Timeline for Boolean Cube of Primes
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2011 at 20:48 | answer | added | jcsp | timeline score: 0 | |
| Sep 13, 2010 at 5:58 | vote | accept | Avishay Tal | ||
| Sep 9, 2010 at 21:49 | answer | added | gowers | timeline score: 2 | |
| Sep 9, 2010 at 19:41 | comment | added | Gerhard Paseman | There may be some value in considering the question for almost-primes and for distinct delta. I think it is possible to construct such a sequence whose members are all relatively prime to some large primorial, and where the system of deltas belongs to a set of small primorials. If you do computations with this set, you may get a good figure for m asympotically. Then you can compare this with what the (in this case m is 2^k) m-prime tuples conjecture says. Gerhard "Ask Me About System Design" Paseman, 2010.09.09 | |
| Sep 9, 2010 at 17:02 | answer | added | M.S | timeline score: 0 | |
| Sep 9, 2010 at 1:04 | comment | added | Hashem sazegar | I think in above question should be $\Delta_1=2$,$\Delta_2=4$ and $\Delta_3=2$ | |
| Sep 5, 2010 at 6:04 | comment | added | dvitek | @Tracy: one may consider an arithmetic progression of primes of length say $k = 2^{m+1}$ with common difference $D$. Then set $\Delta_i = 2^{i-1}D$. Clearly then any choice of $a_i$ gives a prime. | |
| Sep 4, 2010 at 21:03 | history | edited | Avishay Tal | CC BY-SA 2.5 |
added clearer notes
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| Sep 4, 2010 at 19:52 | comment | added | Tracy Hall | As the problem is currently stated, the $2^m$ choices do not necessarily give distinct integers, and so for example an arithmetic progression of length $k$ corresponds to the situation where $m=k-1$ and the $\Delta$ values are equal. (I do assume that each $\Delta_i$ is required to be strictly positive, since otherwise $m$ is trivially unbounded.) If $2^m$ distinct primes are what is asked for, then an arithmetic sequence of length $k$ only gives $m=\lfloor \log_2 k \rfloor$. | |
| Sep 4, 2010 at 18:55 | comment | added | Péter Komjáth | Gil: I do not know Avishay's argument but log log n is the size Szemeredi obtains for a cube in a set of positive density in [1,..,n]. | |
| Sep 4, 2010 at 18:05 | comment | added | Gil Kalai | For a specific choice of Delta's you have here 2^m integers from [n.2n] so you can "estimate" the probability that they are all promes by (1/logn)^{2^m}. If m is fixed maybe this gives the correct asymptotic value of the expected number of such cubes of primes by the recent Green-Tao-Ziegler theory. This gives you also a guess for the correct value of m but I suppose that transforming this guess to lower and upper bounds is beyond present technology. How do you get the log log n/2 result? | |
| Sep 4, 2010 at 17:19 | comment | added | Robin Chapman | The existence of such $p$ and $\Delta_i$ for each $m$ certainly follows from the Green-Tao theorem. I don't know anything about the asymptotics of that though. | |
| Sep 4, 2010 at 17:11 | history | asked | Avishay Tal | CC BY-SA 2.5 |