0
$\begingroup$

Thanks to S. Carnahan for suggestion:

Question: I would like to know if there are results in the literature concerning the prime numbers of the form $$ p = 2q^n-1 $$ where $$ q $$ is an odd prime number.

Moreover, conjectural asymptotics are of special interest.

Furthermore, there seems not to be an entry in the OEIS about these prime numbers.

First terms are: $$ 5,13,17,37,53,61,73,97,\dots $$

$\endgroup$
7
  • 3
    $\begingroup$ They're odd. $\endgroup$ Apr 21, 2011 at 7:07
  • $\begingroup$ Sure.... Observe that $n$ can be even, e.g. $$ 17 = 2 3^2-1 $$ $\endgroup$ Apr 21, 2011 at 7:19
  • $\begingroup$ Or odd...: $$ 53 = 2 \cdot 3^3 -1 $$ So your observation is key for the possible solution... $\endgroup$ Apr 21, 2011 at 7:23
  • 1
    $\begingroup$ Luis, questions of the form "What is known about X?" are almost always unacceptable here, and are likely to be closed. Please add more context to your question, e.g., "I would like to know if there are results in the literature concerning the abundance of such numbers, and conjectural asymptotics are of special interest", or "I would like to know how these numbers might show up in Iwasawa theory". $\endgroup$
    – S. Carnahan
    Apr 21, 2011 at 7:34
  • $\begingroup$ There seems not to be an entry in the OEIS about these prime numbers $\endgroup$ Apr 21, 2011 at 8:33

1 Answer 1

3
$\begingroup$

Each fixed $n$ is a special case of Schinzel's "Hypothesis H" (also known as the Bateman-Horn conjecture), which conjectures that there are infinitely many pairs of primes $p,q$ related by your equation $p=2q^n-1$. (For $n=1$ this is a special case of the Hardy-Littlewood prime $k$-tuples conjecture.) So it's conjectured that there are infinitely many examples for each positive $n$, but it's not known even whether there are infinitely many among all $n$ together.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.