5
$\begingroup$

Looking at a problem in representation theory I ran into a question on small primes in arithmetic progressions.

Let me begin with a short summary of results on small primes in arithmetic progressions. By Linnik's theorem there are constants $c,L$ such that for every $d \geq 2$ and $1\leq a<d$ with $(d,a) = 1$ the least prime $p_{\text{min}}(d,a)$ congruent $a$ modulo $d$ satisfies $$ p_{\text{min}}(d,a) \leq c d^L. $$ Currently the best known value for the exponent is $L=5$ (Xyloris). On the extended Riemann hypothesis or the generalized Riemann hypothesis, we have $L = 2+\epsilon$ for every $\epsilon > 0$. A folklore conjecture (sometimes attributed to Chowla, sometimes to Heath-Brown) states that $L = 1+\epsilon$ for all $\epsilon$.

Fix a prime number $p$ and fix $a$, say $a=1$. For my purposes the only relevant case is when $d$ a power of the fixed prime number $p$. In this case stronger results are known. Let $L(p)$ be defined as $$ L(p) = \limsup_{j \to \infty} \frac{\log(p_{\text{min}}(p^j,1))}{j \log(p)}. $$ In other words $L(p)$ is the infimum over all real numbers $L> 0$ such that $p_{\text{min}}(p^j) \leq c_L p^{jL}$ for some $c_L > 0$ and all $j \geq 1$. Barban, Linnik and Tshudakov proved $L(p) \leq \frac{8}{3}$. Gallagher established $L(p) < 2.5$ and Huxley improved this to $L(p) \leq 2.4$. The best bound I am aware of can be found in a paper of Banks-Shparlinski: $L(p) < 2.1115$.

My question is: what can be said if $\limsup$ is replaced by $\liminf$? Let's define $$ K(p) = \liminf_{j \to \infty} \frac{\log(p_{\text{min}}(p^j,1))}{j \log(p)}.$$ Clearly, $K(p) \leq L(p)$. So according to the strongest conjectures on $L(p)$ one would have $K(p) =1$. However, it seems possible that one can approach $K(p)$ with different methods. Put differently: one "only" needs to show that for infinitely many $j$ there is a small prime in the arithmetic progression $\equiv 1 \bmod p^j$

Question: Are there upper bounds for $K(p)$ which are better than the known bounds for $L(p)$?

(For me the case $a = 1$ is sufficient, but I don't see how this might be useful.)

$\endgroup$

1 Answer 1

2
$\begingroup$

With $p_{\min}(d,a)$ as the OP defines it, let us take

$$p_{\min}(d)=\max_{(a,d)=1}p_{\min}(d,a).$$

Li, Pratt, and Shakan proved (see their Theorem 1.1) that for all $0<\varepsilon<\frac{1}{2}$, there exists $d(\varepsilon)>0$ such that if $d>d(\varepsilon)$ and $d$ has no more than

$$\exp\Big(\Big(\frac{1}{2}-\varepsilon\Big)\frac{(\log\log d)(\log\log\log\log d)}{\log\log\log d}\Big)$$

distinct prime factors, then

$$p_{\min}(d)\gg \varphi(d)\frac{(\log d)(\log\log d)(\log\log\log\log d)}{\log\log\log d}.$$

Take $d=p^j$. Then $d$ has one distinct prime factor, and $\varphi(d)=p^j-p^{j-1}$. From this, it follows that $K(p)\geq 1$.

$\endgroup$
2
  • $\begingroup$ Hi! I don't see the connection. A lower bound for $p_{\text{min}}(d)$ should only give a lower bound for $K(p)$ (here: $K(p) \geq 1$). An upper bound should involve proving the existence of small primes in some arithmetic progressions. $\endgroup$ Apr 9, 2022 at 7:44
  • 1
    $\begingroup$ @SteffenKionke Right, I updated so that $K(p)\geq 1$. You said "My question is: what can be said if limsup is replaced by liminf?" I tried to answer that. $\endgroup$
    – 2734364041
    Apr 9, 2022 at 8:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.