The sequence $(y_n)$ satisfies the relationship ${y_{n1}}{y_{n+1}} + y_n = 1$ for all $n \ge 2$. If $y_1= 1$ and $y_2= 2$, What can you say about the sequence? What happens for other starting values?
closed as offtopic by Alexey Ustinov, Robert Israel, abx, მამუკა ჯიბლაძე, Stefan Kohl Mar 19 '18 at 9:52
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3$\begingroup$ Have you tried computing a few values? $\endgroup$ – Robert Israel Mar 19 '18 at 7:16

$\begingroup$ I am unable to find for values n=7 $\endgroup$ – user542993 Mar 20 '18 at 10:04

$\begingroup$ As I commented to Johann Cigler's answer, $y_7$ is undetermined. $\endgroup$ – Robert Israel Mar 20 '18 at 20:04
For arbitrary Initial values $y_1=a,y_2=b$ (with $ab$ not zero) you get $a,b, \frac{1b}{a},\frac{a+b1}{ab},\frac{1a}{b}$ . Then the sequence is periodic with period 5. Thus for $y_1=1,y_2=2$ the sequence should be $1,2,1,1,0,1,2,1,1,0,\dots.$


1$\begingroup$ In $P^1\times P^1$ $(y_{n1},y_n)\mapsto (y_n,y_{n+1})$ can thus be written as $(a,b)\mapsto (b,(1b)/a)$. This is a Cremona transformation. Cremona transformations of finite order are completely classified; it would be interesting to identify a "minimal" model on which it acts as an automorphism. $\endgroup$ – YCor Mar 19 '18 at 10:08

$\begingroup$ If $(x_n)$ is the sequence defined by $x_n=y_n$ for all $n$, then the relation becomes $x_{n1}x_{n+1}=x_n+1$. So the elements are cluster variables in a cluster algebra of type $A_2$. $\endgroup$ – Philipp Lampe Mar 19 '18 at 12:41

$\begingroup$ Except that when $y_{n2} = 0$, $y_n$ is undefined if $y_{n1} \ne 1$, undetermined if $y_{n1}=1$. So in the case $y_1=1$, $y_2=2$ it should actually be $1,2,1,1,0,1,undetermined,\ldots$. $\endgroup$ – Robert Israel Mar 19 '18 at 18:08