Crazy Iterated Square Roots I was messing around with infinite square root nesting problems like
$$w_1=\sqrt{1+\sqrt{1+\sqrt{...}}}$$
which is an easy example. I decided to try one where the terms inside of the square roots form a geometric sequence:
$$w_2=\sqrt{2^{-1}+\sqrt{2^{-2}+\sqrt{2^{-3}+\sqrt{...}}}}$$
but I can't figure it out. I figured that I probably have to set it equal to $x$ and define $x$ as a function of itself, but I'm not sure how to do that. A similar problem that I did figure out how to do was
$$w_3=\sqrt{2^{-1}+\sqrt{2^{-2}+\sqrt{2^{-4}+\sqrt{2^{-8}+\sqrt{...}}}}}$$
because if you multiply it by $\sqrt{2}$, it collapses to a problem of the first form that I mentioned. Any ideas about that second problem?
 A: If $$G_k(x) = (2x)^{k+1} + \sqrt{G_{k+1}(x)},$$ then $G_0(x)$ is the generating function of sequence A274850 and $$\sqrt{G_0(1/4)} = \sqrt{2^{-1}+\sqrt{2^{-2}+\sqrt{2^{-3}+\sqrt{...}}}}$$ but it is unlikely that this has a closed form value. More generally, let
$$ f(x,q) = x + \sqrt{f(xq,q)}.$$
We can expand it in a power series
$$ f(x,q) = 1 + x\frac{2}{2-q} - x^2\frac{q^2}{(2-q)^2(2-q^2)} + x^3\frac{2q^3}{(2-q)^3(2-q^2)(2-q^3)} + \dots$$
and now $G_0(x) = f(2x,2x).$
A: I haven't answered my exact question, but I have managed to compute the values of some similar "crazy iterated square roots" in which the coefficients inside are in geometric progression. For example,
$$\sqrt{a^2+\sqrt{4a^2+\sqrt{16a^2+\sqrt{64a^2+...}}}}=a+1$$
and
$$\sqrt{1+\sqrt{a^2+a\sqrt{a^4+a^2\sqrt{a^6+a^3\sqrt{a^8+...}}}}}=\frac{a+\sqrt{a^2+4}}{2}$$
Additionally, we have the following involving the Fibonacci numbers, with $F_0=F_1=1$ (not exactly geometric progression, but whatever):
$$\sqrt{1+F_1{-1+F_2\sqrt{1+F_3\sqrt{-1+...}}}}=2$$
