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I've seen the graph for $\varphi(n)$ vs $n$ on Wolfram, and it seems like the $\varphi(n)$ values for primes follow a constant slope, but is there a proof that states that the totient values between two consecutive primes are always smaller than the totient values of those two primes?

EDIT: I got a satisfactory answer for the case where there is a strict inequality. But what is the condition for a relaxed inequality? Is there a proof that establishes $\varphi(n)\le\varphi(p_n)$ and $\varphi(n)\le\varphi(p_{n+1})$

Martin Sleziak
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Arannya Monzur
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1 Answers1

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Let's assume that Oppermann's conjecture is true and for every prime $p_n$ and next prime $p_{n+1}$,$$p_{n+1}<p_n+\sqrt{p_n}$$ EDIT: This inequality has counterexample at small prime numbers. The list of known counterexample $p_n$s are $(3,7,13,23,31,113)$ (see A124129)

Now note that every number $k$ such that $p_n<k<p_{n+1}$ is composite, so it has at least one prime factor $q$ such that $q^2\le k$. Now we get$$\varphi(k)=k\prod_{p|k}\left(1-\frac{1}{p}\right)\le k\left(1-\frac{1}{\sqrt{k}}\right)=k-\sqrt{k}<p_{n+1}-\sqrt{p_n}<p_n$$ and it is proved.

EDIT: One can verify or find counterexamples of the inequality manually for small $k$s around above counterexamples.

If Oppermann's conjecture is not true, and furthermore, there exists a prime $q$ and integer $n$ such that $p_n<q^2-q<q^2<p_{n+1}$, then$$\varphi(q^2)=q^2-q>p_n$$so your conjecture has counterexample.

didgogns
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  • I agree with the answer, but would point out that the question asks if the totient values "are always smaller". $\varphi(3)=\varphi(4)=2$ and $\varphi(7)=\varphi(9)=6$ are counterexamples where the totient value is equal to that for the lower prime. – nickgard Apr 12 '17 at 15:13
  • Thanks, that does help, but I have a question, let's say I relax the strict inequality. Is the hypothesis known to hold with certainty if equality is allowed between two consecutive primes? – Arannya Monzur Apr 12 '17 at 15:50