(a_{n-1}-a)$ with $b_n=1/(u(a_{n-1})+a)$ hence $0\lt b_n\lt1/a$ hence $|a_n-a|\leqslant a^{-n}\,|a_0-a|$. Since $a\gt1$, this shows that $a_n\to a$.
More is true: since $b_n\to b=1/(2a)=1/6$, $a_n-a=b^{n+o(n)}$. In other words, since $a_n\gt a$ for every $n$,
$$
\lim\limits_{n\to\infty}\frac{\log(a_n-a)}n=\log(b)=-\log(6),
$$
and a little more work shows that $a_n-a=c\,b^n\,(1+o(1))$, where $c$ depends on $a_0$ and has no simple explicit form.
Edit: The algebraic trick used above to compute $b$ might hide the fact that $b=u'(a)$, where $a=u(a)$ is the fixed point of $u$.
A: Rewrite it as
$$
a_n^2=a_{n-1}+6\\
a_{n}-a_{n-1}=a_n-a_n^2+6
$$
and approximate it as
$$
a_n'-a_n+a_n^2-6=0,
$$
which is solved by 

 $$a_n = \frac{3 e^{5 n}+2 e^{5 c_1}}{e^{5 n}-e^{5 c_1}},$$ with $c_1= \frac15 \left(5-3 \log(2)+\log(3)\right)\;\;$ such that $a_1=6$. Here's a plot...

A: 
Zegalur: I followed your interesting post but cannot see the final
  closed form. My first question is why to "change direction of element
  index" when the changed recurrence Fn+1=F2n−6 is as hard as the
  original a2n=an−1+6? Has this backward recurrence, say, something to
  do with symmmetries in a further elaboration? Would you please clarify
  a bit?

Strange. I can't write any comments.. so I wrote a new answer.. Anyway...
I take a square of two parts of equation and move $6$ to the left side of equation.
$$
a_n = \sqrt{a_{n-1}+6} \ \ \ => \ \ \ a^2_n - 6 = a_{n-1} \ \ \ => \ \ \ a_{n-1}=a^2_{n} - 6.
$$ 
If we have $a_0$ we now can find $a_{-1}, a_{-2}, ...$ ($a_{-1}=30$).
I've changed direction of index because it's more native to work with something like this:
$$
F_n = Q(F_{n-1}).
$$
(when the next element of sequence is expressed through previous one).
Now $F_{n} = a_{-n}$. If I will find solution in a closed-form $F_{n} = W(n)$ then I can use it to express solution for $a_n$. 
$$
a_n = F_{-n} = W(-n).
$$
It is exactly what I did.

You also write that you can't find a final result. That is because solution is expressed by non elementary function $f(x)$ (see my answer). But you can't even find a Taylor series for it. But that function exist and it is continuous (see for example the plot of it).
Final result is
$$
a_n = F_{-n} = f(X^{2^{-n}})
$$
where $f$ and $X$ are defined in my previous answer.
