least common multiple of $\{1,2,...,n\}$ is bigger than $2^{n-1}$

Question

The least common multiple of $\{1,2,...,n\}$ is greater than $2^{n-1}$ for any $n \ge 3$.

I found this in a MATHEMATICA book, but I don't know how to prove this. Can you help me?

[Edit: This thread has a discussion of an asymptotic stronger result, but that relies on the Prime Number Theorem. What else is known about this? JL]

Answer 1

Here is a neat proof:

Consider the product $$\text{lcm}(1, 2, \dots, 2n + 1)\int_0^1 (x)^n(1 - x)^n \text{d}x$$ This is an integer, because expanding using the binomial theorem and integrating each part gives an integer. Additionally, the integral is less than $$\int_0^1 \left(\frac14\right)^n \text{d}x = \frac{1}{4^n}$$ so $$\text{lcm}(1, 2, \dots, 2n + 1) > 4^n = 2^{2n}$$ which proves the desired inequality for odd values.

To prove it for even values, consider the product $$\text{lcm}(1, 2, \dots, 2n)\int_0^1 (x)^{n}(1-x)^{n-1} \text{d}x$$ This product is also an integer. Additionally, the integral is less than $$\int_0^1 x\left(\frac14\right)^{n-1}\text{d}x = \frac{1}{2\cdot4^{n-1}}$$ so we arrive at $$\text{lcm}(1, 2, \dots, 2n) > 2\cdot4^{n-1} = 2^{2n - 1}$$ which proves the desired result for even values.

Answer 2

This paper proves the identity $$ \operatorname{lcm}(1,2,\dots,n)=n \operatorname{lcm}\left(\binom{n-1}{0}, \binom{n-1}{1},\dots,\binom{n-1}{n-1}\right) $$ by computing the number of factors of $p$ which appear in each expression, for all primes $p$.

From this it follows that $$ \operatorname{lcm}(1,2,\dots,n) \geq n \binom{n-1}{\lfloor (n-1)/2 \rfloor} \geq \sum_{k=0}^{n-1} \binom{n-1}{k}=2^{n-1} $$

Answer 3

(Updated due to lhf's comment).

In a few words,

$$ LCM(1,2,...,n) = e^{\psi(n)}, $$ where $\psi(n)$ is second Chebyshev function (see formula here).

In fact, it is enough to prove that $$ \psi(n) > (n-1)\ln 2 \approx 0.693(n-1).\tag{1} $$

Function $\psi(n)$ has asymptotic $\psi(n) \sim n$. Using lower bound for $\psi(n)$ $$ \psi(n)>0.916n−2.318\tag{2} $$ (see discussion here and paper here, Lemma $2$, p.$179$) we get $(1)$ immediately for $n\ge 8$.

And it remains to check $(1)$ manually for $n=1,2,...,7$. It is shown in this table: $$ \begin{array}{|c|c|c|} \hline n & \psi(n) & (n-1)\ln 2 \\ \hline 2 & 1.79176 & 0.693147 \\ 3 & 2.48491 & 1.38629 \\ 4 & 4.09434 & 2.07944 \\ 5 & 4.09434 & 2.77259 \\ 6 & 6.04025 & 3.46574 \\ 7 & 6.73340 & 4.15888 \\ \end{array} $$

Answer 4

Let $n\in \mathbb{N}^*$.
$\bullet$ If $n=2k+1$, where $k \in \mathbb{N}$. We set $I_n=\sum_{j=0}^{k}\frac{(-1)^j\binom{k}{j}}{k+j+1}$. $k+1,...,2k+1$ are divisors of $\text{lcm}\left( 1,...,n \right)$, so $\text{lcm}\left( 1,...,n \right)I_n \in \mathbb{Z}$. We have $$I_n=\sum_{j=0}^{k}(-1)^j\binom{n}{k}\int_{0}^{1}t^{k+j}dt=\int_{0}^{1}t^k(1-t)^kdt$$ On the othere hand, $I_n>0$, hence $\text{lcm}\left( 1,...,n \right)I_n \ge 1$. Moreover $\sup_{0\le t\le1}t^k(1-t)^k=\frac{1}{2^{2k}}$ so $\text{lcm}\left( 1,...,n \right)\ge \frac{1}{I_n}\ge 2^{2k}=2^{n-1}$.
$\bullet$ If $n=2k$, where $k \in \mathbb{N}^*$. We set $J_n=\sum_{j=0}^{k}\frac{(-1)^j\binom{k}{j}}{k+j}$. $k,...,2k$ are divisors of $\text{lcm}\left( 1,...,n \right)$, so $\text{lcm}\left( 1,...,n \right)J_n \in \mathbb{Z}$. We have $$J_n=\sum_{j=0}^{k}(-1)^j\binom{k}{j}\int_{0}^{1}t^{k-1+j}dt=\int_{0}^{1}t^{k-1}(1-t)^kdt$$ On the othere hand , $J_n>0$, hence $\text{lcm}\left( 1,...,n \right)J_n \ge 1$. Moreover $\sup_{0\le t\le1}t^{k-1}(1-t)^k=\left( \frac{k-1}{2k-1} \right)^{k-1}\left( \frac{k-2}{2k-1} \right)^{k}\le \frac{1}{2^{2k-1}}$ so $\text{lcm}\left( 1,...,n \right)\ge \frac{1}{J_n}\ge 2^{2k-1}=2^{n-1}$.