Division of space by hyper-planes

Let's denote by $S_{k,n}$ the set of possible integers $m$, such that $\mathbb R^k$ can be divided into $m$ regions by $n$ hyperplanes. If we denote by $S^{P}_{k,n}$ the set defined similarly but for the projective space $\mathbb {RP}^k$. We have that $S_{k,n}=S^P_{k,n+1}$ since every affine arrangement can be lifted to a projective arrangement with the same number of regions by adding the hyperplane at infinity. Similarly any projective arrangement gives rise to affine arrangements with the same number of regions by deleting one of the hyperplanes. The following result solves the problem for $k=2$:

Theorem: (N. Manturov, "Classification of arrangements by the number of their cells") We have $m\in S^P_{2,n}$ if and only if there exists some integer $0\le r\le n-2$ such that $$(n-r)(r+1)+\binom{r}{2}-\min\bigg\{ n-r, \binom{r}{2}\bigg\}\le m\le (n-r)(r+1)+\binom{r}{2}.$$

The proof in that paper involves quite a bit of casework and it seems like it wouldn't easily generalize to higher dimensions, however I find one particular aspect very interesting. The way it is shown that all numbers satisfying these inequalities do work is by exhibiting a very simple family of arrangements $\mathbb B^c_{a,b}$ which are obtained as the union of a pencil of $a$ lines (all passing through the same point), a simple arrangement of $b$ lines (no three lines concurrent), such that $c$ of the intersection points of the simple arrangement lie on the first pencil. This leads us to a (possibly too optimistic?) conjecture:

Conjecture: If an arrangement of $n$-hyperplanes in $\mathbb {RP}^k$ has $m$ regions, then there exists an arrangement of $n$-hyperplanes, obtained as the union of $n$ generalized pencils (one for each type), which also has $m$ regions.

By a generalized pencil, I mean taking two disjoint projective plane planes of dimension $r_1,r_2$ in a $(r_1+r_2+1)$-dimensional projective space, picking a generic arrangement of hyperplanes in the first space, and taking the span of each hyperplane with the second projective space. We say that the pencil has type $(r_1,r_2)$. If this conjecture were to be true, there would be some hope of a proof strategy along the lines of starting with an arbitrary arrangement and somehow deforming it into such a form while preserving the number of regions.

The motivation of the question was the observation that the least number of lines in the plane that give five regions is $n=4$ parallel lines. Surprisingly, as long as $m \gt 5$ and $m \leq \binom{n}2+n+1$, the maximum number of regions possible using $n$ lines, there is a division of the plane into $m$ regions using no more than $n$ lines.

I suspected this based on this easy observation (and some computation):

If the plane is divides by $t$ parallel classes of lines with sizes $a_1,\cdots,a_t$, no three lines concurrent, then the plane is split into $\frac{n^2-s}2+n+1$ regions where $n=\sum a_i$ and $s=\sum a_i^2.$

The nice construction by Manturuv does capture some small numbers of regions which this does not, but those attained are enough to establish the result claimed above up to $n=50$ except for $m=5$ and $m=17.$ Naively running over all partitions of $n$ becomes time consuming.

So for $k=2$ then set of attainable number of regions is the union of intervals. Once $n-r \leq \binom{r}2$ we have an interval starting at $$(n-r)(r+1)+\binom{r}{2}-(n-r)=nr-\binom{r+1}2$$ with the previous interval ending at $$(n-r+1)r+\binom{r-1}2=nr-\binom{r+1}2+1.$$ Since this happens near $r=\sqrt{2n},$ we can certainly attain any number of regions $n\sqrt{2n} \leq m \leq n\frac{n+1}2+1$ using exactly $n$ lines.