Nested square root for random variables

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I am looking for the distribution of a random variable $Z$ defined as

$$Z = \sqrt{X_1+\sqrt{X_2+\sqrt{X_3+\cdots}}} .$$

I am looking to find a simple distribution for $X_k$, that results in a simple distribution for the nested square root $Z$. The $X_k$'s are i.i.d. Thus my idea to investigate distributions stable under some particular transformations, see here. But this may not be the easiest way.

I tried a Bernoulli (with parameter $\frac{1}{2}$) for $X_k$, but this leads to some very difficult, nasty stuff, and a distribution on $[1, \frac{1+\sqrt{5}}{2}]$ full of gaps - some really big - for $Z$. So far the most promising result is the following.

Use a discrete distribution for $X_k$, taking on three possible values $0, 1, 2$ with the probabilities

  • $P(X_k = 0) = p_1$
  • $P(X_k = 1) = p_2$
  • $P(X_k = 2) = p_3 = 1-p_1-p_2$.

Now the resulting domain for $Z$'s distribution is $[1, 2]$, and the gaps are eliminated. The resulting distribution is still very wild, unless $p_1, p_2, p_3$ are carefully chosen. Consider

  • $p_1=\sqrt{5\sqrt{2}-1}-2$,
  • $p_2=\sqrt{5\sqrt{3}-1}-\sqrt{5\sqrt{2}-1}$,
  • $p_3=3-\sqrt{5\sqrt{3}-1}$.

I was naively thinking that this would lead to $Z$ being uniform on $[1, 2]$, based on the table featured in my article Number Representation Systems Explained in One Picture (published here, see column labeled "nested square root", with row labeled "digits distribution".) But $Z$ does not appear to be uniform, though it does appear to be well behaved: it looks like $F_Z(z)$ is a polynomial of degree 2 if $z\in [1, 2]$. Then I modified a bit the values of $p_1, p_2, p_3$, removing $0.02$ to $p_1$ and adding $0.02$ to $p_3$. The result for $Z$ looks much closer to uniform on $[1, 2]$ this time.

Question

With appropriate values for $p_1, p_2, p_3$ (and what would these values be?) can we have a simple distribution for $Z$? (uniform or polynomial on $[1,2]$)

Note: With the particular discrete distribution in question, the support domain for $Z$ is $[1, 2]$. Sure, if all $X_k$'s are zero, then $Z=0$ but that happens with probability zero. If all but one of the $X_k$'s is zero, then $Z\geq 1$.