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Sep 22, 2020 at 17:45 vote accept Vincent Granville
Sep 21, 2020 at 17:06 comment added Yuval Peres Suppose $b_1$ and $b_2$ are positive. As Random wrote, by definition $X_{k+2}-b_2 X_{k+1}−b_1X_k$ is an integer between $0$ and $- b_1-b_2$. So that actually gives up to 9 planes in your example. No triple can lie outside these planes.
Sep 21, 2020 at 17:06 history edited Yuval Peres CC BY-SA 4.0
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Sep 20, 2020 at 18:21 vote accept Vincent Granville
Sep 22, 2020 at 7:05
Sep 20, 2020 at 5:37 comment added Random Notice that $X_{k+1} - b_2 X_k - b_1 X_{k-1}$ is by definition an integer, and you can check that the expression can acheive at most $|b_1| + |b_2|$ different values.
Sep 20, 2020 at 2:18 comment added Vincent Granville @ Yuval: how did you come up with number of at must 8 planes? Is it by visual inspection or based on some logic? That number 8 may just be $|b_1|+|b_2|$. Maybe just a coincidence, maybe not. Maybe working with the sequence $X_{3k+1}$ would fix this at least in 3D.
Sep 20, 2020 at 2:16 comment added Vincent Granville @ Yuval: I will look into that in more details. I noticed that for the simple case ($b_1=0$) you have the same problem in 2D, with couples $(X_k, X_{k+1})$ fitting on exactly $b_2$ parallel lines. If you consider the sequence $X_{2k+1}$ instead of $X_k$, the problem is gone. I would expect in the general case (3D issue) the number of planes should depend on $b_1, b_2$. The larger $b_1, b_2$, the more planes, and thus the better. But I haven't checked this yet, just a wild guess at this point.
Sep 20, 2020 at 0:03 comment added Vincent Granville Thank you, I clarified my conjecture according to your comment.
Sep 19, 2020 at 20:47 history answered Yuval Peres CC BY-SA 4.0