Let $\Bbb T$ be the set of all square free integers with ordering derived from $\Bbb N$. Essentially $PNT$ says if you pick $\log N$ integers less than $N$ you can expect one of them to be prime.

  1. What is the analog in $\Bbb U\subseteq\Bbb T$ where $\Bbb U$ contains only those integers in $\Bbb T$ with factors $1\bmod m$ for some fixed integer $m>0$ (that is how many integers we expect to pick in random before we hit a prime)?

  2. What is the analog in $\Bbb U_\alpha\subseteq\Bbb T$ where $\Bbb U$ contains only those integers in $\Bbb T$ with factors $1\bmod m$ for some fixed integer $m>0$ and each integer $N\in\Bbb U_\alpha$ has only $O((\log \log N)^\alpha)$ factors for some fixed $\alpha\in(0,1)$?

  • $\begingroup$ The density of primes congruent to $1 \bmod m$ is known, so to answer (1) it suffices to understand the density of integers constructed from primes congruent to $1 \bmod m$. Similarly for (2), though the density estimate seems to be more annoying. $\endgroup$ – davidlowryduda May 16 '17 at 18:52
  • $\begingroup$ @mixedmath can you post whatever you know? $\endgroup$ – Brout May 16 '17 at 19:19

I describe how to understand (1).

The number of primes congruent to $1 \bmod m$ that are less than $X$ is asymptotically given by $$ \pi(X; 1, m) = (1 + o_m(1))\frac{1}{\varphi(m)} \frac{X}{\log X}.$$ This is a quantitative version of Dirichlet's Theorem on Primes in arithmetic progressions, and generally it is possible to give error terms analogous to the error term in the Prime Number Theorem.

In "Strings of Congruent Primes", Journal of the London Mathematical Society 61.2 (2000): 359-373, Shiu proves that $$ \zeta(s; 1, m) := \prod_{p \equiv 1 \bmod m} \big(1 - \frac{1}{p^s}\big)^{-1} = (s-1)^{-1/\varphi(m)} g(s) $$ where $g(s)$ is a regular, non-zero function in regions of shape $1 - \frac{c}{\log T} \leq \textrm{Re } s \leq 2$, $\textrm{Im } s \leq T$. [This is meant to be similar to the classical zero-free region for $\zeta(s)$].

We can construct a Dirichlet series associated to those $n \in \mathbb{U}$ through the computations $$ D(s) = \sum_{n \in \mathbb{U}} \frac{1}{n^s} = \prod_{p \equiv 1 \bmod m} \big( 1 + \frac{1}{p^s}\big) = \prod_{p \equiv 1 \bmod m} \frac{(1 - p^{-2s})}{(1 - p^{-s})} = \frac{\zeta(s; 1,m)}{\zeta(2s; 1, m)}.$$ Notice that this is essentially similar to the Dirichlet series associated to counting squarefree integers in $\mathbb{Z}$, $$ \sum_{n \; \text{squarefree}} \frac{1}{n^s} = \frac{\zeta(s)}{\zeta(2s)}.$$

Then classical Perron-type estimates and standard Tauberian arguments (similar to those detailed in Shiu's paper) show that the number of elements of $\mathbb{U}$ up to $X$ is given by $$ \# \{n \in \mathbb{U} : n \leq X \} = (1 + o_m(1)) \frac{1}{\zeta(2; 1, m)} \frac{X (\log X)^{1/\varphi(m)}}{\log X}.$$

Thus the density you're looking for is $$ \frac{\pi(X; 1, m)}{\# \{n \in \mathbb{U} : n \leq X \}} \sim \frac{\zeta(2; 1, m)}{\varphi(m)} \frac{1}{(\log X)^{1/\varphi(m)}},$$ where $\zeta(2; 1,m)$ is the constant $$ \zeta(2; 1,m) = \prod_{p \equiv 1 \bmod m} \big(1 - \frac{1}{p^2}\big).$$

Thus you should expect (roughly) a constant out of every $(\log N)^{1/\varphi(m)}$ many random choices up to $N$ to be prime within this set. Note that this also gives the correct estimate for the "degenerate case" corresponding to $m = 1$.

It might be possible to perform some sort of sieving argument to approach your (2), but that is a bit separate from things that I routinely think about.

  • $\begingroup$ should it be '.... a constant out of every $(\log N)^{\varphi(m)}$...'? In any case the density went down since you now have to choose lot more integers before you expect to hit a prime. However you kept $\frac1m$ of all primes but threw out much more than $\frac{m-1}m$ fraction of integers right and so the density of primes should have increased right? $\endgroup$ – Brout May 17 '17 at 2:39
  • $\begingroup$ I think it should be $(\log N)^{\frac1{\varphi(m)}}$ $\endgroup$ – Brout May 17 '17 at 2:45
  • 1
    $\begingroup$ Oh, indeed I mistyped that line. I've corrected that now. $\endgroup$ – davidlowryduda May 17 '17 at 2:45

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