# A conjecture related to primes of the form $\phi(n)$ [closed]

Recently I have been studying primes of various forms (and those forms are number theoretic and special functions). Let $$\pi_{f(x)}(n)$$ denote the number of primes less than $$n$$ which are of the form $$f(x)$$ for some $$x\in\Bbb{Z}$$. I first studied the function $$\pi_{\phi(x)}(n)$$. Few primes of such form occur. Here is a table of $$\pi_{\phi(x)}(n)$$: $$\begin{array}{|c|c|} \hline n&\pi_{\phi(x)}(n)\\ \hline 100&3\\ \hline 1000&3\\ \hline 10000&3\\ \hline 100000&3\\ \hline 1000000&3\\ \hline 10000000&3\\ \hline \end{array}$$ I noticed that for all $$10000000\ge n\ge6$$, $$\pi_{\phi(x)}(n)=3$$. So I conjectured that $$\forall n\geq6,\,\,\pi_{\phi(x)}(n)=3$$ is this conjecture true?
Note: The software I used for these computations is pari/gp.

• Failure to note that phi(n) is even for all $n\geq 3$ indicates that more thought should have been put into this question/conjecture before asking it on MathOverflow – Yemon Choi Nov 25 '20 at 2:28

Assuming $$\phi(x)$$ is referring to Euler's totient function, note $$\phi(1) = \phi(2) = 1$$ and it's always even for all $$x \ge 3$$. This is because of the totient function's multiplicative nature, with all odd primes having an even totient value, and $$2$$ or more factors of $$2$$ also having an even totient function.
Thus, $$\phi(x)$$ can only be a prime $$\ge 2$$ when it's $$2$$. This only happens when $$x$$ has just one factor of $$3$$, a factor of $$3$$ and a factor of $$2$$, or $$2$$ factors of $$2$$. Thus, this means it only occurs for the $$3$$ values of $$x \in \{3, 4, 6\}$$.
However, as the question says it's looking for the number of primes of the form $$\phi(x)$$ which are less than $$n$$, then for all $$n \ge 3$$, using the standard definition of what are prime numbers, the answer would be $$1$$ (i.e., prime $$2$$). If $$1$$ is considered to be a prime (as the OP states in a comment below), the result would then be $$2$$ instead.
• There are only two such primes, $2,3$. So, $\pi_{\phi(x)}(n)$ should be equal to $2$. I think $1$ was considered to be a prime. – Alapan Das Nov 24 '20 at 6:54
• @epic_math You're welcome. As Alapan Das indicated, there is actually only one prime of the form $\phi(x)$, i.e., $2$ (or, if you consider $1$ to be a prime, then $2$). It seems you're actually counting the number of $x$ for which $\phi(x)$ is a prime, which is technically not quite the same thing as what you wrote in your question. – John Omielan Nov 24 '20 at 7:01