2

In other words, are there infinitely many numbers $n$ such that all numbers smaller than $n$ and relatively prime to it are primes ? For example, $ n = 30$ works.

Note that existence of such numbers would trivially imply the conjecture in this question as true.

katana_0
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2 Answers2

2

It is not difficult to see that $\omega(n)\leq 2\log n$, while there exists a constant $A>0$ such that

$$\varphi(n)>A\frac{n}{\log\log n}$$

(see here). Hence,

$$\pi(n)=\varphi(n)-\omega(n)>A\frac{n}{\log\log n}-2\log n$$

but by the PNT $\pi(n)\sim n/\log n$, so for all large $n$, there is an $\epsilon>0$ such that

$$\frac{n}{\log n}-\epsilon<A\frac{n}{\log\log n}-2\log n \qquad\forall n\geq n_0\text{ satisfying the condition},$$

but this is not possible if you take $n$ large enough. Hence there are only finitely many solutions.

user246336
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1

If $n$ is not a multiple of 2 then it has to be less than $2^2$.

If $n$ is not a multiple of 3 then it has to be less than $3^2$.

If $n$ is not a multiple of 5 then it has to be less than $5^2$ etc.

This suggest that the only candidates for $n$ are the primorials. But the next primorial after $30$ is $2 \times 3 \times 5 \times 7 = 210$, and $210 > 11^2$, so that doesn't work. And neither will higher primorials.

gandalf61
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