why $ \rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$

Let $n\ge 7$ be positive integers,show that $$f(n)=\rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$$

Anyone know this problem background?or maybe have best proof or best result?

Solution 1:

the $2^n$ side is answered at $\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$

The governing phenomenon here is that the LCM in question is $e^{\psi(n)},$ where $\psi(n)$ is Chebyshev's second function. So, from the text comments for https://oeis.org/A003418 , we find

The prime number theorem implies that LCM(1,2,...,n) = exp(n(1+o(1))) as n -> infinity. In other words, log(LCM(1,2,...,n))/n -> 1 as n -> infinity. - Jonathan Sondow, Jan 17 2005

We can be very specific with the upper bound, Rosser and Schoenfeld (1962) Theorem 12 on page 71, $$ \psi(x) < 1.03883 \, x $$ for $x > 0.$ As $e^{1.03883} \approx 2.8259087,$ we find $$ \operatorname{lcm} (1,2,3,\ldots,n) < 2.826^n $$ This bound is tightest at $n=113,$ which is where the maximum of $\psi(x)/x$ is obtained.

Apparently, Hanson (1972) showed that $ \operatorname{lcm} (1,2,3,\ldots,n) < 3^n, $ see references in http://arxiv.org/abs/0906.2295

Solution 2:

I have the following euristic arguments.

It is easy to see that

$$f(n)=\prod_{p\in\Bbb P} p^{\lfloor \log_p n\rfloor},$$

where $\Bbb P$ is the set of prime numbers. Subdividing the diapason, we obtain

$$f(n)=\prod_{k=1}^\infty \prod_{p\in\Bbb P_k} p^k,$$

where $$\Bbb P_k=\{p\in\Bbb P: p^k\le n<p^{k+1}\}.$$

Then

$$\prod_{k=1}^\infty n^{|\Bbb P_k|\frac{k}{k+1}}<f(n)\le\prod_{k=1}^\infty n^{|\Bbb P_k|}.$$

Let $\pi(x)$ be the quantity of prime numbers less that $x$. Then

$$|\Bbb P_k|\simeq\pi(n^{\frac 1k})-\pi(n^{\frac 1{k+1}})\simeq$$

$$\frac{n^{\frac 1k}}{\ln n^{\frac 1k}}-\frac{n^{\frac 1{k+1}}}{\ln n^{\frac 1{k+1}}}=$$ $$\frac{kn^{\frac 1k}-(k+1)n^{\frac 1{k+1}}}{\ln n}.$$

Then $$\sum_{k=1}^\infty |\Bbb P_k|\simeq \frac{n}{\ln n},$$

which yields an upper bound $n^{\frac{n}{\ln n}}=e^{\ln n{\frac{n}{\ln n}}}=e^n$ for $f(n)$.

And $$\sum_{k=1}^\infty |\Bbb P_k|\frac{k}{k+1}\ge \frac{n}{2\ln n},$$

which yields a lower bound $e^{n/2}$ for $f(n)$.