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!

One way to solve this is to use the theory of continuous iterations, which is often best tackled using power series around fixed points. Let $g(z) = \frac{z + \sqrt{z^2 + 4}}2$. Then notice
$$
g^{-1}(w) = \frac{w^2 - 1}w
$$
This has no fixed points except at infinity, but we can actually make use of that if we conjugate by $w\to \frac1w$, since $h(w) = \frac1{g^{-1}(1/w)}$ will have a fixed point at $w=0$. We can see the series for $h(w)$ is given by $$
h(w) = \frac{w}{1-w^2} = w + w^3 + w^5 + ...
$$
Formally composing this series with itself is fairly straightforward, for example:
\begin{eqnarray}
h(h(w)) &=& (w + w^3 + w^5 +...) + (w + w^3 + w^5 +...)^3 + (w + w^3 + w^5 +...)^5 + ...
\\ &=& w + 2w^3 + 2w^5 + ...
\end{eqnarray}
In general, we will have
$$
h^n(x) = w + p_3(n)w^3 + p_5(n)w^5 + ...
$$
where $p_i(n)$ satisfy a recurrence given by
\begin{equation}
w + p_3(n+1) w^3 + p_5(n+1)w^5 + ... = h^{n+1}(x) = h(h^n(x)) \\ = (w + p_3(n)w^3 + p_5(n)w^5 + ...) + (w + p_3(n)w^3 + p_5(n)w^5 + ...)^3 + (w + p_3(n)w^3 + p_5(n)w^5 + ...)^5 + ... \\ = w + (p_3(n) + 1) w^3 + (p_5(n) + 3p_3(n) + 1) w^5 + ...
\end{equation}
We get recursions like $p_3(n+1) = p_3(n) + 1$. Clearly, this is solved by $p_3(n) = n$. The $w^5$ coefficient is $p_5(n+1) = p_5(n) + 3n + 1$, which is solved by $p_5(n) = n + \frac32 n(n+1)$. In general, you can actually find a unique polynomial for each $p_i(n)$ by using Faulhaber's formula since each one is defined by a recursion $p_i(n+1) = p_i(n) + q_i(n)$ for some appropriate polynomial $q_i(n)$. Notice this recursion will still be satisfied for noninteger $n$, so we have a function $H(t,w) = w + p_3(t)w^3 + p_5(t)w^5 +...$ satisfying:
$$
 H(t+1,w) = h(H(t,w))
$$
OK so what does this have to do with the original question? Well, we have
\begin{eqnarray}
H(t+1,w) &=& \frac1{g^{-1}(1/H(t,w))} \\
\frac1{H(t+1,w)} &=& g^{-1}\left(\frac1{H(t,w)}\right) \\
g\left(\frac1{H(t+1,w)}\right) &=& \frac1{H(t,w)}
\end{eqnarray}
substituting $-n = t+1$, and recalling the definition of $g$, we find:
\begin{eqnarray}
\frac1{H(-(n+1),w)} &=& g\left(\frac1{H(-n,w)}\right) \\
\frac1{H(-(n+1),w)} &=& \frac{\frac1{H(-n,w)} + \sqrt{\frac1{H(-n,w)^2} + 4}}2
\end{eqnarray}
Then, as you can see, we will have $f(n) = \frac1{H(-n,w)}$, where $w$ is the starting value. I do not expect this function to have a nice closed form.
This formula for $f$ is only helpful on a small scale, but it does demonstrate we can draw a smooth (hence continuous) function through those points. If you want to know about the asymptotic behavior of $f(n)$ as $n\to\infty$, somos has already covered that in his answer.
