Heres my go. Originally I had something that was performing between 2N-1 and N*((N+1)/2), but kaif provided the insight I needed to get it to N log N when he pointed out that you only need to store one path of a given length at any one time.
BTW, I wanted a single routine that could return either the indexes or the actual values, so this is a little clunkier than it need be if you didnt care about the indexes.
Update: I was wrong, the worse case run time for this routine is (N+1)/2*N, which is O(N2).
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