Explaining symmetry in sequences defined by $\frac{a_{n+1}}{n+1}=\frac{a_n}{n}-\frac12$. (Eg, for $a_1=5$, we get $5,9,12,14,15,15,14,12,9,5$) I've noticed recently the following fact: consider ratios $\frac{a_n}{n}$ with $n = 1,\dots,10$. If we require that
$$
\frac{a_{n+1}}{n+1} = \frac{a_n}{n} - \frac12
$$
that is, each successive ratio drops by $\frac12$ in value, then for $a_1 = 5$ we get $a_2 = 9$, $a_3 = 12$, $a_4 = 14$, $a_5 = 15$, $a_6 = 15$, $a_7 = 14$, $a_8 = 12$, $a_9 = 9$ and $a_{10} = 5$. That is, the numerators show a symmetric behvaior $a_{11 - n} = a_n$. This looks pretty random to me, so I wonder whether there's something obvious I've missed why such symmetry may arise, and perhaps a more general case where it happens exists.

$b_n = \dfrac{a_n}{n}\,$ is an arithmetic progression with starting value $\,a_1\,$ and common difference $\,-\dfrac{1}{2}\,$, so:
$$
b_n = \frac{2a_1 - n + 1}{2} \quad\implies\quad a_n = n \cdot \frac{2a_1 - n + 1}{2}
$$
For $\,2a_1 \in \mathbb N\,$, it follows that:
$$
\require{cancel}
\begin{align}
a_{2a_1+1 - n} &= (2a_1+1 - n) \cdot \frac{\cancel{2a_1} - (\cancel{2a_1}+\bcancel{1} - n) + \bcancel{1}}{2}
\\ &= n \cdot \frac{2a_1 - n + 1}{2}
\\ &= a_n
\end{align}
$$
OP's problem is the case $\,a_1=5\,$.

[ EDIT ] More generally, consider the recurrence $\,\dfrac{a_{n+1}}{n+1}=\dfrac{a_n}{n} - a\,$, then $b_n = \dfrac{a_n}{n}\,$ is an arithmetic progression with starting value $\,a_1\,$ and common difference $\,-a\,$, so:
$$
b_n = a_1 - (n - 1)a \quad\implies\quad a_n = n \cdot \left(a_1 - (n - 1)a\right)
$$
For $\,\dfrac{a_1}{a} \in \mathbb N\,$, it follows that:
$$
\require{cancel}
\begin{align}
a_{a_1/a + 1 - n} &= \left(\frac{a_1}{a} + 1 - n\right) \cdot \left(\cancel{a_1} - \left(\cancel{\frac{a_1}{a}} + \bcancel{1} - n - \bcancel{1}\right)a\right)
\\ &= n \cdot \left(a_1 - (n-1) a\right)
\\ &= a_n
\end{align}
$$
This reduces to the previous case when $\,a = \dfrac{1}{2}\,$.
