Timeline for Function that produces primes
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Jun 22, 2023 at 6:20 | comment | added | mathoverflowUser | Observation in Sagemath: sagecell.sagemath.org/… | |
| Mar 24, 2022 at 20:46 | comment | added | Sebastien Palcoux | The sequence overlaps next_prime(n), for n=1,...,15. | |
| Mar 17, 2022 at 12:48 | comment | added | Joachim König | @JohnOmielan I don't know for sure of course, but I don't think there will be counterexamples, and exactly because of the "erratic" (i.e. "random") behavior. Indeed, for reasonably large n, the termination of the iteration (i.e. first prime hit) comes so incredibly early (like, less then 1% of the total range when $n>10^6$) that a counterexample (i.e. no prime hit all the way through) would have to be more than just a bit weird, see also my comment on Sean's answer. | |
| Mar 17, 2022 at 10:46 | comment | added | John Omielan | @Notamathematician FYI, I wrote & ran (for over $13$ hours!) a relatively optimized C++ program. The main optimization is that, similar to what's explained in Sean's answer, if $b = a(m-1,n) + n - m$ is prime, then $a(n-1,n) = b$, so can skip the remaining iterations. I have checked some of my results against PARI/GP output using code similar to yours to help verify my program worked properly. Anyway, no counter-examples up to $5 \times 10^8$ were found. Nonetheless, I suspect there'll be a few eventually due to the "erratic" aspect of the increases stated in Sean's answer. | |
| Mar 17, 2022 at 9:54 | history | edited | Glorfindel | CC BY-SA 4.0 |
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| Mar 16, 2022 at 20:07 | history | became hot network question | |||
| Mar 16, 2022 at 18:51 | answer | added | Sean Eberhard | timeline score: 28 | |
| Mar 16, 2022 at 18:30 | comment | added | Sean Eberhard | @TimothyChow If $2n-1$ is prime then $a(m, n) = n + m$ for all $m$. | |
| Mar 16, 2022 at 17:41 | comment | added | Timothy Chow | @IlyaBogdanov Could you please elaborate on your comment? | |
| Mar 16, 2022 at 17:25 | comment | added | T. Amdeberhan | @TimothyChow: I hope moderators could intervene in these kinds of matters. Sadly, there are some users who jump to "close votes", prematurely. It appears like a pessimistic impulse. | |
| Mar 16, 2022 at 13:50 | comment | added | Timothy Chow | I can't understand why there are two close votes (as of this writing). | |
| Mar 16, 2022 at 12:55 | answer | added | Carlo Beenakker | timeline score: 31 | |
| Mar 16, 2022 at 10:43 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| Mar 16, 2022 at 10:22 | comment | added | Ilya Bogdanov | @HenriCohen Surely, if $2n-1$ is prime, then $a(n-1,n)=2n-1$. | |
| Mar 16, 2022 at 9:08 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| Mar 16, 2022 at 8:58 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| Mar 15, 2022 at 22:54 | comment | added | mathworker21 | Unless I made a mistake, it suffices to show the following for all positive integers $n,d \ge 1$ (it is vacuously true if no such $r$ exists). If $r$ is the largest thing at most $d+1$ that has non-1 gcd with $n+d$, then either $n+d+r-1$ is prime or there's something at most $r$ with non-1 gcd with $n+d+r-1$. | |
| Mar 15, 2022 at 19:53 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Mar 15, 2022 at 19:41 | history | edited | Henri Cohen | CC BY-SA 4.0 |
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| Mar 15, 2022 at 19:40 | comment | added | Henri Cohen | It even seems to give all odd primes, with some repetition. | |
| Mar 15, 2022 at 19:18 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| Mar 15, 2022 at 19:08 | review | Close votes | |||
| Mar 17, 2022 at 18:01 | |||||
| Mar 15, 2022 at 18:13 | history | asked | Notamathematician | CC BY-SA 4.0 |