How to write $[(2+\sqrt{3})^n + (2-\sqrt{3})^n + (2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1}]/6$ to the form $a^2 + 2 b^2$ ($a, b \in \mathbb{N}$).

We know that $ (2+\sqrt{3})^n + (2-\sqrt{3})^n$ is an integer (See hear).

However, we want to write the formula \begin{align} &\frac{3+\sqrt{3}}{6} (2+\sqrt{3})^n + \frac{3-\sqrt{3}}{6} (2-\sqrt{3})^n\ &=\frac{1}{6} \left[(2+\sqrt{3})^{n+1} + (2-\sqrt{3})^{n+1} + (2+\sqrt{3})^n + (2-\sqrt{3})^n\right] \end{align} to the form $ $ a^2 + 2\,b^2,\ (a, b \in \mathbb{N}).$ $


Is the Prouhet-Thue-Morse constant transcendental in any integer base $b>2$?

I have asked the following question on Math.SE some time ago and offered a bounty, yet received no answers nor comments, so I’m posting it here.

The Prouhet-Thue-Morse constant, defined as

$ $ \tau =\sum _{{i=0}}^{{\infty }}{\frac {t_{i}}{2^{{i+1}}}}=0.412454033640\ldots $ $

where the $ t_i$ are elements of the Thue-Morse sequence, is transcendental. But is

$ $ \tau_b =\sum _{{i=0}}^{{\infty }}{\frac {t_{i}}{b^{{i+1}}}} $ $

also transcendental, for $ b>2$ ?