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.

[URL="http://www.factordb.com/index.php?query=2^x-x-2&use=x&x=1&VP=on&VC=on&EV=on&OD=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]Here[/URL] are the first 199 positive terms, with the initial 0.

Not even 2*3-3-2 ?!

[URL="http://www.primenumbers.net/prptop/searchform.php?form=2^39137-%3F&action=Search"]2^39137-39137-2[/URL] is a PRP, known since 2005.
And another one is [URL="http://www.primenumbers.net/prptop/searchform.php?form=2^59819-m&action=Search"]2^59819-59819-2[/URL]
So, no, -- no covering set here.

Fixed. :smile:

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.

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
[B]41674 2 59819 59821 18008 Henri Lifchitz 04/2005[/B]
46849 2 55601 55599 16738 Henri Lifchitz 04/2005
[B]93028 2 39137 39139 11782 Henri Lifchitz 04/2005[/B]
95385 2 38203 38199 11501 Henri Lifchitz 04/2005
112109 2 34656 34655 10433 Henri Lifchitz 04/2005
[/CODE]

I also noticed that if p and p+2 form a twin prime pair, p+2 divides 2^p-p-2.

[QUOTE=Stargate38;418979]I also noticed that if p and p+2 form a twin prime pair, p+2 divides 2^p-p-2.[/QUOTE]
Huh?

[QUOTE=LaurV;419008]Huh?[/QUOTE]

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.

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.

If 2^p+p^2 is prime, than p is a multiple of 3. (congruent to 3 mod 6):smile:

[QUOTE=PawnProver44;428679]If 2^p+p^2 is prime, than p is a multiple of 3. (congruent to 3 mod 6):smile:[/QUOTE]
False.
2^p+p^2 = 3 is prime. p=1 is [B]not[/B] a multiple of 3 (and [B]not[/B] congruent to 3 mod 6).