What is the asymptotic growth rate of the unitary totient function, $\phi^*(n)$?

It appears that $$\phi^*(n)\geq c\frac{n}{\ln n}$$ but I am sure there is a stronger lower bound.

Any linkes to references or resources are greatly appreciated.

  • $\begingroup$ There are some links at the OEIS page. Have you followed them? $\endgroup$ – Gerry Myerson Mar 17 '13 at 23:34
  • $\begingroup$ Yes, I have. Let me know if I missed something, however. $\endgroup$ – pre-kidney Mar 18 '13 at 0:06
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    $\begingroup$ How could we let you know if you missed something if you do not tell us what you found? $\endgroup$ – Mariano Suárez-Álvarez Mar 18 '13 at 4:59

It will take a few minutes for a complete proof. but I already think that Ramanujan's procedure gives exceedingly low values of this at the primorials. In which case

$$ \liminf \; \frac{e^\gamma \phi^\ast(n) \log \log n}{n} = 1. $$

Yep, it works. See both my answers at Is the Euler phi function bounded below? and my answer at Euler's Phi Function Worst Case

However, no separate proof is really necessary. Everything comes from the results of Nicolas on primorials and $\phi$ along with $$ \phi^\ast(n) \geq \phi(n). $$

  • $\begingroup$ What is Ramanujan's procedure? $\endgroup$ – user27126 Mar 18 '13 at 0:04
  • $\begingroup$ @Sanchez, hi. In this case, for any real $0 < \delta < 1,$ you find the number(s) $n,$ just one such value if $\delta$ is rational, where the minimum of $$ \frac{\phi^\ast(n)}{n^\delta} $$ is achieved. In this case, for any $\delta$ you get a primorial $n.$ $\endgroup$ – Will Jagy Mar 18 '13 at 0:08
  • $\begingroup$ Can you justify your statements? I see the general outline of what you're doing ("small values" for $\phi^*(n)$ occur at primorials, and $\phi^*(n)=\phi(n)$ on the primorials). But it still needs justification, which is where I am stuck. $\endgroup$ – pre-kidney Mar 18 '13 at 0:11
  • $\begingroup$ @pre-kidney, please read my full proof for $\phi$ at one of the questions to which I linked, it is the answer with zero votes. I'm not absolutely against writing a separate proof for this, but my guess it you will be able to do that yourself, the proofs are so very similar. Actually, one of two answers(out of three on this topic) with zero votes. Sigh. $\endgroup$ – Will Jagy Mar 18 '13 at 0:14
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    $\begingroup$ And I would prefer to have a pony. But maybe you can read through what Will has done, and come up with your own clear and unified treatment, which you can then post here as a new answer. $\endgroup$ – Gerry Myerson Mar 18 '13 at 4:54

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