<s>[INST] If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8...}}}}$. This can be written more generally as $f(x)=\sqrt{x+(x+1)f(x+2)}$ where $x=1$. This is where the problem arises from. If anybody can find an expression for the nested radical or find $f(x)$ I would be very happy!
 [/INST] Here is a suggested alternate approach.
Consider the recursive sequence $a_k\in\mathbb{R}$
\begin{eqnarray}
a_0&=&\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+...}}}}}\\
a_{k+1}&=&\frac{a_k^2+1}{2(k+1)}-1 \tag{1}
\end{eqnarray}
This gives the increasing unbounded sequence
\begin{eqnarray}
a_1&=&\sqrt{3+4\sqrt{5+6\sqrt{7+8\sqrt{9+10\sqrt{11+...}}}}}\\
a_2&=&\sqrt{5+6\sqrt{7+8\sqrt{9+10\sqrt{11+12\sqrt{13+...}}}}}\\
a_3&=&\sqrt{7+8\sqrt{9+10\sqrt{11+12\sqrt{13+14\sqrt{15+...}}}}}\\
&\vdots&
\end{eqnarray}
This converts the problem of finding $a_0$ into the problem of finding a generating function
\begin{equation}
G(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots
\end{equation} 
and a formula for the general term $a_n$ of the coefficient sequence.
</s>