Timeline for Asymptotics for primality of sum of three consecutive primes
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2011 at 21:17 | vote | accept | Raj | ||
| Aug 10, 2011 at 19:43 | comment | added | Will Jagy | Joriki, it does depend on the problem. I put in a fair amount of effort on something by Kevin O'Bryant, maybe this one mathoverflow.net/questions/28462/… but maybe not, anyway Paul Monsky (I think in offsite email) pointed out that I was doing a limit with a known error term that resembled $$ \frac{1}{\log \log n}$$ which meant, in effect, that the limit was undetectable by computer. As I am using C++ and each program takes effort, I felt I would sit this one out. | |
| Aug 10, 2011 at 6:43 | comment | added | joriki | @Will: That seems like an overvaluation of proofs to me. It can't be proved, but it seems highly likely that it does, and something interesting might be learned from what the limit seems to be quite independent of whether its existence can be proved. | |
| Aug 10, 2011 at 4:10 | comment | added | Will Jagy | In that case, I won't bother programming anything. | |
| Aug 10, 2011 at 3:57 | comment | added | GH from MO | @Will: I doubt it (with current technology). | |
| Aug 10, 2011 at 3:28 | comment | added | Will Jagy | Can it be proved that $$ \frac{R(n)\log n}{n} $$ has a limit? | |
| Aug 10, 2011 at 2:50 | history | edited | GH from MO | CC BY-SA 3.0 |
added 125 characters in body
|
| Aug 10, 2011 at 2:49 | comment | added | GH from MO | @Noam: I believe you, I haven't looked at these densities. Thanks for pointing this out. | |
| Aug 10, 2011 at 2:40 | comment | added | Noam D. Elkies | The factor of $2$ isn't quite right, because even for prime $l>2$ the sum of three consecutive primes will diverge from equidistribution $\bmod l$ even if we believe that the residues of $p_n,p_{n+1},p_{n+2}\bmod l$ are independent. If I did this right, the probability that the sum of $3$ independent elements of $({\bf Z}/l{\bf Z})^*$ is nonzero $\mod l$ is $(l^2-3l+3)/(l-1)^2$, which exceeds $(l-1)/l$ by a factor $1 + 1/(l-1)^3$. So the constant should be $\prod_l \bigl(1+1/(l-1)^3\bigr) = 2.30096+$. This $15\%$ discrepancy should be experimentally detectable though it will take some care. | |
| Aug 10, 2011 at 2:28 | comment | added | GH from MO | @Raj: Even the statement that your $R(n)$ tends to infinity seems hopeless to verify rigorously. | |
| Aug 10, 2011 at 2:27 | comment | added | GH from MO | For example, by such heuristics one can believe that there are infinitely many primes $p$ such that $p+2$ is prime (or $2p+1$ is prime), but this is considered hopeless to prove with current methods. | |
| Aug 10, 2011 at 2:24 | comment | added | GH from MO | @Raj: My response contained vague statements like "rather evenly distributed". It is very far from a rigorous proof, and I would be surprised if anyone could give a rigorous proof. | |
| Aug 10, 2011 at 2:20 | comment | added | Raj | I dont understand how this is not a proof? Assuming everything you stated is known... I understand its not fully rigorous, but my question only asked that it is APPROXIMATELY 2n/ln(n), and you showed just that...? | |
| Aug 10, 2011 at 2:13 | history | answered | GH from MO | CC BY-SA 3.0 |