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Topic: Composites, and neat relation
Replies: 18   Last Post: Sep 14, 2004 3:38 AM

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Dik T. Winter

Posts: 7,899
Registered: 12/6/04
Re: Composites, and neat relation
Posted: Sep 12, 2004 7:24 PM
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In article <3c65f87.0409121043.5988edb8@posting.google.com> jstevh@msn.com (James Harris) writes:
...
> Using that with x = N/2kj, where N, k and j are naturals, and N>=2kj,
> you get
> [(N+2j)/2kj] = [N/kj] - [N/2kj]
> and to test it, I'll use N=13, k=2, j=3, which gives
> [19/12] = [13/6] - [13/12]
> so that works.

Yes, that it works for one set of values *must* mean that it works for all
sets of values... Try k = 6, j = 1, and N any of the values from 18 to 21.

Where is the null test?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; <a href="http://www.cwi.nl/~dik/">http://www.cwi.nl/~dik/</a>




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