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I really don't know where to start here. I have looked at "Rational Difference Equation" on Wikipedia, but found nothing useful.

The recurrence I am trying to solve is $$a_{n+1} = P/(P-1), \text{ where } P_{n-1} := \prod_{i=1}^{n}a_{i},$$ where $a_{1} \neq 1$.

The motivation for this comes from finding solutions in real numbers to $$\sum_{i=1}^{n} \alpha_{i} = \prod_{i=1}^{n} \alpha_{i},$$ where $n \in \mathbb{N}$. Of course $(\alpha, \frac{\alpha}{\alpha - 1})$ is a solution when $n = 2$ ($\alpha \neq 1$). So in general, we can find solutions for any $n$ by choosing $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n-1}$ to satisfy the sum/product relation above, and then letting $$\alpha_{n} = \frac{\alpha_{1}\alpha_{2}\cdots\alpha_{n-1}}{\alpha_{1}\alpha_{2}\cdots\alpha_{n-1} - 1}.$$ Finding explicitly the $\alpha_{i}$ in terms of $\alpha_{1} =: x$ is equivalent to the recurrence relation above (right?).

Edit. As lulu stated, letting $b_{n} := 1/a_{n}$ and $Q_{n} := \prod_{i=1}^{n} b_{i}$, we have $b_{n} = 1 - Q_{n-1}$. Noticing that $b_{n} = Q_{n}/Q_{n-1}$, we get the relation $$Q_{n} = Q_{n-1} - Q_{n-1}^{2} = Q_{n-1}(1 - Q_{n-1}),$$

which seems to be the logistic map. Perhaps there's another way to look at this problem?

amWhy
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Freddie
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  • Do you mean to omit $a_n$ from the right hand? Also, the notation isn't great....$P$ should be $P_n$ or $P_{n-1}$, no? – lulu Feb 21 '18 at 01:15
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    If $b_i=\frac 1{a_i}$ and $Q_{n-1}=\prod_{i=1}^{n-1}b_i$ then $b_{n+1}=1-Q_{n-1}$ which seems a lot more tractable. – lulu Feb 21 '18 at 01:17
  • @lulu Sorry I did not mean to omit $a_{n}$ on the right! So from your second comment, we'd have $b_{n} = 1 - Q_{n-1}$. – Freddie Feb 21 '18 at 01:44
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    That makes more sense. I ran a bunch of examples...didn't see any great pattern emerging. Doesn't prove there isn't one, of course. – lulu Feb 21 '18 at 01:52
  • Well, by your remark that $b_{n} = 1 - Q_{n-1}$, and also noticing that $b_{n} = Q_{n}/Q_{n-1}$, we get the relation $Q_{n} = Q_{n-1} - Q_{n-1}^{2}$. – Freddie Feb 21 '18 at 01:53
  • Well, the logistic recursion hasn't got a simple closed solution in general. – lulu Feb 21 '18 at 01:58
  • Related https://math.stackexchange.com/questions/2492408/sum-and-product-of-n-natural-numbers-is-equal/ – rtybase Feb 22 '18 at 21:14
  • @rtybase Hi. Thanks for the reference. Although helpful, I am trying to solve in real numbers. – Freddie Feb 22 '18 at 21:24
  • Integers are still reals ... anyway, up to you. – rtybase Feb 22 '18 at 21:25
  • @rtybase Right. I would like to find a general solution (I know integers are real numbers...). I will look at your reference in more detail soon. – Freddie Feb 22 '18 at 23:47

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