$w =\operatorname{arcsinh}(1+2\operatorname{arcsinh}(1+2^2\operatorname{arcsinh}(1+2^{2^2}\operatorname{arcsinh}(1+\dotsm$

Question

In the context of positive reals consider$\DeclareMathOperator{\arcsinh}{arcsinh}$

$$ w= \arcsinh( 1 + 2 \arcsinh( 1 + 2^2 \arcsinh ( 1 + 2^{2^2} \arcsinh( 1 + 2^{2^{2^2}} \arcsinh( 1 + \dotsm $$

Now consider a real $A > w$

Then the iterations

$A_0 = A$

$A_1 = (\sinh(A) - 1)/2 $

$A_2 = (\sinh(A_1) - 1)/ 2^2 $

$...$

grow superexponentially ! (Tetration)

Reals $B$ in $[0,w[ $ do not give a sequence $B_n$ ( same iterations as above ) that grows to infinity.

So the interesting things are

1) the value of $w$

2) How fast does the sequence

$w_0 = w $

$w_1 = (\sinh(w_0) - 1 ) / 2 $

$ ...$

grow ?

Could it be slower than superexponential ?

Consider the analogue

$$ 3 = \sqrt{1 + 2 \sqrt{ 1 + 3 \sqrt{ 1 + 4 \sqrt {...}}}} $$

Where the iterations $c_1 = 3, c_n = ( c_{n-1}^2 - 1) / n $ give the sequence $3,4,5,6,7,...$ rather than a double exponential growth ( as expected at first ).

Unfortunately tetration-like ideas tend to give Numbers too large for computation, If we are not careful. So we need a trick or some theory probably.

I tried to compute $w$ and arrived at the estimate

$$ w = 2.613\,022\,592\,281\,8\!\dots $$

This Number might be wrong but this is my guess. Also the number seems familiar but that may be my imagination.

Is there an efficient way to compute $w$ ?

Are my digits correct ?

And the main question again :

How fast does $w_n$ grow ??

Related : Julia set of $x_n = \frac{ x_{n-1}^2 - 1}{n}$

But this is in the context of reals only.