# A question on the prime divisors of p-1

For each positive integer n we may define the convergent sum $$s(n)=\sum_{p}\frac{(n,p-1)}{p^2}$$ where the summation is over primes p and $(a,b)$ denotes the greatest common divisor of a,b.

It is immediate to deduce that s(n) is bounded on average:

Using $\sum \limits_{d|a, d|b}\phi(d)=(a,b)$ and inverting the order of summation we get

$s(n)=\sum_{d|n}\phi(d) a_d$ where $a_d=\sum_{p \equiv 1 (mod \ d)}p^{-2}$

Ignoring the fact that we sum over primes we get the bound $a_d \ll \frac{1}{d^2}$ which leads to $$\sum_{n \leq x}s(n) \ll x \sum_{d \geq 1}\frac{\phi(d)a_d}{d}=O(x)$$ and $$s(n) \ll \exp(\sum_{p|n}1)$$ The last inequality means that $s(n)$ stays bounded if $\omega(n)$ is bounded. Towards the other direction, it seems fair to expect that $s(n)$ grows to infinity if $\omega(n)$ is large in some quantitative sense, say $\omega(n) \geq (1+\epsilon) \log \log n$. Taking into account that the contribution to the sum $s(n)$ of the primes $p$ that satisfy $(p-1,n) \leq \frac{p}{\log p}$ is bounded, since $\sum_{p}\frac{1}{p \log p}$ converges, we see that $s(n)=s'(n)+O(1)$ where $s'(n)=\sum_{ (p-1,n)>\frac{p}{\log p}} \frac{(n,p-1)}{p^2}$ We are therefore led to the question as to whether a condition of the form $\frac{\omega(n)}{\log \log n}-1 \gg 1$ can guarantee that $s'(n) \to +\infty$ Are there any non-trivial techniques that can be used to answer this question ?

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Are you just trying to show $s(n)$ is unbounded? and do you insist on a non-trivial technique? Let $n=m!$; then $s(n)>\sum_{p\lt m}{p-1\over p^2}=\sum_{p\lt m}{1\over p}+O(1)$ and of course the sum diverges.

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(corrected the typo log log n, thanx !) your choice of n shows that $s(n) \gg \log \log \log n$ infinitely often. Choosing $n=\prod_{p \leq m}(p-1)$ we deduce that $s(n) \gg \log \log n$ infinitely often. But my question was of another nature : Is it true that $s(n) \to \infty$ for each sequence of integers n that have a large number of prime divisors (not for just a particular sequence of the form $m!$ or $\prod_{p \leq y}(p-1))$ ? –  Captain Darling Sep 15 '10 at 12:13

Your guess that $s(n)$ gets large if $\omega(n)$ is large is not correct.
It is possible for $n$ to have many primes, and for $s(n)$ still to be small.

This can be seen from some of the work in your question. As you note $s(n) =\sum_{d|n} \phi(d) a_d$ where $a_d =\sum_{p\equiv 1\pmod d} p^{-2} \ll 1/d^2$. Therefore $$s(n) \ll \sum_{d|n} \frac{1}{d} \le \prod_{p|n} \Big(1-\frac 1p\Big)^{-1}.$$ If now every prime factor of $n$ exceeds $\log n$, then (since $\omega(n) \le \log n$ trivially) we have $$s(n) \ll \Big(1-\frac{1}{\log n} \Big)^{-\log n} \ll 1.$$

Thus $n$ can have about $\log n/\log \log n$ prime factors, all larger than $\log n$ and still $s(n)$ would be $\ll 1$.

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A few more ideas: using the chebyshev upper bound, by partial summation we have $\sum_{p>y}p^{-2}=O(\frac{1}{y \log y})$ and therefore we see that $s(n)=\sum_{p \leq \frac{n}{\log n}}\frac{(p-1,n)}{p^2}+O(1).$ Furthermore by the equality $\sum_{p \leq x}p^{-1}=\log \log x +A +(\frac{1}{\log x})$ we get $$s(n)=\sum_{p \leq n^{1/3}}\frac{(p-1,n)}{p^2}+O(1)$$ and one can make this more accurate. There is something in the expression $s(n)=\sum_{d|n}\phi(d)a_d$ that is linked to Linnik's constant(or the Elliot-Halberstam Conjecture). In particular, using $a_d=\sum_{p=1(mod d), p>y} p^{-2} \leq d^{-2}\sum_{m>y/d}m^{-2}$ one can deduce that $$s(n)=\sum_{d|n}\phi(d) a'(d)+O(1)$$ where $$a'(d)=\sum_{p \leq d \log d {\log \log d}^2, p=1(mod d)}\frac{1}{p^2}$$ That seems to suggest that for each n for which $s(n)$ is quite large then for many divisors $d|n$ there might be many primes $p=1 (mod d)$ in the interval $[d,d \log d {\log \log d}^{2}]$ and conversely, but I haven't been able to establish a clear connection between these two facts. To this end we may compute the mean values $\sum_{n \leq x} s^{2k}(n), k \geq 0$ which is quite straightforward. Does all this set-up reminds you anything I could look up?

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