# Does this sequence always give an integer?

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.

• Wrong answer deleted, I read too fast. What pops out of the Somos-6 proof (by Fomin-Zelevinsky) is that the only prime that can occur in the denominator is $2$. Numerical data suggest that the $2$-adic valuation is periodic, repeating 0,0,0,0,1,1,0,0,0,0,2,2,0,0,0,0,2,2,0,0,0,0,1,1 ... I'll try to write a proof of this if noone else does. – David E Speyer Sep 30 '13 at 18:29
• I'm just mentioning here a link of interest to casual readers like me who might be interested in general related work: faculty.uml.edu/jpropp/somos.html – Suvrit Sep 30 '13 at 21:36
• @David: The 2-adic valuation is easy to compute in this case. Just look at the numbers mod 32. The nominator will never be divisible by 32 but by at most 16, while the denominator by at most 4, so we can always compute the new remainder mod 32 and see that we get a periodic sequence. So if your claim is correct, then this proves the question. – domotorp Oct 5 '13 at 7:49
• @domotorp Can you tell me exactly what your inductive statement is? The trouble is that, if we know $(s(n-6), s(n-5), \ldots, s(n-1))$ modulo $32$, and $s(n-6)$ is even, we don't know $s(n)$ modulo $32$. In exchange, I'll write out an explanation of why $2$ is the only troublesome prime. – David E Speyer Oct 5 '13 at 15:18
• I took a liberty to add this sequence to the OEIS as oeis.org/A227999 – Max Alekseyev Dec 4 '13 at 18:19

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

(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.

• Could you please explain why you deleted your answer once and why you wrote it again? I can't find any difference between the two. I'm confused a bit. – mathlove Oct 5 '13 at 16:36
• The version which was there when you wrote the above comment doesn't claim to have a proof, and explains why the only prime that can occur in the denominator is $2$. The current version now claims to prove that $2$ never occurs in a denominator, subject to proving the stated bilinear recursions for $x_{6n+e}$ and $x_{12n+e}$. – David E Speyer Oct 5 '13 at 16:53
• There is also a recursion for $x_{3n+e}$, again with width $8$, which you can use to show that $x_{3n}$ is always odd. – David E Speyer Oct 5 '13 at 16:53
• There are some details in the article of Hone "Analytic solutions and integrability for bilinear recurrences of order six" tandfonline.com/doi/abs/10.1080/00036810903329977#.VSFavfAfO6E – Alexey Ustinov Apr 5 '15 at 15:56

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.)

• Ah, nice. And this explains why I could always find a relation of the form $a \tau_n \tau_{n+8k} + b \tau_{n+k} \tau_{n+7k} + c \tau_{n+2k} \tau_{n+5k} + d \tau_{n+3k} \tau_{n+5k} + e \tau_{n+4k}^2$. I knew what I was doing was computing the kernel of a $5 \times \infty$ matrix, but I didn't know why it always had rank $4$. – David E Speyer Apr 8 '15 at 10:28
• But I am missing something. Are you claiming to have a proof, without relying on the computations I already did, that $(\tau_{m+2} \tau_{m-2} + \tau_{m+4} \tau_{m-4})(\tau_{4+n} \tau_{4-n} + \tau_{5+n} \tau_{5-n})$ is always even? – David E Speyer Apr 8 '15 at 10:31
• Yes, because this product $\equiv\Delta\equiv0\pmod 2$. From you claculations probably follows that each multiple here is even. – Alexey Ustinov Apr 8 '15 at 10:38
• Isn't this circular? If we know that all the $\tau$'s in the matrix except for $\tau_{m+n}$ are integers, then we can deduce that $18 \tau_{m+n} \tau_{m-n} + (\tau_{m+2} \tau_{m-2} + \tau_{m+4} \tau_{m-4})(\tau_{4+n} \tau_{4-n} + \tau_{5+n} \tau_{5-n})$ is an even integer. But how do you rule out that $\tau_{m+n}$ is a half integer and $(\tau_{m+2} \tau_{m-2} + \tau_{m+4} \tau_{m-4})(\tau_{4+n} \tau_{4-n} + \tau_{5+n} \tau_{5-n})$ is odd? – David E Speyer Apr 8 '15 at 10:47
• @David Speyer I've corrected my answer using your approach. – Alexey Ustinov Apr 9 '15 at 3:25

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.

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

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

• Thanks for great information. This does surprise me. But I have two things I would like to say:One is that your page doesn't seem to include my question. The other is that your page doesn't seem to include mathematical proofs. I hope I'm not mistaken. Anyway, thanks. – mathlove Sep 30 '13 at 15:44
• Ah, S. B. Ekhad, that's one of my favourite math authors, although I hear he does all calculations in binary and not decimal... – Per Alexandersson Apr 8 '15 at 15:51