Let the sum to be calculated be
$$s=\sum_{k=1}^{\infty} \frac{1}{a(k)}$$
With the sequence $\{a(k)\} = \{2, 4, 8, 10, 20, 22, 44, 46, 92, ...\}$.
As stated in the OP there is a simple rule to generate this sequence: start with 2, then first multiply by 2, then add 2 to the result, and so on.
This leads to two sequences combined into one.
Now what are the explicit formulas for the terms?
First take the terms with the odd indices 1,3,5,.... These are {2, 8, 20, 44, 92, ...}, and then observe that each term with an even index 2,4,6,... is just 2 more than the previous odd term, i.e. {4,10,22,46, ...}.
Now the first differences of {2, 8, 20, 44, 92, ...}, are {6,12,24,48,...} which is $6\cdot 2^p$. Adjusting $p$ to the counting of terms and adding a constant $(-4)$ to match the initial term we get for the odd terms
$$a(k) = -4 + 3\cdot 2^\frac{k+1}{2}, k=1,3,5,...\tag{1a}$$
and hence, adding $2$, for the even ones
$$a(k) = -2 + 3\cdot 2^\frac{k}{2}, k=2,4,6,...\tag{1b}$$
Letting for odd and even k, respectively, $k=2i-1$ and $k=2 i$ the sum in question can be written as $s = s_o + s_e$ where
$$s_o = \sum_{i=1}^{\infty}\left( \frac{1}{-4+3\cdot 2^{i}} \right)$$
$$s_e = \sum_{i=1}^{\infty}\left( \frac{1}{-2+3\cdot 2^{i}} \right)$$
We shall show below that these can be expressed in terms of the q-polygamma function using
$$ \begin{align}s(a)&:=\sum_{i=1}^{\infty}\left( \frac{1}{-a+3\cdot 2^{i}} \right)\\&=
\frac{1}{a \log(2)}\left(- \psi _2^{(0)}\left(\frac{\log \left(\frac{6}{a}\right)}{\log (2)}\right)- \log (a)+\frac{1}{2} \log (18)\right)\end{align}\tag{2}$$
The numerical value is
$$N(s) = 1.157482445913889917496363751936696728866 ...\tag{3}$$
Short excursion to some higher trancendental function appearing here
It turns out that in order to express the simple looking sum $s(a)$ in terms of know function we have to dive into the deeper waters of the family of the so called q-functions.
The q-polygamma function $\psi _q^{(0)}(z)$ is defined in terms of the q-gamma function $\Gamma_q(z)$ which in turn is defined in terms of the q-Pochhammer symbol $(a;q)_n$.
In particular we have
$$\psi _q^{(0)}(z)= \frac{d}{dz} \log(\Gamma_q(z))$$
$$\Gamma_q(z) = \frac{(q;q)_{\infty}}{(q^z;q)_{\infty}}(1-q)^{1-z}$$
$$(a;q)_{\infty}=\prod_{j\ge 0}(1-a q^j)$$
from which follows that
$$\psi _q^{(0)}(z)= -\log(1-q) + \log(q) \sum_{i \ge 0} \frac{q^{i+z}}{1-q^{i+z}}\tag{4}$$
Here appears a sum with the structure we need to identify our sum $s(a)$.
Starting with the index $i=0$ and writing the summand as
$$\frac{1}{-a+6\cdot 2^{i}}=\frac{1}{a}\left(\frac{1}{-1+\frac{6}{a}\cdot 2^{i}}\right)=\frac{1}{a}\left(\frac{\frac{a}{6}(\frac{1}{2})^i}{1-\frac{a}{6}(\frac{1}{2})^i}\right)=\frac{1}{a}\left(\frac{q^{i+z}}{1-q^{i+z}}\right) $$
where we have set $\frac{a}{6}=q^{z}$ or $z=\frac{\log(\frac{a}{6})}{ \log(q)}$ and $q=\frac{1}{2}$.
we find
$$s(a) =\frac{1}{a}- \frac{1}{a \log(2)}\psi _{\frac{1}{2}}^{(0)}\left(\frac{\log(\frac{6}{a})}{ \log(2)}\right)\tag{5}$$
Notice that here, in contrast to the equivalent expression $(2)$, the parameter of the q-polygamma function is $q=\frac{1}{2}$.
Finally, the complete sum is given by $s = s(2)+s(4)$. This can be simplified, using transformation formulas below to
$$s = -\frac{1}{8 \log(2)} \left( 6 \psi _{2}^{(0)}\left(\frac{\log(3)}{\log(2)}\right)-\log(2) - 6 \log(3) \right)\tag{6}$$
This expression was given by @KStarGamer in a comment.
Transformation formulas for q-polygamma
I found for real $z$ and $q \gt 0$ the following transformation formulas
Shifting $z$
$$\psi _{q}^{(0)}(z+1)=\psi _{q}^{(0)}(z) - \log(q) \frac{q^z}{1-q^z}\tag{7a}$$
letting $q \to \frac{1}{q}$
$$\psi _{q}^{(0)}(z)=\psi _{1/q}^{(0)}(z)+\log(q)(z-\frac{3}{2})\tag{7b} $$
It was easy to derive $(7a)$ from $(4)$, but I had to resort to "experimenting" with the equality of $(2)$ and $(5)$ to find $(7b)$ (I would appreciate a derivation from $(4)$ as well).
