Let G be a cyclic group of order n, where n is odd. What is the infimum of the average of the sum of the element orders in G?
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2$\begingroup$ I don't understand the question. $n = \sum_{d \vert n}\phi(d) d$. $\endgroup$– AndreaCommented Jun 17, 2012 at 8:42
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3$\begingroup$ I don't undertand the comment. I'd assume the phi is Euler totient but then the equality is false. $\endgroup$– user9072Commented Jun 17, 2012 at 10:38
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$\begingroup$ Maybe Andrea meant to write $n = \sum_{d|n} \phi (d)$? $\endgroup$– Patricia HershCommented Jun 17, 2012 at 10:57
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1$\begingroup$ @Patricia Hersh: Maybe, but then it does not seem to answer the question. $\endgroup$– user9072Commented Jun 17, 2012 at 11:01
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$\begingroup$ @quid: good point. Is it easy to see that there is an infinite sequence of odd numbers $n_1 < n_2 < n_3 < \dots $ where this average keeps decreasing in value? For instance, the odd primes would not work. $\endgroup$– Patricia HershCommented Jun 17, 2012 at 11:37
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As (indirectly) commented by Andrea the quantity in question is $$\alpha(n)=\frac{1}{n} \sum_{d \mid n} d \varphi(d) $$ so that in particular $\varphi(n)$ is always a lower bound. And the actual infimum is $1$, attained for $1$.
However, there are various interesting question related to this and (perhaps) since the 'average' was not quite clear in the start actually something else could be meant.
In view of the above it is natural to further scale down by $\varphi(n)$ and study the resulting quantity $\alpha(n)/\varphi(n)$. Its limit inferior is $1$ while the limit superior is $\zeta(2) \zeta(3) / \zeta(6)$.
Or, one could study $\frac{1}{x} \sum_{n \le x} \alpha(n)$. This would then by asymptotic to $\frac{\zeta(3)}{2 \zeta(2)} x$.
These results and more information can be found in a paper by von Zur Gathen et al. (JNTB, 2004)