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I investigated the sequence $Y_k=X_{3k}$, which has far more communal planes, and thus more useful to build a random generator. Of course, choosing large values for $b_1,b_2$ will further drastically improve the generator by adding a lot more planes. I suggest choosing values larger than (say) $2^{30}$ for $b_1,b_2$.

Here $b_1, b_2>0$ are integers. I investigated the sequence $Y_k=X_{3k}$, which has far more communal planes, and thus more useful to build a random generator. Of course, choosing large values for $b_1,b_2$ will further drastically improve the generator by adding a lot more planes. I suggest choosing values larger than (say) $2^{30}$ for $b_1,b_2$.

$$b_1^3\cdot Y_k+b_2(b_2^2+3b_1) \cdot Y_{k+1}-Y_{k+2} =db_1^3.$$

$$b_1^3\cdot Y_k+b_2(b_2^2+3b_1) \cdot Y_{k+1}-Y_{k+2} =d.$$

I have deleted my previous answer and instead posting this one, which is much more interesting. I investigated the sequence $Y_k=X_{3k}$, which has far more communal planes, and thus more useful to build a random generator. Of course, choosing large values for $b_1,b_2$ will further drastically improve the generator by adding a lot more planes. Here $b_1=5, b_2=3$. I suggest choosing values larger than (say) $2^{30}$ for $b_1,b_2$. Also $X_1=\frac{\sqrt{2}}{2}$.

I found 197 communal planes and they all have an equation of the form

$$Y_k+\frac{b_2(b_2^2+3b_1)}{b_1^3} \cdot Y_{k+1}-\frac{1}{b_1^3}\cdot Y_{k+2} =c.$$

The intercept $c$ ranges from $0$ to $196/b_1^3$ by equal increments of $1/b_1^3$. Each plane (identified by $c$) contains a different proportion of triplets $(Y_k,Y_{k+1},Y_{k+2})$. The empirical distribution for these proportions is featured in the histogram below, where the X-axis represents $c$, and the Y-axis the proportion of triplets lying in plan $c$.

Of course it is easy by looking at this chart to guess what the exact theoretical distribution is. I also investigated the case $b_1=3,b_2=7$ and here we have 433 planes. Their equations have the form

$$Y_k+\frac{b_2(b_2^2+3b_1)}{b_1^3} \cdot Y_{k+1}-\frac{1}{b_1^3}\cdot Y_{k+2} =c.$$

The intercept $c$ ranges this time from $0$ to $432/b_1^3$ by equal increments of $1/b_1^3$. To identify these planes, I used the program below.

I investigated the sequence $Y_k=X_{3k}$, which has far more communal planes, and thus more useful to build a random generator. Of course, choosing large values for $b_1,b_2$ will further drastically improve the generator by adding a lot more planes. I suggest choosing values larger than (say) $2^{30}$ for $b_1,b_2$.

There are $M=b_2^3+3b_1b_2+b_1^3$ communal planes and they all have an equation of the form

$$b_1^3\cdot Y_k+b_2(b_2^2+3b_1) \cdot Y_{k+1}-Y_{k+2} =db_1^3.$$

The possible values for $d$ are $0,1,\cdots,M-1.$ Each plane (identified by $d$) contains a different proportion of triplets $(Y_k,Y_{k+1},Y_{k+2})$. The empirical distribution for these proportions is featured in the histogram below (corresponding to $b_1=5,b_2=3$), where the X-axis represents $d$, and the Y-axis the proportion of triplets lying in plan $d$.

Of course it is easy by looking at this chart to guess what the exact theoretical distribution is. To identify these planes, I used the program below and some experimental math.