Does a graded poset on $\mathbb{N}_{>0}$ generated from subtracting factors define a lattice? Consider the partial ordering of positive integers with covering relations $n - \frac np \lessdot n$ for all prime divisors $p \mid n$. This defines a graded poset with $A064097$$(n)+ 1$ rank levels and a unique minimal element, $1$.
I'd like to know a bit more about these posets:


*

*Have these posets been studied?

*Is this poset a lattice? A distributive lattice? A semimodular lattice?

*If this post is a join-semilattice, it appears that $n \vee k$ divides $\operatorname{lcm}(n,k)$. Is this true and are there any nice properties of $$\frac{\operatorname{lcm}(n,k)}{n \vee k}?$$

*Does every interval $[1, n]$ have the Sperner property?

*Does this poset or its intervals have any other nice properties?


Example
An example of a descending saturated chain from $15$ to $1$ is $$
15 \gtrdot \underbrace{15 - \frac{15}{3}}_{12} \gtrdot \underbrace{12 - \frac{12}{2}}_{6} \gtrdot \underbrace{6 - \frac{6}{3}}_3 \gtrdot \underbrace{3 - \frac 31}_{2} \gtrdot \underbrace{2 - \frac 22}_1
$$
More generally, the Hasse diagram of the interval $[1,15]$ is

(Image from Michael De Vlieger. Click the image to see examples of $[1,n]$ for $n \leq 211$.)

Related OEIS Sequences


*

*A333123: Number of descending saturated chains from $n$ to $1$.

*A334184: Size of the rank levels of the poset.

*A332809: Size of the interval $[1, n]$.

 A: Yes, your poset is a lattice, and here is why.
I will denote your poset by $K$ and its order by $\leq_K$. For any prime $p$, let $\lambda_p=\frac{p}{p-1}$. Notice that $p$ appears in $\lambda_p$ but not in any of $\lambda_1,\ldots,\lambda_{p-1}$, so the $\lambda_p$'s are multiplicatively independent.
If we denote by $\Lambda$ the poset whose elements are finite products of $\lambda_p$'s and the covering relations are given by $x \lessdot \lambda_p x$, it follows that $\Lambda$ is isomorphic to the set of finite sequences in $\mathbb N$, with the usual product order ($(u_k) \leq (v_k)$ iff $u_k\leq v_k$ for all $k$), and this last poset is clearly a lattice.
By the definition of $K$, the inclusion $i:K \to \Lambda$ is a poset homomorphism, i.e. $x\leq_K y \Rightarrow x\leq_{\Lambda} y$. But the converse is also true :
Lemma. $i$ is a isomorphism, i.e. $x\leq_{\Lambda} y \Rightarrow x\leq_K y$ when $x$ and $y$ are integers.
Proof of lemma. Suppose that $x\leq_{\Lambda} y$ for $x,y\in K$. Then, there is an increasing sequence $p_1\lt p_2 \lt\ldots \lt p_t$ of primes, and exponents $e_1,\ldots,e_t$ such that $y=\lambda_{p_t}^{e_t}\ldots\lambda_{p_1}^{e_1}x$. If we put $A=(p_t-1)^{e_{t-1}}\ldots(p_1-1)^{e_1}$ and $B=p_{t-1}^{e_{t-1}}\ldots {p_1}^{e_1}$, then $x=\frac{Ay}{{p_t}^{e_t}B}$ but $p_t$ does not divide $A$, so ${p_t}^{e_t}$ must divide $y$. Then $y'=\frac{y}{\lambda_{p_t}^{e_t}}$ is an integer, with $y' \leq_K y$, and the result is now clear by induction on $t$.
Thus $K$ is an induced subposet of $\Lambda$. It also stable by meet and join :
if $x,y\in K$ then $x\vee_{\Lambda} y$, $x\wedge_{\Lambda} y$ are in $K$ (this is because $\prod_{p\in P}\lambda_p^{\min(x_p,y_p)}$  and $\prod_{p\in P}\lambda_p^{\max(x_p,y_p)}$  is an integer if $\prod_{p\in P}\lambda_p^{x_p}$ and $\prod_{p\in P}\lambda_p^{y_p}$ are). This finishes the proof.
