[racket] Generative recursion

Luke, you'd better ignore my third approach and follow the instructions of
Todd.
My method is far more difficult to implement than Todd's methods.
Jos


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From: users-bounces at racket-lang.org [mailto:users-bounces at racket-lang.org]
On Behalf Of Jos Koot
Sent: 05 November 2010 12:06
To: 'Todd O'Bryan'; lukejordan at gmail.com
Cc: users at racket-lang.org
Subject: Re: [racket] Generative recursion


Actually there is one more way. We only have to check numbers 1 up to the
integer-sqrt of n. For each check whose remainder is 0, we immediately have
two divisors, the checking number and the quotient (except when these two
are equal, giving one divisor only -this happens only when n is a square-))
 
Jos
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