OFFSET
1,1
COMMENTS
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.
For all n, a(n+1)<a(n)^(2n), so the series is infinite. [John Tromp, Apr 13 2011]
EXAMPLE
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.
PROG
(Haskell)
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..]), [])]
-- for greater efficiency, replace map reverse (inits[2..]) by
-- [m:[m-d|d<-divisors(m*(m-1)), d<m-1]|m<-[2..]], defining divisors appropriately.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
John Tromp, Apr 08 2011
EXTENSIONS
Finally found a(11); about the square of a(10). I doubt if a(12) will ever be found.
STATUS
approved