If $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$, then is it always divisible by $a_{n-4}$?

Question

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.

Answer 1

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 $n=5,6,7,8$.