In article <email@example.com> firstname.lastname@example.org (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>