<s>[INST] General term of $a_{n}= \sqrt{ a_{n-1}+6}$ What is the general term of  $a_{n}= \sqrt{ a_{n-1}+6}$,$a_{1}=4$?
 [/INST] I doubt if such general closed form exists. 
We have $\sqrt{a_{n-1}+6}$ to be $\sqrt{10}$ in the first step, and $\sqrt{6+\sqrt{10}}$ is hard to simplify as if one assume it has the form $a+\sqrt{b}$, then we must have $a^{2}+b=6,2a\sqrt{b}=\sqrt{10}$. This force $a^{2}b=5/2$, and one has to solve $x^{2}-6x+5/2$. This gives $a^{2}$ or $b$ equal to $\frac{6+\sqrt{36-10}}{2}=3+\frac{\sqrt{23}}{2}$, which is fairly ugly. Another guess might be $\sqrt{c}+\sqrt{d}$, but then we would have $6=c+d$ and $10=4cd$, which ends up to the same thing. Since $a_{2}$ is already such a mess it is hard to come up with general formula for $a_{i}$. 
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