There is a one-to-one mapping between $B_n$ and $X_n$. I studied the patterns in the distribution of successive values of $X_n$ and haven't found any. For instance, unlike other RNG's (see here and follow-up discussion here), the triplets $(X_n,X_{n+1},X_{n+2})$ do not appear to lie in a small number of parallel planes. Successive values of $X_n$ are asymptotically un-correlated. For modern tests (George Marsaglia, 2020) to assess the quality of a RNG, see here and here.
Possible improvementimprovements
Consider a $q$-order recursion $B_{n}=f(B_{n-1},\cdots,B_{n-q})$ instead of a first-order one as here. Then the period can be of the order $2^{Nq}$. Such an example for a Xorshift generator is provided here by G. Marsaglia, with $q=4$. It uses four seeds. In our case, if we were to use $q$ seeds, can can pick up $q$ irrational numbers that are linearly independent over the set of rational numbers. Their digits sequences are independent from each other (see section 1.3 in this article for a proof). An example (with $q=4$) is the first $N$ binary digits of the following numbers: $\log 2, \frac{\pi}{4}, \frac{\sqrt{2}}{2}$ and $\exp(-\frac{3}{5})$.
Of course, instead of choosing $\sqrt{2}/2$, one might choose an irrational number impossible to guess, for instance
$$\alpha=\zeta(\sqrt{31}\log 5)\cdot\Gamma(e^{73 \sin 7})+\psi_2\Big(5e^{-11\cos 19}\log(53\pi+\sin 101)\Big)$$
Further improvement is obtained by using $N$ digits of $\alpha$ or $\sqrt{2}/2$ starting at position $M$ in their binary expansion, with $M$ very large and kept secret, rather than $M=0$ as in the code below. If you work with $q$ seeds, choose a different $M$ for each seed.
It also computes the period. If the period is larger than Niter (in the code) it will return $-1$ for the period: you need to increase Niter accordingly. Use for values of $N$ smaller than 4545; to eliminate this problem, get the digits of the seed from a table or use a tool such as this one to get millions of digits for the seed.
