Does 2^n-n-2 have a covering set?

[Stargate38]

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.

[Batalov]

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.

[Stargate38]

Fixed.

[Stargate38]

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.

[Batalov]

Henri Lifchitz had found many 2^n-n PRPs over years:

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

[Stargate38]

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

[LaurV]

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?

[science_man_88]

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.

[LaurV]

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.

[PawnProver44]

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

[Batalov]

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).