hihi, all - look at the equation eric angelini wrote for sliding numbers 1/a + 1/b = (a+b)/10^k the left-hand side is (a+b)/ab, and we can cancel out (a+b), so we are looking for all (a+b) with a and b positive integers, such that a*b is a (non-negative integral) power of 10, sorted by (a+b) it should be easy to prove some things about them (to guarantee that after a certain point, there will be no new smaller ones) for example, for a given k, the smallest (a+b) with that k is 2*sqrt(10^k) (equality when k i even, unless we also require a!=b) more soon, cal