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.
