Somos 3 Sequence Consider the Somos 3 sequence $a_{n+3}a_n = a_{n+1}a_{n+2}$ with $a_1 = \alpha$, $a_2 = \beta$ and
$a_3 = \gamma$ where $\alpha$, $\beta$ and $\gamma$ are all integers.
(a) Try different values of the initial data to see whether it generates integer
sequences.
(b) From your observation, formulate the conditions on $\alpha$, $\beta$ and $\gamma$ which guarantee that the sequences generated are integers.
(c) Prove your statement formulated in (b).
 A: $$\frac{a_{n+3}}{a_{n+1}}=\frac{a_{n+2}}{a_{n}}$$
Let  $b_n=\frac{a_{n+2}}{a_{n}}$,
$$b_{n+1}=b_n=\cdots=b_1=\frac{a_{3}}{a_{1}}=\frac {\gamma}{\alpha}$$
$$\implies \frac{a_{n+2}}{a_{n}}=\frac {\gamma}{\alpha}$$
So, $a_{n+2}=\frac {\gamma}{\alpha}a_n=\cdots=\left(\frac {\gamma}{\alpha}\right)^{\frac{n+2-r}2} a_r$
If $n$ is even,$=2m$(say) $a_{2m+2}=\left(\frac {\gamma}{\alpha}\right)^{\frac{2m+2-r}2} a_r=\left(\frac {\gamma}{\alpha}\right)^ma_2=\left(\frac {\gamma}{\alpha}\right)^m\beta$
So, $(\alpha)^m\mid (\gamma)^m\beta$ to make $a_{2m+2}$ integer.
If $n$ is odd,$=2m+1$(say) $a_{2m+3}=\left(\frac {\gamma}{\alpha}\right)^{\frac{2m+3-r}2} a_r=\left(\frac {\gamma}{\alpha}\right)^ma_3=\left(\frac {\gamma}{\alpha}\right)^m\gamma$
So, $(\alpha)^m\mid (\gamma)^{m+1}$ to make $a_{2m+3}$ integer.
Combining $\alpha^m\mid (\gamma^{m+1},\gamma^m\beta)\iff\alpha^m\mid (\gamma^m(\beta,\gamma))$ to keep $a_r$ integer for integer $r\ge 4$
A: It's not too difficult to write out a few terms of your sequence. Let us fix $a_1,\ a_2,\ a_3$. Then your sequence begins
$$a_1,\ a_2,\ a_3,\ a_4=\frac{a_3a_2}{a_1},\ a_5=\frac{a_3^2}{a_1},\ a_6=\frac{a_3^2a_2}{a_1^2},\ a_7=\frac{a_3^3}{a_1^2},\ a_8=\frac{a_3^3a_2}{a_1^3},\ a_9=\frac{a_3^4}{a_1^3}$$
The pattern should be rather clear. I will leave it to you to verify that the sequence continues in the expected manner. A necessary and sufficient condition is then
$$\forall n\in\mathbb{N}:\ a_1^n\mid a_3^{n+1}\ \land\  a_1^n\mid a_3^na_2$$
I claim that $\forall n\in\mathbb{N}:\ a_1^n\mid a_3^{n+1}$ necessarily implies $a_1\mid a_3$. To see this, suppose otherwise. 
Clearly the prime factors of $a_1$ is a subset of $a_3$'s. Therefore if $a_1\nmid a_3$ it must be because there is a prime factor occurring with larger exponent in $a_1$ than in $a_3$. Suppose that $p$ is such a prime with $p^k\|a_1$ and $p^\ell\|a_3$ such that $k > \ell$. But this is absurd, since
$$a_1^n \mid a_3^{n+1} \implies p^{nk} \mid p^{(n+1)\ell} \implies nk < (n+1)\ell \implies \frac{k}{\ell} < 1 + \frac{1}{n}\ \ \forall n\in\mathbb{N}^+$$
The rational $\frac{k}{\ell}$ is bounded away from $1$ while the latter sequence converges towards $1$. This is clearly impossible.
Therefore it is necessary and sufficient that $a_1\mid a_3$.
A: Consider any prime $p$ and denote by $x_n$ the exponent with which $p$ is present in $a_n$. The recursion for the $a_n$ then implies
$$x_{n+3} - x_{n+2}-x_{n+1}+x_n=0 \qquad(n\geq1)\ .\qquad(*)$$
The characteristic polynomial of this difference equation is $$\chi(\lambda)=\lambda^3-\lambda^2-\lambda+1=(\lambda-1)^2(\lambda+1)\ .$$
Therefore the general solution of $(*)$ is
$$x_n=A \ n+B+ C(-1)^n\qquad (n\geq1)\ ,$$
whereby the coefficients $A$, $B$, $C$ have to be determined from the initial conditions. 
If we want the $a_n$ to be integers for all $n\geq1$ then the  exponents $x_1$, $x_2$, $x_3$ with which $p$ appears in the initial values $\alpha$, $\beta$, $\gamma$ have to be such that $x_n\geq0$ is guaranteed for all $n\geq 1$, and this for every prime $p$ individually.
