20160401, 22:23  #1 
Apr 2016
1_{16} Posts 
Largest known prime p such that 6p1 and 6p+1 are prime?
What is the largest known prime p such that 6p1 and 6p+1 are twin primes? Some primes of this form are listed in the OEIS at http://oeis.org/A060212.

20160402, 01:30  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,281 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 6p1 and 6p+1 regardless of their form. 
20160402, 05:40  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,281 Posts 
Here is a small one for you, for starters:
9975332*997#/35+1 = p (421 digits) 59851992*997#/35+5 = 6p1 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*2^66661 twin pair = 3*251637551*2^66675, 3*251637551*2^66677 Last fiddled with by Batalov on 20160403 at 02:36 
20160405, 18:58  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,281 Posts 
2280196563*2^9982+1 = p (3015 digits)
and twin pair = 3*2280196563*2^9983+5 = 6p1 3*2280196563*2^9983+7 = 6p+1 
20160406, 18:10  #5 
"Matthew Anderson"
Dec 2010
Oregon, USA
1001010100_{2} 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 
20160406, 18:22  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
21644_{8} Posts 
For all of the values in that table, is prime? (So that the twins would be 6x1 and 6x+1, you know?)
Hint: no. 
20160406, 19:27  #7 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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 6p1 have to be composite.

20160408, 17:48  #8 
"NOT A TROLL"
Mar 2016
California
305_{8} Posts 
I believe there are infinitely many primes of the form 6q+1, 6q1, for prime q. Don't know how to prove this.
Likewise, there should be infinitely many primes q such that 15*q4, 15*q2, 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: 2q1, q, and (2q+1)/3 are prime. (2q1)/3, q and 2q+1 are prime. 6q1 and 6q+1 are prime. Anyone please find an example for this. Last fiddled with by PawnProver44 on 20160408 at 17:48 
20160408, 18:12  #9  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
q=2,3,6x1,6x+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=6x1: produces : 90x19, 90x17,90x13,90x11 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 6x1 or both infinitely often. x must already be of a certain form for 6x1 or 6x+1 to be composite so prove infinitely often that these forms are not met ? Last fiddled with by science_man_88 on 20160408 at 18:15 

20160408, 20:26  #10  
Aug 2006
1011100011111_{2} Posts 
Quote:
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. 

20160408, 20:52  #11  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,281 Posts 
Quote:
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 "nonA"):  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. "nonA") gives you no information about its complement's infinitude. 

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