TITLE: What's the first non-planar graph of the first $n$ numbers where edges show divisibility?
QUESTION [6 upvotes]: Let $G_n$
be a graph with vertices $v_1, v_2, v_3,\dots, v_n$,
where there is an edge between $v_i$
and $v_j$
if and only if either $v_i$
divides $v_j$
or vice versa.
Is there a value $n$
such that $G_n$
is non-planar,
ie cannot be drawn in the plane without intersecting edges?
If so, what is the least value of $n$
for which this happens?
By way of example,
the drawing below shows that
$G_{12}$ is planar.

REPLY [5 votes]: $G_{14}$ is planar: From your $G_{12}$ sketch, relocate $7$ into the $1{-}2{-}6$ triangle. This allows you to add $14$ with edges to $1$, $2$, an $7$. And the prime $13$ can be placed anywhere in the outer region.
One readily sees that the five numbers $1,2,4,8,16$ form a $K_5$, hence $G_{16}$ is not planar.
So what about $G_{15}$? It turns out, we would have a $K_{3,3}$ with vertices $1,2,3$ and $6,12,m$ if there were another common multiple of $2 $ and $3$ available. While there is no such $m$, we have $5$, which is indirectly linked to $3$ via $15$ and to $2$ via $10$.  Therefore $G_{15}$ is not planar.