Solve the recurrence relation $n(n-1)a_n - (n-2)^2a_{n-2}= 0$, $a_0=0$, $a_1=1$

Question

Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.

I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.

Answer 1

As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1\cdot 3^2\cdot 5^2\cdot...\cdot (n-2)^2\over n!}$$ or $$a_n = {((n-2)!!)^2\over n!}$$

You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).

Answer 2

$a_n=\frac{(n-2)^2a_{n-2}}{n(n-1)}, n\geq2;\ a_0=0,\ a_1=1$

Clearly, all the even-index terms of the recurrence are $0$. For the odd-index terms, you could do this:

$a_3=\frac{1^2a_1}{2\cdot3}\\ a_5=\frac{3^2a_3}{4\cdot5}\\ a_7=\frac{5^2a_5}{6\cdot7}\\ \ \ \ \cdot\\ \ \ \ \cdot\\ \ \ \ \cdot\\ a_{2n+1}=\frac{(2n-1)^2a_{2n-1}}{(2n)\cdot(2n+1)}, n\geq1\\ $

Multiplying all the terms $\thinspace\thinspace \implies a_3\cdot a_5\cdot a_7 \cdot \cdot \cdot a_{2n+1}=\frac{1^2\cdot3^2\cdot5^2\cdot\cdot\cdot(2n-1)^2}{(2n+1)!}a_1\cdot a_3\cdot a_5\cdot\cdot\cdot a_{2n-1}\\ \implies a_{2n+1}=\frac{(1\cdot3\cdot5\cdot\cdot\cdot(2n-1))^2}{(2n+1)!}a_1\\ \implies a_{2n+1}=\frac{(2n-1)!^2}{2^{2n-2}\ (n-1)!^2\ (2n+1)!}\\ \implies a_{2n+1}=\frac{(2n)!}{2^{2n}\ (2n+1)\ n!^2};\ n\geq1 $

Answer 3

The recurrence gives $a_{n + 2}$ in terms of $a_n$:

$\begin{equation*} (n + 2) (n + 1) a_{n + 2} -n^2 a_n = 0 \end{equation*}$

so it is really a first order linear recurrence in disguise. From $a_0 = 0$ you easily get $a_{2 n} = 0$ for all $n$. For odd indices, call $b_n = a_{2 n + 1}$, so that:

$\begin{align*} \left(\frac{n - 1}{2} + 2\right) \left(\frac{n - 1}{2} + 1\right) b_{n + 1} - \left(\frac{n - 1}{2}^2\right) b_n &= 0 \\ (n + 3) (n + 1) b_{n + 1} - (n - 1)^2 b_n &= 0 \\ b_{n + 1} &= \frac{(n - 1)^2}{(n + 1)(n + 2)} b_n \end{align*}$

The solution is simple:

$\begin{align*} b_n &= \frac{((n - 1)!)^2}{(2 n)!} b_0 \\ a_{2 n} &= 0 \\ a_{2 n + 1} &= \frac{((n - 1)!)^2}{(2 n)!} a_1 \end{align*}$