[seqfan] 2^(2^n-1)-1 as de Polignac numbers for n > 2

[Tomasz Ordowski]

Dear number lovers!

Crocker (1971) proved that,
if n > 2 and a <> b, then 2^(2^n)-1 - 2^a - 2^b is not prime.

If a = 2^n-1, then b < a, so for n > 2, 2^(2^n-1)-1 is a de Polignac number.
Note that 2^(2^n-1)-1 - 2^m is divisible by a prime factor of 2^(2^n)-1.

Conjecture: if n > 5, then |2^(2^n-1)-1 - 2^m| is not prime for every m >
0.
If so, then by the dual Riesel conjecture, 2^(2^n-1)-1 is a Riesel number,
i.e. if n > 5, then (2^(2^n-1)-1)2^m-1 is composite for every integer m >
0.
For example, the Mersenne prime 2^127-1 may be a dual Riesel number.

Similarly, if n > 6, then 2^(2^n+1)-1 + 2^m is composite for every natural
m > 0.
If so, then by the dual Sierpinski conjecture, 2^(2^n+1)-1 is a Sierpinski
number,
i.e. if n > 6, then (2^(2^n+1)-1)2^m+1 is composite for every positive
integer m.
For example, the Mersenne number 2^257-1 may be a dual Sierpinski number.

Covering sets contain the prime factors of 2^(2^n)-1 and some other odd
primes.

Best,

Tom Ordo