Let $\pi(x)$ the prime-counting function and $\varphi(n)$ denotes the Euler's totient function.

I would like to know if next arithmetic function, that I've defined searching a comparison of such arithmetic function, tends to zero as $N\to\infty$


When $N=10^D$, with $D=0,1,2,3$ one gets that $f(N)$ is calculated as, respectively, $1,\approx0.917,\approx0.511$ and $\approx0.370$. Thus our positive function is decreasing in this segment.

Question. Prove or refute $$f(N)\to 0$$ as $N\to\infty$. Many thanks.

I presume that the solution needs the prime number theorem and inequalities for the Euler's totient function.

  • 2
    $\begingroup$ Your sum seems rather arbitrary. In what context did it arise? $\endgroup$ – Antonio Vargas Jan 18 '18 at 22:56
  • $\begingroup$ I accept your words, many thanks @AntonioVargas . My intention was present a comparison between these counting functions in a question, showing their composition in the quotients $\frac{\pi(\varphi(k)+N)}{\varphi(\pi(k)+N)}$, after I've take the sum. Finally with the help of a CAS I have adjusted the factor $\frac{1}{N}$ (here I know that it, this last factor has mathematical meaning). $\endgroup$ – user243301 Jan 19 '18 at 10:26
  • $\begingroup$ If you want @AntonioVargas you can study in you home what happens with $\frac{1}{N}\sum_{1\leq n\leq N}\frac{\sigma(\pi(\varphi(n)+N))}{\varphi(\sigma(\pi(n)+N))}$, as $N\to\infty$, where $\sigma(m)$ denotes the sum of divisors function. As I said I don't know if it was in the literature or these combinations are interestings. I don't know if this last is more difficult/complicated or arbitrary than wich was solved. With a CAS I can deduce that this positive sequence seems decreasing, $N=100$ as $\approx 0.98$, $N=1000$ is $\approx 0.65$ and my last experiment with $N=2000$ is $\approx0.63$ $\endgroup$ – user243301 Jan 23 '18 at 13:40

Since $$C_{2}\frac{n}{\log\left(n\right)}\leq\pi\left(n\right)\leq C_{1}\frac{n}{\log\left(n\right)}$$ where $C_{1},C_{2}>0$ are suitable constants and $$\frac{n}{e^{\gamma}\log\left(\log\left(n\right)\right)+\frac{3}{\log\left(\log\left(n\right)\right)}}<\phi\left(n\right)<n$$ where $\gamma$ is the Euler Mascheroni constant, we have $$\pi\left(\phi\left(k\right)+N\right)\ll\frac{\phi\left(k\right)+N}{\log\left(\phi\left(k\right)+N\right)}\ll\frac{2N}{\log\left(N\right)}$$ and $$\frac{1}{\phi\left(\pi\left(k\right)+N\right)}\ll\frac{\log\left(\log\left(\pi\left(k\right)+N\right)\right)}{\pi\left(k\right)+N}\ll\frac{\log\left(\log\left(2N\right)\right)}{N}$$ so $$\sum_{k\leq N}\frac{\pi\left(\phi\left(k\right)+N\right)}{\phi\left(\pi\left(k\right)+N\right)}\ll\frac{\log\left(\log\left(2N\right)\right)}{\log\left(N\right)}\sum_{k\leq N}1\ll\frac{\log\left(\log\left(2N\right)\right)}{\log\left(N\right)}N$$ then the claim.

  • $\begingroup$ Many thanks I am going to study your nice answer this morning. $\endgroup$ – user243301 Jan 19 '18 at 10:15

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