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While solving a differential equation I ended up with this recurrence

$$(n+1)(n+2)a_{n+2}=a_{n-1}$$ With $n=1,2,3,...$

I know that $a_{3k+2}=0$ for all $k$. When it comes to finding series for $a_{3k}$ I have a problem expressing it correctly:

$n=1$ $$(2)(3)a_3=a_0$$ $$\therefore a_3=\frac{a_0}{(2)(3)}$$ $n=4$ $$(5)(6)a_6=a_3$$ $$\therefore a_6=\frac{a_0}{(5)(6)(2)(3)}$$ So on and so on. So we end up with : $$a_{3k}=\frac{a_0}{(3k)(3k-1)(3k-3)(3k-4)...(3)(2)}$$ I don't know how to express that product as a closed product. I thought about rearranging it: $$(3k*(3k-3)(3k-6)...*3)*((3k-1)(3k-4)...*2)$$ And whilst the first bracket I found easy to solve :

$$3^k*3!$$ I can't do similar "trick" with the second one. Any help is appreciated

1 Answers 1

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$$(n+1)(n+2)a_{n+2}=a_{n-1}$$ without losing generality using the following substitution: $$ a_k=\frac{c_k}{k!} $$ leads to: $$(n+1)(n+2)\frac{c_{n+2}}{(n+2)!}=\frac{c_{n-1}}{(n-1)!}$$ and remembering that $$(n+2)!=(n+2)(n+1)\cdot n!$$ leads to this simplification: $$c_{n+2}=\frac{n!}{(n-1)!}c_{n-1}=n\,c_{n-1}$$ now all that is left is solving $$ c_{n+2}=n\,c_{n-1} \iff c_{n+3}=(n+1)c_n $$ that is solved using the $\Gamma$ function which has this fundamental property $$\Gamma(n+1)=n\Gamma(n)$$ Again without loss of generality set: $$ c_n=q_n \Gamma(\alpha n+ \beta) $$ substituting this leads to $$ q_{n+3}\Gamma(\alpha n +3\alpha +\beta)=(n+1)q_n\Gamma(\alpha n + \beta) $$ Applying the property of $\Gamma$ to the LHS: $$ q_{n+3}(\alpha n +3\alpha +\beta-1)\Gamma(\alpha n +3\alpha +\beta-1)=(n+1)q_n\Gamma(\alpha n + \beta) $$ now in order to simplify the $\Gamma$ on both sides: $$ \Gamma(\alpha n +3\alpha +\beta-1)=\Gamma(\alpha n + \beta)\iff \alpha=\frac{1}{3} $$ let's substitute and see what comes next: $$ q_{n+3}(\frac{n}{3} +\beta)=(n+1)q_n $$ setting $\beta=\frac{1}{3}$ leads to: $$ q_{n+3}=3 q_n $$ now all the linearly independent solutions of this last recurrence are obtained using: $$ q_n=K^n $$ substituting gives the characteristic equation: $$ K^3=3 \iff K_{i}=\sqrt[3]{3},\,\,\sqrt[3]{3}e^{i2\pi/3},\,\,\sqrt[3]{3}e^{-i2\pi/3} $$ this means that the most general $q_n$ is: $$ q_n=\lambda_0 K_0^n+\lambda_1 K_1^n+\lambda_2 K_2^n=3^{n/3}\left(\lambda_0+\lambda_1 e^{i2n\pi/3}+\lambda_2 e^{-i2n\pi/3}\right) $$ so putting it all together: $$ a_n=\frac{1}{n!}c_n=\frac{1}{n!}\Gamma\left(\frac{n+1}{3}\right)q_n= $$ $$ =\frac{1}{n!}\Gamma\left(\frac{n+1}{3}\right)3^{n/3}\left(\lambda_0+\lambda_1 e^{i2n\pi/3}+\lambda_2 e^{-i2n\pi/3}\right)$$ where the $\lambda$ are arbitrary complex numbers.