I can answer in the bounded case, and the bounded case only.
If you construct the bounded analytic hyper-operators you get the following sequence of analytic functions:
\[
\begin{align}
\alpha \in (1,\eta)\,\, & n \ge 1\,\,\Re(z) > 0\\
\alpha \uparrow^n z &: \mathbb{C}_{\Re(z)> 0} \to \mathbb{C}_{\Re(z) > 0}\\
\alpha \uparrow^1 z &= \alpha^z\\
\alpha \uparrow^n 1 &= \alpha\\
\alpha \uparrow^n ( \alpha \uparrow^{n+1} z) &= \alpha \uparrow^{n+1} (z+1)\\
\end{align}
\]
Then, I believe you are asking for the fixed points of tetration not pentation, because the formula you wrote is for the fixed points of exponentiation, or the infinite power tower.
To begin, in this schema:
\[
\lim_{k\to\infty} \alpha \uparrow^2 ...(k\,\text{times})...\uparrow^2 \alpha = \omega_2\\
\]
And additionally:
\[
\lim_{x \to \infty} \alpha \uparrow^3 x \to \omega_2\\
\]
There are ways to evaluate the \(\omega_2\) constant more effectively, but they're probably not what you are looking for. The Lambert W identity for the infinite tetration is more of a coincidence than a rule of law for hyper operators. But:
\[
\alpha \uparrow^2 \omega_2(\alpha) = \omega_2(\alpha)\\
\]
This is a geometrically attracting fixed point for \(\alpha \in (1,\eta)\), and this is enough to construct an implicit solution. Which honestly would be the fastest route, as you'd just be solving for the taylor expansion in \(\alpha\) about a point of your desire.
There is no real quick way, personally I would use the iterative limit, as it's fast and simple. But if that's not up your alley, then I'd suggest you expand the Schroder function:
\[
\alpha \uparrow^2 z = \Psi_\alpha^{-1}\left(\Psi_\alpha(1) (\log(\alpha)\omega_1)^z\right)\\
\]
Where \(\omega_1\) is your Lambert W expression above. Now you are asking for a point:
\[
\Psi^{-1}_\alpha\left(\Psi_\alpha(1) (\log(\alpha)\omega_1(\alpha))^{\omega_2(\alpha)}\right) = \omega_2(\alpha)\\
\]
Discovering a taylor expansion from here is difficult, but should be doable. This reduces into, \(\lambda(\alpha) = \log(\alpha)\omega_1(\alpha)\) and \(a_k(\alpha) > 0\):
\[
\alpha \uparrow^2 z = \omega_1 - \sum_{k=1}^\infty a_k(\alpha) \lambda(\alpha)^{kz}\\
\]
Which then isn't too hard to run a an algo to get:
\[
\omega_1 - \sum_{k=0}^\infty a_k(\alpha) \lambda(\alpha)^{k\omega_2(\alpha)} = \omega_2(\alpha)
\]
It's especially fast to construct this function too because:
\[
\alpha < .... <\omega_n(\alpha) <....<\omega_2(\alpha)< \omega_{1}(\alpha) < e
\]
If you construct the bounded analytic hyper-operators you get the following sequence of analytic functions:
\[
\begin{align}
\alpha \in (1,\eta)\,\, & n \ge 1\,\,\Re(z) > 0\\
\alpha \uparrow^n z &: \mathbb{C}_{\Re(z)> 0} \to \mathbb{C}_{\Re(z) > 0}\\
\alpha \uparrow^1 z &= \alpha^z\\
\alpha \uparrow^n 1 &= \alpha\\
\alpha \uparrow^n ( \alpha \uparrow^{n+1} z) &= \alpha \uparrow^{n+1} (z+1)\\
\end{align}
\]
Then, I believe you are asking for the fixed points of tetration not pentation, because the formula you wrote is for the fixed points of exponentiation, or the infinite power tower.
To begin, in this schema:
\[
\lim_{k\to\infty} \alpha \uparrow^2 ...(k\,\text{times})...\uparrow^2 \alpha = \omega_2\\
\]
And additionally:
\[
\lim_{x \to \infty} \alpha \uparrow^3 x \to \omega_2\\
\]
There are ways to evaluate the \(\omega_2\) constant more effectively, but they're probably not what you are looking for. The Lambert W identity for the infinite tetration is more of a coincidence than a rule of law for hyper operators. But:
\[
\alpha \uparrow^2 \omega_2(\alpha) = \omega_2(\alpha)\\
\]
This is a geometrically attracting fixed point for \(\alpha \in (1,\eta)\), and this is enough to construct an implicit solution. Which honestly would be the fastest route, as you'd just be solving for the taylor expansion in \(\alpha\) about a point of your desire.
There is no real quick way, personally I would use the iterative limit, as it's fast and simple. But if that's not up your alley, then I'd suggest you expand the Schroder function:
\[
\alpha \uparrow^2 z = \Psi_\alpha^{-1}\left(\Psi_\alpha(1) (\log(\alpha)\omega_1)^z\right)\\
\]
Where \(\omega_1\) is your Lambert W expression above. Now you are asking for a point:
\[
\Psi^{-1}_\alpha\left(\Psi_\alpha(1) (\log(\alpha)\omega_1(\alpha))^{\omega_2(\alpha)}\right) = \omega_2(\alpha)\\
\]
Discovering a taylor expansion from here is difficult, but should be doable. This reduces into, \(\lambda(\alpha) = \log(\alpha)\omega_1(\alpha)\) and \(a_k(\alpha) > 0\):
\[
\alpha \uparrow^2 z = \omega_1 - \sum_{k=1}^\infty a_k(\alpha) \lambda(\alpha)^{kz}\\
\]
Which then isn't too hard to run a an algo to get:
\[
\omega_1 - \sum_{k=0}^\infty a_k(\alpha) \lambda(\alpha)^{k\omega_2(\alpha)} = \omega_2(\alpha)
\]
It's especially fast to construct this function too because:
\[
\alpha < .... <\omega_n(\alpha) <....<\omega_2(\alpha)< \omega_{1}(\alpha) < e
\]