Primes q = (p - 2^m)2^m + 1 with 1 < 2^m < p prime

[Tomasz Ordowski]

Hello to everyone interested! 

   Every odd natural number n > 1 

has a unique representation of the form 

n = (k - 2^m)*2^m + 1, where 1 < 2^m < k odd. 

Cf. A387016, see https://oeis.org/A387016.
   Notice that if this odd k is divisible by 3, 

then n = (k - 2^m)*2^m + 1 is as well.

   Question: how to prove that 

there are infinitely many primes of the form 

q = (p - 2^m)*2^m + 1, where p is an odd prime ?
   The numbers q are primes in A332075: 

see https://oeis.org/A332075.
The odd primes that are not of this form are A332078: 

see https://oeis.org/A332078. 

   Proving this would be nontrivial, because 

there exist odd primes p such that all numbers 

(p - 2^m)*2^m + 1 are composite for 1 < 2^m < p.
These primes p are 229, 509, 523, 1181,1481, ... 

(and there are probably infinitely many of them). 

Since, although p-2^m is not a Sierpiński number, 

by the dual his conjecture; but the above form would 

have to be expanded as follows: q = (p - 2^m)*2^M + 1, 

so that for every 1 < 2^m < p there exists a number M > 0 

(and not necessarily M = m) such that the above q is prime.  

At the other end are A388054, see https://oeis.org/A388054 

Only look at the odd numbers (odd primes) in this sequence.

Best, 

Tom Ordo 

PS. Embarrassing question (on a more basic matter): 

are there infinitely many primes of the form p*2^m + 1, 

  where p is an odd prime (over all p and all m > 0) ?  

It is enough that such p are not Sierpiński numbers. 

However, it has not been proven (as far as I know) 

that there are infinitely many primes p 

that are not Sierpinski numbers.

That they are, also not. Well. 

[Max Alekseyev]

For a fixed m, you question would follow from Dickson's conjecture

https://en.wikipedia.org/wiki/Dickson%27s_conjecture

which asserts that p and 2^m p - 2^(2m) + 1 are simultaneously prime for infinitely many values of p.

Regards,

Max

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[Tomasz Ordowski]

Thanks Max! 

Yes, for a fixed m > 0 it is only hypothetical,
  2p-3, 4p-15, 8p-63, 16p-255, 32p-1023, ...  
But for all integers m > 0 it should be provable.

   To prove:
There are infinitely many primes of the form
  q = p2^m - 4^m + 1 with 1 < 2^m < p prime.

  For infinitely many odd primes p,
there should be at least one integer m > 0
such that q = (p - 2^m)2^m + 1 is a prime. 

I am asking for har proof.

Good luck!

- Tom  

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[Tomasz Ordowski]

PS. I'm asking for (hard) proof (and sorry for the typo).

[Tomasz Ordowski]

These primes q = (p - 2^m)2^m + 1 with 1 < 2^m < p prime

are primes q such that (q + 4^m - 1)/2^m is an odd prime. 

Note that odd (2n+1 + 4^m - 1)/2^m = A386220(n). Thus: 

q is such a prime if and only if A386220((q-1)/2) is prime. 

Look https://oeis.org/A386220 (to complete this proof)... 

[Tomasz Ordowski]

Odd primes q such that, for some m > 0, 

(q - 1)/2^m + 2^m = p is an odd prime.

Equivalently, (q + 4^m - 1)/2^m = p. 

Or, A386220((q-1)/2) = p. OEIS,

see https://oeis.org/A386220 

Primes of A332075. Look

 https://oeis.org/A332075

Complement of A332078, 

 https://oeis.org/A332078 

 to all odd primes. END.