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.
Best Answer
This is the special case $(p,q,r)=(1,2,3)$ of the $3$-term Gale-Robinson recurrence: $$x_{n+p+q+r} x_n = x_{n+p} x_{n+q+r} + x_{n+q} x_{n+p+r} + x_{n+p+q} x_{n+r}$$
Fomin and Zelevinsky proved that, treating the initial values as formal variables, all the $x_n$ are Laurent polynomials with integer coefficients in $(x_0, x_1, \cdots x_{p+q+r-1})$. Therefore, any prime which occurs in the denomintator of some $x_n$ must divide one of the initial values. In your case, that shows that the only prime which can occur in the denominator is $2$.
The sequence seems to have a lot of periodicity modulo powers of $2$, but I haven't yet found a specific statement which I can inductively prove to make sure the denominator never has a $2$ in it.
Here is a proof subject to an unproved relation. According to my laptop, for $n \leq 100$, we have $$a x_n x_{n+48} + b x_{n+6} x_{n+42} + c x_{n+12} x_{n+36} + d x_{n+18} x_{n+30} + e x_{n+24}^2 =0$$ where $(a,b,c,d,e)$ is
The only thing you need to see here is that $(a,b,c,d,e) \equiv (1,1,0,1,1) \mod 2$.
This shows inductively, if $x_{6n}$ is odd for $0 \leq n \leq 7$ (and it is), then $x_{6n}$ is odd for all $n$. Similarly, $x_{6n+1}$, $x_{6n+2}$ and $x_{6n+3}$ are always odd.
The remaining two residue classes are harder, but I found them by doubling my spacing: $$f x_n x_{n+96} + g x_{n+12} x_{n+84} + h x_{n+24} x_{n+72} + i x_{n+36} x_{n+60} + j x_{n+48}^2 =0$$ with $(f,g,h,i,j)$ equal to
(Checked for $n \leq 200$.) Again, all that matters is that $(f,g,h,i,j)$ are $(1,0,0,0,1) \bmod 2$. This lets us show inductively that, if the first $7$ values of $x_{12n+4}$ are always $2 \bmod 4$, then they all are. Similarly, $x_{12n+5}$ is always $2 \bmod 4$, $x_{12n+10}$ and $x_{12n+11}$ are always $4 \bmod 8$.
There is obviously a lot of mystery going on with this bilinear relations. I know that they have something to do with $\theta$ functions for abelian surfaces, but the details don't really seem to be recorded anywhere.