[seqfan] Is 2^(2^5)+1 + 2^n composite for every n?

[Tomasz Ordowski]

Dear number enthusiasts!

If the dual Sierpinski conjecture is true,
then 78557 is the smallest Sierpinski number.
Cf. A033919 and A067760 (see last comment).*

Let's go ahead and define odd numbers k > 1 such that
both k+2^n and k*2^n+1 are composite for every n < k.
This is a subsequence of A033919. I found the first term,
namely 7013. Finding the next one will be difficult. Try it.
The keywords "hard" and "more" are appropriate here!
Of course, all (provable) Sierpinski numbers (A076336)
are in this sequence, by the dual Sierpinski conjecture.

There are odd numbers k > 1 such that
both |k+-2^n| and both k*2^n+-1 are composite
for every natural n with 2^n < k. See A256163.
7913, 8923, 24943, 34009, 35437, 42533, ...
Also, "... for every n < k.", hard, but less than
the Brier numbers A076335 (a subsequence).

On the other hand...
Let us define the numbers m > 1 such that
2^m+1 + 2^n is composite for every natural n < m.
I found: 3, 8, 25, 32, ... The number 2^32+1 is noteworthy!
F(5) = 2^(2^5)+1 is the smallest composite Fermat number.
F(5) = 641*6700417 (the prime 641 is my house number).
The Fermat number F(5) is not a dual Sierpinski number,
because F(5)*2^m+1 is prime for m = 63 (find more).
Is there a number n such that F(5)+2^n is prime?
By the dual Sierpinski conjecture, it must exist.
If it exists, then n > 2^16. If not, then what?!
(*) Is F(5) a further term of A033919 ?
The keyword "more" is missing there.

Let's define the following two sequences to see this in general.
Let a(n) be the smallest k such that F(n)*2^k+1 is prime. We have:
a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 279, a(4) = 287, a(5) = 63, a(6) =
225.
Let b(n) be the smallest k such that F(n)+2^k is prime.** We have:
b(0) = 1, b(1) = 1, b(2) = 1, b(3) = 9, b(4) = 1, b(5) = ? , b(6) = 2241.
Conjecture: if n > 6, then both a(n) and b(n) do not exist.
If so, then for n > 6, F(n) is a dual Sierpinski number.
The case F(5) was considered earlier.
Continued (for those interested) in PS.

Any comments are welcome!

Best,

Thomas
____________________________
(*) https://oeis.org/A033919 (more)
and https://oeis.org/A067760 (zeros).
See https://oeis.org/A076336 (provable Sierpinski numbers)
and https://oeis.org/A101036 (Riesel number with covering).
See also https://oeis.org/A076335 (sample Brier numbers).
____________________________
Post Scriptum (for those interested).
*****************
Note that 2^(2^5)-1 = 3*5*17*257*65537.
Is {3, 5, 17, 257, 65537} a covering set for F(5)+2^n?
If such a covering set exists at all, this one is incomplete.

F(5)^4 = (2^32+1)^4 is not a dual Sierpinski number,
because F(5)^4*2^n+1 is prime for n = 924, 1308, ...
Is F(5)^4+2^n composite for every natural n?

I noticed that (2^31-1)(2^32+1) is a de Polignac number.
Is M(M(5))F(5) = (2^31-1)(2^32+1) a dual Riesel number?

(**) 2^(2^5)-1 - 2^n is prime for n = 2, 4, 6, 23, 24, 25, 29, and 31.
These primes are not de Polignac numbers only for n = 23 and 24,
which is consistent with the assumption a <> b of Crocker's theorem:
if m > 2 and a <> b, then all positive 2^(2^m)-1 - 2^a - 2^b are not prime.
Cf. https://oeis.org/history/view?seq=A232565&v=35 (see my comment).
Positive 2^(2^2)-1 - 2^n is prime and 2^(2^2)-1 + 2^n is prime for n < 7.
Also, 2^(2^2)-1 + 2^(2^n) is prime for n < 6.  Abundance of primes!

The number 2^(2^5)-1 + 2^n is prime for n = 4, 12, 16, 18, 24, 34, 51, ...
It seems that if such an n < 34, then 2^(2^5)-1 + 2^n + 2^m are composite.
In addition, if n > 4, then 2^(2^n)-1 + 2^n + 2^m are composite for every
m.

Thanks for your attention and please provide counterexamples!