What is this function related with continued fractions? Playing with continued fractions, I came with the idea of iterating the limit of the simplest one:
$$1 + \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}}}\ = \Phi$$
And then I thoght about iterating the result:
$$\Phi + \cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cfrac{1}{\Phi+\cdots}}}}}\ = ?$$
And after that keep iterating again and again. Essentially, what I am doing is solving this:
$$x = a + \frac{1}{x}$$
for different values of $a$. Whose solution is:
$$ x = \frac{a \pm \sqrt{a^2 + 4}}{2}$$
And I'm just staying with the positive solution. At the end, I'm just exploring this sequence defined recursively:
$$ f(1) = \Phi \\
f(n+1) = \frac{f(n) + \sqrt{f(n)^2 + 4}}{2}
$$
When you see the plot of the terms of the sequence, you get this picture:

I would like to know which is the function that generates that curve. It really seems to be a continuos function (notice that the picture is a set of points so close, that may seem to be continuous, but it's not).
So... What's the function underlying this curve and how can we find it?
Many thanks in advance!

I think you can set up a differential equation and solve it.
\begin{eqnarray}
2f'(n) &= -f(n)+\sqrt{f(n)^2+4}\\
\text{d}n&=\frac{2\, \text{d}\! f}{-f+\sqrt{f^2+4}} \\
n &= \frac{1}{4} (f (f + \sqrt{4 + f^2} + 4 \sinh^{-1}\left(\frac{f}{2}\right)
\end{eqnarray}
Solving for $f$ should result in the function you wanted.
Wolfram plot
