If $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$, then is it always divisible by $a_{n-4}$? The Somos Sequences are neat because they are integral yet contain a division operation. 
If we consider the numerator $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$ of the Somos-4 sequence, $a_1 = a_2 = a_3 = 1$, then is it always divisible by $a_{n-4}$?
Note that this is not obvious, since integrality in the orignal Somos-4 sequence implies that $a_{n-4}$ divides the numerator, $a_{n-4}$ and the numerator each built with respect to that sequence's previous elements and not with respect to the simpler sequence above.

No. Working out the first few terms of the recurrence $a_n = a_{n-1} a_{n-3} + a_{n-2}^2$, we get
$$
\begin{array}{c|c}
n & a_n \\ \hline
1 & 1\\
2 & 1\\
3 & 1\\
4 & 2\\
5 & 3\\
6 & 7\\
7 & 23\\
8 & 118\\
9 & 1355\\
\end{array}
$$
from which we see $a_{5} = 3$ and $a_9 = 1355$, but $3 \nmid 1355$.
In fact for $n \leq 50$, $a_{n-4} \mid a_n$ only for $n=5,6,7,8$.
