Consider numbers N of the form:
$(2^a\cdot 3^b)-(2^{(a-3)}\cdot 3^{(b-3)})+1$, with a, b integers >3.
Is it known if there are infinitely many N which are prime?
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2$\begingroup$ I doubt the answer is known. The set of numbers of this form is quite sparse, way beyond the capabilities of current methods. Also let me note that the way the title is phrased it seems to be asking a different question, which may be easier $\endgroup$– WojowuAug 19, 2020 at 15:38
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1$\begingroup$ So these are numbers of the form $215k+1$ where $k$ is 3-smooth. Have you done any calculations? $\endgroup$– Gerry MyersonAug 20, 2020 at 1:30
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$\begingroup$ No I did not do any calculation $\endgroup$– Enzo CretiAug 20, 2020 at 6:18
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1 Answer
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Heuristically speaking, there should be infinitely many such primes, by the same sort of heuristic as one would use to expect there to be infinitely many Mersenne primes. That is, by the Prime Number Theorem, the "chance" that a given number is prime is about 1/log n, and 1/log (s) where s ranges over all of the numbers in your series diverges. So we should expect that there are infinitely many such primes. Your set is in fact denser than the Mersenne primes, since for a given x, there are about (log x)^2 numbers in your set which are at most x, while there are about only (log x) powers of 2 less than x.
However, proving that such a thin set has infinitely many primes is in general beyond us at this time for any set which isn't more or less constructed to have infinitely many primes explicitly. The best state of the art now is results is much weaker. For example, Iwaniec proved there are infinitely many numbers of the form $n^2+1$ which have at most two prime factors. See here. Similarly, it has been proven that there are infinitely many primes of the form $x^4 +y^2$. See here. Both of the corresponding sets are much denser than the set of numbers in your question and are close to what the state of the art can do.
