20151211, 21:01  #1 
"Daniel Jackson"
May 2011
14285714285714285714
1001111001_{2} Posts 
Does 2^nn2 have a covering set?
Does \(x=2^nn2\) have a covering set? If not, are there any primes of that form, other than 3? I've checked n=432768 with no results.
EDIT: I noticed that nx if n is prime, and (n+1)x if n+1 is prime. Here are the first 199 positive terms, with the initial 0. Last fiddled with by Stargate38 on 20151211 at 21:16 Reason: n=3 is prime. 
20151211, 21:07  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
47×197 Posts 
Not even 2*332 ?!
2^39137391372 is a PRP, known since 2005. And another one is 2^59819598192 So, no,  no covering set here. Last fiddled with by Batalov on 20151211 at 21:25 Reason: at least three primes are known 
20151211, 21:12  #3 
"Daniel Jackson"
May 2011
14285714285714285714
3×211 Posts 
Fixed.
Last fiddled with by Stargate38 on 20151211 at 21:13 
20151211, 21:22  #4 
"Daniel Jackson"
May 2011
14285714285714285714
3×211 Posts 
I see why they are so rare. For prime p, p divides $2^pp2$ and ${{2}^{p1}}p3$. If n is divisible by m and n+2 is of the form k^m, $2^nn2$ will have algebraic factors.
Last fiddled with by Stargate38 on 20151211 at 21:29 
20151211, 21:33  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9259_{10} Posts 
Henri Lifchitz had found many 2^nn PRPs over years:
Code:
83 2 1061095 1061095 319422 Henri Lifchitz 10/2013 91 2 1015321 1015321 305643 Henri Lifchitz 08/2013 701 2 505431 505431 152150 Henri Lifchitz 04/2005 714 2 500899 500899 150786 Henri Lifchitz 04/2005 775 2 481801 481801 145037 Henri Lifchitz 03/2005 4348 2 228271 228271 68717 Henri Lifchitz 11/2004 6096 2 182451 182451 54924 Henri Lifchitz 10/2004 12872 2 108049 108049 32526 Henri Lifchitz 11/2001 40325 2 61011 61011 18367 Henri Lifchitz 09/2001 40375 2 60975 60975 18356 Henri Lifchitz 09/2001 79771 2 44169 44169 13297 Henri Lifchitz 09/2001 and many 2^nm where m is near n 25405 2 75329 75325 22677 Henri Lifchitz 04/2005 28285 2 70866 70865 21333 Henri Lifchitz 04/2005 29847 2 69510 69507 20925 Henri Lifchitz 04/2005 37182 2 65597 65593 19747 Henri Lifchitz 04/2005 37662 2 64764 64763 19496 Henri Lifchitz 04/2005 41674 2 59819 59821 18008 Henri Lifchitz 04/2005 46849 2 55601 55599 16738 Henri Lifchitz 04/2005 93028 2 39137 39139 11782 Henri Lifchitz 04/2005 95385 2 38203 38199 11501 Henri Lifchitz 04/2005 112109 2 34656 34655 10433 Henri Lifchitz 04/2005 
20151211, 21:50  #6 
"Daniel Jackson"
May 2011
14285714285714285714
3·211 Posts 
I also noticed that if p and p+2 form a twin prime pair, p+2 divides 2^pp2.

20151212, 02:16  #7 
Romulan Interpreter
Jun 2011
Thailand
3×3,049 Posts 

20151212, 03:03  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
I think I figured it out 2^pp2 = 2^p(p+2); eliminate the (p+2) and we get the statement of (p+2)(2^p) now if p+2 is prime (p+2)2^(p+2)2 aka (p+2)2*(2^(p+1)1); eliminate the 2 it can't divide and you get (p+2)(2^(p+1)1) which of course doesn't lead to it, clearly I'm with Laurv on this one, as the statement made is equivalent to (p+2)(2^(p+1)) a contradiction is formed.

20151212, 08:50  #9 
Romulan Interpreter
Jun 2011
Thailand
21673_{8} Posts 
I didn't make any calculus, from Fermat 2^p=2 (mod p) and therefore 2^pp2 is divisible by p. If divisible by p+2, they are prime eachother, it would mean is divisible by p^2+2p  etc, total confusion. From which I spotted immediately in my mind that 3252=25 is not divisible by 7, neither by 35. That was it.
Last fiddled with by LaurV on 20151212 at 08:51 
20160311, 01:05  #10 
"NOT A TROLL"
Mar 2016
California
305_{8} Posts 
If 2^p+p^2 is prime, than p is a multiple of 3. (congruent to 3 mod 6)

20160311, 03:02  #11 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
47×197 Posts 

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