$\DeclareMathOperator{\ord}{ord}$Artin's conjecture stipulates that $\ord_p(2) = p -1$ for infinitely many primes $p$, where $\ord_p(2)$ denotes the multiplicative order of $2$ modulo $p$. More generally one expects that $\ord_p(2)$ is often quite large. I'm looking for a weakened version of this, namely:

   Does the sum $\displaystyle\sum_{p \leq x} \frac{1}{\ord_p(2)^2}$ converge as $x\to\infty$?

I would prefer unconditional results, but results conditional on e.g. GRH are still welcome.

  • 6
    $\begingroup$ This doesn't answer the question, but in case this is useful for something: since $p|2^n-1$ if and only if $\mathrm{ord}_p(2)=d$ for some $d|n$, your (infinite) sum is equal to $\zeta(2)^{-1}\sum_{n\geq 1}\frac{\omega(2^n-1)}{n^2}$. $\endgroup$ Jul 30 at 20:27
  • 1
    $\begingroup$ The following paper might be helpful mast.queensu.ca/~murty/erdos-ram.pdf $\endgroup$
    – P. Koymans
    Jul 31 at 12:29
  • $\begingroup$ Two other papers which might be useful: My own paper "Upper bounds for the number of primitive ray class characters with conductor below a given bound" in Acta Arithmetica which showed that for any fixed $a$, and any $\alpha$, $\sum_{n \leq x, (a,n)=1} \frac{n}{ord_n(a)} = O(\frac{x^2}{(\log x)^{\alpha}})$, and Sungjin Kim's "Average reciprocals of the order of a modulo n" which obtains a tighter bound using a somewhat different technique. $\endgroup$
    – JoshuaZ
    Aug 2 at 13:04

Since Joe Silverman raised the possibility of variations of the question, I want to point out (as a long comment) that even a mild variation

$$ \sum_{p } \frac{1}{ \operatorname{ord}_p(2)^2 \log ( \operatorname{ord}_p(2))^\epsilon}$$

provably converges, since (following an argument given by Julian Rosen in the comments) $\operatorname{ord}_p(2)=d$ only if $p$ divides $2^d-1$, and the number of such is $$\omega(2^d-1) = O \left( \frac{\log(2^d-1)}{ \log (\log(2^d-1))} \right) = O \left( \frac{d}{ \log d} \right)$$


$$ \sum_{p } \frac{1}{ \operatorname{ord}_p(2)^2 \log ( \operatorname{ord}_p(2))^\epsilon} \leq \sum_d \frac{1}{ d^2 \log(d)^\epsilon} O \left( \frac{d}{ \log d} \right) = O \left( \sum_d \frac{1}{ d( \log d)^{1+\epsilon}}\right) <\infty.$$

One can try to improve this argument by taking advantage of the fact that one $p$ can't divide $2^d-1$ for too many different $d$s, but I don't think you will solve it with such reasoning. A bad scenario you would need to rule out is that for each $d$ there are $ \sim c_1 d / \log d$ primes, all of size $\sim d^{c_2}$, that divide $2^{d}-1$, for some arbitrary constants $c_1,c_2$ with $c_2 > 3$ and $c_1 c_2 < \log 2 $.

We want $c_2>3$ so the number of primes between $d^{c_2}$ and $(d+1)^{c_2}$ that are congruent to $1$ modulo $d$ is still at least $\sim d/\log d$, and we want $c_1 c_2 < \log 2$ so that the product of all these primes is still less than $2^{d}-1$.

  • 3
    $\begingroup$ One can of course quantify your first observation by noting that if $d$ is the smallest exponent so that $p$ divides $2^d-1$, then $v_p(2^m-1)$ is $0$ if $p\nmid m$, and it is equal to $v_p(2^d-1)+v_p(m/d)$ if $p\mid m$. But ruling out $2^d-1$ from having a lot of small prime factors of approximately equal size does sound difficult. $\endgroup$ Jul 30 at 21:39
  • $\begingroup$ @JoeSilverman And the fact that we can take the factors to be very sparse among the primes and still make the sum diverge means that even GRH is unlikely to be helpful. $\endgroup$
    – Will Sawin
    Jul 30 at 21:59

Not quite what you've asked for, but in case it helps in whatever application you have in mind: $$ \sum_{p~\text{prime}} \frac{\log p}{p \operatorname{ord}_p(a)^\epsilon} \le \log\log a + \frac{2}{\epsilon} + C $$ for all $\epsilon>0$ and an absolute constant $C$. Here $a\in\mathbb Z$ is any integer with $|a|\ge2$. This will at least tell you that $\operatorname{ord}_p(a)$ cannot be too small, too often. The proof, which is fairly elementary, is in Variations on a theme of Romanoff, Internat. J. Math. 7 (1996), 373-391 [MR1395936].

  • $\begingroup$ Thanks I'll take a look! Any idea if the method can be adapted to my question? $\endgroup$ Jul 30 at 20:35
  • $\begingroup$ My first guess is that no, it's not going to answer your question. Indeed, I'm not sure it can even be adapted to prove that there exists a $K>0$ and an $\epsilon>0$ so that $$\sum_p \frac{1}{p^{1-\epsilon}\cdot\operatorname{ord}_p(a)^K}$$ converges. $\endgroup$ Jul 30 at 20:43

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