How is Ramanujan's recurrence relation for his nested radical solved? The Wikipedia article, here, describes in some detail the derivation of Ramanujan's famous nested radical, $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}.$$ In the Wikipedia article it provides a generalized expression: $$F(x)=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\cdots}}}$$ where the first expression is defined for $a=0$, $n=1$, $x=2$. Taking the square root of both sides allows for the recurrence relation, $$F(x)^2=ax+(n+a)^2+xF(x+n).$$  The article then solves the relation, simply saying, "it can then be shown that" $$F(x)=x+n+a.$$ How is this kind of recurrence relation solved? I don't want to see Ramanujan's method. I want to know exactly what steps were taken to get from recurrence to no recurrence.
More importantly, can a recurrence relation be solved such as $$F(x)^2=ax+(n+a)^2+g(x)F(x+n)?$$ By all means, let $a=0$ and $n=1$ for simplification.
 A: I'm not very familiar with functional equations, but I was able to get some seemingly meaningful results.  Someone feel free to correct if wrong.  So, clearly $$F(0)=a+n$$
Then, we can rewrite the equation as
$$F(x)=g(x)+a+n$$
With $g(0)=0$.
Now, after a substitution with $F(0)$, we find
$$F(-n)=a$$
And then plugging in again,
$$F(-n)=g(-n)+a+n=a$$
So, $g(-n)=-n$.  A fixed point! Then,
$$F(x)=x+n+a$$
Of course plugging in will confirm the results.
As for the other part of the question, I seriously doubt that this functional equation can be solved in general.  Notice how solving it depended greatly on the form of the equation.
A: This may be of interest to you, given the pattern to your questions, I had invented a generalization of this:
http://vixra.org/pdf/1310.0177v1.pdf
See bottom of page 1 to page 2. 
The idea is that if you encounter a continuous increase between the coefficients in front of the radicals.
Then a simple test can be done to the floating additive constants (in this case the 1's) that allows you to verify if it has a good closed form.
Ramanujan's radical falls out of here as a special case, but this method doesn't directly line up with his, so it's better to say that I have a different mechanism of finding the same results as his.
