1

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, $B_1(n) = n$, $B_2(n) = n^2$.

Michal Ferov
  • 370
  • 1
  • 10
  • 2
    Does this answer your question? Maximal order of an element in a symmetric group – Václav Mordvinov Nov 11 '20 at 16:05
  • In the question I linked, you can find that this $\mathrm{lcm}$ is bounded by $\exp(n/e)$. – Václav Mordvinov Nov 11 '20 at 16:06
  • @VáclavMordvinov it does and at the same time it does not. It is certainly a valid upper bound, but it is bit more general than what I asked. I edited the question to make it less vague. – Michal Ferov Nov 11 '20 at 16:26
  • The question is are there any known upper bounds and the answer is yes, e.g. a trivial upper bound such as $n!$. If you ask for a tight upper bound, then the answer is again yes: just define $B_k(n)$ as the $\max$ of the $\operatorname{lcm}$ for all such $k$-tuples (there are only finitely many of them). – WhatsUp Nov 11 '20 at 16:47
  • 2
    From your example of $B_2(n)$, it seems that you only want some upper bound that is tight up to a big O constant: it is clear that $B_2(n)$ can be improved to $\lceil n/2\rceil^2$. If this is the intention, then you may want to add it to your question. – WhatsUp Nov 11 '20 at 16:49
  • 1
    @WhatsUp Indeed, I'm interested in upper bounds up to coarse equivalence. I added it to the question. Originally I did not want to overload the question by being overly specific with the notation, but that turned out to be rather counterproductive - my bad. – Michal Ferov Nov 11 '20 at 17:16

1 Answers1

2

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.

Daniel Fischer
  • 206,697