<s>[INST] What is this number: $x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\approx 4.14031...$? I was watching this Mathologer video about Ramanujan's nested radical identity
$$
3 = \sqrt{1 + 2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\, ,
$$
and I decided to look at this for different sets of "radical coefficients". Using powers of $2$ yields the quantity
$$
x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, .
$$ 
I can't figure out much about the nature of the number $x$. I can't even prove that it exists, let alone anything analytically about its value. Nevertheless, by looking at the truncation of this expression at different levels in Mathematica, it seems to be the case that:
1) The expression converges relatively quickly.
2) $x \approx 4.14031456214125981180937290...$
I tried feeding this number into an Inverse Symbolic Calculator... to no avail, although $x$ does match about 8 digits of
$$
\frac{5}{4}\, \frac{\sqrt{3} + \sqrt{5}\ln(5)}{\ln(5)}\, .
$$
Can anyone tell me anything else about this quantity?
Edited to add:
I note that the apparently simpler Nested Radical Constant
$$
\sqrt{1 + \sqrt{2 + \sqrt{3 + ...}}}
$$
has no closed form, so we clearly have no reason to expect my quantity $x$ to have a closed form either...
Edit 2:
There was an answer that someone posted, and then deleted for some reason, which gave a method for getting a lower bound on this quantity. A variation of that method leads to:
\begin{align}
x &= \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\\
&> \sqrt{1+2^1\sqrt{0+2^2\sqrt{0+2^3\sqrt{0+...}}}}\\
&= \sqrt{1+2^1 \times 2^{2/2}\times 2^{3/4}\times 2^{4/8}\times 2^{5/16}...}\\
&= \sqrt{1 + 2^{\sum_{m=0}^\infty \frac{m+1}{2^m}}}\\
&= \sqrt{1 + 2^4} = \sqrt{17} \approx 4.1231...
\end{align}
This is apparently a relatively tight lower bound.
 [/INST] Interpreting the sequence as mvw, we have $x_1<\sqrt{2^1+2^1}=2$, $x_2<\sqrt{2^1\sqrt{1+2^2}+2^1\sqrt{1+2^2}}=2\sqrt[4]{1+2^2}<2^{1+3/4}$ and similarly $$x_n<2\sqrt[4]{2^2\sqrt{\cdots\sqrt{2^n+2^n}} +2^2\sqrt{\cdots\sqrt{2^n+2^n}}}=2^{2/2+3/2^2+\cdots+(n+1)/2^n}<2^3.$$ The modus operandi is to turn each $1$ into the term that stands at the right of it, which is certainly bigger than $1$. Since $x_n$ is also increasing, its limit exists. One might as well note that the analogous sequence with powers of any fixed $m>1$ does converge, and if $m=1$ it converges to the golden ratio.
I guess my method can be refined, or there is another one, to yield a better bound.
UPDATE
I thought it wouldn't be easy to majorize effectively and more sharply the sequence in a similar fashion, but I was wrong. It suffices to turn each $1$ into half the term that stands at the right of it. We thus get, for $n>1$, \begin{align} x_n &<\sqrt{(2^0+2^1)\sqrt{(2^1+2^2)\sqrt{\cdots\sqrt{2^{n-1}+2^n}}}} \\ &=2^{1/4+2/8+\cdots+(n-1)/2^n}\cdot3^{1/2+1/4+\cdots+1/2^n}<6.  \end{align}
UPDATE 2 Finally nailed it! Let $P_m=\prod_{n=1}^\infty(1+2^{n+m})^{1/2^n}$. Then we find $$x<\sqrt{(1+2^1)\sqrt{(1+2^2)\sqrt{\cdots}}}=P_0<5.284,$$ $$x<\sqrt{1+2\sqrt{(1+2^2)\sqrt{(1+2^3)\sqrt{\cdots}}}}=\sqrt{1+2P_1}<4.429,$$ $$x<\sqrt{1+2\sqrt{1+2^2\sqrt{(1+2^3)\sqrt{(1+2^4)\sqrt{\cdots}}}}}=\sqrt{1+2\sqrt{1+4P_2}}<4.215 $$ etc.
</s>