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A prime number of the form $k\times 2^n+1$ with $n\ge 1\ ,\ k<2^n$ is called a Proth-prime.

Is it known whether the number of Proth-primes is infinite ?

It seems to be almost surely true that there are infinite many proth-primes, but I do not know a proof.

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Yes, there is an infinite number of Proth primes (see the discussion here). The essence of the proof is to use the fact, that every arithmetic progression $a+bm$ with $gcd (a,b)=1$ has infinitely many primes.

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    $\begingroup$ In the discussion, I did not find the restriction $k<2^n$, making a prime of the form $k\times 2^n+1$ a Proth-prime. Without this ristriction, it is clear that even the number of primes of the form $k\times2^n+1$ with a fixed $n$ is infinite. $\endgroup$ – Peter Oct 3 '15 at 10:13
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    $\begingroup$ In fact, without the restriction $k<2^n$, every odd prime is of the form $k\times 2^n+1$, and the number of odd primes is, of course, infinite. $\endgroup$ – Peter Oct 3 '15 at 10:16
  • $\begingroup$ Yes, but the Dirichlet's theorem works without any restriction of $k$. If you define your arithmetic progression $a + bm$ as $a = k \times 2^n+1$ and $b = k \times 2^n$, it will assure the existence of infinitely many primes in the sequence $k \times 2^n + 1$ for any value of $k$. $\endgroup$ – mamuteek Oct 3 '15 at 10:43
  • $\begingroup$ If $k$ is fixed, it is not clear, whether $k\times 2^n+1$ has infinite many primes. $k\times 2^n+1$ is not an arithmetic progression, if $n$ runs. $\endgroup$ – Peter Oct 3 '15 at 12:06
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    $\begingroup$ @Peter: See zyx's answer on the linked post, where it explicitly says that there are no known values of k for which infinitely many primes of the form $k\cdot2^n+1$ are known to exist. $\endgroup$ – Lucian Oct 3 '15 at 12:37

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