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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 $$b_6=5,b_7=11,b_8=25,b_9=97,b_{10}=220,b_{11}=1396,b_{12}=6053,b_{13}=30467$$ $$b_{14}=249431,b_{15}=1381913,b_{16}=19850884,b_{17}=160799404$$ $$b_{18}=1942868797,b_{19}=36133524445, \cdots.$$

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.

Update : I crossposted to MO.

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Can't you just pick arbitrary $a_1,...,a_5$ and then choose $a_0$ not to be a divisor of $a_5\cdot a_1+a_4\cdot a_2+a_3^2$ to provide a counterexample? –  Christoph Sep 26 '13 at 12:34
    
In the "similar sequence defined", what is $\;a_n\;$ ?? –  DonAntonio Sep 26 '13 at 13:23
    
I edited the question. I hope this is better. –  mathlove Sep 26 '13 at 14:21
    
The Somos sequences don't give integers for longer recursions than $k=8$ I think, according to the site you mention –  coffeemath Sep 26 '13 at 20:43
    
@coffeemath: You are right. Thanks. –  mathlove Sep 27 '13 at 14:15

1 Answer 1

This is not a solution. This is a reply to William's above comment, which said

"Have you already tried going through the proof of the original theorem, to see what might change if you replace 1 by 2 in the initial conditions?"

In order to make it easier to show the proof of the $6$-Somos sequence, I'm going to show that the $4$-Somos sequence always gives an integer.

The $4$-Somos sequence $\{c_n\}$ is defined as $$\begin{align}c_{n+4}=\frac{c_{n+3}\cdot c_{n+1}+{c_{n+2}}^2}{c_{n}}\ \ (n\ge0)\qquad(1)\end{align}$$ $$c_0=c_1=c_2=c_3=1.$$

Note that it is sufficient to prove the following theorem :

Theorem : Let $c_0=w,c_1=x,c_2=y,c_3=z$ be variables, and let $\frac{p_n}{q_n}$ be the rational expression of $c_n$ with irreducible representation by $(1)$. Then, the denominator of $q_n$ is always a monomial expression about $w,x,y,z$ whose coefficient is $1$.

Proof : The $n\le 7$ cases are obvious. The point is the $n=8$ case. Let $A$ be $p_4=xz+y^2$. Noting that $q_5=wx,q_6=w^2xy,q_7=w^3x^2yz$, we get $$p_5=Ay+z^2w, p_6=A^2x+Ayzw+z^3w^2,$$$$ p_7=A^3(x^2+yw)+A^2z^2w^2+Ay^2z^2w^2+yz^4w^3.$$

Hence, we know that $p_4$ is coprime to each of $p_5,p_6,p_7$ (as a polynomial). Since the constant term of the numerator of $c_8$ is $y^2z^6w^4+xz^7w^4=z^6w^4A,$ we know that the numerator of $c_8$, as a whole, can be divided by $A=p_4$. By induction, treating $c_9$ as $c_8$ which starts from $x,y,z,a=\frac{A}{w}$, by the same argument above, we know that the denominator is a monomial expression only with $w,x,y,z$ and so on. Now the proof is completed.

Now, I'm going to show the proof of the $6$-Somos sequence briefly. It is known that this proof is completed by using Macsyma. (by Dean Hickerson)

The way of this proof is the same as above.

See $a_n$ as the rational function about $$a_0=u, a_1=v, a_2=w, a_3=x, a_4=y, a_5=z,$$

and show that the denominator is always a monomial expression about $u,v,w,x,y,z$ whose coefficient is $1$ with irreducible representation. The point is to show every numerator can be represented as a polynomial of $B=vz+wy+x^2$. The difficulty lies in showing the constant term of the numerator of $a_{12}$, which is a polynomial with $194$ terms, can be represented by $B$.

(In my opinion, this way is something like "caluculations tells us this is true". As far as I know, the mystery that "why these can be divided?" still remains unsolved.)

Anyway, what I would like to say is that the above theorem (the way of thinking), which is all I know, is not sufficient to solve my question.

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