5
$\begingroup$

Clearly if $p$ is prime, the sequence $\frac{\phi(p)}{p} \rightarrow 1$. In general, however, if $s_n \in S \subseteq \mathbb{N}$, we are not even guaranteed of the existence of: $\displaystyle \lim_{n \to \infty} \frac{\phi(s_n)}{s_n}$.

My question is this: Does there exist an infinite sequence $S \subseteq \mathbb{N}$ such that $\displaystyle \lim_{n \to \infty} \frac{\phi(s_n)}{s_n}=1$ and at most finitely many $s_i$ are prime?

My intuition tells me no, but I'm not sure. Having the above limit exist for some sequence alone is quite a strong statement, so having it equal one and contain finitely many primes is pretty restrictive. Admittedly, that isn't an argument and I've been having trouble finding one. Any thoughts would be appreciated.

EDIT: It just occurred to me that if $s_i=p_i^2$ for prime $p_i$, then the above holds. Reformulating, is there such an $S$ with finitely many prime powers?

| cite | improve this question | |
$\endgroup$
8
$\begingroup$

Certainly. Let $\{p_i\}_{i\in\mathbb{N}}$ enumerate the prime numbers, and take $s_n=p_np_{n+1}$. We have $$\lim_{n\rightarrow\infty}\frac{\phi(p_np_{n+1})}{p_np_{n+1}}=\lim_{n\rightarrow\infty}\frac{p_np_{n+1}-p_n-p_{n+1}+1}{p_np_{n+1}}=1.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Awesome. Thanks. $\endgroup$ – M10687 Apr 14 '16 at 6:18
  • $\begingroup$ And informally, if $n$ has "only large factors" then $\phi(n)$ is "approximately $n$". A product of consecutive primes is a good way to ensure a limited number of large factors while avoiding prime powers. $\endgroup$ – Steve Jessop Apr 14 '16 at 9:47
  • $\begingroup$ I am reminded of the Q of showing that $\lim \sup_{n\to \infty}\phi (n+1)/\phi(n)=1$ and $\lim \inf_{n\to \infty}\phi (n+1)/\phi (n)=0.$.... Let $Q_n$ be the product of the first $n$ primes. Using $n/p_n\to 0,$ where $p_n$ is the $n$th prime, and using $\prod_{j=1}^n(1-1/p_j)\to 0,$ we can show that $\phi (Q_n)/Q_n \to 0$ and that $\phi ( Q_n\pm 1)/( Q_n\pm 1)\to 1.$ $\endgroup$ – DanielWainfleet Apr 14 '16 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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