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Let's $\pi(n)$ denote the prime-counting function. The prime number theorem famously states that $\pi(n) \underset{n\to+\infty}\sim \dfrac{n}{\log n}$. I'm interested in rougher but easier estimates. I will use the $\ll$ symbol instead of the $O$ notation.

  1. Most proofs of the infiniteness of prime numbers give some sort of lower bound. For example, Euclid's well-known proof can be used to get the very weak $\pi(n) \gg \log \log n$.
  2. One of my favourite proofs is the following: every number $k \leq n$ can be factored out as a squarefree number (which is a product of distinct primes, all $\leq n$) and a perfect square (and there are $\leq \sqrt n$ perfect squares which are $\leq n$). That gives $n \leq 2^{\pi(n)}\,\sqrt n$, so $\pi(n) \gg \log n$. Euler-type proofs also give this sort of bounds (and, conversely, this argument is the basis of Erdős's proof that $\sum_p \dfrac 1p = +\infty$).
  3. There are some clever, relatively easy proofs of the Čebyšëv estimates $\dfrac{n}{\log n} \ll \pi(n) \ll \dfrac{n}{\log n}$.

My question is: is there something nontrivial between 2. and 3? Specifically, I'm looking for a simple argument (simpler than all proofs of the Čebyšëv estimates) which would give a significant improvement on the logarithmic bound. Ideally, I would love a proof that I can remember almost as easily as 2. but which would give a lower bound $\pi(n) \gg n^\epsilon$, for some $\epsilon > 0$.

PseudoNeo
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    The Legendre conjecture (open but believed to be true) that there is always a prime between $n^2$ and $(n+1)^2$ would imply $n^{0.5} \ll \pi(n)$ and the subsequent work on other exponents provides proofs for other $n^\epsilon$. – Henry Jun 11 '25 at 13:39
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    It's fairly easy to show $\pi(x) \gg (\log x)^C$ for every $C > 0$ using Euler-type arguments and $\pi(x) = o(x)$. But of course that's still a far cry from $\pi(x) \gg x^{\varepsilon}$. – Dermot Craddock Jun 11 '25 at 14:03
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    In other words, you'd like to give a polynomial upper bound on $p_n$. So this and this are related, though they don't answer your question because the arguments are all through Chebyshev estimates or something harder. (although I'll make the unsolicited remark that Zagier's presentation of the lower bound is really stunningly simple - I think it's so simple that it makes your question almost unanswerable!) – Izaak van Dongen Jun 11 '25 at 17:56
  • Thanks a lot for the links. I agree that the answer I'm dreaming of cannot exist if a very simple proof of Čebyšëv's estimates exists, so I'm gonna look at that presentation of Zagier's. – PseudoNeo Jun 11 '25 at 19:25
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    After looking at Zagier's article, I think the question is still relevant. His presentation is concise, but it is the same clever, short proof you can find in textbooks (with Legendre's formula for $v_p(n!)$ and some telescoping arguments: he just skips details). Definitely a jewel but I think there is enough room for a easier, weaker result. – PseudoNeo Jun 12 '25 at 19:45
  • Fair enough! Not sure what the telescoping argument is - the way I would prove $p^{\nu_p\binom nk} \le n$ if someone put a gun to my head is to use the fact that $\lfloor a + b \rfloor \le \lfloor a \rfloor + \lfloor b \rfloor + 1$ for any $a, b$, which I find quite intuitive (wlog $a, b \in [0, 1)$: all it says is $a + b < 2$), then apply that to Legendre's formula which I guess I've internalised pretty well (let $a = k/p^i, b = (n - k)/p^i$, and sum over $i$). I guess that's the "key step" to memorise. A good "reason" and aide memoire for this why this fact is true is Kummer's theorem. – Izaak van Dongen Jun 13 '25 at 17:10
  • Are the two Čebyšëv estimates meant to be identical? – J.G. Jun 13 '25 at 18:52
  • Yes @J.G., here $f(n) \ll g(n)$ is a shorthand for $f(n) = O(g(n))$ - see the first paragraph. It is an atypical meaning for that symbol! – Izaak van Dongen Jun 13 '25 at 19:07
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    This meaning of $\ll$ is very typical in analytic number theory. – Greg Martin Jun 13 '25 at 19:45
  • Use ${\pi(x)\gg\log{x}}$ with Bertrand's Postulate "there is at least a prime ${\in[n,2n]}$" to get ${\pi(x)\gg\log{_{_2}}x}\approx 1.443\log{x}$ – Hazem Orabi Jun 14 '25 at 18:13

2 Answers2

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Here's a quick argument, in the line of Chebyshev's arguments, which gives a quasi-polynomial lower bound $\pi(n)\geq e^{c\sqrt{\log(n)}}$ for some constant $c$. It's a bit annoying to carry through, but the sketch is pretty simple.

The starting point is Legendre's formula $$\nu_p(n!)=\sum_{k=1}^\infty\left\lfloor\frac n{p^k}\right\rfloor\leq\frac n{p-1}.$$ This implies $$\log n\ll\frac 1n\log(n!)=\frac1n\sum_{p\leq n}\nu_p(n!)\log p\leq\sum_{p\leq n}\frac{\log p}{p-1}.$$ Note that $x\mapsto\frac{\log x}{x-1}$ is nonincreasing, so $$\log n\ll\sum_{r=2}^{\pi(n)+1}\frac{\log r}{r-1}\ll(\log\pi(n))^2$$ The result follows.

Carl Schildkraut
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3

You can take the repeated difference of $1,\tfrac13,\tfrac15,\ldots,\tfrac1{2n+1}$; namely, let $$ D = \bigg( \Big(\big(1-\tfrac13\big)-\big(\tfrac13-\tfrac15\big)\Big)- \Big(\big(\tfrac13-\tfrac15\big)-\big(\tfrac15-\tfrac17\big)\Big)-\ldots\bigg) = \sum_{k=0}^n\dfrac{(-1)^k\binom{n}{k}}{2k+1}. $$

By an easy induction, $D=\dfrac{2^nn!}{(2n+1)!!}$.

On the other hand, $D=\dfrac{K}{lcm(1,3,5,\ldots,2n+1)}$ with some integer $K$.

Updated: The denominators are odd, so $K$ is divisible by $2^n$ $$ \dfrac{1}{lcm(1,3,5,\ldots,2n+1)} \le \dfrac{K/{\color{red}{2^n}}}{lcm(1,3,5,\ldots,2n+1)} \le \dfrac{n!}{(2n+1)!!} < (\tfrac23)^{n-1} $$ $$ lcm(1,3,5,\ldots,2n+1) > (\tfrac32)^{n-1}. $$ The lcm on the left-hand side is the product of $\pi(2n+1)-1$ maximal odd prime powers, so $$ (2n+1)^{\pi(2n+1)-1} \ge lcm(1,3,5,\ldots,2n+1) > (\tfrac32)^{n-1} $$ Therefore $$ \pi(2n+1) > \dfrac{n}{\log_{3/2}(2n+1)}. $$

G.Kós
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