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Does every sequence $4^nx_0+\dfrac{4^n-1}{3}$ contain a prime?

Fix some odd positive integer $x_0$ and define the sequence $x_0,x_1,x_2,\ldots$ as follows:

$x_n=4^nx_0+\frac{4^n-1}{3}$

So for example choosing $x_0=1$ gives the sequence:

$1,5,21,85,\ldots$

And choosing $x_0=3$ gives the sequence:

$3,13,53,214,\ldots$

Does every such sequence contain a prime number?


Dirichlet's Theorem states that there are infinitely many primes in every arithmetic progression:

$a, a+d, a+2d,\ldots$

where $a,d$ are coprime. But I'm unaware of similar results for other progressions. I found this paper (which is beyond me) but my rudimentary understanding suggests this is saying the result is unknown for geometric progressions.

The progressions I'm asking about are exponential in nature. These are in fact the canonical set of linear combinations of the Lucas Sequences $U_1(5,4)$ and $V_1(5,4)$ over the odd integers.

Every successor $x_n:n\geq0$ is of course $\equiv 5\mod 8$ while every sequence for which $x_0\not\equiv\{1,3,7\}\mod 8$ is a subsequence of a longer sequence (with a smaller starting number), which does start with $x_0\equiv\{1,3,7\}\mod 8$.

So the set of all odd integers can be canonically indexed according to membership of the sequences starting with $x_0\equiv\{1,3,7\}\mod 8$.

I suspect these sequences have a close relationship with the kernel of the 2-adic logarithm, although I don't know a lot about that.

EDIT So far we have found that $5$ is the only prime in $1,5,21,85,...$ so not every sequence has infinitely many primes.

UPDATE Peter posted this related question for a candidate for a sequence with no primes and Paolo Leonetti has elegantly verified it has no primes, plus provided a class of such sequences which contain no primes; those for which $3x_0+1$ is a square number.

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    $\begingroup$ Look up Dirichlet's theorem $\endgroup$ – steven gregory Mar 23 '18 at 23:12
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    $\begingroup$ Does Dirichlet's theorem have a case that covers exponentially distributed sequences? $\endgroup$ – Valborg Mar 23 '18 at 23:16
  • $\begingroup$ @stevengregory thanks. I was aware of that... And can see it's related. But does it decide this? These sequences are neither arithmetic not geometric progressions. $\endgroup$ – user334732 Mar 23 '18 at 23:16
  • $\begingroup$ @Valborg precisely $\endgroup$ – user334732 Mar 23 '18 at 23:17
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    $\begingroup$ $4^{72}\cdot 8+\frac{4^{72}-1}{3}$ is a Fermat-pseudoprime to bases $3$ and $5$ , but composite. $x_0=8$ is a hard case as well. $\endgroup$ – Peter Mar 24 '18 at 21:10

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