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
_____ From: users-boun...@racket-lang.org [mailto:users-boun...@racket-lang.org] On Behalf Of Jos Koot Sent: 05 November 2010 12:06 To: 'Todd O'Bryan'; lukejor...@gmail.com Cc: users@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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