20190430, 09:30  #1  
Jun 2003
Oxford, UK
3·5^{4} 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 ktuple permissable patterns and prime constellations that fit these patterns.
To advance this idea requires us to define subgroups of ktuple 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:
I seems to me that we can easily find first instance twin prime constellations that fit known ktuple positions up to perhaps 10length. To do so we need to generate the ktuples with leading numbers of twin prime positions, but I don't know where to source the list of all possible efficient ktuples. The leading work of Engelsma does not actually list the ktuples see http://www.opertech.com/primes/ktuples.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 20190430 at 10:50 

20190430, 10:39  #2 
Jun 2003
Oxford, UK
3·5^{4} 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 20190430 at 10:52 
20190430, 12:27  #3 
Jun 2003
Oxford, UK
3·5^{4} 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. 
20190430, 16:52  #4 
Einyen
Dec 2003
Denmark
2×3×5^{2}×19 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 15n4, 15n2, 15n+2, 15n+4.
There are 3 options for 0 (mod 7): 15n8 (and 15n+6), 15n6 (and 15n+8), and 15n. If the gap is 2*30 = 4 (mod 7) then 15n8 becomes 15n4, 15n6 becomes 15n2, 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 15n8 becomes 15n+2, 15n6 becomes 15n+4 and 15n becomes 15n+10 which is 15n4, 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 
20190430, 17:39  #5 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
5^{2}×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? 
20190430, 18:01  #6  
Jun 2003
Oxford, UK
3×5^{4} Posts 
Quote:
Thank you ATH for the very clear explanation of the mods around quads. 

20190430, 19:36  #7  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
5^{2}×227 Posts 
Quote:
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 20190430 at 19:37 

20190430, 21:06  #8  
Jun 2015
Vallejo, CA/.
2^{2}×239 Posts 
Quote:
2  5 ======= 9 12 16 19 23 26 30 33 37 40 ... ... 702 705 

20190430, 22:28  #9  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT)
5675_{10} Posts 
Quote:
I think I am being a victim here of the numbers getting larger and less likely to be prime as I search deeper. 

20190501, 07:34  #10 
Jun 2003
Oxford, UK
3·5^{4} 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 20190501 at 07:35 
20190501, 16:28  #11  
"Dana Jacobsen"
Feb 2011
Bangkok, TH
1606_{8} Posts 
Quote:
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 

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