Question 862879
<pre>
{{{1}}}, {{{1/2}}}, {{{1/2}}}, {{{3/4}}}, {{{3/2}}}, ...

1st term = 1, then
multiply the 1st term, {{{1}}}, by {{{1/2}}} to get the 2nd term, {{{1/2}}}, 
multiply the 2nd term, {{{1/2}}}, by {{{1}}} to get the 3rd term, {{{1/2}}}, 
multiply the 3rd term, {{{1/2}}}, by {{{3/2}}} to get the 4th term, {{{3/4}}}, 
multiply the 4th term, {{{3/4}}}, by {{{2}}} to get the 5th term, {{{3/2}}} 

Notice that if we think of 1 as {{{2/2}}} and 2 as {{{4/2}}}, we have:

1st term = 1, then we
multiply the 1st term, {{{1}}}, by {{{1/2}}} to get the 2nd term, {{{1/2}}}, 
multiply the 2nd term, {{{1/2}}}, by {{{1/2}}} to get the 3rd term, {{{1/2}}}, 
multiply the 3rd term, {{{1/2}}}, by {{{3/2}}} to get the 4th term, {{{3/4}}}, 
multiply the 4th term, {{{3/4}}}, by {{{4/2}}} to get the 5th term, {{{3/2}}} 

So we multiply the nth term by {{{n/2}}} to get the (n+1)st term. 

So the recursive formuna for a<sub>n</sub> is

a<sub>1</sub> = 1, a<sub>n+1</sub> = {{{n/2}}}·a<sub>n</sub>

Edwin</pre>