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Let $f$ be defined by $$f(n) = \frac{\phi(n)}{n}.$$ Then define a sequence $(n_k)$ by $$n_1 = 1, \mbox{and for } k \geq 2,$$ $$n_k = \ \mbox{smallest integer }n\ > n_{k-1} \ \mbox{with}\ f(n) > f(n_k)$$ for any $n < n_k$. Deduce a formula for $n_k$ and $f(n_k)$, with proof.

Sol. I try to calaulate some values of the $n_k$ as follows : $$\begin{array}{c|ccccccccccccccccc} n & 1 &2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17 \\ \hline \phi(n) &1&1&2&2&4&2&6&4&6&4&10&4&12&6&8&8&16\\ \hline f(n)&1&\frac{1}{2}&\frac{2}{3}&\frac{2}{4}&\frac{4}{5}&\frac{2}{6}&\frac{6}{7}&\frac{4}{8}&\frac{6}{9}&\frac{4}{10}&\frac{10}{11}&\frac{4}{12}&\frac{12}{13}&\frac{6}{14}&\frac{8}{15}&\frac{8}{16}&\frac{16}{17} \end{array}$$

and so $n_1 = 1, n_2 = 2$,and $n_3 = 6$

To see what $n_k$ should be, it requires computations, and I still do not see potential formula.

Can anyone please suggest the formula, or a more effective way to analyse this problem ?

Nathaniel B
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user117375
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    Try to get some more insight into how to compute $\phi(n)/n$: do you know how to do it easily knowing the prime factors of $n$? Also, you know that $n_4 > 13$ so this helps you to guess that the sequence grows pretty quickly. Does $1,2,6$ match any fast-growing sequence you are familiar with? – Erick Wong Sep 06 '16 at 16:01
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    Yes, I think $$\phi(n)/n = \prod_{p|n} (1 - 1/p).$$ Oh, okay, so something connects to number of distinct prime dividing $n$, but might be more complicated. – user117375 Sep 06 '16 at 16:04
  • That's correct, the value of $\phi(n)/n$ comes from the product of terms that are $<1$ coming from prime factors of $n$. Now think about the following: what's the smallest possible value of $\phi(n)/n$ when $n$ has 1 prime factor (is prime)? What's the smallest possible value with 2 prime factors? What about 3? – Erick Wong Sep 06 '16 at 17:44

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Erick is working this out with you, good. Eventually, you want two items, PLANAT which is commentary on NICOLAS. The Nicolas paper appeared in the Journal of Number Theory, volume 17 (1983), pages 375-388 and has an English abstract. Here we go, NICOLAS 2012 is an update, better estimates, in English.

In an earlier problem, I proved that Ramanujan's method, the same that gives the Superior Highly Composite numbers and the Colossally Abundant numbers, gives the same sequence that Erick is hinting at in comments. Is the Euler phi function bounded below?

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Will Jagy
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