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%I #17 Mar 31 2012 10:27:20
%S 2,3,4,6,13,28,64,756,126225,157366665
%N Minimal largest k in set of n fractions of the form (k-1)/k all of whose ratios (smaller fraction / larger fraction) are also of that form.
%C These ratios, together with 1, can be the constant speeds of cyclists going forever around a circular track while only allowed to pass each other at a single point.
%C For all n, a(n+1)<a(n)^(2n), so the series is infinite. [John Tromp, Apr 13 2011]
%e All pairwise ratios in the set {5/6,4/5,3/4,2/3} are also of the form (k-1)/k, while for the only "lesser" size-4 set {4/5,3/4,2/3,1/2}, we have (1/2)/(4/5)=5/8 is not of that form. So a(4)=6.
%o (Haskell)
%o let ext (c,l) = [(tails.filter(\b->a*(a-1)`mod`(b-a)==0)$r,a:l) | (a:r)<-c] in map (last.snd.head) . tail . iterate (>>= ext) $ [(map reverse (inits[2..]),[])]
%o -- for greater efficiency, replace map reverse (inits[2..]) by
%o -- [m:[m-d|d<-divisors(m*(m-1)),d<m-1]|m<-[2..]], defining divisors appropriately.
%K nonn,more
%O 1,1
%A _John Tromp_, Apr 08 2011
%E Finally found a(11); about the square of a(10). I doubt if a(12) will ever be found.
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