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$3,0,2,3,2,5$ and I know the $8$th term is $10$.

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@Thomas: Is this the Perrin Sequence? If so, the site shows the recurrence. Regards –  Amzoti Jan 17 '13 at 17:22
    
seems it is –  Santosh Linkha Jan 17 '13 at 17:25
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Do you really mean THE recursive sequence? Of what order? –  HSN Jan 17 '13 at 17:26

2 Answers 2

http://oeis.org/A001608 starts this way and has $10$ as the $9$th term if that helps. It starts counting at $0$, so $10$ is the term corresponding to $8$.

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If we know that the recursive sequence is a simple linear recurrence, then an $n^{th}$ order linear recurrence needs $2n$ known values - the initial $n$ terms and the coefficients.

You have seven knowns, so we can hypothesize that the recurrence has the form

$$ a_n=\alpha_1a_{n-1}+\alpha_2a_{n-2}+\alpha_3a_{n-3} $$

with initial terms

$$ \begin{align} a_0 &= 3 \\ a_1 &= 0 \\ a_2 &= 2. \end{align} $$

With the next three terms we can create a set of linear equations:

$$ \begin{align} a_3 &= 3 &= \alpha_1(2)+\alpha_2(0)+\alpha_3(3) \\ a_4 &= 2 &= \alpha_1(3)+\alpha_2(2)+\alpha_3(0) \\ a_5 &= 5 &= \alpha_1(2)+\alpha_2(3)+\alpha_3(2). \end{align} $$

Solving this yields

$$ \begin{align} \alpha_1 &= 0 \\ \alpha_2 &= 1 \\ \alpha_3 &= 1. \end{align} $$

Now we can calculate all terms, specifically $a_6=5$, $a_7=7$, and $a_8=10$. So the Perrin Sequence as mentioned is the only third-order simple linear recurrence given your constraints.

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