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It is known that the Somos-$k$ sequences for $k\ge 8$ do not give integers. But the first terms of Somos-8 sequence $s_n=a_n/b_n$ $$1, 1, 1, 1, 1, 1, 1, 1, 4, 7, 13, 25, 61, 187, 775, 5827, 14815,\frac{420514}{7}, \frac{28670773}{91}$$ defined by $s_1=s_2=s_3=s_4=s_5=s_6=s_7=s_8=1$, $$s_{n+8}s_n=s_{n+7}s_{n+1}+s_{n+6}s_{n+2}+s_{n+5}s_{n+3}+s_{n+4}^2\qquad(n\ge1)$$ have only odd denominators $b_n$. Morever $s_n$ has even numerator $a_n$ only for $n=9k$. It was checked for $n\le 67$. First terms of $s_n\mod 8:=a_n\cdot b_n^{-1}\mod 8$ are $$\begin{array}{l} 1, 1, 1, 1, 1, 1, 1, 1, {\bf 4},\\ 7, 5, 1, 5, 3, 7, 3, 7, {\bf 6},\\ 7, 7, 5, 7, 5, 3, 1, 1, {\bf 6},\\ 5, 1, 5, 5, 3, 3, 7, 1, {\bf 4},\\ 7, 7, 3, 3, 7, 7, 3, 1, {\bf 2},\\ 1, 3, 1, 3, 7, 5, 3, 5, {\bf 2},\\ 5, 7, 3, 7, 7, 3, 7, 3, {\bf 0},\\ 1, 3, 7, 3,\ldots\end{array}$$

Is it possible to prove that the sequence $s_n\mod 2$ is periodic?

EDT. It was found by მამუკა ჯიბლაძე that numerator of $s_{71}$ is even while that of $s_{72}$ is odd. The last line of the table above is $$1, 3, 7, 3, 1, 5, 1, {\bf 2}, 5$$ So the conjecture about numerators of $s_n$ is wrong. But the qustion about periodicity is still valid.

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  • $\begingroup$ There might be something in Robinson, Periodicity of Somos sequences, Proc Amer Math Soc 116 (1992) no. 3, 613-619, MR1140672 (93a:11012). The review by Peter Kiss says, in part, "For some sequences related to the Somos($k$) sequences, the author shows that they are periodic modulo $m$ for every $m$, although the periodicity problem remains open for the original sequences. Some observations are also made concerning the prime divisors of the terms and the rate of growth of certain sequences. $\endgroup$
    – Gerry Myerson
    Commented Jun 9, 2017 at 6:54
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    $\begingroup$ Believe it or not but actually the numerator of $s_{71}$ is even while that of $s_{72}$ is odd (unless there is a bug in Mathematica 11, that is). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 9, 2017 at 7:51
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    $\begingroup$ Doing some further calculations with 2-adics; although I am not sure I am doing it right but seems like $s_{95}$ has even denominator... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 9, 2017 at 8:51
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    $\begingroup$ Still have to re-check it but I switched to Pari and computed in 2-adics (with $2^{100000}$ precision), seemingly $s_{103}$ has even denominator. Up to that, results with Mathematica and Pari are identical, and also coincide with results for exact rationals (these I am currently running, they are at 82 so far). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 9, 2017 at 11:17
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    $\begingroup$ Alas, run out of memory on 85 with exact rationals. There might be other ways to find out about parity of the denominator in $s_{103}$ though... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jun 10, 2017 at 9:31

1 Answer 1

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I confirm the observation of @მამუკაჯიბლაძე that $\nu_2(s_{103})=-1$. That is, $s_n\bmod 2$ is not well-defined at first place, which invalidates the question.

Moreover, for $n\geq 133$, $\nu_2(s_n)$ seems to form a strictly decreasing function, i.e., $s_n$ accumulates larger and larger powers of $2$ in the denominators.

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