# $\mathrm{lcm}(1, 2, 3, \ldots, n)$?

I want to find $\mathrm{lcm}(1, 2, 3, \ldots, n)$ where $2 \le n \le 10^8$ .

I'm trying to find a formula .

I claim that

$${\rm lcm}(1,2,3,\cdots,n)=\prod_p p^{\large\lfloor\log_p n\rfloor}.$$

Let's review what an LCM is: we say $\ell$ is an LCM of the elements of a finite set $S$ precisely if it has the following "universal property" (the GCD similarly has such a definition):

$$s\mid m~{\rm for~all~}s\in S ~~\iff~ \ell\mid m.$$

Which is to say that if we create a partial order on the set of things which are simultaneously multiples of everything in $S$, then $\ell$ is minimal in this collection. Now, call the product I defined earlier $\Pi$. Suppose $s\mid m$ for each $1\le s\le n$ and some integer $m$. Then in particular, for each prime number $p$, the highest power of $p$ that is $\le n$ (which is $p^{\lfloor\log_pn\rfloor}$ - check) will divide $m$, and hence (since all these things are coprime), their product, which is $\Pi$, will divide $m$. Thus

$$s\mid m{\rm ~for~all~}s\in S\implies \Pi\mid m.$$

Now we only need the reverse direction to show that $\Pi$ is an LCM. Thus, suppose $\Pi\mid m$ and let $1\le s\le n$. Then $s=\prod_p p^{e_p}$ has a unique prime factorization, and $e_p\le \lfloor\log_pn\rfloor$, so that in fact $p^{e_p}\mid p^{\lfloor\log_pn\rfloor}$ for each $p$, and hence $s\mid\Pi\mid m\implies s\mid m$. This finishes the proof.

Another possible formula is the exponential of the summatory Mangoldt function,

$${\rm lcm}(1,2,3,\cdots,n)=\exp\psi(n).$$

The reason this might be interesting is because of the connection between $\psi$ and the PNT.

This is OEIS A03418 where the formula $$a(n)=\prod p^{\lfloor(\log n/\log p\rfloor}$$ where p runs through primes not exceeding n is given, also

Recursive formula useful for computations: $a(0)=1; a(1)=1; a(n)=\operatorname{lcm}(n,a(n-1))$