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Fix positive a, b, c. Define a sequence of real numbers by $x_0 = a$, $x_1 = b$ and $x_{n+1} = cx_nx_{n-1}$. Find formula for $x_n$.

Exodd
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naseefo
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  • Have you tried writing down the first few terms to see if you can spot a pattern? – Daniel Fischer Mar 01 '17 at 13:55
  • well, you know that $x_n=c^{\gamma_n} a^{\alpha_n}b^{\beta_n}$, so you can derive a recurrence formula for $\alpha_n$, $\beta_n$ and $\gamma_n$ – Exodd Mar 01 '17 at 14:07
  • $\alpha_i = 1, 0, 1, 1, 2, 4, 6...$ $\beta_i = 0 1 1 2 3 5 8$ ---> Fibinocci series $\gamma_i = 0 0 1 2 4 7 12$ – naseefo Mar 01 '17 at 14:17
  • Your $\alpha_i$ are incorrect. The exponents are all closely related to the Fibonacci sequence. Perhaps it's easier to see when you consider $y_n = cx_n$. – Daniel Fischer Mar 01 '17 at 14:25
  • $\alpha_i = 1, 0, 1, 1, 2, 4, 6...$ ---> I can't understand this series $\beta_i = 0, 1, 1, 2, 3, 5, 8,...$ ---> Fibinocci series $\gamma_i = 0, 0, 1, 2, 4, 7, 12,...$ I can't understand this series – naseefo Mar 01 '17 at 14:39

1 Answers1

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The first few terms of sequence is $a, b, cab, c^2ab^2, c^4a^2b^3, c^7a^3b^5, c^12a^5b^8 ...$

If we look at the exponents of a, b, c, we have:

$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & expression \\ \hline fib\_n & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & \\ \hline a & 1 & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & fib\_\{n-1\} \\ \hline b & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & fib\_n \\ \hline c & 0 & 0 & 1 & 2 & 4 & 7 & 12 & 20 & 33 & fib\_\{n+1\}-1 \\ \hline \end{array} $

Assuming you have closed form for fib, just join everything together to get the closed form for $x_n$.

eatfood
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