well posed problem?
Franklin T. Adams-Watters
franktaw at netscape.net
Sat Mar 12 01:20:02 CET 2005
First, shouldn't these numbers be one higher? The empty set always sums to zero.
Second, why not say the nth roots of 1, instead of -1? The values are the same (just rotate by pi/n), and this formulation is simpler.
Third, I think some of these values must be wrong. Since the entire set of roots always sums to 0 for n > 1, a set and its complement must either both or neither sum to 0; hence (for n > 1, and after adding 1 as in my first point) the number of sets summing to 0 must be even.
"wouter meeussen" <wouter.meeussen at pandora.be> wrote:
>of all subsets of the n-th roots of -1, n= 1 to 20, how many add to zero?
>??
>0, 1, 1, 3, 1, 6, 1, 11, 4, 8, 1, 50, 1, 10, 9, 107, 1, 240, 1, 316
>??
>where are the 'groupies' when you need them?
>
>W.
>
>
>
>
--
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645
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