Erdős Problem 482

Define a sequence by $a_1=1$ and $$a_{n+1}=\lfloor\sqrt{2}(a_n+1/2)\rfloor$$ for $n\geq 1$. The difference $a_{2n+1}-2a_{2n-1}$ is the $n$th digit in the binary expansion of $\sqrt{2}$.

Find similar results for $\theta=\sqrt{m}$, and other algebraic numbers.

The result for $\sqrt{2}$ was obtained by Graham and Pollak [GrPo70]. The problem statement is open-ended, but presumably Erdős and Graham would have been satisfied with the wide-ranging generalisations of Stoll ([St05] and [St06]).

The binary expansion is $\sqrt{2} = 1.0110101\ldots$, and the $n$th digit counts the leading $1$ as digit $1$. It is stated with Mathlib's Real.digits: the $n$th digit of $\sqrt{2}$ is digit $n - 1$ of $\sqrt{2}/2 = 0.10110101\ldots$, that is $\lfloor \sqrt{2} \cdot 2^{n-1} \rfloor \bmod 2$. The second conjunct records that these digits are the binary expansion: they reconstruct $\sqrt{2}/2$.