Timeline for Mersenne Prime Sequences
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2017 at 12:04 | answer | added | Gottfried Helms | timeline score: 2 | |
| Sep 29, 2017 at 11:58 | comment | added | Gottfried Helms | This looks even more drastical for higher iterates. Let's denote $n_0$ a variable having any positive integer value, $n_1 = 2^{n_0}-1$, $n_2 = 2^{n_1}-1$ and so on. Then $n_4$ can have at most two primefactors below $10,000$, namely $2879$ and $4703$ or at most $5$ primefactors below $100,000$ and the likeliness of being prime, based on subsequent statistical considerations only, should be much larger that that of a non-iterated Mersennenumber of the same size. | |
| Sep 29, 2017 at 10:56 | comment | added | Gottfried Helms | @FrançoisBrunault : Mersenne-numbers and iterated Mersenne-numbers of the same size have the significant difference, that the iterated ones cannot have "small" prime-factors while for the basic Mersenne numbers there is no such restriction. For instance a two-time iterated Mersenne number cannot have primefactors $3,5,11, ... $ and only $7,23,..$ can be primefactors of such a number. So the possible number-of-primefactors for highly iterated Mersenne-numbers is much smaller than a naive expectation. | |
| Sep 28, 2017 at 14:34 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 30, 2013 at 3:38 | comment | added | Thomas | On the wikipedia page: en.wikipedia.org/wiki/Double_Mersenne_number, it says that M(M(13)), M(M(17), M(M(19)), and M(M(31))=M(M(M(5))) are all composite, and factors have been found. The next possibility for your B1 is 61. | |
| Jun 17, 2013 at 16:30 | answer | added | alias | timeline score: 0 | |
| Jun 15, 2013 at 20:45 | comment | added | Dietrich Burde | @François: I agree. Thank you for the interesting link (I am still reading...). | |
| Jun 15, 2013 at 19:40 | comment | added | François Brunault | @Dietrich : Eugène Catalan's footnote is here archive.org/stream/nouvellecorresp01mansgoog#page/n353/mode/2up In fact, he states this as an "empirical theorem" which holds for all terms "up to a certain limit". To me this seems far from conjecturing that all terms are prime. | |
| Jun 15, 2013 at 19:38 | comment | added | François Brunault | @Barakman : The point is that there is no obvious bias towards primality arising from belonging to $A_n$. The exponents of the numbers in $A_n$ are very large, and I see no reason why they should be more prime than the Mersenne numbers of comparable size. So their primality becomes soon unlikely. | |
| Jun 15, 2013 at 19:32 | comment | added | Stefan Kohl♦ | Based on the heuristics mentioned by Francois, it seems rather reasonable to conjecture that $A_n$ is composite for all $n \geq 6$. | |
| Jun 15, 2013 at 19:27 | comment | added | barak manos | Thank you Francois! $2^{p}-1$ is a Mersenne number. I think that the probability of a Mersenne number to be in $A{n}$ is smaller than the probability of a Mersenne number to be a prime. So if I'm correct, then your probabilistic argument cannot be used in order to conclude that it is unlikely for $A{n}$ to contain only prime numbers. | |
| Jun 15, 2013 at 19:03 | comment | added | Dietrich Burde | Yes, for a discussion see also en.wikipedia.org/wiki/Double_Mersenne_number. Whether Catalan really conjectured that all of $A_n$ are prime is not clear to me. | |
| Jun 15, 2013 at 19:00 | comment | added | François Brunault | The Wagstaff heuristics primes.utm.edu/mersenne/heuristic.html assert that for large prime $p$, the probability of $2^p-1$ being prime is about $(\log p)/p$ (up to some multiplicative constant). So it seems unlike to me that $A_n$ contains only prime numbers. I would rather conjecture that any such sequence will contain a composite number. | |
| Jun 15, 2013 at 18:43 | history | asked | barak manos | CC BY-SA 3.0 |