Hooley proves in Applications of Sieves to the Theory of Numbers that there are only $o(x)$ numbers $n\le x$ such that $n\cdot2^n+1$ is a (Cullen) prime. The proof generalizes to forms $n\cdot2^{n+a}+b$.

It seems evident that the same result would hold without the $n$ out front, that is, for $a,b,c$ with $a>0$ and $b>1$ there are only $o(x)$ primes of the form $a\cdot b^n+c$ with $n\le x.$ Has this been proved?

This is essentially the same as a question asked on math.se which was never answered.


Unfortunately, proving such a result for a sequence like $2^n-3$ (say) seems very difficult. The reason one can handle $n\cdot2^n+1$ by sieve methods is because that sequence is equidistributed modulo each odd number $m$. (This equidistribution statement is maybe due to Rieger; I think Hooley uses a weaker result.)

In the same tract, Hooley actually discusses the very question you ask about, maybe in a special case like $2^n-3$. But he is only able to get the $o(x)$ result under the assumption of GRH and another auxiliary ad-hoc assumption.

  • $\begingroup$ Thanks! So no major revisions in our knowledge since the Hooley tome? $\endgroup$ – Charles Jan 9 '14 at 4:03
  • 2
    $\begingroup$ Not as far as I know. I would love to be wrong and would even be happy to see a proof that was conditional on just GRH. $\endgroup$ – so-called friend Don Jan 9 '14 at 4:09

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