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This afternoon I am trying to get variations of sequences inspired from the inequality that defines the so-called strong primes, see the definition of this inequality in number theory from this Wikipedia. To me seems a very interesting sequence (in this post as I've said I am going to get different variations using different arithmetic functions in such inequality, but maybe it can be interesting to explore variations of this definition when our sequence is restricted to different prime constellations).

To provide context to this question here is next easy example that I can get using a proof by cases.

Example. If $\pi(x)$ is the prime-counting function then an integer $n$ satisfies $$\pi(n+1)>\frac{\pi(n)+\pi(n+2)}{2}$$

if and only if $n$ is the $k$th odd prime number minus $1$ (that is, the sequence A006093 from the OEIS with the exception of the first term).$\square$

After I wanted to explore a different variation, now using the Euler's totient function $\varphi(n)$, and I wondered next question.

Question. Let $\varphi(n)$ the Euler's totient function. Prove or refute that there exist infinitely many even numbers $n\geq 2$ such that satisfy $$\varphi(n+1)\leq\frac{\varphi(n)+\varphi(n+2)}{2}.\tag{1}$$ Many thanks.

Thus the sequence of even and positive integers satisfying $(1)$ starts as

$$104,164,314,524,584,734,944,\ldots\tag{2}$$

and I would like to know if it is possible to prove that these solutions are infinitely many.

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  • $\begingroup$ And about strong primes, I emphasize that it looks like an interesting sequence. $\endgroup$ – user243301 Mar 17 '18 at 14:39
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    $\begingroup$ Besides the formulation of the problem, what are your attempts? Naively, one might try to construct $n+1$ such that $n+1$ is the product of many primes and $n,n+2$ are the products of much fewer. Since $\frac{\varphi(n)}{n}\approx\exp\left(-\sum_{p\mid n}\frac{1}{p}\right)$, it looks to me one might prove much more, i.e. that the set of $(n+1)$s fulfilling such inequality has a positive density. Still I fail to see why such inequality is relevant. $\endgroup$ – Jack D'Aurizio Mar 17 '18 at 14:52
  • $\begingroup$ I didn't attempts of solution since my thoughts is that it is very difficult to prove to me. On the other hand I don't think that this inequality is relevant of course. Many thanks for your remarks, if you want feel free to add it as draft of your answer @JackD'Aurizio $\endgroup$ – user243301 Mar 17 '18 at 14:55
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Suppose your conjecture is false, then there is some integer $N$, such that: $$\varphi(n+1)>\frac12(\varphi(n)+\varphi(n+2))$$ for all $n\ge N$.

Let $p>N$ be prime. Now, $\varphi(p)=p-1>\frac{p+1}{2}\ge\varphi(p+1)$, so $\varphi(p)>\varphi(p+1)$. We now have: $$\varphi(p+1)>2\varphi(p+1)-\varphi(p)>\varphi(p+2)$$ from which it follows that: $$\varphi(p+2)>2\varphi(p+2)-\varphi(p+1)>\varphi(p+3)$$ and so on, so the sequence $\varphi(p),\varphi(p+1),\varphi(p+2),\ldots$ is strictly decreasing. However, since $\varphi(p)$ is finite and $\varphi$ can only take on integer values, this means that $\varphi(p+k)$ will eventually be negative for large $k$, which is impossible.

Therefore, your conjecture is true.

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    $\begingroup$ Many thanks, this afternoon I am going to study your clear answer. Any case I encourage to other users if they want to do more contributions. $\endgroup$ – user243301 Mar 17 '18 at 15:10

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