[seqfan] Mersenne numbers that are de Polignac numbers

[Tomasz Ordowski]

Dear number lovers,

I have something interesting!

Let's define (notice the obvious binary interpretation of this definition):
Numbers m > 1 such that 2^m-1 - 2^n is composite for every natural n < m.
7, 15, 23, 27, 31, 37, 39, 43, 55, 58, 63, 71, 79, 82, 91, 95, 111, 123,
127, ...

All integers m = 2^k-1 for k > 2 are in this sequence (by Crocker theorem).*

The following property is more general.
For many terms m, 2m+1 is also in this sequence (iteration).
So maybe b(n) = (m+1)2^n-1 for n >= 0 is an infinite subsequence.
For which m can this be true? For m = 7, 15, 31, .. this is already proven.*
It seems that (27+1)2^n-1 is a subsequence: 27, 55, 111, 223, 447, ...

How to prove that there are infinitely many primes in this sequence?

Unfortunately, not all numbers n == -1 (mod 8) are in this sequence.
However, it should be noted that many terms m == -1 (mod 8).
See https://oeis.org/A004771 (a(n) = 8n+7 for n >= 0).

Cf. https://oeis.org/A006285 (de Polignac numbers)
and https://oeis.org/A000225 (Mersenne numbers).

Best,

Tom Ordo
______________
(*) Crocker (1971) proved that,
if m > 2 and a <> b, then 2^(2^m)-1 - 2^a - 2^b is not prime.
If a = 2^m-1, then b < a, so for m > 2, 2^(2^m-1)-1 is a de Polignac number.
Note that 2^(2^m-1)-1 - 2^n is divisible by some prime factor of 2^(2^m)-1.

_____Post Scriptum__________
Note that 127^7 and (2^127-1)^3 are de Polignac numbers.
Difficult question: is (2^127-1)^4 a dual Sierpinski number?