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I'm having a difficult time figuring out the functional form of this sequence:

$$\{9,144,3600,129600,6350400,...\}$$

I'm trying to determine the recursive relationship for a differential equation using power series solutions.

Any help or recommendations are appreciated.

Thank you!

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$$a(n)=\left(\frac{(n+2)!}{2}\right)^2$$ or $$a_{n+1}=a_n(n+3)^2$$

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  • $\begingroup$ Oh geez, that makes my comment above seem stupid, ha! Thanks! $\endgroup$ – Cody_15243 Mar 19 at 20:27
  • $\begingroup$ I've been thinking about group theory recently, so I'd frame it as $|A_{n+2}|^2$, but it's the same answer as above. $\endgroup$ – Robert Shore Mar 19 at 20:50
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We have $$\Big\{9\quad\Big|\quad9\times 16\quad\Big|\quad9\times 16\times 25\quad\Big|\quad 9\times 16\times 25\times 36\quad\Big|\quad 9\times 16\times 25\times 36\times 49\quad\Big|\quad \text{and so on ...}\Big\}$$ therefore$$a_1=9\\a_n={\Big[(n+2)!\Big]^2\over 4}$$

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  • $\begingroup$ I'd argue that specifying $a_1 = 9$ is superfluous here. $\endgroup$ – Peiffap Mar 19 at 23:04
  • $\begingroup$ Yes that's right. I was just calculating a recurrence relation in my mind that time :) $\endgroup$ – Mostafa Ayaz Mar 19 at 23:05
  • $\begingroup$ @Peiffap Well, yes and no, if we consider $a_0=9$ then the formula becomes $(n+3)!^2/4$. Since OP didn't fix the origin index, it somehow has to be explicited. $\endgroup$ – zwim Mar 19 at 23:11
  • $\begingroup$ @zwim. That's true as well. $\endgroup$ – Peiffap Mar 19 at 23:33
  • $\begingroup$ That's a good point! My apologies. The sequence begins at $n=0$ $\endgroup$ – Cody_15243 Mar 20 at 2:39

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