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.

  • 3
    $\begingroup$ 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? $\endgroup$
    – Christoph
    Sep 26, 2013 at 12:34
  • $\begingroup$ In the "similar sequence defined", what is $\;a_n\;$ ?? $\endgroup$
    – DonAntonio
    Sep 26, 2013 at 13:23
  • $\begingroup$ I edited the question. I hope this is better. $\endgroup$
    – mathlove
    Sep 26, 2013 at 14:21
  • $\begingroup$ The Somos sequences don't give integers for longer recursions than $k=8$ I think, according to the site you mention $\endgroup$
    – coffeemath
    Sep 26, 2013 at 20:43
  • $\begingroup$ @coffeemath: You are right. Thanks. $\endgroup$
    – mathlove
    Sep 27, 2013 at 14:15

2 Answers 2


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.


With Yuri Fedorov we have found an explicit analytic description of a general Somos-6 sequence in terms of genus 2 curves; see http://arxiv.org/abs/1512.00056. This might provide an answer to your question, but it is not entirely straightforward to check, as in general one must look at relations of higher order. In fact, Melanie de Boeck did a project with me in her master's, in which she identified a 3-parameter family of Somos-6 sequences which appear to consist entirely of integers, and checked hundreds of terms in many different cases, but it is not yet obvious why this should be so.

Hickerson's proof is essentially the same as proving the Laurent property for Somos-6 (see Fomin and Zelevinsky's paper http://arxiv.org/abs/math/0104241), which is enough to show that any Somos-6 recurrence with arbitrary integer coefficients on the r.h.s. and 6 initial values all either +1 or -1 gives an integer sequence (as long as it doesn't hit a 0 term). Yet in fact Somos-6 should satisfy a stronger property than the ordinary Laurent property, meaning that it generates many more integer sequences than just these.

If I can say more about your example I'll post it here soon.


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