<table cellspacing="0" cellpadding="0" border="0" ><tr><td valign="top" style="font: inherit;">I had made a previous post on this subject and have spent a few days trying to work out this problem but I feel like I have hit a dead end, and I am hoping I can get some advice from a few people with more experience than I. The code I have managed to create so far is: (define(find-div n i)   (cond     [(=(remainder n i)0)i]     [else(find-div n(sub1 i))])) ;(find-div 20 3) (define(tabulate-div n)   (cond     [(= n 1)(list 1)]     [else(append           (cons n empty)           (tabulate-div(find-div n(sub1 n))))])) (tabulate-div 20) I believe that I understand how generative recursion, and I have spent a lot of time attempting to figure this problem out in a much more complex way. The above code seems much easier to read and understand, but, it suffers similar problems as the other things I have tried. Mainly, it doesn't produced the wanted answer, it's close, but not quite there. (tabulate-div 20) should equal (list 1 2 4 5 10 20) However, with this code the answer given is (list 20 10 5 1). I have figured out that if I somehow divide 20 by each number in the list I get (list(/ 20 20)(/ 20 10)(/ 20 5)(/ 20 1))=(list 1 2 4 20) which seems a little closer, but still not quite there. Although this would only be close and not all the way there, I feel its a step in the right direction. My problem is that if I add a local definition of: (local((define q(/ n (find-div n n)))) it produces an incorrect answer. This would give me(list(/ 20 20)(/ 10 10)(/ 5 5)(/ 1 1)) Maybe I don't understand how generative recursion works exactly. I could really use some advice, this problem is starting to aggravate me. </td></tr></table>