<s>[INST] 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.
 [/INST] 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 $n=5,6,7,8$.
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