This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010
are on their own paths, but the first branch contains 3 curves and the second one 2 curves. If you want the same result for all pairs of curves you'd need to split up the first and second branches, so that all curves are on their own branch.…
Added by Lars Renklint at 4:33am on September 6, 2009