What is the value of the nested radical $\sqrt[3]{1+2\sqrt[3]{1+3\sqrt[3]{1+4\sqrt[3]{1+\dots}}}}$? The closed-forms of the first three are well-known,
$$x_1=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}\tag1$$
$$x_2=\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\dots}}}}\tag2$$
$$x_3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots}}}}\tag3$$
$$x_4=\sqrt[3]{1+2\sqrt[3]{1+3\sqrt[3]{1+4\sqrt[3]{1+\dots}}}}=\;???\tag4$$
with $x_1$ the golden ratio, $x_2$ the plastic constant, and $x_3=3\,$ (by Ramanujan). 

Questions: 



*

*Trying to generalize $x_3$, what is the value of $x_4$ to a $100$ or more decimal places? (The Inverse Symbolic Calculator may then come in handy to figure out its closed-form, if any.)

*What is the Mathematica command to compute $x_4$?


P.S. This other post is related but  only asks for its closed-form which resulted in speculation in the comments. (A method/code to compute $x_4$, and a verifiable numerical value is more desirable.)

To answer this question we need to find the system that is error resilient.
We can write the equation as
$$y(x)=x\sqrt[3]{1+(x+1)\sqrt[3]{1+(x+2)\sqrt[3]{1+...}}}$$
from where we have
$$y(x)=x\sqrt[3]{1+y(x+1)}$$
or
$$y(x-1)=(x-1)\sqrt[3]{1+y(x)}$$
Now we need to estimate how this function behaves and we can easily see that
$$y(x) \sim x^{\frac{3}{2}}$$
because
$$(x^{\frac{3}{2}})^3\sim(x-1)\sqrt[3]{1+x^{\frac{3}{2}}}$$
From here we have an algorithm. Take large $N$ start with $y(N)=N^{\frac{3}{2}}$ and go backwards using
$$y(k-1)=(k-1)\sqrt[3]{1+y(k)}$$
With, for example, 
$N=20$ we have $y(1)=1.70221913267155$
$N=50$ we have $y(1)=1.70221913269546$ already fixing 14 digits which is obvious 
from $N=100$ $y(1)=1.70221913269546$
It is not difficult to estimate the error terms. If we have missed initial value by $\Delta x$ the error term exponentially diminishes until we reach $y(1)$.
Extracting Mathematica or any other command or language from this is rather trivial.
