Solving a recurrence relation in 2 variables

Question

Given this sequence $Q_1(x)=x$, $Q_{n+1}(x)={Q_n(x+1)\over Q_n(x)}$, with $n>=1$, how can I get the explicit n-th term relation?

More precisely, $Q_n(x)=$ ? (when $n>=0$)

I'm eager to learn a method for expliciting this multi-variable recurrences in the future.

Answer 1

For every $n\geqslant0$ and $k\geqslant1$, $$ Q_{n+k}(x)=\prod_{i=0}^nQ_k(x+i)^{(-1)^{n-i}{n\choose i}}, $$ hence, for every $n\geqslant0$, $$ Q_{n+1}(x)=\prod_{i=0}^{n}(x+i)^{(-1)^{n-i}{n\choose i}}=\frac{(x+n)}{(x+n-1)^{{n\choose1}}}\cdot\frac{(x+n-2)^{{n\choose 2}}}{(x+n-3)^{{n\choose3}}}\cdots $$

Answer 2

Just try for small $n.$

$$Q_1(x) = x$$

$$Q_2(x) = Q_1(x+1)/Q_1(x) = (x+1)/x.$$

$$Q_3(x) = Q_2(x+1)/Q_2(x) = ((x+2)/(x+1))/((x+1)/x) = (x+2)x/(x+1)^2.$$

$$Q_4(x) = Q_3(x+1)/Q_3(x) = ((x+3)(x+1)/(x+2)^2)/((x+2)x)/(x+1)^2) = (x+3)(x+1)^3/((x+1)^2(x+2)^2 x).$$

It does not look like there will be any simple expression.