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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.

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  • $\begingroup$ Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even. $\endgroup$ – José Carlos Santos Nov 24 '18 at 21:05
  • $\begingroup$ @JoséCarlosSantos Yes, that is what I was given. $\endgroup$ – cosmicbrownie Nov 24 '18 at 21:16
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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).

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

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 $

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