Size of the least common multiple Suppose $k \in \mathbb{N}$ is fixed and that $n, n_1, \dots, n_k \in \mathbb{N}$ are given such that $n = n_1 + \dots + n_k$.
Are there any known upper bounds for $\text{lcm}(n_1, \dots, n_k)$ in terms of n, that take $k$ into account as well?
I care up to coarse equivalence. Given two non-decreasing functions $f, g \colon \mathbb{N} \to \mathbb{N}$, we say that $g$ coarsely dominates $f$, write $f \preceq g$, if there exists $C>0$ such that $f \leq C g(Cn)$ for all $n \in \mathbb{N}$. We say that $f,g$ are coarsely equivalent, write $f \approx g$ if $f \preceq g$ and $g \preceq f$. Note that this notion is really coarse, for example $n2^n \preceq 2^n$.
EDIT:Landau's function does work, but it is rather general, as it allows any number of summands.
Ideally, for each $k \geq 1$ I would like a to have a function $B_k(n)$ given by a closed form formula that coarsely dominates $\text{lcm}(n_1, \dots, n_k)$ and is tight up to coarse equivalence. For example, $B_1(n) = n$, $B_2(n) = n^2$.
 A: With $k$ fixed, an asymptotically tight upper bound is
$$B_k(n) = \frac{n^k}{k^k}\,.$$
First, it is immediate that $\operatorname{lcm}(n_1, \dotsc, n_k) \leqslant n_1\cdot \ldots \cdot n_k$, and
$$\prod_{j = 1}^{k} x_j$$
with the constraints $x_j \geqslant 0$ and $\sum x_j = n$ is maximised for $x_j = \frac{n}{k}$, $1 \leqslant j \leqslant k$. Thus $B_k(n)$ as given above is an upper bound.
For $k = 1$ it is trivial that this is not only asymptotically tight, but in fact an equality. Thus consider $k \geqslant 2$. Let $\varepsilon > 0$ be small. For sufficiently large $n$, we can find at least $k-1$ primes between $e^{-\varepsilon}\cdot \frac{n}{k}$ and $\frac{n}{k}$. Pick the largest $k-1$ of these primes as $n_1, \dotsc, n_{k-1}$ and put $n_k = n - n_1 - \ldots - n_{k-1}$. Then
$$\frac{n}{k} \leqslant n_k \leqslant n\cdot\biggl(1 - e^{-\varepsilon}\frac{k-1}{k}\biggr) = e^{-\varepsilon}\cdot \frac{n}{k}\cdot \bigl(1 + k(e^{\varepsilon}-1)\bigr) < 2e^{-\varepsilon}\cdot \frac{n}{k}$$
(we need $e^{\varepsilon} < 1 + \frac{1}{k}$ for this), and hence the $n_j$ are coprime. Therefore
$$\operatorname{lcm}(n_1,\dotsc,n_k) = \prod_{j = 1}^{k} n_j \geqslant e^{-(k-1)\varepsilon}\cdot \frac{n^k}{k^k}\,.$$
Since we can choose $\varepsilon > 0$ arbitrarily small, it follows that $B_k(n)$ is asymptotically tight.
