Infinite nested radical and infinite continued fractions If $$a = \sqrt{k_0+\sqrt{k_1+\sqrt{k_2+\sqrt{k_3+\sqrt{\cdots}}}}}$$
and  $$b = \cfrac{1}{k_0+\cfrac{1}{k_1+\cfrac{1}{k_2+\cfrac{1}{\cdots}}}}$$
what is the relation between $a$ and $b$. What function always satisfies $a = f(b)$? Would $f(x)$ be bijective, injective, or niether?

Edit: all values in the sequence $k_n$ are whole numbers.
 A: This is not a complete answer, but it aims to put the problem in a formally better defined form.
As was mentioned by @MichaelHardy in the comments, we must assume that the $\{k_n\}_n$ are integers for the function $f$ to make sense. In truth, we have two functions
$$\mathbb{R}\stackrel{g}\longleftarrow U \stackrel{h}\longrightarrow\mathbb{R},$$
where $g(\{k_n\}) = a$ and $h(\{k_n\}) = b$. Here, $U\subset \mathbb{N}^\mathbb{N}$ is the set of sequences $\{k_n\}$ such that $\forall n\in\mathbb{N},\exists m>n:k_m\neq0$ and such that $g(\{k_n\})$ gives a well defined, finite number (this is not always the case, take $k_n = 2^n$...) Then $h$ is bijective on its image, and $f = g\circ h^{-1}:h(U)\to\mathbb{R}$. The problem is reduced to understanding what the second condition on $U$ is explicitly, what is the image of $U$ under $h$, and what properties does $g$ satisfies as a function $U\to\mathbb{R}$.
A: I'm going to provide the most basic example of your function to show how complicated it can get.
Let's take $\{k_n\}$ to be a constant sequence $\{p\}$. Then we find $a$ and $b$ explicitly:
$$a=\sqrt{p+a}$$
$$a=\frac{1+\sqrt{1+4p}}{2}$$
$$b=\frac{1}{p+b}$$
$$b=\frac{\sqrt{p^2+4}-p}{2}$$
Using $p=a^2-a$, it is easier to find $b(a)$ which will be:
$$b=\frac{1}{2} (a-a^2+\sqrt{a^4-2a^3+a^2+4})$$
But remember, since $k_n$ can only be whole, if follows that $p > 0$, or one of the $a,b$ would not converge. So, the smallest possible $p=1$, with $a(1)=\phi$ and $b(1)=\phi-1$, where $\phi$ is the Golden ratio.
Here is the plot of the function $b(a)$, where we correctly obtain that $b(p)$ is decreasing, while $a(p)$ is increasing. Only the orange curve is allowed, so the functions $b(a)$ and $a(b)$ are both single valued.

However, because of your condition that $k_n$ are whole, the function only exists for $p=1,2,3,4, \dots$, with $a,b$ values determined by the above expressions.
--
If we take even a little bit more complicated sequence with two parameters ${p,q}$, we get a quartic equation for $a$:
$$a^4-2pa^2-a+p^2-q=0$$
For $b$ the equation stays quadratic (every periodic simple continued fraction converges to a quadratic irrational as far as I know):
$$pb^2+pqb-q=0$$
Because of the quartic and two parameters it is extremely hard even to plot $a(b)$ or $b(a)$. I'm leaving it up to you, if you want.

As another example, we can compare the continued fraction constant:
$$\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cdots}}}}=\frac{I_1(2)}{I_0(2)}=0.697774657964$$ 
to the nested radical constant, which has no known closed form:
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}=1.757932756618$$
The sum, difference, product or quotient of these two numbers make no known constant (as far as I was able to check).
