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In the course of some research computations I have been doing, I run up against a recursion $$ a_{n+3} = a_{n+2}a_{n+1} - a_n $$

I've tried to find out if it's possible to solve recursions of this form, but can't find much since it's nonlinear. Does anyone know of methods that might be applicable; or failing that, if there are any assumptions on the initial conditions which might make it solvable?

Hope this is clear; I don't have any background in number theory or discrete math. Thanks.

ec92
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    If the first three terms are each $0$ or $1$ then $a_{n+12}=a_n$ – Henry Jul 13 '12 at 04:05
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    Personaly, when I run into things like this, I do the following. Compute a few terms from $a_0$. By a few terms I mean, perhaps, more than 25. Then, Look at what you've produced, and see if you can find a pattern. If you can find a pattern, then you've gotten lucky, and may be able to do soemthing with induction. – RougeSegwayUser Jul 13 '12 at 04:05
  • @ChrisDugale: right, i do the same thing; i think most of us do. – Aang Jul 13 '12 at 04:59
  • I just type the recursion into Wolfram Alpha. It tells you a lot about the sequence straight away, and sometimes gives a closed form. Sometimes it doesn't give one when it is possible to find one, but that's not so often. – Ragib Zaman Jul 13 '12 at 05:13
  • If the first few terms are increasing, this is going to grow really fast. – Alex Becker Jul 13 '12 at 05:15
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    @AlexBecker Indeed, this grows slower than $b_{n+3} = b_{n+2} b_{n+1} $ (simply omitting the subtracted term) and that has solutions of the form $ b_n = \exp ( c_1 F_n + c_2 L_n)$ where $F_n$ and $L_n$ are the Fibonacci and Lucas numbers. – Ragib Zaman Jul 13 '12 at 05:17
  • You could also calculate some terms and ask OEIS... – draks ... Jul 13 '12 at 05:27
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    @Henry: Wow! (With period 12 when at least two ones in the starting word $a_0a_1a_2$, period 6 when exactly one one, and period 1 when no one.) Why is that so? – Did Jul 13 '12 at 09:19
  • Thanks for all the input. I've been working with formal variables as initial conditions, so I can't really look for patterns in the terms. This discussion gives my an idea of some simple cases that might be manageable, though. – ec92 Jul 13 '12 at 22:52
  • Maybe you are interested in this: Solving Recurrence $T_n = T_{n-1}*T_{n-2} + T_{n-3}$ – draks ... Jul 16 '12 at 17:58

1 Answers1

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Let $F_1,F_2,\dots$ be a Fibonacci-like sequence, such that $F_{n}=F_{n-1} + F_{n-2}$, and let $$a_{n} = e^{F_{n}} + e^{-F_{n}} = 2\cosh F_n.$$ Then $$ \begin{eqnarray} a_{n-1}a_{n-2} &=& \left(e^{F_{n-1}} + e^{-F_{n-1}}\right) \left(e^{F_{n-2}} + e^{-F_{n-2}}\right) \\ &=& e^{F_{n-1} + F_{n-2}} + e^{F_{n-1} - F_{n-2}} + e^{-F_{n-1} + F_{n-2}} + e^{-F_{n-1} - F_{n-2}} \\ &=& e^{F_{n}} + e^{F_{n-3}} + e^{-F_{n-3}} + e^{-F_{n}} \\ &=& a_{n} + a_{n-3}, \end{eqnarray} $$ which is exactly your recursion. This family of solutions, parametrized by $(F_1, F_2)$, only covers part of the full range of initial conditions $(a_1, a_2, a_3)$. In particular, it will cover those cases where $$ a_3 = \frac{1}{2} a_1 a_2 \pm \frac{1}{2}\sqrt{\left(a_1^2-4\right)\left(a_2^2-4\right)}. $$

mjqxxxx
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