8
$\begingroup$

In connection with this question and its follow-up.

Suppose that $a\ge 2$ and $b\ne 0$ are integers, and $f$ is a monotonically increasing function such that $f(2)>1$, $f(p)\to\infty$, and the series $\sum_p 1/f(p)$ (extended onto all rational primes) diverges. Suppose further that for any integer $K>0$, there exist infinitely many integers $n\ge 1$ with $\gcd(a^n+b,K)=1$.

Is it true that, under the stated assumptions, every point of the interval $[0,1]$ is a limit point of the sequence with the $n$th term $$ \prod_{p\mid a^n+b}\Big(1-\frac1{f(p)}\Big)? $$ In particular, is it true that every point of $[0,1]$ is a limit point of the sequence $\varphi(2^n-1)/(2^n-1)$?

$\endgroup$

1 Answer 1

5
$\begingroup$

This is an answer to the part of the question where the OP is asking if the image of the function $$R_{2,\varphi}: \mathbf N^+ \to \mathbf R: n \mapsto \frac{\varphi(2^n - 1)}{2^n-1}$$ is dense in the interval $[0,1]$. The short answer is yes, some more details follow.


This week, I met Carlo Sanna in Turin during a meeting of number theory and mentioned the problem to him. He remembered to have read about the same question in a paper of Florian Luca. Today, I got an e-mail from Carlo with the full reference:

F. Luca, On the sum of divisors of the Mersenne numbers, Math. Slovaca 53, No. 5 (2003), 457-466 (EuDML link).

More precisely, see point ii) of the theorem on the bottom of p. 458.

By the way, Luca points out on p. 459 that:

  • his method works for any multiplicative function $f: \mathbf N^+ \to \mathbf R$ for which "there exist [...] $c > 0$ and $\lambda > 1$ so that $f(p^a) = 1 + \frac{c}{p} + O(p^{-\lambda})$ holds for all prime numbers $p$ and all positive integers $a$", on condition that $[0,1]$ is replaced with the interval $[\liminf_n f(n), \limsup_n f(n)]$ (the question in the OP corresponds to the case when $f = R_{2,\varphi}$);

  • the conclusion remains true if the sequence of Mersenne numbers is replaced with a Lucas sequence subjected to some technical conditions.

Among other things, Luca's proof makes use of the Siegel-Walfisz theorem.

$\endgroup$

Your Answer

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

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