It is clear from Dirichlet's theorem on arithmetic progressions that for a fixed $n$, there are infinitely many primes of the form $k\cdot 2^n+1$ for a fixed $n$ and $k=1,2,3,..$. However, what if we keep $k$ odd?
From Arturo's comment on Are there infinitely many primes of the form $k\cdot 2^n +1$?, he writes:
And if you want $k$ odd, start with $a=2^n+1$, $b=2^n\cdot 2$, and look at $a+bm$.
Can someone explain whether this is correct and why, and whether this proved there are infinitely many primes of the form $k\cdot 2^n+1$ for a fixed $n$ and odd $k$.
Best Answer
That's still an arithmetic progression: $k = 2m + 1$ for some integer $m$, and your form is then $2^{n+1}m + (2^n + 1)$, which is still an arithmetic progression satisfying the hypotheses of Dirichlet's Theorem, so there are still infinitely many primes of that form.