[seqfan] Numbers k such that both k -/+ 2^(2^m) are prime

[Tomasz Ordowski]

Dear OEIS Community!

I will remind you of the numbers
https://oeis.org/history/view?seq=A370523&v=26
that I have already written about on the SeqFan forum:
http://list.seqfan.eu/pipermail/seqfan/2024-February/075022.html
Let's go ahead and define
(stronger) numbers k > 2 such that
both k - 2^(2^m) and k + 2^(2^m) are prime
for every integer m >= 0 with 2^(2^m) < k.
I only found {9, 15, 195}. Seek and ye shall find no more!
This set of numbers k is probably finite and appears complete.
Amiram Eldar checked that there are no more terms k < 10^8.
Is this provable* at all? This is the question (not only for Robert)!
If k is a term, then both (k-4,k-2) and (k+2,k+4) are twin prime pairs.
So, for k > 9, such a number k must be divisible by 15, as you can see.

The slightly weakened definition (to eliminate all twin prime pairs):
numbers k > 4 such that
both k - 2^(2^n) and k + 2^(2^n) are prime
for every natural n > 0 with 2^(2^n) < k.
It gives the set of numbers k that is probably infinite:
{7, 9, 15, 27, 57, 63, 195, 267, 363, 405, 483, 603, 1197, 1233, 1443,
1737, 2715, 4257, 5403, 6117, 21855, 22287, 26817, 40755, 63777,
260007, 617253, 986733, 1151655, 1167837, 1174503, 1199373, ... }.
Thanks to Amiram, I know that there are many more of them.
If k > 7 is such a number, then it is odd and divisible by 3.
   These numbers deserve a new sequence on the OEIS.
The previous numbers have a weak data section, I think.
Their mysterious rarity requires formal hard proof first.*
Can anyone solve the problem of that finite set?

This is a good example of how a slight change in definition
leads from a (probably) finite set to a (probably) infinite set.

Best regards,

Thomas Ordowski (3/3).
_______________
Cf. https://oeis.org/A039669
See: https://oeis.org/A129613 and also this
https://oeis.org/draft/A370523 (mentioned draft).
I invite you to discuss (any comments are welcome)!
________________
(*) Draft of the proof (to be completed).

Let k - 2 = p and k + 2 = P.
Let k - 4 = q and k + 4 = Q.
Let k - 16 = r and k + 16 = R.
Where r < q < p < k < P < Q < R
with all r, q, p, P, Q, R being primes
and k being the number you are looking for.

We have k = p + 2 = P - 2 = q + 4 = Q - 4 = r + 16 = R - 16.
Hence 2k = p + P = q + Q = r + R, so the differences of primes
Q - P = p - q = 2 and R - Q = q - r = 12 and R - P = p - r  =`14.
Important note: not all of them have to be consecutive primes.

It is necessary to prove that the only solution to this system of equations
is
the seven numbers (r, q, p, k, P, Q, R) = (179, 191, 193, 195, 197, 199,
211),
where the number k = 195 is the largest element of the set {9, 15, 195},
QED.

But it probably won't be that easy. If there are other solutions for larger
k,
then another pair of equations must be added, e.t.c. (it may never end).
So an appropriate conjecture should be used (in a conditional proof).
Cf. https://en.wikipedia.org/wiki/Prime_k-tuple#Prime_constellations
See also: https://mathworld.wolfram.com/k-TupleConjecture.html
However, perhaps there is a trivial divisibility condition that prevents
the considered constellation of primes (with the same differences)
from repeating itself infinitely often. This should be checked first.
Does anyone know how to do this? If so, let him finish my proof.
Unless someone has a better idea for proof to show us.

Thanks for your attention!

Tom Ordo