5

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.)

1 Answers1

2

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 $K(p)\geq 1$.

2734364041
  • 5,059
  • 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. – Steffen Kionke Apr 09 '22 at 07:44
  • 1
    @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. – 2734364041 Apr 09 '22 at 08:22