Timeline for The maximum lengthed sequence of prime numbers with certain conditions (denizens)
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 8, 2015 at 13:57 | comment | added | Brad Graham | @Gerhard you might be interested in this open question mathoverflow.net/questions/217756/… | |
| Sep 6, 2015 at 4:52 | comment | added | Gerhard Paseman | To name, no. However, you can use Westzynthius's construction to build one. It will require lots of primes. Current calculations suggest you will need x > 50. Gerhard "Check Out Thomas Hagedorn's Paper" Paseman, 2015.09.05 | |
| Sep 6, 2015 at 3:07 | comment | added | Brad Graham | @Gerhard do you know of any $p_x$-denizen longer than $4p_x$? | |
| Sep 3, 2015 at 19:56 | comment | added | Brad Graham | Smallest interval that contains a totative of a primorial? Thats my next approach in the future. | |
| Sep 3, 2015 at 19:55 | comment | added | Brad Graham | I've considered splitting, the interval of integers $I=[0, p_x \# !)$, into intervals with $p_y \in \mathbb{P}$ length and that are complete residues of $p_y$, where $p_y \leq p_x$. In each of these smaller intervals; $(p_y -1)/p_y$ numbers are not divisble by $p_y$. From this, its easy to show that atleast $1/p_x$ of the numbers in $I$ are totatives. From this, if we half the interval I, we can say there are half as many numbers not divisible by any prime upto $p_x$ except $2$. There are just under or over a half the amount of even numbers in I. So following this approach, can we find a small | |
| Sep 3, 2015 at 19:37 | comment | added | Brad Graham | @Gerhard Its strange because for the smallest prime numbers (upto 23), the denizen centred at a primorial is the largest... So what happens after 23? What the best explanation in mathematical terms? And yeah it makes sense to look at the totatives, but describing their distribution naturally is difficult when we cant even say how far apart they are at most! | |
| Sep 3, 2015 at 19:26 | comment | added | Gerhard Paseman | Which is as I suspected, a pattern of least prime factors of integers n with central number a multiple of a primorial. While some of the large prime gaps are near multiples of large primorials, many do not have such multiples near the center. I show my bias when I say it makes more sense to study the distribution of totatives to a primorial, and see where those large gaps appear. Gerhard "Sensing A Philosophical Clash Here" Paseman, 2015.09.03 | |
| Sep 3, 2015 at 19:18 | comment | added | Brad Graham | @Gerhard Consider the denizen $2,5,2,3,2,7,0,11,2,3,2,5,2$. I have used $0$ to represent its centre (although it should be a 2). If in this sequence $0$ has position $k$, we can see a $2$ at position $k \pm 2$, a $3$ at position $k \pm 3$ and a $5$ at position $k \pm 5$. So we can $0$ has a symmetric depth of $5$. Now, as the symmetric depth of all other members in this sequence is less than $5$, then this denizen has a symmetric depth of $5$. I initially thought the longest denizen consisting of prime numbers upto $p_x$ was the $p_x$ denizen with symmetric depth of $x-2$. | |
| Sep 3, 2015 at 19:05 | comment | added | Gerhard Paseman | I think you will find nonsymmetric patterns will abound with large values of x. You might consider looking at recent papers on large prime gaps. They are implicitly constructing denizens using a log's worth of small primes and interesting sprinklings of large primes, starting with a base of medium primes. I think the symmetric depth for these will be pretty small, and I don't even understand the definition of symmetric depth. Gerhard "Check Out Rankin And Pomerance" Paseman, 2015.09.03 | |
| Aug 31, 2015 at 3:00 | comment | added | Brad Graham | @GerhardPaseman Now to try and explain why there exists a denizen$\lbrace a_k \rbrace^{23}_5$ larger than $\lbrace a_k \rbrace^{23}_17$ i'll look up Westzynthius thank you. | |
| Aug 31, 2015 at 2:56 | comment | added | Brad Graham | Excellent so there are larger denizens! | |
| Aug 31, 2015 at 2:49 | comment | added | Gerhard Paseman | 23252a232b2_23272_232_25232a2723252b232 . Length 39. a and b represent the primes 11 and 13. Fill in the three blanks with 17, 19, and 23. Use Chinese Remainder Theorem to calculate the numbers corresponding to the digits, or lookup Westzynthius. Gerhard "And There Are Larger Examples" Paseman, 2015.08.30 | |
| Aug 29, 2015 at 18:59 | comment | added | Brad Graham | Sorry but thats not a counter example! | |
| Aug 29, 2015 at 18:56 | comment | added | Gerhard Paseman | There is likely an example computed already (I know Westzynthius has one in his paper for primes up to 23). The best link I have right now is mathoverflow.net/questions/49400/question-in-prime-numbers . Gerhard "Yes, It's Been Done Before" Paseman, 2015.08.29 | |
| Aug 29, 2015 at 18:52 | comment | added | Brad Graham | @GerhardPaseman i would love to see a counter example. | |
| Aug 29, 2015 at 18:49 | comment | added | Brad Graham | Well the key insight is that all prime numbers less than some $p_{x}^2$ are those numbers not divisible by a prime less than $p_{x}$ except for those primes themselves. In effect we are creating maximum gaps relative to squared primes by creating sieves corresponding to prime numbers less the prime squared. | |
| Aug 29, 2015 at 18:45 | comment | added | Gerhard Paseman | Others have thought that was the upper bound, and were wrong. It may not be the best way, but it is (according to Terry Tao and others in their December ArXiv preprint) the way used by most researchers on lower bounds of prime gaps. If you edit the question to reflect the viewpoint I suggested above, it might be closed as a duplicate on this forum. If you edit it to show clearly a key idea or insight using your definition, I am willing to consider it. Gerhard "Always Looking For New Ideas" Paseman, 2015.08.29 | |
| Aug 29, 2015 at 18:44 | comment | added | Brad Graham | I have yet to find a denizen consisting of prime numbers upto $p_x$ that has a length greater than $2p_{x-1}, and i suspect that is an upper bound. | |
| Aug 29, 2015 at 18:42 | comment | added | Brad Graham | @GerhardPaseman thank you. Yeah that's pretty much it, and I'm trying to maximise the length of the denizen consisting of only prime numbers up to some $p_x$. The definition tries to capture the axioms that restrict the sequence you mentioned, so that we don't need to know "a" "a+1" etc. I also came across Erdos' attempts at a similar problem using arithmetic sequences, and i realised that perhaps this in not the best approach to finding a maximum distance between two prime numbers. | |
| Aug 29, 2015 at 18:42 | comment | added | Gerhard Paseman | Oh, and I suspect the maximum length is bounded from above (for all but finitely many $a$) by $p_a(\log p_a)^2$, and more likely by something not much larger than $p_a\log p_a$. Gerhard "Speaking From Somewhat Limited Experience" Paseman, 2015.08.29 | |
| Aug 29, 2015 at 18:33 | comment | added | Gerhard Paseman | I am still having trouble understanding your definition of denizen. To me it looks like you are looking at sequences of the form L(a+1),L(a+2),...,L(a+m) where a and m are positive integers and L() is the least prime factor. If so, there are better ways of characterising denizens, and there is literature and some computation involving longer denizens. Search on this forum for "Westzynthius" for some detail. Gerhard "Good To Be Back Home" Paseman, 2015.08.29 | |
| Aug 25, 2015 at 2:41 | comment | added | Gerry Myerson | There's a theorem that says you can't partition the integers into finitely many arithmetic progressions with different common differences. An IRDCS is a partition of a finite segment of the integers into arithmetic progressions with different common differences, so it shows you can't weaken the theorem hypotheses too much. The numbers were chosen because they work: there is no IRDCS shorter than length 11, and the only one of length 11 is the one I gave, and its reverse. Try it! | |
| Aug 24, 2015 at 23:46 | comment | added | Brad Graham | I'm curious about what the purpose for creating the concept of an IRDCS was and why the numbers $6,9,3,4,5$ were chosen? Essentially a placement condition is one of the requirements of a denizen; a prime number will appear in the $m$th position if it is the lowest prime factor of $m$. I note that an IRDCS allows two consecutive terms to be odd so it doesn't look like it was created to represent a section of the natural number line. Denizens were created to represent the sequence of composite numbers between two consecutive primes. | |
| Aug 24, 2015 at 23:19 | comment | added | Gerry Myerson | Distantly related: some years ago, my co-authors and I defined the concept of an IRDCS of length $n$. The smallest example, of length 11, is $6,9,3,4,5,3,6,4,3,5,9$. Note that for each $m$ appearing in the list, $m$ appears at every $m$th position, and only at every $m$th position; also, every $m$ that appears appears at least twice. | |
| Aug 24, 2015 at 17:07 | history | edited | Brad Graham | CC BY-SA 3.0 |
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| Aug 24, 2015 at 15:16 | history | edited | Brad Graham | CC BY-SA 3.0 |
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| Aug 24, 2015 at 6:45 | comment | added | Gerry Myerson | Question also posted, without notice to either site, to m.se: math.stackexchange.com/questions/1404429/… | |
| Aug 23, 2015 at 18:15 | history | edited | Brad Graham | CC BY-SA 3.0 |
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| Aug 21, 2015 at 0:14 | review | First posts | |||
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| Aug 21, 2015 at 0:04 | history | asked | Brad Graham | CC BY-SA 3.0 |