Does this sequence always give an integer?

Question

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.

Answer 1

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.

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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

(42872600952532756413944577, -7642585197866180463969286501605177115683023, -14777777125160439954108773045163128226074672889387272080, 148964391693661992923680078954077756067110081304751719212599407, 50595833510742832041116346653564092895564724512883353187577334991)

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

(275482421676870371359371463998680435538426963076376380974139366712401528564856518381424556318432563385462892296746457695, -1204187018838917569168734117714049614241185115821559510667269181326087284012251290418681275712354771003853097159487786429076362441774104293746291761783054990301019370687256368668064571321856, 46237948454612900518472923159014977519269442452459902128288590632859589486362269029696379416299719982558060325802294250784058152579910008058460895113571344637868831455176995514194104478127008847575580360960143927556981822085371274654718852, -25258589788169445344223776170236779359054408212184860769933304831364012583819797167161677185288240579806555892962118768351982581424384932133901083260133870630044033809364594884555608989561773335221097351699363330231907242444968619795482350223356786592137691094080623104, 102927547672248207711100989742092219264928069372556751802909240396120983835548385841382536622225530269264900739652796985313302402437855010196494257440627546507583982533406361509714125293160643210579263127894875120075097659020181644866881731502319992829790755348150678034508425757)

(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.

Answer 2

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.)

Answer 3

It is not an answer just a piece of fun.

I've made a mistake in my program and calculated the sequence with wrong recurrence $$a_{n+6}=\frac{a_{n+5}\cdot a_{n+1}+a_{n+4}\cdot a_{n+2}\cdot {a_{n+3}}^2}{a_n}\ \ (n\ge0).$$ In both cases (1) $a_0=a_1=a_2=a_3=a_4=a_5=1$ and (2) $a_0=a_1=a_2=a_3=1$, $a_4=a_5=2$ it gives at least $40$ integer values: $$\tag{1}1, 1, 1, 1, 1, 1, 2, 3, 5, 17, 107, 1489, 79541, 96735414,\ldots$$ $$\tag{2}1, 1, 1, 1, 2, 2, 4, 12, 44, 472, 13144, 5509040, 32227528976,\ldots$$

EDT: These sequences are really integers. It can be proved with simple (George Bergman's) argument described in David Gale's The strange and surprising saga of the Somos sequences, see Tracking the automatic ant and other mathematical explorations.

Answer 4

The following article by Ekhad and Zeilberger might be of interest.

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/somos.html