From: "Meeussen Wouter (bkarnd)" <wouter.meeussen@vandemoortele.com>
it is a small surprise that, for w = Exp[ 2 Pi I /30 ] alias 1^(1/30) , w+w^7+w^8+w^18+w^19+w^25 = 0 no cancellation of symmetric pairs, triples or quintuples in sight, but w+w^7+w^19+w^25 = -w^13 (* four out of five *) and w^8+w^18 = -w^28 = +w^13 (* two out of three *)
could be called "cancellation by missing parts" ;-))
As (a) {0,6,12,18,24} cancels, (b) {0,10,20} cancels, and (c) {0,15} cancels. {5,15,25} cancels, so {5,6,12,18,24,25} cancels. The largest gap (mod 30) between terms is that between 25 and 5, so {0,1,7,13,19,20} is the symmetry-less set with minimal largest member that cancels. You've exploited the gap of 6 between 18 and 24 instead. 2*p*q is easily so minimised - any takers for zeta_210? Phil When inserting a CD, hold down shift to stop the AutoRun feature In the Device Manager, disable the SbcpHid device. http://www.cs.princeton.edu/~jhalderm/cd3/ __________________________________ Do you Yahoo!? Yahoo! Small Business - Try our new resources site! http://smallbusiness.yahoo.com/resources/