Finding the limit of a pseudo geometric recurring sequence $$
a_{n+1}=\frac{1}{2^{a_n}}
$$
Whilst practising to find the n-th term of a sequence, I haven't been able to successfully generalize this recurring sequence. I've called it pseudo geometric but honestly I got no idea what it's called, I would love to learn!
Things I've tried:

*

*In my calculator, this sequence converges to a value around $0.641185744$.

*We can assume the sequence to be geometric, so $k^{n+1}=\frac{1}{2^{k^n}}$, then $\log_2(k^{n+1})=\frac{1}{k^n}$ but I believe this is a dead end.

 A: The $n$-th term depends on the first term of the sequence. However this sequence do converge to the value you found. This value satisfies:
$$L = \frac1{2^L}, \text{ or } L\times 2^L = 1$$
A sequence of its decimal expansion is at https://oeis.org/A104748.
Its closed form involves the Lambert W Function, which is non-elementary:
$$L\times 2^L = 1 \iff (L\ln 2)e^{L\ln 2} = \ln 2 \iff L \ln 2= W(\ln2)\iff L = \frac {W(\ln 2)}{\ln 2}$$
so the only way to obtain its value is by numerical approximation.
EDIT: The recurrence relation given is equivalent to:
$$a_{n+1} = \left(\frac12\right)^{a_n}$$
Given that $a_1=1$, we have the sequence:
$$\left\{1,\frac12, \left(\frac12\right)^\frac12, \left(\frac12\right)^{\left(\frac12\right)^\frac12},\left(\frac12\right)^{\left(\frac12\right)^{\left(\frac12\right)^\frac12}}, \dots\right\}$$
or, in Tetration notation:
$$\left\{^0\left(\frac12\right),^1\left(\frac12\right),^2\left(\frac12\right),^3\left(\frac12\right)\dots\right\}$$
giving the "closed form" $a_n = ^{n-1}\left(\frac12\right)$.
