Riccati difference equation

Remarks

Example: $x_{n+1} = 1/(1 + x_n)$ has $a = 0$, $b = c = d = 1$, so $c \ne 0$ and $ad - bc = -1 \ne 0$. Non-example: $x_{n+1} = (2 x_n + 1)/3$ has $c = 0$; it is an affine linear recurrence, not a Riccati equation. The condition $ad - bc \ne 0$ excludes the degenerate case where the right side is a constant. Linearization: if $(y_n)$ satisfies $y_{n+2} = (a + d)\, y_{n+1} - (ad - bc)\, y_n$ with all $y_n \ne 0$, then $x_n = (y_{n+1}/y_n - d)/c$ solves the equation. To check, put $w_n = c x_n + d$; the equation becomes $w_{n+1} = (a + d) - (ad - bc)/w_n$, and $w_n = y_{n+1}/y_n$ turns this into the linear recurrence.