[racket] 26.1.1 tabulate-div

From: Ken Hegeland (hegek87 at yahoo.com)
Date: Thu Nov 4 21:16:44 EDT 2010

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)
    [(=(remainder n i)0)i]
    [else(find-div n(sub1 i))]))
;(find-div 20 3)

(define(tabulate-div n)
    [(= n 1)(list 1)]
          (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.

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