Let $\{a_n\}_{n\geq 0}$ be a sequence of integers given by the rule $a_{n+1}=2a_n+1$, Does there exists a value for $a_0$ for which the sequence consists entirely of prime numbers.
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If $a_0=p$ is an odd prime, then $a_{p-1}$ is divisible by $p$ by FLT. Remaining case $a_0=2$ is simple. – Sil Apr 25 '20 at 00:29
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See here for a proof of a more general result (replace $10x+1$ by $2x+1$) – Bill Dubuque Apr 25 '20 at 00:43
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No. Mod any odd prime $p$, the sequence is periodic with period at most $p-1$. If $p$ is a prime divisor of $a_1$, $a_n$ will again be divisible by $p$ (and thus not prime) for some $n$ with $2 \le n \le p$.
Robert Israel
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