Timeline for Recursive random number generator based on irrational numbers
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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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 |