For each natural number $x\ge2$ , we have $x\le2^{\pi(x)}~?$

Question

Is the following statement true ?

For each natural number $x\ge2$ , we have $x\le2^{\pi(x)}$, where $\pi(x)$ is the prime-counting function .

It seems true for $x$ is small. As $x$ getting larger then the distance between two prime numbers possibly getting larger. So is it true for all natural number $x\ge 2$ ? or is there a counterexample ?

Thanks for patient reading and any comment or answer I will be grateful.

Answer 1

I am very bad with prime numbers; so I shall use algebra.

Consider the function $$f(x)=\log \left(2^{\pi (x)}\right)-\log(x)$$ and we know that, for any $x \ge 17$, $\pi (x) > \frac {x}{\log(x)}$ $$f(x) >\frac {x\log(2)}{\log(x)}- \log(x)$$ Consider now the function $$g(x)=\frac {x\log(2)}{\log(x)}- \log(x)$$ Its derivative $$g'(x)=-\frac{1}{x}-\frac{\log (2)}{\log ^2(x)}+\frac{\log (2)}{\log (x)}$$ cancels for $x \approx 5.87$ (numerical method) and for this value $g(x) \approx 0.529$ and the second derivative test shows that this is a minimum.

Is this any good for you ?

Answer 2

Let $p_1,p_2,p_3,\ldots$ be the increasing sequence of all prime natural numbers. From Bertrand's Postulate, we can show that $p_n< 2^{n-1}$ for every integer $n\geq 4$.

For an integer $x$ such that $1\leq x \leq 4$, we can easily see that $x\leq 2^{\pi(x)}$ with equality cases $x=1$, $x=2$, and $x=4$. Now, suppose that $x\geq 5$ is an integer. Then, $\pi(x)\geq 3$. Thus, $p_{\pi(x)+1}<2^{\pi(x)}$ from the paragraph above. That is, $$p_{\pi(x)}\leq x< p_{\pi(x)+1} < 2^{\pi(x)}\,.$$ Thus, $$x<2^{\pi(x)}\text{ for every integer }x\geq 5$$ as required. (In deed, $x<2^{\pi(x)}$ holds for all real numbers $x\geq 5$.)

In fact, due to a result by Jitsuro Nagura, there exists a prime between $n$ and $\frac{6}{5}n$ for every $n\geq 25$. We can then show that, for every integer $n\geq 27$, $p_{n} <\left(\frac{6}{5}\right)^{n-1}$. This proves that $$x< \left(\frac65\right)^{\pi(x)}$$ for every $x\geq p_{26}=101$.