mersenneforum.org Twin Prime Constellations
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

2019-04-30, 09:30   #1
robert44444uk

Jun 2003
Oxford, UK

3·54 Posts
Twin Prime Constellations

rudy235 has pm'd me on the topic of twin prime constellations and it got me thinking that a lot of work has been carried out on k-tuple permissable patterns and prime constellations that fit these patterns.

To advance this idea requires us to define sub-groups of k-tuple permissible patterns that contain a preponderance of twins.

For example from Tony Forbes paper at https://sites.google.com/site/anthon...attredirects=0

we can see a pattern of 6 positions twins in a width of 81 integers ( I don't know if this is the most dense)
Quote:
 k=20 s=80 B={0 2 8 12 14 18 24 30 32 38 42 44 50 54 60 68 72 74 78 80} 1 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 The only other known example of this pattern is 3941119827895253385301920029 + d, d = 0, 2, 8, 12, 14, 18, 24, 30, 32, 38, 42, 44, 50, 54, 60, 68, 72, 74, 78, 80 (28 digits, 24 June 2014, Raanan Chermoni & Jaroslaw Wroblewski) k=20 s=80 B={0 2 6 8 12 20 26 30 36 38 42 48 50 56 62 66 68 72 78 80} The smallest known example of this pattern is 14374153072440029138813893241 + d, d = 0, 2, 6, 8, 12, 20, 26, 30, 36, 38, 42, 48, 50, 56, 62, 66, 68, 72, 78, 80 (29 digits, October 6, 2014, Raanan Chermoni & Jaroslaw Wroblewski)
rudy235 points out there are Hargrave Primes, where there are 5 twin positions in a width of 39, presumably 0,2,6,8,18,20,30,32,36,38

I seems to me that we can easily find first instance twin prime constellations that fit known k-tuple positions up to perhaps 10-length. To do so we need to generate the k-tuples with leading numbers of twin prime positions, but I don't know where to source the list of all possible efficient k-tuples. The leading work of Engelsma does not actually list the k-tuples see http://www.opertech.com/primes/k-tuples.html

Then maybe this work has already been carried out, in which case my Google skills aren't what they used to be.

Last fiddled with by robert44444uk on 2019-04-30 at 10:50

 2019-04-30, 10:39 #2 robert44444uk     Jun 2003 Oxford, UK 187510 Posts I just ran up a quick program to check for 5 twins in a range of 51, a subset of 14 prime spots, of which the twins are 0,2..6,8..18,20..30,32..48,50 (not the most efficient constellation) The first primes in the first few such sets are 11, 325267931, 905119331, 2013256361, 3066212111, 10816172201, 22194295811, 23641113911 Last fiddled with by robert44444uk on 2019-04-30 at 10:52
 2019-04-30, 12:27 #3 robert44444uk     Jun 2003 Oxford, UK 3×54 Posts Here's an interesting fact, looking at quads, (i.e. two sets of twins in a space of 9), the difference between two sets of quads must be a multiple of 30. However, differences cannot be 30*x, when x= 2mod7 or 5mod7, for some reason that a mod specialist will be able to explain. The run of quads in the range 5 to 1e9 produces x of values between 1 to 13184.
 2019-04-30, 16:52 #4 ATH Einyen     Dec 2003 Denmark 285010 Posts The middle point between the 4 primes must be 15*n, because it must be 0 (mod 3) and 0 (mod 5) at the same time. So the primes are 15n-4, 15n-2, 15n+2, 15n+4. There are 3 options for 0 (mod 7): 15n-8 (and 15n+6), 15n-6 (and 15n+8), and 15n. If the gap is 2*30 = 4 (mod 7) then 15n-8 becomes 15n-4, 15n-6 becomes 15n-2, and 15n becomes 15n+4, so at least 1 of the prime spots are 0 (mod 7) If the gap is 5*30 = 10 (mod 7) then 15n-8 becomes 15n+2, 15n-6 becomes 15n+4 and 15n becomes 15n+10 which is 15n-4, and again at least 1 prime spot is 0 (mod 7). Here is a "schematic" looking only at the odd numbers: Code:  15n P P P P mod 3: 0 2 1 0 2 1 0 2 1 0 2 1 0 mod 5: 3 0 2 4 1 3 0 2 4 1 3 0 2 mod 7: 3 5 0 2 4 6 1 3 5 0 2 4 6 mod 7: 1 3 5 0 2 4 6 1 3 5 0 2 4 mod 7: 2 4 6 1 3 5 0 2 4 6 1 3 5
 2019-04-30, 17:39 #5 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT) 52×227 Posts Finding sets of 6 twins within a range of 81 is easy. The first is 2595051759329+c with c = 0 2 12 14 30 32 42 44 72 74 78 80 There are several more that are easy to find with polysieve. I am ignoring all other values of c so there could be more than 6 sets of twins assuming a range of 81 is not the minimum. 0 2 6 8 18 20 30 32 36 38 48 50 is possible. I strongly suspect this is minimal for 6 twins. It is only necessary to search 1451 mod 11#=2310 + c for this form. 1256522812841 = 11#*543949269+1451 + c; c= 0 2 6 8 18 20 30 32 36 38 48 50 No proof that this is optimal for 7 but it won't be far off: 11#*491403340492+1451 + c; c = 0 2 6 8 18 20 30 32 36 38 48 50 60 62 I have started a search for 0 2 6 8 18 20 30 32 36 38 48 50 60 62 78 80. I think this will take a while though. Can anyone prove this is optimal?
2019-04-30, 18:01   #6
robert44444uk

Jun 2003
Oxford, UK

3·54 Posts

Quote:
 Originally Posted by henryzz Finding sets of 6 twins within a range of 81 is easy. The first is 2595051759329+c with c = 0 2 12 14 30 32 42 44 72 74 78 80 There are several more that are easy to find with polysieve. I am ignoring all other values of c so there could be more than 6 sets of twins assuming a range of 81 is not the minimum. 0 2 6 8 18 20 30 32 36 38 48 50 is possible. I strongly suspect this is minimal for 6 twins. It is only necessary to search 1451 mod 11#=2310 + c for this form. 1256522812841 = 11#*543949269+1451 + c; c= 0 2 6 8 18 20 30 32 36 38 48 50 No proof that this is optimal for 7 but it won't be far off: 11#*491403340492+1451 + c; c = 0 2 6 8 18 20 30 32 36 38 48 50 60 62 I have started a search for 0 2 6 8 18 20 30 32 36 38 48 50 60 62 78 80. I think this will take a while though. Can anyone prove this is optimal?
henryzz, this is great!
• Where are you getting the sets of possible positions?
• How do you derive the primorial and the offset?

Thank you ATH for the very clear explanation of the mods around quads.

2019-04-30, 19:36   #7
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT)

52×227 Posts

Quote:
 Originally Posted by robert44444uk henryzz, this is great! Where are you getting the sets of possible positions? How do you derive the primorial and the offset? Thank you ATH for the very clear explanation of the mods around quads.
I created the sets of possible positions by hand. After 0 2 and 6 8, 3 can't divide 0 mod 6 or 2 mod 6 and it is clear that 4 mod 6 is divisible by 3 so they can be excluded. After choosing the next twin pair 18 20, 4 mod 10 was the only even number mod 10 that hadn't been seen already so 4 mod 10 must be divisible by 5. I made two lists of numbers excluded by these two criteria, used them to work out a list of possible numbers and then excluded cases that weren't twins. It helped that I could try a case by putting it into polysieve to see if it was possible. I am fairly confident I have optimal solutions but there was a couple of places where I could have chosen other twins.

In this case the offset was calculated by polysieve which uses a wheel sieve. It basically crossed out all offsets <= 11# where one number in the tuple would be divisible by 2 then 3 etc upto 11. In this case there was only one remaining. Polysieve actually continues doing this in order to gain more speed. There are only two possible offsets modulo 13# which leaves 2/13 of the sieve size if I ignored that. I am currently testing 45900 possible offsets for the 8 twins case calculated using primes upto 31 in a modified version of polysieve.

I can go into more detail if necessary.

Last fiddled with by henryzz on 2019-04-30 at 19:37

2019-04-30, 21:06   #8
rudy235

Jun 2015
Vallejo, CA/.

22·239 Posts

Quote:
 Originally Posted by ATH The middle point between the 4 primes must be 15*n, because it must be 0 (mod 3) and 0 (mod 5) at the same time. So the primes are 15n-4, 15n-2, 15n+2, 15n+4. There are 3 options for 0 (mod 7): 15n-8 (and 15n+6), 15n-6 (and 15n+8), and 15n. If the gap is 2*30 = 4 (mod 7) then 15n-8 becomes 15n-4, 15n-6 becomes 15n-2, and 15n becomes 15n+4, so at least 1 of the prime spots are 0 (mod 7) If the gap is 5*30 = 10 (mod 7) then 15n-8 becomes 15n+2, 15n-6 becomes 15n+4 and 15n becomes 15n+10 which is 15n-4, and again at least 1 prime spot is 0 (mod 7). Here is a "schematic" looking only at the odd numbers: Code:  15n P P P P mod 3: 0 2 1 0 2 1 0 2 1 0 2 1 0 mod 5: 3 0 2 4 1 3 0 2 4 1 3 0 2 mod 7: 3 5 0 2 4 6 1 3 5 0 2 4 6 mod 7: 1 3 5 0 2 4 6 1 3 5 0 2 4 mod 7: 2 4 6 1 3 5 0 2 4 6 1 3 5
If the number is 2 or 5 modulo 7 then the gap is non-existent.

2 | 5
=======
9 12
16 19
23 26
30 33
37 40
... ...
702 705

2019-04-30, 22:28   #9
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT)

162B16 Posts

Quote:
 Originally Posted by henryzz I have started a search for 0 2 6 8 18 20 30 32 36 38 48 50 60 62 78 80. I think this will take a while though. Can anyone prove this is optimal?
I was correct about this taking a while. I have searched 11#*x + 1451 with x upto 1e14 in about 3 hours with no results(best was upto 48). Started a run upto 1e15.
I think I am being a victim here of the numbers getting larger and less likely to be prime as I search deeper.

 2019-05-01, 07:34 #10 robert44444uk     Jun 2003 Oxford, UK 3×54 Posts For my 6 twins in 81 combination (0,2,6,8,36,38,48,50,66,68,78,80) I found no examples testing up to 6.5e12 For the other 6 in 81 combination (0,2,12,14,30,32,42,44,72,74,78,80) I found 3 examples (29, 2,595,051,759,329, and 4,073,652,347,549) testing up to 6.37e12 Last fiddled with by robert44444uk on 2019-05-01 at 07:35
2019-05-01, 16:28   #11
danaj

"Dana Jacobsen"
Feb 2011
Bangkok, TH

90210 Posts

Quote:
 Originally Posted by robert44444uk For my 6 twins in 81 combination (0,2,6,8,36,38,48,50,66,68,78,80) I found no examples testing up to 6.5e12 For the other 6 in 81 combination (0,2,12,14,30,32,42,44,72,74,78,80) I found 3 examples (29, 2,595,051,759,329, and 4,073,652,347,549) testing up to 6.37e12
For your second example, it takes 40 seconds to test to 10^14 using sieve_prime_cluster in serial on my macbook. Use the ktuplet-threads.pl example script to run in parallel.

29
2595051759329
4073652347549
15351703501919
20051437920179
75726824820389

For the first:
Code:
\$ time perl -Mntheory=:all -E 'prime_set_config(verbose=>2); say for sieve_prime_cluster(0,1e14,2,6,8,36,38,48,50,66,68,78,80);'
cluster sieve found 1 residues mod 30
cluster sieve found 1 residues mod 210
cluster sieve found 3 residues mod 2310
cluster sieve found 9 residues mod 30030
cluster sieve found 63 residues mod 510510
cluster sieve found 567 residues mod 9699690
cluster sieve found 6804 residues mod 223092870
cluster sieve using 122472 residues mod 6469693230
cluster sieve ran 800074 MR and 48 Lucas tests
15204559856741
32799902384231
56052330360071
91598391480641

real	0m41.310s
user	0m39.732s
sys	0m0.343s

 Thread Tools

 Similar Threads Thread Thread Starter Forum Replies Last Post hal1se Miscellaneous Math 13 2018-11-05 16:34 MattcAnderson MattcAnderson 25 2018-07-31 23:11 MattcAnderson MattcAnderson 119 2018-03-14 20:22 CRGreathouse Software 10 2017-07-14 09:45 cuBerBruce Puzzles 3 2014-12-01 18:15

All times are UTC. The time now is 07:57.

Sun Jul 5 07:57:13 UTC 2020 up 102 days, 5:30, 1 user, load averages: 1.04, 1.05, 1.13

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.