Timeline for Automorphisms of a certain digraph defined on the set of primes? [Edited]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2012 at 6:09 | comment | added | David Feldman | Another thought: given $n$, then for $0<i<n$ one could give color $c_i$ to edges such that $p^i|q-1$ but not $p^{i+1}|q-1$ with color $c_n$ for all other edges. Doesn't the same logic still suggest an uncountable automorphism group? Then these automorphism groups would give a filtration by subgroups as $n$ increases. Not sure where to go from there, but this filtered groups would seem to encode a great deal of arithmetic. | |
| Jul 24, 2012 at 4:14 | comment | added | David Feldman | It would be interesting to know (asymptotically, as a function of $x$) the number of distinct Pratt DAGs that occur for primes less than $x$. | |
| Jul 23, 2012 at 22:11 | comment | added | Gjergji Zaimi | @David, one can obviously distinguish between primes which have non-isomorphic Pratt trees, but on the other hand, given the "infinite outdegrees" hypothesis, the Pratt tree (or Pratt DAG more precisely :)) is the only distinguishing factor, so your comment above seems absolutely right. | |
| Jul 23, 2012 at 17:22 | comment | added | David Feldman | Dirichlet's theorem does guarantee infinitely many primes q of the form pk+1. Must think about your question. BTW, scare quotes, because "Pratt trees" are not generally trees, but rather quotients of trees. | |
| Jul 23, 2012 at 14:42 | comment | added | joro | You don't mention the infinitely many outgoing edges. Can you prove it for p,q in N? | |
| Jul 23, 2012 at 8:11 | comment | added | David Feldman | And you expect two primes in the same orbit if their "Pratt trees" have the same topology, right? | |
| Jul 23, 2012 at 8:09 | comment | added | David Feldman | So if I understand, you expect the automorphism group to look like a countably-iterated wreath product, right? | |
| Jul 23, 2012 at 7:25 | comment | added | Erick Wong | @David The same heuristic does suggest infinitely many primes of the form $2^m 3^n + 1$, though. These are called Pierpont primes. | |
| Jul 23, 2012 at 6:41 | comment | added | David Feldman | If one views the potential Fermat primes merely as typical numbers of there size, the prime number theorem would lead you to guess only finitely many. | |
| Jul 23, 2012 at 6:02 | history | answered | Gjergji Zaimi | CC BY-SA 3.0 |