Question 838454
<pre>
1/1, 1/1, 1/2, 1/3, 1/4, 1/9...

Sequence of denominators b<sub>n</sub>: 1,1,2,3,4,9,...

5(b<sub>3</sub>)-6(b<sub>1</sub>) = 5(2)-6(1) = 10-6 = 4 = b<sub>5</sub>

5(b<sub>4</sub>)-6(b<sub>2</sub>) = 5(3)-6(1) = 15-6 = 9 = b<sub>6</sub>

So a recursion equation for the sequence of denominators is:

b<sub>1</sub>=1,b<sub>2</sub>=1,b<sub>3</sub>=2,b<sub>n</sub>=5b<sub>n-2</sub>-6b<sub>n-4</sub>

Therefore a recursion equation for the given sequence a<sub>n</sub> is

a<sub>1</sub>=1,a<sub>2</sub>=1,a<sub>3</sub>=1/2,a<sub>n</sub>=5/a<sub>n-2</sub>-6/a<sub>n-4</sub>    

Edwin</pre>