I believe I see that $a_n = 2^n(a_0+1) - 1$ but am somewhat uncertain where to proceed afterwards. I am a complete beginner at number theory and would appreciate it if someone could point me in the right direction--surely there is some obvious argument I am missing.



General term in your sequence is $a_n = 2^n(a+1) - 1$, where, by assumption, $a$ is prime.

There is some sufficiently large $n$ such that $2^n \equiv 1 \pmod{a}$, or in other words $2^n = k a + 1$. But then $a_n$ is divisible by $a$, hence not prime.

EDIT: As Ben Frankel rightly notes, there is a special case when $a = 2$. Argument above fails, but $a_5 = 95$ is composite, so we are in good shape regardless.

Even more general statement is true: If $p$ is a prime, $p \neq 2$, and $p$ divides one of $a_n$, then $p$ divides infinitely many $a_n$'s. In particular, if $a_n$ is an odd prime, then $a_n \mid a_m$ for infinitely many $m$, and these $a_m$'s are composite.

  • $\begingroup$ Thank you very much -- I understand now. $\endgroup$ – Marcus Emilsson Oct 14 '14 at 9:48
  • 5
    $\begingroup$ Unless $a = 2$, but then $a_5 = 95$ which is composite. $\endgroup$ – Ben Frankel Oct 14 '14 at 9:48

This is equivalent to find an integer $k.2^{n_0}$ such that $k$ is odd and $p_n=k.2^{n+n_0}-1$ is a prime for all $n\ge 0.$
If $k=1,$ then $p_n$ is not a prime for composite $n+n_0,$ As

enter image description here

If $k=2m+1$ where $m\ge1,$ then we can find a large $n$ such that
$p_n=k(2^{n+n_0}-1)+(k-1)$ is not a prime.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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