Please explain the features of this plot of the squareness ratio $r(n)$ versus $n$.
I define the squareness of a natural number $n$ to be the closest its factors can be partitioned into a balanced ratio of $1$. A perfect square has squareness $1$. A prime $p$ has squareness $1/p$. In a sense, the squareness measures how close is $n$ to a perfect square.
Example. The squareness ratios for the first ten number $n=1,2,\ldots,10$ are $$1,\frac{1}{2},\frac{1}{3} ,1,\frac{1}{5},\frac{2}{3},\frac{1}{7},\frac{1}{2},1,\frac {2}{5}$$
Example. $n=1032 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 43$. One can partition its $5$ factors into two parts which have products whose ratios are $$ \left\{\frac{1}{1032},\frac{1}{2 58},\frac{3}{344},\frac{2}{12 9},\frac{3}{86},\frac{8}{129} ,\frac{6}{43},\frac{24}{43}\right\} $$ with $\frac{24}{43} \approx 0.558$ the largest ratio, its squareness.
Example. For $n=12600=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7$, the largest ratio is $$\frac{3 \cdot 5 \cdot 7}{2^3 \cdot 3 \cdot 5}=\frac{7}{8}=0.875 \;.$$
Among this plot's evident features are: straight rays from the origin, hyperbolas, discernable density change at $r=\frac{1}{2}$, interesting patterns near $r=1$. There is more structure here than I anticipated.
(Some detail is lost converting the image for posting.)
Added. Riffing on PattuX's idea, for a prime, $n=p$, all the numbers $k n$ for $k=1,2,\ldots,p$ have squareness ratios $k/p$. For example, for $n=17$, $$n = 17,34,51,68,85,102,119,136,153 ,170,187,204,221,238,255,272, 289$$ have squareness $$\frac{1}{17},\frac{2}{17} ,\frac{3}{17},\frac{4}{17},\frac{5}{17},\frac{6}{17}, \frac{7}{17},\frac{8}{17},\frac{9} {17},\frac{10}{17},\frac{11}{ 17},\frac{12}{17},\frac{13}{1 7},\frac{14}{17},\frac{15}{17 },\frac{16}{17},1 $$ and so all lie on a line through the origin.
Amazing stuff, very interesting.
If we say $r(x) = \frac{p}{q} < \frac{1}{a}$ and $a>b$, then $r(bx) = \frac{bp}{q}\space \space \space \space \space \space \space \space \space \space (a,b \in \mathbb{N})$
Proof: Assume $p=n_1 n_2 ... n_k, \space q=m_1 m_2 ... m_l$, so that $\frac{p}{q} \le1$ is maximal.
Then $\forall i,j:n_i<m_j \iff \frac{n_i}{m_j} < \frac{1}{a^{2}}$ because otherwise swapping $n_i$ and $m_j$ would result in $\frac{p}{q}$ incresing, but still be less than 1.
If we now say $p=bn_1 n_2 ... n_k, \space q=m_1 m_2 ... m_l$, we have to show that this is still optimal.
Assume $r(bx)\ne \frac{bp}{q} \implies \exists j: b<m_j, \frac{b}{m_j} \ge \frac{1}{a^{2}} \iff ba^{2}\ge m_j $
Swapping $b$ and $m_j$ effectivelty multiplies the fraction by $\frac{m_j^{2}}{b^{2}}$, therefore $\frac{m_j^{2}}{b^{2}} < a \implies m_j^{2}<b^{2}a$
Multiplying those together: $m_j^{3} < (ab)^{3} \implies m_j < ab < a^{2}$.
But we said that $\frac{n_i}{m_j} < \frac{1}{a^{2}}$. We can say $n_i = 1$ (as we didn't assume it is prime factorization), which yields $\frac{1}{m_j} < \frac{1}{a^{2}} \implies m_j > a^{2}$
[I'm not 100% sure this proof is correct, so please confirm this for yourself. However, even if it is not true for all numbers, it is true for most numbers, which is sufficient to create a pattern]
So $r(x)<\frac{1}{a} \implies r(bx) = b*r(x) \space \space \space (a>b)$. This may explain the partial linearity.