mersenneforum.org Does 2^n-n-2 have a covering set?
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 2015-12-11, 21:01 #1 Stargate38     "Daniel Jackson" May 2011 14285714285714285714 23×34 Posts Does 2^n-n-2 have a covering set? Does $$x=2^n-n-2$$ have a covering set? If not, are there any primes of that form, other than 3? I've checked n=4-32768 with no results. EDIT: I noticed that n|x 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 2015-12-11 at 21:16 Reason: n=3 is prime.
 2015-12-11, 21:07 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23×52×47 Posts Not even 2*3-3-2 ?! 2^39137-39137-2 is a PRP, known since 2005. And another one is 2^59819-59819-2 So, no, -- no covering set here. Last fiddled with by Batalov on 2015-12-11 at 21:25 Reason: at least three primes are known
 2015-12-11, 21:12 #3 Stargate38     "Daniel Jackson" May 2011 14285714285714285714 23×34 Posts Fixed. Last fiddled with by Stargate38 on 2015-12-11 at 21:13
 2015-12-11, 21:22 #4 Stargate38     "Daniel Jackson" May 2011 14285714285714285714 23·34 Posts I see why they are so rare. For prime p, p divides $2^p-p-2$ and ${{2}^{p-1}}-p-3$. If n is divisible by m and n+2 is of the form k^m, $2^n-n-2$ will have algebraic factors. Last fiddled with by Stargate38 on 2015-12-11 at 21:29
 2015-12-11, 21:33 #5 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23·52·47 Posts Henri Lifchitz had found many 2^n-n 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^n-m 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
 2015-12-11, 21:50 #6 Stargate38     "Daniel Jackson" May 2011 14285714285714285714 64810 Posts I also noticed that if p and p+2 form a twin prime pair, p+2 divides 2^p-p-2.
2015-12-12, 02:16   #7
LaurV
Romulan Interpreter

Jun 2011
Thailand

24AE16 Posts

Quote:
 Originally Posted by Stargate38 I also noticed that if p and p+2 form a twin prime pair, p+2 divides 2^p-p-2.
Huh?

2015-12-12, 03:03   #8
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by LaurV Huh?
I think I figured it out 2^p-p-2 = 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.

 2015-12-12, 08:50 #9 LaurV Romulan Interpreter     Jun 2011 Thailand 24AE16 Posts I didn't make any calculus, from Fermat 2^p=2 (mod p) and therefore 2^p-p-2 is divisible by p. If divisible by p+2, they are prime each-other, it would mean is divisible by p^2+2p - etc, total confusion. From which I spotted immediately in my mind that 32-5-2=25 is not divisible by 7, neither by 35. That was it. Last fiddled with by LaurV on 2015-12-12 at 08:51
 2016-03-11, 01:05 #10 PawnProver44     "NOT A TROLL" Mar 2016 California 197 Posts If 2^p+p^2 is prime, than p is a multiple of 3. (congruent to 3 mod 6)
2016-03-11, 03:02   #11
Batalov

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

23×52×47 Posts

Quote:
 Originally Posted by PawnProver44 If 2^p+p^2 is prime, than p is a multiple of 3. (congruent to 3 mod 6)
False.
2^p+p^2 = 3 is prime. p=1 is not a multiple of 3 (and not congruent to 3 mod 6).

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