Timeline for Prime numbers from permutation
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 25, 2022 at 8:57 | comment | added | Ilya Bogdanov | @Notamathematician In this case, the claim about the permutation goes with no doubt, so I do not think it is necessary (neither I think MO is the best place to collect such proofs). Moreover, the claim is very easy: applying the $k$th operation simultaneously to the permutation of integers (and removing zeroes), you still get a permutation, so the result is surely injective. Moreover, a number $n$ (which exists after any operation) is stable after $k$ exceeds $n$, so it is present in the final state. | |
| Feb 24, 2022 at 14:53 | comment | added | Notamathematician | @IlyaBogdanov, today I discovered that A066937 is a permutation of the positive integers. Do you have ideas why is it so? If so, then it would be nice to ask a separate question about it where you could put your answer. | |
| Feb 15, 2022 at 21:29 | comment | added | Ilya Bogdanov | @YaakovBaruch Already approved | |
| Feb 15, 2022 at 19:47 | comment | added | Yaakov Baruch | @IlyaBogdanov: I'd like to suggest that you, or someone, post an edit at OEIS, with a link to your answer. | |
| Feb 15, 2022 at 16:11 | comment | added | Yaakov Baruch | @Notamathematician. I haven't done it the smart way, so my limit was just $n\le50000$ so up to $217$ in your list. In any case, I wouldn't call this "conjecture", just a verifiable calculation. I could have speculated on a conjecture about record breakers, hadn't it been for $a(16493)=147$. Right now I'm more curious about what $\lim_{n\rightarrow\infty} \big(2\sum_{i\le n} a(i)\big)/n^2$ might be. | |
| Feb 15, 2022 at 10:37 | comment | added | Yaakov Baruch | @Notamathematician I was was doing brute computation of $a(n)$ for all $n$ up to a large number, and reached my limit that way. But it's actually much cheaper to compute $n$ for all $a(n)$ up to a much smaller limit, and thus verify your extra terms. | |
| Feb 15, 2022 at 9:23 | comment | added | Yaakov Baruch | Very neat, and much simpler than I was expecting! As for the record breakers, I used a much more natural metric: sorted all pairs $(n,a(n))$ by $a(n)$ and printed those with $n$ higher than all preceding ones. I got the same result, including $(16493,147)$. I think one could probably get out of your proof a bound on $n$ as a function of $a(n)$. | |
| Feb 15, 2022 at 8:09 | vote | accept | Notamathematician | ||
| Feb 15, 2022 at 7:32 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
deleted 4 characters in body
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| Feb 15, 2022 at 7:15 | comment | added | Ilya Bogdanov | I've added a proof that the result is a permutation. It would be interesting to analyze whether this proof may show something about the records listed by @YaakovBaruch. | |
| Feb 15, 2022 at 7:13 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
A proof that the result is a permutation is added.
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| Feb 14, 2022 at 21:40 | comment | added | Yaakov Baruch | On further thought, the $(16403,147)$ interloper is probably a result of $n/a(n)$ not necessarily being quite the "correct" metric to define the records. | |
| Feb 14, 2022 at 21:18 | comment | added | Yaakov Baruch | Barring mistakes, up to $n=30000$ the sequence of record breaking pairs $(n,a(n))$ with respect to the ratio $n/a(n)$ is this: $(2,2), (5,3), (14,5), (41,8), (122,14), (365,21), (1094,32), (3281,56), (9842,93), (16403,147), (29525,152)\dots$ and notice that $n_{i+1}=3n_i-1$ provided we ignore the second to last pair! No similar regularity seems apparent in the $a(n_i)$ sequence. | |
| Feb 14, 2022 at 16:40 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |