TITLE: Does this sequence always give an integer?
QUESTION [23 upvotes]: It is known that the $k$-Somos sequences always give integers for $2\le k\le 7$.
For example, the $6$-Somos sequence is defined as the following : 
$$a_{n+6}=\frac{a_{n+5}\cdot a_{n+1}+a_{n+4}\cdot a_{n+2}+{a_{n+3}}^2}{a_n}\ \ (n\ge0)$$
where $a_0=a_1=a_2=a_3=a_4=a_5=1$.
Then, here is my question.
Question : If a sequence $\{b_n\}$ is difined as 
$$b_{n+6}=\frac{b_{n+5}\cdot b_{n+1}+b_{n+4}\cdot b_{n+2}+{b_{n+3}}^2}{b_n}\ \ (n\ge0)$$$$b_0=b_1=b_2=b_3=1, b_4=b_5=2,$$
then is $b_n$ an integer for any $n$?
We can see 
$$b_6=5,b_7=11,b_8=25,b_9=97,b_{10}=220,b_{11}=1396,b_{12}=6053,b_{13}=30467$$
$$b_{14}=249431,b_{15}=1381913,b_{16}=19850884,b_{17}=160799404$$
$$b_{18}=1942868797,b_{19}=36133524445, \cdots.$$
Remark : This question has been asked previously on math.SE without receiving any answers.
Motivation : I've been interested in seeing what happens when we change the first few terms. It seems true, but I can neither find any counterexample nor prove that the sequence always gives an integer. As far as I know, it seems that this question cannot be solved in the way which proved that the $6$-Somos sequence always gives an integer.

REPLY [7 votes]: Andrew Hone in the articles Analytic solutions and integrability for bilinear recurrences of order six and Sigma-function solution to the general Somos-6 recurrence via hyperelliptic Prym varieties (with Yuri N. Fedorov) considered   general Somos-6 recurrence
$$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$
with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form
$$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$  From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907))
$$\dfrac{\sigma(\mathbf{u}+\mathbf{v})
\sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}=
\wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+
\wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$
follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$
It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes.
Let's apply this theory to the given sequence. (From this point I'll follow David Speyer's solution.)
Taking two $5$-tuples of $m$'s and $n$'s $(m,22,21,20,19)$ and $(24,18,12,6,0)$ from 
$$\left|
\begin{array}{ccccc}
 \tau_{m-24} \tau_{m+24} & \tau_{m-18} \tau_{m+18} & \tau_{m-12} \tau_{m+12} & \tau_{m-6} \tau_{m+6} & \tau_m^2 \\
 \tau_{22-24} \tau_{22+24} & \tau_{22-18} \tau_{22+18} & \tau_{22-12} \tau_{22+12} & \tau_{22-6} \tau_{22+6} & \tau_{22}^2 \\
 \tau_{21-24} \tau_{21+24} & \tau_{21-18} \tau_{21+18} & \tau_{21-12} \tau_{21+12} & \tau_{21-6} \tau_{21+6} & \tau_{21}^2 \\
 \tau_{20-24} \tau_{20+24} & \tau_{20-18} \tau_{20+18} & \tau_{20-12} \tau_{20+12} & \tau_{20-6} \tau_{20+6} & \tau_{20}^2 \\
 \tau_{19-24} \tau_{19+24} & \tau_{19-18} \tau_{19+18} & \tau_{19-12} \tau_{19+12} & \tau_{19-6} \tau_{19+6} & \tau_{19}^2 \\
\end{array}
\right|=0$$
we get his first recurrence which proves that  $\tau_{6n}$, $\tau_{6n+1}$, $\tau_{6n+2}$, $\tau_{6n+3}$ are always odd.
Taking two $5$-tuples of $m$'s and $n$'s as $(m,45,44,43,42)$ and $(48,36,24,12,0)$ we get his second formula which proves that $2$-adic valuation has period $24$. 
(Thanks to David Speyer who found a gap in first version of this answer.)