mersenneforum.org Largest known prime p such that 6p-1 and 6p+1 are prime?
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 2016-04-01, 22:23 #1 pythonic   Apr 2016 18 Posts Largest known prime p such that 6p-1 and 6p+1 are prime? What is the largest known prime p such that 6p-1 and 6p+1 are twin primes? Some primes of this form are listed in the OEIS at http://oeis.org/A060212.
 2016-04-02, 01:30 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×37×131 Posts This is easy to make with a modified NewPgen. There is a sieve mode for triple sets, but you need to change the multiplier to 6 in the code (then use mode). It is likely that you can easily find a p with at least 10,000 or 20,000 digits. The proof of course will not be a problem, because p will be a helper for 6p-1 and 6p+1 regardless of their form.
 2016-04-02, 05:40 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·37·131 Posts Here is a small one for you, for starters: 9975332*997#/35+1 = p (421 digits) 59851992*997#/35+5 = 6p-1 59851992*997#/35+7 = 6p+1 _____________________________ Then, twice larger... 14411087*2003#/35+1 (852 digits) twin pair = 86466522*2003#/35+5, 86466522*2003#/35+7 _____________________________ And over 1000+ digits: 547561666*3001#/210+1 (1284 digits) twin pair = 547561666*3001#/35+5, 547561666*3001#/35+7 _____________________________ And over 2000 digits: p = 251637551*2^6666-1 twin pair = 3*251637551*2^6667-5, 3*251637551*2^6667-7 Last fiddled with by Batalov on 2016-04-03 at 02:36
 2016-04-05, 18:58 #4 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×37×131 Posts 2280196563*2^9982+1 = p (3015 digits) and twin pair = 3*2280196563*2^9983+5 = 6p-1 3*2280196563*2^9983+7 = 6p+1
 2016-04-06, 18:10 #5 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 24·32·7 Posts There is a table at the University of Tennessee Martin webpage with largest twin primes known - https://primes.utm.edu/top20/page.php?id=1 The largest is 3756801695685 · 2666669 - 1 with 200700 digits. Regards Matt
 2016-04-06, 18:22 #6 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×37×131 Posts For all of the values in that table, is $x = {{p_{TWIN} + 1} \over 6}$ prime? (So that the twins would be 6x-1 and 6x+1, you know?) Hint: no.
2016-04-06, 19:27   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Batalov For all of the values in that table, is $x = {{p_{TWIN} + 1} \over 6}$ prime? (So that the twins would be 6x-1 and 6x+1, you know?) Hint: no.
that's one way to look at it (aka are those twin primes known, not can the twin primes be proven to exist) the other way to look at is are any of the largest known primes (including those primes) not of the form such that 6p+1 or 6p-1 have to be composite.

 2016-04-08, 17:48 #8 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts I believe there are infinitely many primes of the form 6q+1, 6q-1, for prime q. Don't know how to prove this. Likewise, there should be infinitely many primes q such that 15*q-4, 15*q-2, 15*q+2, 15*q+4, are all prime. Still don't know how to prove this. In fact, This is probable the case of minimum sets, where n consecuative integers have the n smallest factors. For your problem this is equivalent that there are infinitely many primes q such that: 2q-1, q, and (2q+1)/3 are prime. (2q-1)/3, q and 2q+1 are prime. 6q-1 and 6q+1 are prime. Anyone please find an example for this. Last fiddled with by PawnProver44 on 2016-04-08 at 17:48
2016-04-08, 18:12   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts

Quote:
 Originally Posted by PawnProver44 Likewise, there should be infinitely many primes q such that 15*q-4, 15*q-2, 15*q+2, 15*q+4, are all prime. Still don't know how to prove this.
well a possible start would be to show all the cases possible

q=2,3,6x-1,6x+1 and what each is equivalent to.

case q=2:

all parts are even so the result would be even so q=2 fails to meet the requirements

q=3:

produces 49 for the last one so q=3 is out.

q=6x-1:

produces : 90x-19, 90x-17,90x-13,90x-11

q=6x+1

produces: 90x+11,90x+13,90x+17,90x+19

now you need to show that for any x values that these are all prime create primes 6x+1 or 6x-1 or both infinitely often. x must already be of a certain form for 6x-1 or 6x+1 to be composite so prove infinitely often that these forms are not met ?

Last fiddled with by science_man_88 on 2016-04-08 at 18:15

2016-04-08, 20:26   #10
CRGreathouse

Aug 2006

3·1,993 Posts

Quote:
 Originally Posted by PawnProver44 I believe there are infinitely many primes of the form 6q+1, 6q-1, for prime q. Don't know how to prove this.
You shouldn't be able to -- this is a hard problem. Lots of progress has been made toward it in recent years
http://arxiv.org/abs/math/0606088
http://arxiv.org/abs/1009.3998
http://arxiv.org/abs/1409.1327
http://arxiv.org/abs/1511.04468
http://arxiv.org/abs/1603.07817
but we're still far from proving Dickson's conjecture.

2016-04-08, 20:52   #11
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

100101110111102 Posts

Quote:
 Originally Posted by science_man_88 ... x must already be of a certain form for 6x-1 or 6x+1 to be composite so prove infinitely often that these forms are not met ?
In addition to what Charles wrote, infinitude of composite members of any sequence has nothing to do with the infinitude of prime members.

sm88, you need to start thinking at least about the basic properties of infinities: that is, (even for complementary sets, e.g. a set "A" and a set of all others "non-A"):
- a sum of a finite set and and infinite set is an infinite set;
- a sum of an infinite set and and infinite set is an infinite set, so
- knowing that one subset is infinite (e.g. "non-A") gives you no information about its complement's infinitude.

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