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Let $\nu (n)$ be the number of primes less than or equal to $n$ where $n$ is a positive integer. Prove that $\dfrac{n}{\nu(n)}=k$ has a solution for every integer $k \geq 2$.

I was thinking of using the Prime Number Theorem which says $\nu(n) \sim \frac{n}{\log{n}}$, but I don't see how to apply it to the question. Is there a simpler way to solve this?

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  • $\begingroup$ The usual notation is $\pi(n)$. $\endgroup$ – lhf Jun 11 '16 at 1:56
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    $\begingroup$ FWIW the sequence of $n$ such that $n/\pi(n)$ is an integer is in OEIS under A057809. $\endgroup$ – daniel Jun 11 '16 at 13:18
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The PNT shows that the expression on the left will eventually exceed every $k$. Let $n_0$ be the largest $n$ for which the expression on the left is less than $k$. Then $n_0+1$ must be composite. So $n_0 < \nu (n_0) k$, but $n_0 + 1\geq \nu ( n_0 +1)k =\nu (n_0) k$. It follows that we have equality, so we're done.

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  • $\begingroup$ Why must $n_0+1$ be composite? $\endgroup$ – user19405892 Jun 11 '16 at 1:21
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    $\begingroup$ Because as you increase $n$, the expression $n/\nu(n)$ will decrease when you hit a prime $n$ since the fraction exceeds 1 and you are increasing both numerator and denominator by 1 $\endgroup$ – Barry Smith Jun 11 '16 at 1:37
  • $\begingroup$ This answer seems correct but personally I would like to see "$n_0+1$ must be composite" and "[i]t follows that we have equality" explained in the answer itself. Even in the comment above the contradiction (that there would be a larger $n_0$ such that the expression on the left is less than $k$) is not stated explicitly . +1 $\endgroup$ – daniel Jun 12 '16 at 8:22

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