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

[PawnProver44]

That is the only exception:

2^p+p^2 is prime, so p is an odd multiple of 3:

Case 1: p = 1 (mod 3):

2^p+p^2 = 0 (mod 3); 2 (mod 3) + 1 (mod 3) = 0 (mod 3)

Case 2: p = 2 (mod 3):

2^p+p^2 = 0 (mod 3); 2 (mod 3) + 1 (mod 3) = 0 (mod 3)

Case 3: p = 0 (mod 3):

2^p+p^2 = 2 (mod 3); 2 (mod 3) + 0 (mod 3) = 2 (mod 3)

See, I was correct.

[Batalov]

Originally Posted by PawnProver44

See, I was correct.

Another false statement. Gee, you are racking them up. Are you proud of being wrong all the time? Like this is some kind of an achievement?

Goodbye. On the "ignore list" you go.

[sweety439]

See http://www.primenumbers.net/Henri/us/NouvP1us.htm, there are also no known primes of these forms:

(2^n-k)*2^n+1, for k = 7, 34, 74, 79, ... and is 3 the only n such that (2^n-k)*2^n+1 is prime for k=6? (if (2^n-k)*2^n+1 is prime for k=1, then this n must be 3-smooth)

(2^n+k)*2^n+1 for k = 15. (if (2^n+k)*2^n+1 is prime for k=1, then this k must be a power of 3)

Besides, for (2^n-k)*2^n-1 and (2^n+k)*2^n-1, is there any reserving for them?