What's the first non-planar graph of the first $n$ numbers where edges show divisibility?

Question

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.

Answer 1

$G_{14}$ 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.