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Re: Composites, and neat relation
Posted:
Sep 12, 2004 7:24 PM


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>



