\section{Numerology}
\label{section-numerology}

We first utilize some facts about graded free modules in order to give constraints on potential splitting types.  Given a graded free module
\[ M = R(e_1) \oplus ... \oplus R(e_r) \]
one can observe that the Hilbert polynomial $H_M$ is given by
\[H_M(m) = rm + e_1 + ... + e_r + r \]
Let $\varphi$ denote a free morphism into $X$.  Noting that the map $\varphi: R(-d)^{\oplus n+1}_m \longrightarrow R_m$ is surjective for $m \gg 0$, then for $m$ sufficiently large,
\begin{align*}
H_{\Omega(\varphi)}(m) &= \dim_\mathbb{K}\left(\ker(R(-d)^{\oplus n+1}_m \longrightarrow R_m)\right) \\
&= (n+1)(-d + m + 1) - (m+1) \\
&= nm + -d(n+1) + n
\end{align*}
A similar calculation shows that,
\[H_{E_X(\varphi)}(m) = nm + d(n+1-e) + n\]
We continue to refer to the splitting type components of $\Omega(\varphi)$ respectively $E_X(\varphi)$ as $e_i$ respectively $f_i$.  In both cases $n=r=5$, so combining these two equations with the general form for the Hilbert polynomial of a free graded module, we obtain our first constraints:
\begin{align}
e_1 + e_2 + e_3 + e_4 + e_5 &= -6d \\
f_1 + f_2 + f_3 + f_4 + f_5 &= d
\end{align}
The curve is free respectively very free if for each $i$, $f_i \geq 0$ respectively $f_i > 0$.  Since $f_i = 4e_i + 5d$, it follows that
\begin{align}
e_i \geq -\frac{5d}{4}
\end{align}
where strict inequality yields a very free curve.  With these two bounds, we can quickly observe a few facts about curves of different degrees.

\begin{remark}
\label{charlie}
\begin{enumerate}
\item
There exist no free curves in degrees $1,2,3,6,$ and $7$.
\item
There are no free curves that are not very free in degrees not divisible by $4$.
\item
There are no very free curves in degrees $4$ or $8$.
\item
The splitting type of a free curve of degree $4$ must be $(e_1, e_2, e_3, e_4, e_5) = (-5,-5,-5,-5,-4)$.
\item
The splitting type of a very free curve of degree $5$ must be $(e_1, e_2, e_3, e_4, e_5) = (-6,-6,-6,-6,-6)$.
\end{enumerate} 
\end{remark}

\noindent All of these observation follow directly from the two constraints.  For example, in degree $6$, $e_1 + e_2 + e_3 +e_4 + e_5 = -6d = -36$.  However, each $e_i \geq \frac{-30}{4} = -7.5$.  So even if each $e_i$ is at best $-7$, the $e_i$ cannot sum to $-35$.  The rest of the remarks follow in a similar manner.  Note that one can glean even more information about these curves from the constraints, but the ones listed above are sufficient for our purposes.
