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.

Solution 1:

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 $u$.

Solution 2:

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...