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Given the other post Somos 3 Sequence i just wanted some insight to how the answer came around,

The Somos 3 sequence is:

$a_{n+3}a_n = a_{n+1}a_{n+2}$

Given $a_1=\alpha$ $a_2 = \beta$ and $a_3=\gamma$

How do you write the $a_n$ in terms of fixing it to $a_1,\ a_2,\ a_3$

And one of the answers wrote this:

$a_1,\ a_2,\ a_3,\ a_4=\frac{a_3a_2}{a_1},\ a_5=\frac{a_3^2}{a_1},\ a_6=\frac{a_3^2a_2}{a_1^2},\ a_7=\frac{a_3^3}{a_1^2},\ a_8=\frac{a_3^3a_2}{a_1^3},\ a_9=\frac{a_3^4}{a_1^3}$

And i was wondering how this came about.

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It's just repeatedly substituting the expressions and simplifying. Any recursion can be written in terms of the seed values. This one just happens to be rather simple. – EuYu Nov 19 '12 at 3:00
up vote 3 down vote accepted

Your recursive relation is given as $$\frac{a_{n+3}}{a_{n+1}} = \frac{a_{n+2}}{a_{n}}$$ Notice that this allows us to repeatedly reduce the index by one $$\frac{a_{n+3}}{a_{n+1}} = \frac{a_{n+2}}{a_{n}} = \frac{a_{n+1}}{a_{n-1}} =\cdots = \frac{a_3}{a_1}$$ This gives the simplified recursion $$a_{n+2} = \frac{a_3}{a_1}a_{n}$$ From this you can easily see how the sequence develops.

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+1 this is better than mine. – Amr Nov 19 '12 at 3:21

Rewrite this as $a_{n+3}/a_{n+2}=a_{n+1}/a_{n}$

Now multiply from n=1 to n=m: $$(a_4/a_3)(a_5/a_4)...(a_{m+3}/a_{m+2})=(a_2/a_1)(a_3/a_2)...(a_{m+1}/a_{m})$$ $$a_{m+3}/a_3=a_{m+1}/a_2$$ $$a_{m+3}/a_{m+1}=a_3/a_2$$

Now do the same thing again.

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