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Prove that for every $ϵ>0$ there exist infinitely many natural numbers $n$, such that $$\tau(n)\geq 2^{(1-ϵ)\frac{\log(n)}{\log\log(n)}}$$

It is obvious to look at squarefree naturals of the form $n=p_1 p_2\cdots p_k$ where $p_1,p_2,\ldots,p_k$ are first $k$ prime numbers. Such naturals have $\tau(n)=2^k$.

So I have to prove that for k large enough, $k>(1-ϵ)\frac{\log(n)}{\log(\log(n))}$. I thought about putting exp on both sides so we have to prove $\exp(k)>n^{\frac{1-ϵ}{\log(\log(n))}}$.

There must be a chain of inequalities starting from $n^{\frac{1-ϵ}{\log(\log(n))}}<\cdots<\cdots<\cdots<\exp(k)$. I was trying to do it yesterday but my inequalities were tnot proper. Of course I can write "$1$" instead of "$1-ϵ$" or write $\log(\log(2))$ or $\log(\log(p_k))$ instead of $\log(\log(n))$ in denominator but that directs me to nothing. Later all I can do with $n$ is saying that $n<p_k^k$ but this is too strong - because this side will be bigger than $\exp(k)$.

Any hints?

rtybase
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robin3210
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  • @Gary, You edited my question badly. log(n)/log(log(n)) is multiplied by (1-epsilon) in power of two. It is not ike you edited. – robin3210 Apr 15 '20 at 08:47
  • Sorry, but it looked more like that from your original post. Btw, what is $\tau$? – Gary Apr 15 '20 at 08:49
  • @Gary tau is the number of positive divisord of n. – robin3210 Apr 15 '20 at 09:34
  • Hint: ${\displaystyle \tau(p_{k})=2}<2^{(1-\varepsilon)\frac{\log{p_k}}{\log{\log{p_k}}}} \iff \frac{1}{1-\varepsilon}<\frac{\log{p_k}}{\log{\log{p_k}}}$ for large enough primes $p_k$, since $\frac{x}{\log{x}}$ is ascending for $x\geq 3$. – rtybase Apr 15 '20 at 09:43
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    @rtybase ok, I see where I made a mistake. My numbers need to be squarefree numbers but no starting form 2,3,5... but all prime divisors must be large enough so holds what You wrote. Now I know what to write :) – robin3210 Apr 15 '20 at 10:09
  • @rtybase no, Im not right at that moment. I have to also prove that log(n)/log(log(n)) is < sum from 1 to k log(pk)/log(log(pk). – robin3210 Apr 15 '20 at 10:15
  • @rtybase that Gary changed the quality badly during edit. t should be >= not <=.. – robin3210 Apr 15 '20 at 10:16
  • Are you trying to prove this? – rtybase Apr 15 '20 at 11:40
  • @rtybase i think we can say so. – robin3210 Apr 15 '20 at 13:28
  • What do you think about the 2nd part of the answer below? – rtybase May 02 '20 at 18:04

1 Answers1

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You could use Wigert's result $$\limsup_{n\to\infty}\frac{\log \tau(n)}{\log n/\log\log n}=\log2 \Rightarrow \\ \limsup_{n\to\infty}\frac{\log \tau(n)}{\log2 \cdot \log n/\log\log n}=1 \Rightarrow \\ \exists n \text{ s.t. } \frac{\log \tau(n)}{\log2 \cdot \log n/\log\log n}> 1-\varepsilon \Rightarrow \\ \tau(n)>2^{(1-\varepsilon)\frac{\log{n}}{\log{\log{n}}}}$$

Or use Chebyshev function and the fact that $\forall \varepsilon>0$, $\exists x(\varepsilon)>0$ s.t. $\forall x > x(\varepsilon)$ $$e^{(1-\varepsilon)x}<\prod\limits_{p\leq x}p< e^{(1+\varepsilon)x} \Rightarrow $$ $$(1-\varepsilon)x<\log{\prod\limits_{p\leq x}p}< (1+\varepsilon)x \Rightarrow \color{blue}{x > \frac{\log{\prod\limits_{p\leq x}p}}{1+\varepsilon}} \tag{1}$$ Use PNT or better $$\color{blue}{\pi(x)>\frac{x}{\log{x}}} \tag{2}$$ Note that $\frac{x}{\log{x}}$ is ascending for large enough $x$'s: $$\frac{x}{\log{x}}\overset{\color{red}{(1)}}{>} \frac{\frac{\log{\prod\limits_{p\leq x}p}}{1+\varepsilon}}{\log{\frac{\log{\prod\limits_{p\leq x}p}}{1+\varepsilon}}}\tag{3}$$

Then, by noting $n_x=\prod\limits_{p\leq x}p$ and considering that $2^x$ is ascending: $$\tau(n_x)=\tau\left(\prod\limits_{p\leq x}p\right)= 2^{\pi(x)}\overset{\color{red}{(2)}}{>} 2^{\frac{x}{\log{x}}} \overset{\color{red}{(3)}}{>} \\ 2^{\frac{1}{1+\varepsilon}\frac{\log{\prod\limits_{p\leq x}p}}{\log{\left(\log{\prod\limits_{p\leq x}p}\right)-\log{(1+\varepsilon)}}}}= 2^{\frac{1}{1+\varepsilon}\frac{\log{n_x}}{\log{\log{n_x}}-\log{(1+\varepsilon)}}}> 2^{(1-\varepsilon)\frac{\log{n_x}}{\log{\log{n_x}}}}$$ because $1>1-\varepsilon^2=(1+\varepsilon)(1-\varepsilon) \Rightarrow \frac{1}{1+\varepsilon}>1-\varepsilon$. We can build an infinite sequence of such $n_x$'s.

rtybase
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