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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.

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No.

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.

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

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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.

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