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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?

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 $a_d=\sum_{p=1(\operatorname{mod} d), p>y} p^{-2} \leq d^{-2}\sum_{m>y/d}m^{-2}$, one can deduce that $$s(n)=\sum_{d|n}\phi(d) a'(d)+O(1),$$ where $$a'(d)=\sum_{p \leq d \log d {\log \log d}^2,\, p \equiv 1(\operatorname{mod} d)}\frac{1}{p^2}.$$ That seems to suggest that if $n$ is such that $s(n)$ is large, then for many divisors $d|n$ there might be many primes $p \equiv 1 (\operatorname{mod} d)$ 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 remind you of anything I could look up?

A few more ideas: using the chebyshev upper bound in the form $\sum_{p>y}p^{-2} \ll 1/y \log y$ 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?

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?

A few more ideas: using the chebyshev upper bound in the form $\sum_{p>y}p^{-2} \ll 1/y \log y$ 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 such that if $s(n)$ 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?

A few more ideas: using the chebyshev upper bound in the form $\sum_{p>y}p^{-2} \ll 1/y \log y$ 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?