Veit Elser kindly noted errors in:
On the other hand, we've already seen examples S(ka,kb,kc) < S(a,b,c) in the
The correct statement: We've seen an example S(ka,kb,kc,kd) > S(a,b,c,d), namely Michael Reid's example 1/3 = S(4k,5k,6k,7k) > S(4,5,6,7) for k > 1.
don't worry, i'll soon have a 3 parameter example. :-) on another note: ) lowest terms. But I expect the more interesting and "universal" projective ) quantity is ) ) S^*(r_1,r_2) = infimum_{k>0} S(k*m, k*m*r_1, k*m*r_2). ) ) It's surely the case, for given a<b<c, that S^*(b/a,c/a) = S(ka,kb,kc) for ) some finite k. I conjecture there's a single finite K_3 which achieves certainly it's plausible that S(ka, kb, kc) is (eventually) periodic in k , with period larger than 1 . whether or not this is *likely* i do not know. but i cannot rule it out. perhaps you are predicting this does not happen; but if it does happen then your S^*(r, s) function does not tell the whole story. mike